Hugh D. Young - Instructor's Solution Manuals To College Physics (2011, Pearson).pdf 3e2a6b

INSTRUCTOR SOLUTIONS MANUAL

SEARS & ZEMANSKY’S

COLLEGE PHYSICS 9TH EDITION

HUGH D. YOUNG

Forrest Newman Sacramento City College

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Copyright © 2012, 2007 Pearson Education, Inc., publishing as Addison-Wesley, 1301 Sansome Street, San Francisco, CA 94111. All rights reserved. Manufactured in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, 1900 E. Lake Ave., Glenview, IL 60025. For information regarding permissions, call (847) 486-2635. Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps.

ISBN 10: 0-321-69665-4 ISBN 13: 978-0-321-69665-6

CONTENTS

Preface.......................................................................................................................................v Chapter 0

Mathematics Review ..................................................................................... 0-1

Mechanics Chapter 1

Models, Measurements, and Vectors ............................................................ 1-1

Chapter 2

Motion along a Straight Line ........................................................................ 2-1

Chapter 3

Motion in a Plane .......................................................................................... 3-1

Chapter 4

Newton’s Laws of Motion ............................................................................ 4-1

Chapter 5

Applications of Newton’s Laws.................................................................... 5-1

Chapter 6

Circular Motion and Gravitation ................................................................... 6-1

Chapter 7

Work and Energy .......................................................................................... 7-1

Chapter 8

Momentum .................................................................................................... 8-1

Chapter 9

Rotational Motion ......................................................................................... 9-1

Chapter 10

Dynamics of Rotational Motion.................................................................. 10-1

Periodic Motion, Waves, and Fluids Chapter 11

Elasticity and Periodic Motion.................................................................... 11-1

Chapter 12

Mechanical Waves and Sound .................................................................... 12-1

Chapter 13

Fluid Mechanics .......................................................................................... 13-1

Thermodynamics Chapter 14

Temperature and Heat ................................................................................. 14-1

Chapter 15

Thermal Properties of Matter ...................................................................... 15-1

Chapter 16

The Second Law of Thermodynamics ........................................................ 16-1

© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

iii

iv

Contents

Electricity and Magnetism Chapter 17

Electric Charge and Electric Field .............................................................. 17-1

Chapter 18

Electric Potential and Capacitance.............................................................. 18-1

Chapter 19

Current, Resistance, and Direct-Current Circuits ....................................... 19-1

Chapter 20

Magnetic Field and Magnetic Forces.......................................................... 20-1

Magnetic Forces Chapter 21

Electromagnetic Induction .......................................................................... 21-1

Chapter 22

Alternating Current ..................................................................................... 22-1

Chapter 23

Electromagnetic Waves .............................................................................. 23-1

Light and Optics Chapter 24

Geometric Optics ........................................................................................ 24-1

Chapter 25

Optical Instruments..................................................................................... 25-1

Chapter 26

Interference and Diffraction........................................................................ 26-1

Modern Physics Chapter 27

Relativity..................................................................................................... 27-1

Chapter 28

Photons, Electrons, and Atoms ................................................................... 28-1

Chapter 29

Atoms, Molecules, and Solids .................................................................... 29-1

Chapter 30

Nuclear and High-Energy Physics .............................................................. 30-1


PREFACE

This Instructor Solutions Manual contains detailed solutions to all end-of-chapter problems. Solutions are done in the Set Up/Solve/Reflect framework used in the textbook. In most cases rounding was done in intermediate steps, so you may obtain slightly different results if you handle the rounding differently. We have made every effort to be accurate and correct in the solutions, but if you find errors or ambiguities it would be very helpful if you would point these out to the publisher.


v

0

MATHEMATICS REVIEW

Solutions to Problems 0.1. Set Up and Solve: (−3x 4 y 2 ) 2 = (−3) 2 ( x 4 ) 2 ( y 2 )2 = 9 x8 y 4 . Reflect: The square of any real expression must be positive.

0.2. Set Up and Solve:

(2344 ) 2 (8) 4

=

[23 (22 ) 4 ]2 (23 )4

=

(2328 ) 2 212

=

(211 ) 2 212

=

222 212

= 2(22 −12) = 210 = 1024.

Reflect: Since we are evaluating a numerical expression you could simply use your calculator to obtain the result without any algebraic manipulation. 2

⎛ 4 x2 ⎞ 42 ( x 2 ) 2 16 x 4 = 2 3 2= = 4 x 4 y −6 . 3⎟ 6 ⎟ 2 (y ) 4y ⎝ 2y ⎠

*0.3. Set Up and Solve: ⎜⎜

Reflect: The answer could also be written as

4x4

y6

.

3

⎛ x2 y4 ⎞ ( x 2 )3 ( y 4 )3 x 6 y12 = (−1)3 3 22 3 = 2 3 −6 = − x (6 −3) y (12+ 6) = − x3 y18 22 ⎟ ⎟ x (y ) x y ⎝ xy ⎠ Reflect: The cube of any real expression has the same sign as the original expression.

0.4. Set Up and Solve: ⎜⎜ −

0.5. Set Up and Solve: The decimal point must be moved 5 places to the left to change 475000 into a number between 1 and 10. Thus, we have 475000 = 4.75 × 105. Reflect: When written in scientific notation, numbers larger than 1 will have positive exponents and numbers smaller than 1 will have negative exponents for their power of ten.

0.6. Set Up and Solve: The decimal point must be moved 6 places to the right to change 0.00000472 into a number between 1 and 10. Thus, we have 0.00000472 = 4.72 × 10−6. Reflect: When written in scientific notation, numbers larger than 1 will have positive exponents and numbers smaller than 1 will have negative exponents for their power of ten.

*0.7. Set Up and Solve: The decimal point must be moved 2 places to the left to change 123 into a number between 1 and 10. Thus, we have 123 × 10−6 = 1.23 × 102 × 1026 = 1.23 × 1024. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

0-1

0-2

Chapter 0

Reflect: Alternatively, we could have written our original number in decimal form by moving the decimal point 6

places to the left to obtain 123 × 1026 = 0.000123. Finally we could convert the result into scientific notation by moving the decimal point 4 places to the right to obtain 0.000123 = 1.23 × 1024.

0.8. Set Up and Solve:

8.3 × 105 7.8 × 102

=

8.3 × 10(5− 2) = 1.1 × 103 , where we have rounded the decimal number to the 7.8

nearest tenth. Reflect: You can make a quick estimate, to check your result, by rounding each number to the nearest power of ten.

Thus, we have the estimate

8.3 × 105 7.8 × 10

2

≈

106 103

= 103 , which can be done without a calculator.

0.9. Set Up and Solve: First we subtract 9 x from both sides to obtain −5 x + 6 = −14. Next we subtract 6 from both sides to get −5 x = −20. Finally, we divide both sides of the equation by −5 to get the solution x = 4. Reflect: There is no need to keep x on the left side of the equation. By adding −4 x + 14 to both sides of the equation we get 20 = 5 x, which reduces to 4 = x. 9 5

0.10. Set Up and Solve: First subtract 32 from both sides to obtain F − 32 = C. Next multiply both sides by obtain

5 to 9

5 5 ( F − 32) = C. Finally, reflect the equation to obtain C = ( F − 32). 9 9

Reflect: The symmetric property of equality allows us to write y = x as x = y.

*0.11. Set Up and Solve: Add −3x 2 − 6 to both sides to obtain x 2 = 12. Next, take the positive and negative square root of both sides to obtain x = ± 12 = ± (3)(4) = ±2 3. Reflect: Since this quadratic has no containing x, it is possible to solve it without factoring or using the quadratic formula.

0.12. Set Up and Solve: Reflect the equation to obtain −9.8t 2 = −196. Next divide both sides by −9.8 to obtain t2 =

2196 = 20. Finally, take the positive and negative square root of both sides to obtain t = ± 20 = 29.8

± (4)(5) = ±2 5. Reflect: Since this quadratic has no containing t, it is possible to solve it without factoring or using the quadratic formula.

*0.13. Set Up and Solve: Notice that −2 + 23 = 25 and (−2)(−3) = 6. Thus, we can factor the equation as x 2 − 5 x + 6 = ( x − 3)( x − 2) = 0. The two roots are x = 3 and x = 2. Reflect: Alternatively, we can use the quadratic formula with a = 1, b = −5, and c = 6 to obtain x =

−b ± b2 − 4ac = 2a

5 ± (−5) 2 − 4(1)(6) 5 ± 1 = = 3 or 2. 2(1) 2

0.14. Set Up and Solve: Use the quadratic formula with a = 1, b = 1, and c = −1 to obtain x =

−b ± b 2 − 4ac = 2a

−1 ± 12 − 4(1)(−1) 21 ± 5 = . 2(1) 2 Reflect: The factorization of this equation is not obvious, so we resort to the quadratic formula.


Mathematics Review

0.15. Set Up and Solve: Use the quadratic formula with a = 4.9, b = 2, and c = 220 to obtain t =

0-3

−b ± b 2 − 4ac = 2a

−2 ± 22 − 4(4.9)(−20) −2 ± 396 = = 22.2 or 1.8. The decimal answers are accurate to the nearest tenth. 2(4.9) 9.8 Reflect: In physics many numbers are based on approximate measurements. Therefore it is usually okay to write

answers in approximate decimal form. For example, we may write

2 as 1.4 or 1.41, depending on the situation.

0.16. Set Up and Solve: Solve the first equation for x to obtain x =

4y +1 and substitute this expression for x in 5

⎛ 4y +1⎞ the second equation to obtain 6 y = 10 ⎜ ⎟ − 4 = 8 y − 2. Solving for y we obtain −2 y = −2 or y = 1. Thus, we ⎝ 5 ⎠ 4y +1 4 +1 have x = = = 1. The solution is x = 1, y = 1. 5 5 Reflect: We could also solve this system by multiplying the first equation by 2 and adding the result to the second equation.

*0.17. Set Up and Solve: Divide the second equation by 3 to obtain

2x y 1 − = . Add this result to the first 3 3 3

1 ⎛ x y ⎞ ⎛ 2x y ⎞ − ⎟ = 2 + . The y cancel and we can clear the fractions by multiplying equation to obtain ⎜ + ⎟ + ⎜ 2 3 3 3 3 ⎝ ⎠ ⎝ ⎠ both sides by 6. Thus, we obtain 3x + 4 x = 12 + 2. Solving for x we obtain x = 2. By substituting this value for x into the second equation we obtain 2(2) − y = 1, so y = 3. Thus, the solution is x = 2, y = 3.

Reflect: We could also solve this equation by solving the second equation for y and substituting the result into the first equation.

0.18. Set Up and Solve: Since x ∝ y we have

x 2 1 8 = or x = y. Substituting y = 8 we obtain x = = 1.6. y 10 5 5

Reflect: In physics, variables often represent physical measurements and so it is common to represent fractions as decimals.

0.19. Set Up and Solve: Let F be the force required to compress the spring by a distance x. Thus, F ∝ x and we F 80 N = or F = (4000 N/m) x. Substituting x = 0.05 m gives F = 200 N. x 0.02 m F Reflect: The newton (N) is a unit of force and the constant k = is measured in N/m. Stiff springs have large x values of k and are difficult to stretch.

have

0.20. Set Up and Solve: Let T be the temperature of the gas (in kelvins) and V be its volume. Thus, we have V 4.0 L = or V = (0.040 L/K)T . Thus, when T = 300 K we have T 100 K V = (0.040 L/K)T = (0.040 L/K)(300 K) = 12 liters.

V ∝ T , when the pressure is fixed, and we have

Reflect: Since the temperature increases by a factor of 3, the volume must also increase by a factor of 3.

*0.21. Set Up and Solve: Let A be the amplitude of the sound and d the distance from the source. Thus, A ∝ and we have Ad = (4.8 × 1026 m)(1.0 m) or A =

1 d

(4.8 × 1026 m 2 ) . Thus, when d = 4.0 m we have A = 1.2 × 1026 m. d


0-4

Chapter 0

Reflect: Since the distance is increased by a factor of 4, the amplitude is decreased by a factor of 4 (i.e., multiplied 1 by ). 4

0.22. Set Up and Solve: Use the given values to solve for k: k = Ir 2 = (60.0 lux)(0.40 m)2 . Use this value of k to find the intensity at a distance of 1.80 meters: I =

k r2

=

(60.0 lux)(0.40 m) 2 (1.80 m) 2

2

⎛ 0.40 m ⎞ = (60.0 lux) ⎜ ⎟ = 3.0 lux ⎝ 1.80 m ⎠

Reflect: Larger distances lead to lower intensities. We could also solve this problem using the ratio equation

I 2 ⎛ d1 ⎞ =⎜ ⎟ I1 ⎝ d 2 ⎠

2

*0.23. Set Up and Solve: Let F be the force of gravity on an object that is a distance d from the earth’s center. Thus, we have F ∝

1 d

2

or Fd 2 = k . Also, when d = 6.38 × 106 m we know that F = 700 N so we can find the

value of k. When the astronaut is 6000 km ( 6 × 106 m ) from the earth’s surface he is 6.38 × 106 + 6.00 × 106 = 1.238 × 107 m from the center of the earth. Thus, we have F =

k d

2

=

(700 N)(6.38 × 106 m) 2 (1.238 × 107 m) 2

= 186 N.

2

F2 ⎛ d1 ⎞ = ⎜ ⎟ . Since the astronaut nearly doubles F1 ⎝ d 2 ⎠ his distance from the earth’s center as he moves from the surface into orbit, his weight decreases by nearly a factor of 1 4 (i.e., 2 ). 2 Reflect: We could also solve this problem using the ratio equation

0.24. Set Up and Solve: (a) log( x 4 y 2 z8 ) = log( x 4 ) + log( y 2 ) + log( z8 ) = 4log x + 2log y + 8log z. 1 1 3 7 (b) log x3 y 7 = log( x3 y 7 ) = ⎡⎣log( x3 ) + log( y 7 ) ⎤⎦ = log x + log y. 2 2 2 2 (c) log 3

x2 y6 z

3

2 = log( x 2/3 y 2 z21 ) = log( x 2/3 ) + log( y 2 ) + log( z21 ) = log x + 2log y − log z. 3

Reflect: There are multiple ways to expand each of these expressions. For example, we could expand log

a as b

1 1 1 −1 ⎛ a⎞ 1 log ⎜ ⎟ = log a − log b or as log(a1/ 2b −1/ 2 ) = log(a1/ 2 ) + log(b −1/ 2 ) = log a + log b. ⎝ b⎠ 2 2 2 2 2

⎡ x4 y ⎤ 3⎥ ⎣⎢ ( x + y ) ⎦⎥

*0.25. Set Up and Solve: (a) 4log x + log y − 3log( x + y ) = log( x 4 ) + log y − log[( x + y )3 ] = log ⎢ ⎛ x( y + x) ⎞ 2 (b) log( xy + x 2 ) − log( xz + yz ) + 2log z = log ⎜ ⎟ + log( z ) = log( xz ) ⎝ z( x + y) ⎠ Reflect: There are multiple ways to combine each of these expressions.

⎛ 10−4 ⎞ = 10log(108 ) = 80. −12 ⎟ ⎟ 10 ⎝ ⎠

0.26. Set Up and Solve: We have β = 10log ⎜⎜

Reflect: It is easy to find common logarithms of powers of ten—no calculator is required.


Mathematics Review

0-5

⎛ I ⎞ ⎛ I ⎞ = 60 so log ⎜ −12 ⎟ = 6. Taking the antilog of both sides we ⎝ 10−12 ⎟⎠ ⎝ 10 ⎠

0.27. Set Up and Solve: We have β = 10log ⎜

I = 106 so I = 106 ⋅ 10−12 = 10−6. 10212 Reflect: It is easy to find the antilog of an integer—no calculator is required. obtain

0.28. Set Up and Solve: We must solve the equation N = N 0e−( 0.210 )t for t. Divide both sides of the equation by N 0 to obtain

N = e − (0.210)t . Next, take the natural logarithm of both sides of the equation and solve for t to obtain N0

⎛ 2.50 × 104 ⎞ ⎛ N ⎞ 1 ln ⎜ ln ⎜ ⎟ = 20.9 ⎟ = −(0.210)t so t = 2 0.210 ⎜⎝ 2.00 × 106 ⎟⎠ ⎝ N0 ⎠ Reflect: Since we are using base-e rather than base-10, we must use the natural logarithm (ln) rather than the common logarithm (log).

*0.29. Set Up and Solve: (a) The circumference of the circle is given by C = 2π r = 2π (0.12 m) = 0.75 m and its area is given by A = π r 2 = π (0.12 m) 2 = 0.045 m 2 . (b) The surface area of the sphere is given by A = 4π r 2 = 4π (0.21 m) 2 = 0.55 m 2 and its volume is given by 4 3 4 π r = π (0.21 m)3 = 0.039 m3. 3 3 (c) The total surface area of the rectangular solid (which consists of three identical pairs of rectangular faces) is 2lw + 2lh + 2 wh = 2[(0.18 m)(0.15 m) + (0.18 m)(0.8 m) + (0.15 m)(0.8 m)] = 0.6 m 2 (rounding to the nearest tenth).

Its volume is V = lwh = (0.18 m)(0.15 m)(0.8 m) = 0.02 m3 (rounding to the nearest hundredth). (d) The total surface area of the cylinder consists of its lateral (side) surface area and the area of its two circular endcaps: A = 2π rh + 2π r 2 = 2π [(0.18 m)(0.33 m) + (0.18 m) 2 ] = 0.58 m 2 . The volume of the cylinder is given by V = π r 2 h = π (0.18 m) 2 (0.33 m) = 0.034 m3. Reflect: When the lateral surface of a cylinder of radius r and height h is unrolled, it forms a rectangle with

dimensions 2π r by h, so it has a surface area of A = 2π rh. The area of each of its two circular end-caps is π r 2 , so

the total surface area of a cylinder is A = 2π rh + 2π r 2 .

0.30. Set Up and Solve: Using the Pythagorean theorem, a 2 + b2 = c 2 , we can find the length of the unknown side: a = c 2 − b 2 = (20 cm) 2 − (16 cm) 2 = 12 cm. The angle, θ , opposite side a can be found from tan θ =

θ = tan −1

12 , so 16

12 = 37D. Since all three angles add to 180D , we have that the other angle must be 180D − 90D − 37D = 16

90D − 37D = 53D. Reflect: This triangle is four times larger than the common 3-4-5 triangle.

*0.31. Set Up and Solve: Assume that each step forms a right angle so that a line drawn between the corners of adjacent steps forms a right triangle with a base of 30 cm and a height of h. The angle between the hypotenuse and h the base is 36D. Thus, tan 36D = , or h = (30 cm) tan 36D = 22 cm. 30 cm 22 Reflect: A line drawn from the base of the stairs to the top of the stairs has a slope of , which is also equal to 30 tan 36D.


1

MODELS, MEASUREMENTS, AND VECTORS

Answers to Multiple-Choice Problems 1. B

2. A

3. C

4. A

5. B

6. A

7. B

8. B

9. C

10. C

11. D

12. B

13. D

Solutions to Problems *1.1. Set Up: We know the following equalities: 1 mg = 1023 g; 1 μ g = 1026 g; 1 kilohms = 1000 ohms; and

1 milliamp = 1023 amp. Solve: In each case multiply the quantity to be converted by unity, expressed in different units. Construct an expression for unity so that the units to be changed cancel and we are left with the new desired units. ⎛ 1023 g ⎞ = 2.40 g/day b 2 − 4ac (a) (2400 mg/day) ⎜ ⎜ 1 mg ⎟⎟ ⎝ ⎠ 2 6 ⎛ 10 g ⎞ = 1.20 × 1024 g/day (b) (120 μ g/day) ⎜ ⎜ 1 μ g ⎟⎟ ⎝ ⎠ ⎛ 1023 g ⎞ = 0.500 g/day (c) (500 mg/day) ⎜ ⎜ 1 mg ⎟⎟ ⎝ ⎠ ⎛ 1 kilohm ⎞ (d) (1500 ohms) ⎜ 3 ⎟ = 1.50 kilohms ⎝ 10 ohms ⎠ ⎛ 1 milliamp ⎞ (e) (0.020 amp) = ⎜⎜ 23 ⎟⎟ = 20 milliamps ⎝ 10 amp ⎠ Reflect: In each case, the number representing the quantity is larger when expressed in the smaller unit. For example, it takes more milligrams to express a mass than to express the mass in grams.

1.2. Set Up: From the meaning of the metric prefixes we know: 1 megaohm = 1 MΩ = 106 ohms; 1 picofarad = 1 pf = 10212 farad; 1 gigameter = 1 Gm = 109 m; 1 nanometer = 1 nm = 1029 m; and 1 femtometer = 1 fm = 10215 m. ⎛ 106 ohms ⎞ Solve: (a) (7.85 megohms) ⎜ = 7.85 × 106 ohms ⎜ 1 megaohm ⎟⎟ ⎝ ⎠ ⎛ 10−12 farad ⎞ (b) (5 picofarads) ⎜ = 5 × 10−12 picofarad ⎜ 1 picofarad ⎟⎟ ⎝ ⎠ © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-1

1-2

Chapter 1

⎛ 1 gigameter ⎞ (c) (3.00 × 108 m/s) ⎜ ⎟ = 0.300 gigameter/s ⎝ 109 m ⎠ ⎛ 1029 meter ⎞ ⎛ 1029 meter ⎞ (d) (400 nm) ⎜ = 4.00 × 10−7 meter; (700 nm) ⎜ = 7.00 × 10−7 meter ⎜ 1 nm ⎟⎟ ⎜ 1 nm ⎟⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 10215 meter ⎞ (e) (2 femtometer) ⎜ = 2 × 10−15 meter ⎜ 1 femtometer ⎟⎟ ⎝ ⎠

1.3. Set Up: We know the equalities: 1 mg = 1023 g, 1 μg = 10−6 g, and 1 kg = 103 g. ⎛ 10−3 g ⎞⎛ 1 μ g ⎞ Solve: (a) (410 mg/day) ⎜ = 4.10 × 105 μ g/day ⎜ 1 mg ⎟⎜ ⎟ ⎜ 1026 g ⎟⎟ ⎠ ⎝ ⎠⎝ ⎛ 10−3 g ⎞ (b) (12 mg/kg)(75 kg) = (900 mg) ⎜ = 0.900 g ⎜ 1 mg ⎟⎟ ⎝ ⎠ (c) The mass of each tablet is ⎛ 10−3 g ⎞ = 2.0 × 10−3 g. (2.0 mg) ⎜ ⎜ 1 mg ⎟⎟ ⎝ ⎠

The number of tablets required each day is the number of grams recommended per day divided by the number of grams per tablet: 0.0030 g/day = 1.5 tablet/day. 2.0 × 10−3 g/tablet Take 2 tablets each day. ⎛ 1 mg ⎞ (d) (0.000070 g/day) ⎜⎜ 23 ⎟⎟ = 0.070 mg/day ⎝ 10 g ⎠

1.4. Set Up: In part (a), we need to solve an equation of the form ⎛ 2.54 cm ⎞ ⎜ ⎟ x(1.0 mi) = y, ⎝ 1.00 in. ⎠ where x is a series of conversion equalities and y is the unknown number of kilometers. For part (b), we must prove (1.00 mL) x = 1.00 cm3 using the equalities 1 L = 1000 cm3 and 103 mL = 1 L. Part (c) requires the application of

the equality 1 mL = 1 cm3 . ⎡⎛ 2.54 cm ⎞⎛ 12.0 in. ⎞⎛ 5280 ft ⎞ ⎤ ⎡⎛ 1.00 m ⎞⎛ 1.00 km ⎞ ⎤ Solve: (a) ⎢⎜ ⎟⎜ ⎟⎜ ⎟ ⎥ (1.00 mi) ⎢⎜ ⎟⎜ ⎟ ⎥ = y; y = 1.61 km 3 ⎣⎝ 1.00 in. ⎠⎝ 1.00 ft ⎠⎝ 1.00 mi ⎠ ⎦ ⎣⎝ 100 cm ⎠⎝ 10 m ⎠ ⎦ 3 ⎛ 1.00 L ⎞ ⎛ 1000 cm ⎞ 3 (b) (1.00 mL) ⎜ 3 ⎜ ⎟ = 1.00 cm ⎟ ⎝ 10 mL ⎠ ⎝⎜ 1.00 L ⎠⎟

⎛ 103 mL ⎞⎛ 1.00 cm3 ⎞ = 1000 cm3 (c) (1.00 L) ⎜ ⎜ 1.00 L ⎟⎜ ⎟⎜ 1.00 mL ⎟⎟ ⎝ ⎠⎝ ⎠

*1.5. Set Up: We need to apply the following conversion equalities: 1000 g = 1.00 kg, 100 cm = 1.00 m, and 1.00 L = 1000 cm3 . 3

⎛ 1.00 kg ⎞ ⎛ 100 cm ⎞ 3 Solve: (a) (1.00 g/cm3 ) ⎜ ⎟⎜ ⎟ = 1000 kg/m ⎝ 1000 g ⎠ ⎝ 1.00 m ⎠


Models, Measurements, and Vectors

1-3

3

⎛ 1000 g ⎞ ⎛ 1.00 m ⎞ 3 (b) (1050 kg/m3 ) ⎜ ⎟⎜ ⎟ = 1.05 g/cm ⎝ 1.00 kg ⎠ ⎝ 1000 cm ⎠ ⎛ 1000 cm3 ⎞ ⎛ 1.00 g ⎞ ⎛ 1.00 kg ⎞ ⎛ 2.205 lb ⎞ (c) (1.00 L) ⎜ = 1.00 kg; (1.00 kg) ⎜ ⎟ = 2.20 lb ⎜ 1.00 L ⎟⎟ ⎝⎜ 1.00 cm3 ⎠⎟ ⎝⎜ 1000 g ⎠⎟ ⎝ 1.00 kg ⎠ ⎝ ⎠ Reflect: We could express the density of water as 1.00 kg/L.

1.6. Set Up: From the appendices we find the conversion from kilometers into miles, and the radius and period of the earth’s orbit: 1 km = 0.6214 mi; r = 1.50 × 1011 m; T = 365.3 days. Since the earth’s orbit is nearly circular, its circumference is C = 2π r . Solve: distance speed = time ⎛ 2π (1.50 × 1011 m) ⎞ ⎛ 1 day ⎞⎛ 1 h ⎞ 4 =⎜ ⎟ = 2.99 × 10 m/s ⎜ ⎟⎟ ⎜⎝ 24 h ⎟⎜ 365.3 days 3600 s ⎠⎝ ⎠ ⎝ ⎠ ⎛ 2π (1.50 × 1011 m) ⎞ ⎛ 1 day ⎞⎛ 1 km ⎞ 5 5 =⎜ ⎟⎟ ⎜ ⎟⎜ ⎟ = 1.075 × 10 km/h = 1.08 × 10 km/h ⎜ 365.3 days 24 h 1000 m ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎛ 0.6214 mi ⎞ 4 = (1.075 × 105 km/h) ⎜ ⎟ = 6.68 × 10 mi/h ⎝ 1 km ⎠ Reflect: To minimize rounding errors, we should avoid rounding in intermediate calculations: this is why we used 1.075 × 105 km/h rather than 1.08 × 105 km/h to calculate our final answer. The significant digits are underlined.

1.7. Set Up: The speed of light is v = 3.00 × 108 m/s; 1 ft = 0.3048 m; 1 s = 109 ns; t = d /v. Solve: t =

0.3048 m 3.00 × 108 m/s

= 1.02 × 1029 s = 1.02 ns

Reflect: In 1.00 s light travels 3.00 × 108 m = 3.00 × 105 km = 1.86 × 105 mi.

1.8. Set Up: We know: 1 in = 2.54 cm; 1 cm = 1022 m; and 1 mm = 1023 m. Solve: (a)

⎛

22

⎞

⎞ 10 m ⎛ 1 mm ⎞ ⎟⎜ ( 83 in.) ⎜⎝⎛ 2.154incm ⎟⎜ ⎟ = 9.5 mm . ⎠ ⎜ 1 cm ⎟ ⎝ 1023 m ⎠ ⎝

⎠

⎛ 10 m ⎞ ⎛ 1 cm ⎞⎛ 1 in. ⎞ (b) (12 mm) ⎜ ⎟ = 0.47 in. ⎜ 1 mm ⎟⎟ ⎜⎝ 1022 m ⎟⎜ ⎠⎝ 2.54 cm ⎠ ⎝ ⎠ 23

(c) Express and compare the English wrench sizes as decimal values: 83 ′′ = 0.375 in.,

The

1 -inch 2

4 ′′ 8

= 0.500 in., and 85 ′′ = 0.625 in.

wrench is closest to 12 mm.

*1.9. Set Up: We apply the equalities of 1 km = 0.6214 mi and 1 gal = 3.788 L. ⎛ 1 km ⎞⎛ 1 gal ⎞ Solve: (37.5 mi/gal) ⎜ ⎟⎜ ⎟ = 15.9 km/L ⎝ 0.6214 mi ⎠⎝ 3.788 L ⎠ Reflect: Note how the unit conversion strategy, of cancellation of units, automatically tells us whether to multiply or divide by the conversion factor.

1.10. Set Up: Apply the given conversion factors, 1 furlong = 0.1250 mi and 1 fortnight = 14 days, along with 1 day = 24 h.


1-4

Chapter 1

⎛ 0.125 mi ⎞⎛ 1 fortnight ⎞ ⎛ 1 day ⎞ Solve: (180,000 furlongs/fortnight) ⎜ ⎟⎜ ⎟⎜ ⎟ = 67 mi/h ⎝ 1 furlong ⎠⎝ 14 days ⎠ ⎝ 24 h ⎠

1.11. Set Up: We know: 1 euro = $1.25; and 1 gal = 3.788 L. ⎛ $1.25 ⎞ ⎛ 3.788 L ⎞ Solve: (1.35 euros/L) ⎜ ⎟ = $6.39 per gallon. Currently, in 2005, gasoline in the U.S. costs about ⎟⎜ ⎝ 1 euro ⎠ ⎝ 1 gal ⎠ $2 per gallon so the price in Europe is about three times higher.

1.12. Set Up: From Appendix A, the volume V of a sphere is given in of its radius as V = 43 π r 3 while its surface area A is given as A = 4π r 2 . Also, by definition, the radius is one-half the diameter or r = d /2 = 1.0 μ m. Finally, the necessary equalities for this problem are: 1 μ m = 1026 m; 1 cm = 1022 m; and 1 mm = 1023 m. 3

⎛ 10−6 m ⎞ ⎛ 1 cm ⎞3 Solve: V = 43 π r 3 = 43 π (1.0 μ m)3 ⎜ = 4.2 × 10−12 cm3 ⎜ 1 μ m ⎟⎟ ⎜⎝ 10−2 m ⎟⎠ ⎝ ⎠ and 2

⎛ 10−6 m ⎞ ⎛ 1 mm ⎞2 = 1.3 × 10−5 mm 2 A = 4π r 2 = 4π (1.0 μ m)2 ⎜ ⎜ 1 μ m ⎟⎟ ⎜⎝ 10−3 m ⎟⎠ ⎝ ⎠

*1.13. Set Up: We apply the basic time relations of 1 h = 60 min and 1 min = 60 s in part (a) and use the result of (a) in part (b). Similarly, apply the result of part (b) in solving part (c). ⎛ 60 min ⎞⎛ 60 s ⎞ Solve: (a) (1 h) ⎜ ⎟⎜ ⎟ = 3600 s ⎝ 1 h ⎠⎝ 1 min ⎠ (b) (24 h/day) [ (3600 s)/(1 h) ] = 86.4 × 103 s/day (c) (365 day/yr)[(86.4 × 103 s)/(1 day)] = 31,536,000 s/yr = 3.15 × 107 s/yr

1.14. Set Up: (a) The appropriate equalities required are: 1 mi = 1609 m; 1 h = 3600 s; 1 mph = 0.4470 m/s; 1 ft = 0.3048 m; and 1 mi = 5280 ft. ⎛ 1 mi ⎞ 5 Solve: (a) (3.00 × 108 m/s) ⎜ ⎟ = 1.86 × 10 mi/s; ⎝ 1609 m ⎠

⎛ 1 mi ⎞⎛ 3600 s ⎞ 8 (3.00 × 108 m/s) ⎜ ⎟⎜ ⎟ = 6.71 × 10 mph 1609 m 1 h ⎝ ⎠⎝ ⎠ ⎛ 0.3048 m ⎞ ⎛ 1 mi ⎞⎛ 3600 s ⎞ (b) (1100 ft/s) ⎜ ⎟ = 335 m/s; (1100 ft/s) ⎜ ⎟⎜ ⎟ = 750 mph ⎝ 1 ft ⎠ ⎝ 5280 ft ⎠⎝ 1 h ⎠ ⎛ 5280 ft ⎞ ⎛ 1 h ⎞ (c) (60 mi/h) ⎜ ⎟ = 88 ft/s ⎟⎜ ⎝ 1 mi ⎠ ⎝ 3600s ⎠ ⎛ 1 ft ⎞ 2 (d) (9.8 m/s 2 ) ⎜ ⎟ = 32 ft/s ⎝ 0.3048 m ⎠

1.15. Set Up: In each case, round the last significant figure. Solve: (a) 3.14, 3.1416, 3.1415927 (b) 2.72, 2.7183, 2.7182818 (c) 3.61, 3.6056, 3.6055513 Reflect: All of these representations of the quantities are imprecise, but become more precise as additional significant figures are retained.

1.16. Set Up: Use a calculator to calculate the decimal equivalent of each fraction and then round the numeral to the specified number of significant figures. Compare to π rounded to the same number of significant figures. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1-5

Solve: (a) 3.14286 (b) 3.14159 (c) The exact value of π rounded to six significant figures is 3.14159. Since the fraction 22/7 differs in the fourth significant figure, it is accurate to only three significant figures. The fraction 355/113 and π agree when expressed to six figures and thus agree to this precision.

*1.17. Set Up: Calculate the trig function with the maximum and minimum possible values of the angle. Solve: (a) cos 3.5D = 0.9981 and cos 4.5D = 0.9969, so the range is 0.9969 to 0.9981. (b) sin 3.5D = 0.061 and sin 4.5D = 0.078, so the range is 0.061 to 0.078. (c) tan 3.5D = 0.061 and tan 4.5D = 0.079, so the range is 0.061 to 0.079.

1.18. Set Up: We know the relations: mass = density × volume; g and 1.00 L = 1.00 × 103 cm3 . Solve: Water m = (1.00 g/cm3 )(1.00 × 103 cm3 ) = 1.00 × 103 g; Blood: m = (1.05 g/cm3 )(1.00 × 103 cm3 ) = 1.05 × 103 g; Seawater: m = (1.03 g/cm3 )(1.00 × 103 cm3 ) = 1.03 × 103 g.

*1.19. Set Up: We are given the relation density = mass/volume = m/V where V = 43 π r 3 for a sphere. From Appendix F, the earth has mass of m = 5.97 × 1024 kg and a radius of r = 6.38 × 106 m whereas for the sun at the end of its lifetime, m = 1.99 × 1030 kg and r = 7500 km = 7.5 × 106 m. The star possesses a radius of r = 10 km = 1.0 × 104 m and a mass of m = 1.99 × 1030 kg.

Solve: (a) The earth has volume V = 43 π r 3 = 43 π (6.38 × 106 m)3 = 1.088 × 1021 m3 .

density =

3 3 m 5.97 × 1024 kg 3 3 ⎛ 10 g ⎞ ⎛ 1 m ⎞ = = . × = 5.49 g/cm3 (5 49 10 kg/m ) ⎜ ⎟ ⎜ ⎟ ⎜ 1 kg ⎟ ⎝ 102 cm ⎠ V 1.088 × 1021 m3 ⎝ ⎠

(b) V = 43 π r 3 = 43 π (7.5 × 106 m)3 = 1.77 × 1021 m3

density =

3 ⎞ m 1.99 × 1030 kg 9 3 ⎛ 1 g/cm = = (1 . 1 × 10 kg/m ) = 1.1 × 106 g/cm3 ⎜ ⎜ 1000 kg/m3 ⎟⎟ V 1.77 × 1021 m3 ⎝ ⎠

(c) V = 43 π r 3 = 43 π (1.0 × 104 m)3 = 4.19 × 1012 m3

density =

3 ⎞ m 1.99 × 1030 kg 17 3 ⎛ 1 g/cm = = (4 . 7 × 10 kg/m ) = 4.7 × 1014 g/cm3 ⎜ ⎜ 1000 kg/m3 ⎟⎟ V 4.19 × 1012 m3 ⎝ ⎠

Reflect: For a fixed mass, the density scales as 1/r 3 . Thus, the answer to (c) can also be obtained from (b) as 3

⎛ 7.50 × 106 m ⎞ (1.1 × 10 g/cm ) ⎜ = 4.7 × 1014 g/cm3 . ⎜ 1.0 × 104 m ⎟⎟ ⎝ ⎠ 6

3

1.20. Set Up: To calculate the densities, we need to find the spherical volume, V = 43 π r 3 , and the mass of the atom or nucleus as the sum of the masses of its constituent particles. For the atom, m = 2( mp + mn + me ) while for the nucleus m = 2( mp + mn ). We thus need mass data from Appendix F: 1.675 × 10

−27

kg; and me = 9.109 × 10

−31

mp = 1.673 × 10−27 kg;

3

mn =

3

kg. The unit conversion factor 1 g/cm = 1000 kg/m is also needed.

Solve: (a) Given r = 0.050 nm = 0.050 × 10−9 m, V = 43 π r 3 = 43 π (0.050 × 10−9 m)3 = 5.24 × 10−31 m3 .

m = 2(mp + mn + me ) = 6.70 × 10−27 kg density =

m 6.70 × 10−27 kg = = 1.3 × 104 kg/m3 = 13 g/cm3 V 5.24 × 10−31 m3


1-6

Chapter 1

The density of the helium atom is 13 times larger than the density of pure water. (b) Given r = 1.0 fm = 1.0 × 10−15 m, V = 43 π r 3 = 43 π (1.0 × 10−15 m)3 = 4.19 × 10−45 m3 . m = 2(mp + mn ) = 6.70 × 10−27 kg

density =

m 6.70 × 10−27 kg = = 1.6 × 1018 kg/m3 = 1.6 × 1015 g/cm3 V 4.19 × 10−45 m3

In Problem 1.19 we found the density of a neutron star to be 4.7 × 1014 g/cm3 . By comparison, the density of the helium nucleus is 3 times larger than the density of a neutron star.

1.21. Set Up: We know the density and mass; thus we can find the volume using the relation density = mass/volume = m/V and the given data, density = 19.5 g/cm3 and mcritical = 60.0 kg. The radius is then found from the volume

equation for a sphere, V = 43 π r 3 , and the result for volume. ⎛ 60.0 kg ⎞ ⎛ 1000 g ⎞ 3 Solve: V = mcritical /density = ⎜⎜ 3⎟ ⎟ ⎜ 1.0 kg ⎟ = 3080 cm 19 5 g/cm . ⎠ ⎝ ⎠⎝ r=

3

3V = 4π

3

3 (3080 cm3 ) = 9.0 cm 4π

1.22. Set Up: From Appendix A, a thin spherical shell has volume V = At , where A = 4π r 2 is the surface area of the shell and t is its thickness. We are given r = 1.0 μ m = 1.0 × 10−6 m and t = 50.0 nm = 50.0 × 10−9 m, and we know the unit conversions 1 m3 = 106 cm3 and 1 mg = 10−3 g. Solve: V = 4π r 2t = 4π (1.0 × 10−6 m) 2 (50.0 × 10−9 m) = 6.28 × 10−19 m3 = 6.28 × 10−13 cm3

mass = (density)(volume) = (1.0 g/cm3 )(6.28 × 10−13 cm3 ) = 6.28 × 10−13 g = 6.28 × 10−10 mg

*1.23. Set Up: The mass can be calculated as the product of the density and volume. The volume of the washer is the volume Vd of a solid disk of radius rd minus the volume Vh of the disk-shaped hole of radius rh . In general, the volume of a disk of radius r and thickness t is π r 2t . We also need to apply the unit conversions 1 m3 = 106 cm3 and 1 g/cm3 = 103 kg/m3 . Solve: The volume of the washer is:

V = Vd − Vh = π ( rd 2− rh 2)t = π [(2.25 cm)2 − (0.625 cm) 2 ](0.150 cm) = 2.20 cm3

The density of the washer material is 8600 kg/m3[(1 g/cm3 )/(103 kg/m3 )] = 8.60 g/cm3 . Finally, the mass of the washer is: mass = (density)(volume) = (8.60 g/cm3 )(2.20 cm3 ) = 18.9 g. Reflect: This mass corresponds to a weight of about 0.7 oz, a reasonable value for a washer.

1.24. Set Up: The world population when this solution is being written is about 6.4 × 109 people. Estimate an average mass of a person to be about 50 kg, which corresponds to a weight of about 110 lbs. Solve: The total mass of all the people is about (50 kg/person)(6.4 × 109 persons) = 3 × 1011 kg

1.25. Set Up: Estimate the number of pages and the number of words per page: There are approximately a thousand pages, and each page has between 500 and 1000 words (counting captions and the smaller print, such as the end-of-chapter exercises and problems). Solve: An estimate for the number of words is about 106. Reflect: We can expect that this estimate is accurate to within a factor of 10. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1-7

1.26. Set Up: Assume that the density of a cell is 1000 kg/m3 , the same as for water. Also assume a mass of 70 kg (a weight of about 150 lbs) for a typical person. The volume of a sphere is V = 43 π r 3 . Solve: The volume is: V = 43 π r 3 = 43 π (1.0 × 10−6 m)3 = 4.2 × 10−18 m3 . The mass of a cell is: mass = (density)(volume) = (1000 kg/m3 )(4.2 × 10−18 m3 ) = 4 × 10−15 kg

The number of cells in the typical person is then (70 kg)/(4 × 10215 kg/cell) = 2 × 1016 cells.

*1.27. Set Up: Estimate that we blink 10 times per minute and a typical lifetime is 80 years. Convert this into blinks per lifetime using 1 y = 365 days, 1 day = 24 h, 1 h = 60 min.

⎛ 60 min ⎞ ⎛ 24 h ⎞⎛ 365 days ⎞ 8 Solve: The number of blinks is (10 per min) ⎜ ⎟⎜ ⎟ (80 y/lifetime) = 4 × 10 ⎟⎜ 1 h 1 day 1 y ⎝ ⎠⎝ ⎠⎝ ⎠ Reflect: Our estimate of the number of blinks per minute can be off by a factor of two but our calculation is surely accurate to a power of 10.

1.28. Set Up: Estimate the diameter of a drop to be d = 2 mm. The volume of a spherical drop is V = 43 πr 3 = 16 πd 3 and 103 cm3 = 1 L . Solve: V = 16 π (0.2 cm)3 = 4 × 10−3 cm3. The number of drops in 1.0 L is

1000 cm3 4 × 10−3 cm3

= 2 × 105

Reflect: Since V ∝ d 3 , if our estimate of the diameter of a drop is off by a factor of 2 then our estimate of the number of drops is off by a factor of 8.

1.29. Set Up: Estimate the thickness of a dollar bill by measuring a short stack, say ten, and dividing the measurement by the total number of bills. I obtain a thickness of roughly 1 mm. From Appendix F, the distance from the earth to the moon is 3.8 × 108 m. The number of bills is simply this distance divided by the thickness of one bill. ⎛ 3.8 × 108 m ⎞⎛ 103 mm ⎞ = 3.8 × 1012 bills ≈ 4 × 1012 bills Solve: N bills = ⎜ ⎜ 0.1 mm/bill ⎟⎜ ⎟⎜ 1 m ⎟⎟ ⎝ ⎠⎝ ⎠ Reflect: This answer represents 4 trillion dollars! The cost of a single space shuttle mission in 2005 was significantly less—roughly 1 billion dollars.

1.30. Set Up: For part (a), estimate that a person takes 12 breaths per minute. In part (b), use the relation for the volume of a sphere, V = 43 π r 3 , to calculate the radius required as r = [3V /(4π )] 1/ 3 . Solve: (a) The estimated number of breaths in two weeks equals: ⎛ 60 min ⎞ ⎛ 24 h ⎞ ⎛ 7 days ⎞ 5 (12 breaths/min) ⎜ ⎟⎜ ⎟⎜ ⎟ (2 weeks) = 2.4 × 10 breaths 1 h 1 day 1 week ⎝ ⎠⎝ ⎠ ⎠⎝

The total amount of air breathed by one person in two weeks is: V=

⎛

23

( 12 L/breath ) (2.4 ×105 breaths) ⎜⎜ 101 Lm ⎝

3

⎞ 2 3 ⎟⎟ = 1 × 10 m ⎠

(b) For a spherical vehicle, the radius required is: 1/3

1/3

⎡ 3(1 × 102 m3 ) ⎤ =⎢ ⎥ = 3 m. 4π ⎢⎣ ⎥⎦ The diameter of the space station needs to be about 6 m to contain the air for one person for two weeks. ⎛ 3V ⎞ r=⎜ ⎟ ⎝ 4π ⎠

*1.31. Set Up: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years. The volume of blood pumped during this interval is then the volume per beat multiplied by the total beats.


1-8

Chapter 1

⎛ 60 min ⎞ ⎛ 24 h ⎞⎛ 365 days ⎞⎛ 80 yr ⎞ 9 Solve: N beats = (75 beats/min) ⎜ ⎟⎜ ⎟⎜ ⎟ = 3 × 10 beats/lifespan ⎟⎜ yr ⎝ 1 h ⎠ ⎝ 1 day ⎠⎝ ⎠⎝ lifespan ⎠ 9 ⎛ 1L ⎞⎛ 1 gal ⎞ ⎛ 3 × 10 beats ⎞ 7 Vblood = (50 cm3 /beat) ⎜ ⎜ ⎟ = 4 × 10 gal/lifespan ⎟⎜ ⎟ ⎝ 1000 cm3 ⎠⎝ 3.788 L ⎠ ⎜⎝ lifespan ⎟⎠

1.32. Set Up: Estimate that one step is

1 3

m. Estimate that you walk 100 m in 1 minute. The distance to the Moon

is 3.8 × 108 m. Solve: The time it takes is

⎛ 1 minute ⎞ (3.8 × 108 m) = 3.8 × 106 minutes, t =⎜ ⎝ 100 m ⎟⎠

which is about 7 years. The number of steps would be: 3.8 × 108 m = 1 × 109 steps. 1 m/step 3

*1.33. Set Up: The cost would equal the number of dollar bills required; the surface area of the U.S. divided by the surface area of a single dollar bill. By drawing a rectangle on a map of the U.S., the approximate area is 2600 mi by 1300 mi or 3,380,000 mi 2 . This estimate is within 10 percent of the actual area, 3,794,083 mi 2 . The population is roughly 3.0 × 108 while the area of a dollar bill, as measured with a ruler, is approximately 6 18 in. by 2 85 in. Solve:

AU.S. = (3,380,000 mi 2 )[(5280 ft)/(1 mi)]2 [(12 in.)/(1 ft)]2 = 1.4 × 1016 in.2 Abill = (6.125 in.)(2.625 in.) = 16.1 in.2 Total cost = N bills = AU.S. /Abill = (1.4 × 1016 in.2 )/(16.1 in.2 /bill) = 9 × 1014 bills

Cost per person = (9 × 1014 dollars)/(3.0 × 108 persons) = 3 × 106 dollars/person Reflect: The actual cost would be somewhat larger, because the land isn’t flat.

1.34. Set Up: Convert the radian values to degrees as: π D

π

rad = 60 ; and rad = 30D. 3 6 Solve: The vectors are shown in the figure below.

*1.35. Set Up: The sum with the largest magnitude is when the two displacements are parallel and the sum with the smallest magnitude is when the two displacements are antiparallel. Solve: The orientations of the displacements that give the desired sum are shown in the figure below.



1-9

Reflect: The orientations of the two displacements can be chosen such that the sum has any value between 0.6 m and 4.2 m.

1.36. Set Up: The displacement vector is directed from the initial position of the object to the final position. Solve: (a) The initial and final positions of the bug are shown in Figure (a) below. The curved path of the bug is shown in Figure (b) and the bug’s displacement vector is shown in Figure (c).

(b) The magnitude of the displacement vector is the length of the line that connects points 1 and 2 and is equal to the radius of the turntable, 6 inches. The direction of the displacement vector is shown in Figure (c) above, with φ = 45D.

1.37. Set Up: Draw the vectors to scale on graph paper, using the tip to tail addition method. For part (a), simply JG JG JG JG JG draw B so that its tail lies at the tip of A. Then draw the vector R from the tail of A to the tip of B. For (b), add JG JG JG JG JG JG JG JG − B to A by drawing − B in the opposite direction to B. For (c), add − A to − B. For (d), add − A to B. Solve: The vector sums and differences are shown in the figures below.


1-10

Chapter 1

JG JG JG JG JG JG JG JG Reflect: − A − B = −( A + B ) so it has the same magnitude and opposite direction as A + B. Similarly, B − A = JG JG JG JG JG JG −( A − B ) and B − A and A − B have equal magnitudes and opposite directions. 1.38. Set Up: The two forces applied by the cables add to give a resultant force that is vertical and has magnitude 2.25 N. Draw the vector addition diagram with the two forces head to tail. Solve: (a) The vector addition diagram is drawn in the figure below.

(b) Careful measurement of the length of F1 and F2 gives F1 = F2 = 1.24 N.

*1.39. Set Up: Use components to add the two forces. Take the + x -direction to be forward and the + y -direction to be upward. Solve: The second force has components F2 x = F2 cos32.4° = 433 N and F2 y = F2 sin 32.4° = 275 N. The first force has components F1x = 725 N and F1 y = 0.

Fx = F1x + F2 x = 1158 N and Fy = F1 y + F2 y = 275 N The resultant force is 1190 N in the direction 13.4° above the forward direction. Reflect: Since the two forces are not in the same direction the magnitude of their vector sum is less than the sum of their magnitudes.

1.40. Set Up: In each case the direction of the vector is specified by the angle θ measured counterclockwise from the +x axis. The components are calculated as: Ax = A cos θ ; Ay = A sin θ . Solve: (a) Ax = (50.0 N)(cos 60D ) = + 25.0 N; Ay = (50.0 N)(sin 60D ) = + 43.3 N. The sketch of the vector and its

components in Figure (a) below shows that both components are positive and that q Ay q > q Axq.



1-11

(b) θ = (5π /6 rad)(180D /π rad) = 150D ; Ax = (75 m/s)(cos 150D ) = −65.0 m/s; Ay = (75 m/s)(sin 150D ) = +37.5 m/s.

Figure (b) above shows that Ax is negative, Ay is positive, and q Ax q > q Ay q. (c) Ax = (254 lb)(cos 325D ) = 208 lb; Ay = (254 lb)(sin 325D ) = −146 lb. Figure (c) above shows that Ax is positive,

Ay is negative, and q Ax q > q Ay q. (d) θ = (1.1π rad)(180D /π rad) = 198D ; Ax = (69 km)(cos 198D ) = − 66 km; Ay = (69 km)(sin 198D ) = −21 km. Figure

(d) above shows that both Ax and Ay are negative and q Ax q > q Ay q.

*1.41. Set Up: In each case, create a sketch (see the figure below) showing the components and the resultant to JG determine the quadrant in which the resultant vector A lies. The component vectors add to give the resultant.


1-12

Chapter 1

Solve: (a) A = Ax 2 + Ay 2 = (4.0 m) 2 + (5.0 m) 2 = 6.4 m; tanθ = (b) A = (−3.0 km) 2 + (−6.0 km) 2 = 6.7 km; tan θ =

Ay Ax

=

5.0 m and θ = 51D. 4.0 m

−6.0 m and θ = 243D. −3.0 m

(My calculator gives φ = 63D and θ = φ + 180D.) (c) A = (9.0 m/s) 2 + ( −17 m/s) 2 = 19 m/s; tan θ =

−17 m/s and θ = 298D. 9 m/s

(My calculator gives φ = −62D and θ = φ + 360D.) (d) A = (−8.0 N) 2 + (12 N) 2 = 14 N. tanθ =

12 N and θ = 124D. −8.0 N

(My calculator gives φ = 256D and θ = φ + 180D.) Reflect: The signs of the components determine the quadrant in which the resultant lies.

JJG

1.42. Set Up: Use coordinates for which the +x axis is horizontal and the +y direction is upward. The force F and its x and y components are shown in the figure below.

Solve: (a) Fx = F (cos 37D ) = (20.0 N)(cos 37D ) = 16.0 N (b) Fy = F sin 37D = (20.0 N)(sin 37D ) = 12.0 N

*1.43. Set Up: In each case, use a sketch (see the figure below) showing the components and the resultant in order JG to determine the quadrant in which the resultant vector A lies. The component vectors add to give the resultant.



Solve: (a) A = Ax 2 + Ay 2 = (8.0 lb) 2 + (6.0 lb) 2 = 10.0 lb; tanθ = (b) A = (−24 m/s) 2 + ( −31 m/s) 2 = 39 m/s; tan θ =

Ay Ax

=

1-13

6.0 lb and θ = 37D. 8.0 lb

−31 m/s and θ = 232D −24 m/s

(My calculator gives φ = 52D and θ = φ + 180D.) (c) A = (−1500 km)2 + (2000 km) 2 = 2500 km; tan θ =

2000 km and θ = 127D −1500 km

(My calculator gives φ = 253D and θ = φ + 180D.) (d) A = (71.3 N)2 + ( −54.7 N) 2 = 89.9 N; tan θ =

−54.7 N and θ = 323D 71.3 N

(My calculator gives φ = −37D and θ = φ + 360D.)

1.44. Set Up: For each vector, use the relations Rx = R cos θ and Ry = R sin θ .

JG Solve: For vector A : Ax = (12.0 m) cos (90D − 37D ) = 7.2 m; Ay = (12.0 m) sin (90D − 37D ) = 9.6 m. JG JG For vector B : Bx = (15.0 m) cos (320D ) = 11.5 m; B y = (15.0 m) sin (320D ) = − 9.6 m. For vector C :

C x = (6.0 m) cos (240D ) = − 3.0 m; C y = (6.0 m) sin (240D ) = −5.2 m.

1.45. Set Up: For parts (a) and (c), apply the appropriate signs to the relations Rx = Ax + Bx and Ry = Ay + By . For (b) and (d), find the magnitude as R = Rx 2 + R y 2 and the direction as θ = tan −1( Ry /Rx ). Solve: (a) Rx = Ax + Bx = 1.30 cm + 4.10 cm = 5.40 cm; Ry = Ay + By = 2.25 cm + (−3.75 cm) = −1.50 cm (b) R = Rx 2 + Ry 2 = (5.40 cm) 2 + ( −1.50 cm) 2 = 5.60 cm; θ = tan −1 [ ( −1.50 cm)/(5.40 cm) ] − 15.5D. The resultant

vector thus makes an angle of 344.5D counterclockwise from the +x axis. (c) Rx = Bx + (− Ax ) = 4.10 cm + (−1.30 cm) = 2.80 cm; R y = By + (− Ay ) = −3.75 cm + ( −2.25 cm) = − 6.00 cm (d) R = (2.80 cm) 2 + (− 6.00 cm)2 = 6.62 cm; θ = tan −1[(−6.00 cm)/(2.80 cm)] = −65.0D. The resultant vector thus

makes an angle of 295.0D counterclockwise from the +x axis. JG JG JG JG Reflect: Note that B − A has a larger magnitude than B + A. Vector addition is very different from addition of scalars.


1-14

Chapter 1

JG

JG

1.46. Set Up: Use coordinates where +x is east and +y is north. A and B are the two given displacements (see the JG Figure (a) below). Displacement B makes an angle of θ = 312D counterclockwise from the +x axis. Calculate JG JG JG R = A + B.

Solve: (a) Ax = A cos 22D = (85 mi)(cos 22D ) = 78.8 mi;

Ay = A sin 22D = (85 mi)(sin 22D ) = 31.8 mi. Bx = B cos 312D = (115 mi)(cos 312D ) = 77.0 mi;

B y = B sin 312D = (115 mi)(sin 312D ) = −85.5 mi Rx = Ax + Bx = 78.8 mi + 77.0 mi = 155.8 mi;

Ry = Ay + By = 31.8 mi + (−85.5 mi) = −53.7 mi R = Rx 2+ R y 2 = (155.8 mi) 2 + ( −53.7 mi) 2 = 165 mi; tanθ =

−53.7 mi and θ = 341D ; 155.8 mi

or φ = 19D south of east, as shown in Figure (b) above. JG JG JG JG (b) The vector addition diagram for R = A + B is given in Figure (c) above. The magnitude and direction of R in this diagram agrees with our calculation using components.



JG

1-15

JG

*1.47. Set Up: We know that the two force vectors, A and B, have the same magnitude ( A = B ) and form a right angle; thus, the two forces and their resultant form an isosceles right triangle. Use coordinates for which the resultant force is parallel to the +y-direction and find the magnitude of A and B by setting the y-component of the resultant equal to 620 N. Solve: (a) The two forces and their resultant are shown below.

(b) R y = Ay + B y = A sin 45D + B sin135D = 2 A sin 45D = 620 N. Thus, A =

620 N 2sin 45D

= 440 N.

Reflect: Note that Rx = Ax + Bx = A cos 45D + B cos135D = 0.

1.48. Set Up: The counterclockwise angles each vector makes with the +x axis are: θ A = 30D , θ B = 120D , and θC = 233D. The components of each vector are shown in Figure (a) below.


1-16

Chapter 1

Solve: (a) Ax = A cos 30D = 87 N; Ay = A sin 30D = 50 N; Bx = B cos 120D = −40 N; B y = B sin 120D = 69 N; C x =

C cos 233D = −24 N; C y = C sin 233D = −32 N. JG JG JG JG (b) R = A + B + C is the resultant pull. Rx = Ax + Bx + C x = 87 N + ( −40 N) + ( −24 N) = + 23 N R y = Ay + B y + C y = 50 N + 69 N + ( − 32 N) = + 87 N

JG (c) Rx , R y , and R are shown in Figure (b) above. R = Rx 2 + R y 2 = 90 N and tanθ =

87 N so θ = 75D 23 N

JG (d) The vector addition diagram is given in Figure (c) above. Careful measurement gives an R value that agrees with our results using components.

*1.49. Set Up: Use coordinates for which + x is east and + y is north. Each of the professor’s displacement vectors make an angle of 0º or 180º with one of these axes. The components of his total displacement can thus be calculated directly from Rx = Ax + Bx + C x and R y = Ay + B y + C y . Solve: (a) Rx = Ax + Bx + C x = 0 + (− 4.75 km) + 0 = 24.75 km = 4.75 km west; R y = Ay + B y + C y = 3.25 km + 0

+ ( −1.50 km) = 1.75 km = 1.75 km north; R = Rx 2 + R y 2 = 5.06 km; θ = tan −1 ( Ry /Rx ) = tan21[(+1.75)/( − 4.75)] = − 20.2D ; φ = 180D − 20.2D = 69.8D west of north (b) From the scaled sketch in the figure below, the graphical sum agrees with the calculated values.

Reflect: The magnitude of his resultant displacement is very different from the distance he traveled, which is 159.50 km.

1.50. Set Up: Use coordinates for which + x is east and + y is north. The driver’s vector displacements are: JG JG JG A = 2.6 km, 0D of north; B = 4.0 km, 0D of east; C = 3.1 km, 45D north of east.



1-17

Solve: Rx = Ax + Bx + C x = 0 + 4.0 km + (3.1 km) cos (45D ) = 6.2 km;

R y = Ay + B y + C y = 2.6 km + 0 + (3.1 km)(sin 45D ) = 4.8 km; R = Rx 2 + Ry 2 = 7.8 km; JG θ = tan −1[(4.8 km) / (6.2 km)] = 38D ; R = 7.8 km, 38º north of east. This result is confirmed by the figure below.

1.51. Set Up: We know that an object of mass 1000 g weighs 2.205 lbs and 16 oz = 1 lb. ⎛ 1 lb ⎞⎛ 1000 g ⎞ ⎛ 1 lb ⎞⎛ 1000 g ⎞ Solve: (5.00 oz) ⎜ ⎟⎜ ⎟ = 142 g and (5.25 oz) ⎜ ⎟⎜ ⎟ = 149 g ⎝ 16 oz ⎠⎝ 2.205 lb ⎠ ⎝ 16 oz ⎠⎝ 2.205 lb ⎠ The acceptable limits are 142 g to 149 g. Reflect: The range in acceptable weight is 14 oz. Since 14 oz = 7 g, the range in mass of 7 g is consistent. As we

will see in a later chapter, mass has units of kg or g and is a different physical quantity than weight, which can have units of ounces. But for objects close to the surface of the earth, mass and weight are proportional; we can therefore say that a certain weight is equivalent to a certain mass.

1.52. Set Up: The cookies are roughly cylindrical in shape: the volume of a cylinder with a radius r (or diameter d ) 1 and a height h is V = π r 2h = π d 2h . 4 1 Solve: (a) V = π (8.50 cm) 2 (0.050 cm) = 2.8 cm3 4 (b)

d 8.50 cm = = 1.7 × 102 h 0.050 cm

Reflect: The accuracy of our answer is limited by the thickness of the cookie, which is only known to two significant figures. Also, the answer to (b) is dimensionless. Why is it better to write the answer to (b) as 1.7 × 102 rather than 170?

*1.53. Set Up: Estimate 12 breaths per minute. We know 1 day = 24 h, 1 h = 60 min, and 1000 L = 1 m3 . ⎛ 60 min ⎞ ⎛ 24 h ⎞ Solve: (a) (12 breaths/min) ⎜ ⎟ = 17,280 breaths/day. The volume of air breathed in one day is ⎟⎜ ⎝ 1 h ⎠ ⎝ 1 day ⎠

( 12 L/breath ) (17,280 breaths/day) = 8640 L = 8.64 m3.

The mass of air breathed in one day is the density of air times the volume of air breathed: m = (1.29 kg/m3 )(8.64 m3 ) = 11.1 kg. As 20% of this quantity is oxygen, the mass of oxygen breathed in 1 day is (0.20)(11.1 kg) = 2.2 kg = 2200 g. (b) 8.64 m3 and V = l 3 , so l = V 1/ 3 = 2.1 m.


1-18

Chapter 1

Reflect: A person could not survive one day in a closed tank of this size because the exhaled air is breathed back into the tank and thus reduces the percent of oxygen in the air in the tank. That is, a person cannot extract all of the oxygen from the air in an enclosed space.

1.54. Set Up: To determine the number of breaths in the atmosphere we can divide the mass of the atmosphere by the mass of a single breath. The mass of a single breath is equal to the density of air times the volume of a single breath. −3 3 ⎛ 1 ⎞ ⎛ 10 m ⎞ ⎛ 1.2 kg ⎞ Solve: The mass of a single breath is ⎜ L ⎟ ⎜ = 4 × 10−4 kg . Thus, the approximate number of ⎟⎟ ⎜ 3 ⎟ ⎜ 3 L ⎝ ⎠⎝ ⎠⎝ m ⎠ ⎛ 1000 kg ⎞ ⎛ 1 breath ⎞ 22 breaths in the atmosphere is (5 × 1015 metric tonnes) ⎜ ⎟⎟ = 1 × 10 breaths ⎟ ⎜⎜ 24 metric tonne ⎝ ⎠ ⎝ 4 × 10 kg ⎠ Since this number is roughly the same as the number of atoms in a single breath, there is a good chance that each breath we take contains at least one molecule that was previously inhaled by any given person. Reflect: The answer is only accurate to one significant figure—but that is good enough to show a rather surprising result! *1.55. Set Up: The volume V of blood pumped during each heartbeat is the total volume of blood in the body divided by the number of heartbeats in 1.0 min. We will need to apply 1 L = 1000 cm3 . Solve: The number of heartbeats in 1.0 min is 75. The volume of blood is thus: V=

5. 0 L = 6.7 × 10−2 L = 67 cm3 . 75 heartbeats

1.56. Set Up: The x axis is shown to lie along the axis of the bone. Let F = 2.75 N be the magnitude of the force applied by each tendon. The angles that the tendons make with the x axis are 20.0D , 30.0D , 40.0D , 50.0D , and 60.0D. Solve: (a) Rx = F1x + F2 x + F3 x + F4 x + F5 x Rx = (2.75 N)(cos 20.0D + cos 30.0D + cos 40.0D + cos 50.0D + cos 60.0D ) = 10.2 N R y = F1 y + F2 y + F3 y + F4 y + F5 y D

R y = (2.75 N)(sin 20.0 + sin 30.0D + sin 40.0D + sin 50.0D + sin 60.0D ) = 8.57 N R = Rx 2 + R y 2 = 13.3 N; tan φ =

8.57 N so φ = 40.0D 10.2 N

JG The resultant R and its components are shown in Figure (a) below.



1-19

(b) The vector addition diagram is shown in Figure (b) above. Careful measurements in the diagram give results that agree with what we calculated using components.

1.57. Set Up: Estimate 2 ft per step and 100 steps per minute. Use: 1 mi = 5280 ft. Solve: (a) The distance is (2160 mi)(5280 ft/mi) = 1.14 × 107 ft. The number of steps is calculated as this distance

divided by the distance per step: 1.14 × 107 ft = 5.7 × 106 steps. 2 ft/step (b) The time in minutes is the number of steps required divided by the number of steps per minute 5.7 × 106 steps = 5.7 × 104 min = 950 h. 100 steps/min At 8 h per day it takes about 120 days.

1.58. Set Up: The number of atoms in your body is your mass divided by the mass of one atom. Assume a 70-kg person and that the human body is mostly water. Use the appendix to find the mass of one H2O molecule (by adding the masses for two hydrogen atoms and one oxygen atom): 18.015 u × 1.661 × 10−27 kg/u = 2.992 × 10−26 kg/molecule. Solve: (70 kg)/(2.992 × 10−26 kg/molecule) = 2.34 × 1027 molecules. Each H 2O molecule has 3 atoms, so there are

about 6 × 1027 atoms. Reflect: Assuming carbon to be the most common atom gives 3 × 1027 molecules, which is a result of the same order of magnitude.

*1.59. Set Up: Estimate the volume of each object. The mass, m, of an object is equal to its density times its volume: the volume of a sphere of radius r is V = 43 πr 3 . The volume of a cylinder of radius r and length l is V = πr 2l . The density of water is 1000 kg/m3 . Solve: (a) Estimate the volume as that of a sphere of diameter 10 cm: V = 5.2 × 1024 m3 . m = (0.98)(1000 kg/m3 )(5.2 × 10−4 m3 ) = 0.5 kg .

(b) Approximate as a sphere of radius r = 0.25 μ m (probably an overestimate): V = 6.5 × 10220 m3. So m = (0.98)(1000 kg/m3 )(6.5 × 10−20 m3 ) = 6 × 10−17 kg = 6 × 10−14 g .

(c) Estimate the volume as that of a cylinder of length 1 cm and radius 3 mm: V = πr 2l = 2.8 × 1027 m3. So

m = (0.98)(1000 kg/m3 )(2.8 × 10−7 m3 ) = 3 × 10−4 kg = 0.3 g. Reflect: The mass is directly proportional to the volume.

1.60. Set Up: Assume a weight of 190 lbs, which corresponds to a mass of 86 kg. I estimate the volume V of my body by approximating it as a rectangular solid with dimensions: 67 in.× 12 in.× 8 in. ⎛ 10−3 m3 ⎞ Solve: (a) V = (67 in.)(12 in.)(8 in.) = (6.4 × 103 in.3 ) ⎜ = 0.10 m3 ⎜ 61.02in.3 ⎟⎟ ⎝ ⎠ mass 86 kg = = 860 kg/m3 average density = volume 0.10 m3 (b) According to this estimate, my average density is somewhat less than the density of water. This is sensible; I know I float in water because my body contains air cavities.

*1.61. Set Up: Use coordinates for which + x is east and + y is north. The spelunker’s vector displacements are: JG JG JG JJG A = 180 m, 0º of west; B = 210 m, 45º east of south; C = 280 m, 30º east of north; and the unknown displacement D.

The vector sum of these four displacements is zero.


1-20

Chapter 1

Solve: Ax + Bx + C x + Dx = − 180 m + (210 m)(sin 45D ) + (280 m)(sin 30D ) + Dx = 0 and Dx = − 108 m.

Ay + B y + C y + D y = (−210 m)(cos 45D ) + (280 m)(cos 30D ) + D y = 0 and D y = − 94 m. JJG D = Dx 2 + D y 2 = 143 m; θ = tan −1[(−94 m)/( −108)] = 41D ; D = 143 m, 41º south of west. This result is confirmed by the figure below.

Reflect: We always add vectors by separately adding their x and y components. JG

1.62. Set Up: Use coordinates for which + x is east and + y is north. The vector displacements are: A = 2.00 km,

JG JG 0º of east; B = 3.50 m, 45º south of east; and R = 5.80 m, 0º east

Solve: C x = Rx − Ax − Bx = 5.80 km − (2.00 km) − (3.50 km)(cos 45D ) = 1.33 km; C y = Ry − Ay − By = 0 km − 0 km − (−3.50 km)(sin 45D ) = 2.47 km;

C = (1.33 km)2 + (2.47 km) 2 = 2.81 km; θ = tan −1[(2.47 km)/(1.33 km)] = 61.7D north of east. The vector addition diagram in the figure below shows good qualitative agreement with these values.

1.63. Set Up: Use coordinates having a horizontal + x axis and an upward + y axis. Then Rx = 5.60 N. Solve: Ax + Bx = Rx and A cos 32D + B sin 32D = Rx . Since A = B, 2 A cos 32D = Rx and A = Rx /[(2)(cos 32D )] = 3.30 N. G G G G G G 1.64. Set Up: The four displacements return her to her starting point, so D = −( A + B + C ) , where A, B, and G G C are in the three given displacements and D is the displacement for her return. Let + x be east and + y be north. Solve: (a) Dx = 2[(147 km)sin 85° + (106 km)sin167° + (166 km)sin 235°] = −34.3 km.

D y = 2[(147 km)cos85° + (106 km)cos167° + (166 km)cos 235°] = +185.7 km.

D = (−34.3 km)2 + (185.7 km)2 = 189 km.



1-21

JJG ⎛ 34.3 km ⎞ (b) The direction relative to north is φ = arctan ⎜ ⎟ = 10.5°. Since Dx < 0 and Dy > 0, the direction of D is ⎝ 185.7 km ⎠ 10.5° west of north. Reflect: The four displacements add to zero.

*1.65. Set Up: Referring to the vector diagram in the figure below, the resultant weight of the forearm and the

JG JG weight, R, is 132.5 N upward while the biceps’ force, B, acts at 43º from the + y axis. The force of the elbow is JJG JG JG thus found as E = R − B.

Solve: The components are: E x = Rx − Bx = 0 − ( −232 N)(sin 43D ) = 158 N;

E y = R y − By = 132.5 N − (232 N)(cos 43D ) = − 37 N. The elbow force is thus

E = (158 N) 2 + ( −37 N) 2 = 162 N and acts at θ = tan −1[(−37)/158] = − 13D or 13º below the horizontal. Reflect: The force exerted by the elbow is larger than the total weight of the arm and the object it is carrying.

1.66. Set Up: Approximate the mass of an atom by the mass mp of a proton, where mp = 1.67 × 10−27 kg. The mass of the sun is 1.99 × 1030 kg. Solve: (a) 10100 represents the numeral “1” followed by one hundred zeroes. (b) The number of atoms in the sun is about

1.99 × 1030 kg 1.67 × 10

−27

kg

= 1 × 1057 atoms.

⎛ 1 googol ⎞ −43 1 × 1057 ⎜ ⎟ = 1 × 10 googol ⎝ 10100 ⎠ (c) 1.0 × 101....0 , where 1....0 is the numeral “1” followed by one hundred zeroes. In standard notation, a googolplex is “1” followed by a googol of zeroes.

Solutions to age Problems *1.67. Set Up: The total volume is equal to the number of alveoli times the average volume of a single alveolus. Note that 1 μ m = 1026 m and 1 m3 = 103 L.

⎛ 4.2 × 106 μ m3 ⎞⎛ 10−6 m ⎞ Solve: (480 × 10 alveoli) ⎜ ⎜ alveolus ⎟⎜ ⎟⎜ 1 μ m ⎟⎟ ⎝ ⎠⎝ ⎠ 6

3

⎛ 103 L ⎞ = 2.0 L. The correct answer is C. ⎜⎜ 3 ⎟ ⎟ ⎝ 1m ⎠


1-22

Chapter 1

4 3 Solve: Solve for r to obtain the diameter of a typical alveolus:

1.68. Set Up: The volume of a sphere is V = π r 3. Note that 1 μ m = 10−6 m = 10−3 mm. 1/ 3

⎛ 3V ⎞ d = 2r = 2 ⎜ ⎟ ⎝ 4π ⎠

1/ 3

⎛ 3(4.2 × 106 μ m3 ) ⎞ = 2⎜ ⎟⎟ ⎜ 4π ⎝ ⎠

⎛ 1023 mm ⎞ ⎜⎜ ⎟⎟ = 0.20 mm. ⎝ 1 μm ⎠

The correct answer is A.

*1.69. Set Up: The graph shows a nearly horizontal line, with no systematic upward or downward trend. Thus, the average volume of the alveoli is not dependent on the total lung volume. Solve: The total lung volume must be equal to the average volume of the alveoli times the number of alveoli; thus, as the total volume of the lungs increases, the volume of the individual alveoli remains constant and the number of alveoli increases. The correct answer is C.


2

MOTION ALONG A STRAIGHT LINE

Answers to Multiple-Choice Problems 1. C, D 2. C 3. C, D 13. B, D 14. B 15. C

4. A, B

5. D

6. C

7. A, D

8. A

9. D

10. C

11. A

12. A, C, D

Solutions to Problems *2.1. Set Up: Let the + x direction be to the right in the figure. Solve: (a) The lengths of the segments determine the distance of each point from O: x A = − 5 cm, xB = + 45 cm, xC = + 15 cm, and xD = − 5 cm. (b) The displacement is Δx; the sign of Δx indicates its direction. The distance is always positive. (i) A to B: Δx = xB − x A = + 45 cm − ( −5 cm) = + 50 cm. Distance is 50 cm.

(ii) B to C: Δx = xC − xB = + 15 cm − 45 cm = − 30 cm. Distance is 30 cm. (iii) C to D: Δx = xD − xC = − 5 cm − 15 cm = − 20 cm. Distance is 20 cm. (iv) A to D: Δx = xD − x A = 0. Distance = 2( AB ) = 100 cm. Reflect: When the motion is always in the same direction during the interval the magnitude of the displacement and the distance traveled are the same. In (iv) the ant travels to the right and then to the left and the magnitude of the displacement is less than the distance traveled.

2.2. Set Up: From the graph the position xt at each time t is: x1 = 1.0 m, x2 = 0, x3 = − 1.0 m, x4 = 0, x8 = 6.0 m, and x10 = 6.0 m. Solve: (a) The displacement is Δx. (i) Δx = x10 − x1 = +5.0 m; (ii) Δx = x10 − x3 = + 7.0 m; (iii) Δx = x3 − x2 = − 1.0 m;

(iv) Δx = x4 − x2 = 0. (b) (i) 3.0 m + 1.0 m = 4.0 m; 90D (ii) 1.0 m + 1.0 m = 2.0 m; (iii) zero (stays at x = 6.0 m )

2.3. Set Up: Let the + x direction be to the right. x A = 2.0 m, xB = 7.0 m, xC = 6.0 m. Solve: Average velocity is

Δx xC − x A +6.0 m − 2.0 m = = = 1.3 m/s. Δt Δt 3.0 s distance 4.0 m + 1.0 m + 1.0 m Average speed = = = 2.0 m/s time 3.0 s Reflect: The average speed is greater than the magnitude of the average velocity. Δx 2.4. Set Up: The average velocity is vav-x = , 13.5 days = 1.166 × 106 s, and at the release point x = +5.150 × 106 m. Δt

υav-x =


2-1

2-2

Chapter 2

x2 − x1 0 − 5.150 × 106 m = = −4.42 m/s Δt 1.166 × 106 s (b) For the round-trip, x2 = x1 and Δx = 0. The average velocity is zero. Solve: (a) vav-x =

Reflect: The average velocity for the trip from the nest to the release point is positive.

*2.5. Set Up: x A = 0, xB = 3.0 m, xC = 9.0 m. t A = 0, t B = 1.0 s, tC = 5.0 s. Solve: (a) υav-x =

Δx Δt A to B: υav-x =

Δx xB − x A 3.0 m = = = 3.0 m/s Δt tB − t A 1.0 s

B to C: υav-x =

xC − xB 6.0 m = = 1.5 m/s tC − t B 4.0 s

xC − x A 9.0 m = = 1.8 m/s tC − t A 5.0 s (b) The velocity is always in the same direction (+x-direction), so the distance traveled is equal to the displacement in each case, and the average speed is the same as the magnitude of the average velocity. Reflect: The average speed is different for different time intervals. A to C: υav-x =

2.6. Set Up: t A = 0, t B = 3.0 s, tC = 6.0 s. x A = 0, xB = 25.0 m, xC = 0. Solve: (a) υav { x =

Δx Δt A to B: υav-x =

Δx xB − x A 25.0 m = = = 8.3 m/s Δt tB − t A 3.0 s

B to C: υav-x =

xC − xB 225.0 m = = − 8.3 m/s tC − t B 3.0 s

xC − x A =0 tC − t A (b) For A to B and for B to C the distance traveled equals the magnitude of the displacement and the average speed equals the magnitude of the average velocity. For A to C the displacement is zero. Thus, the average velocity is zero but the distance traveled is not zero so the average speed is not zero. For the motion A to B and for B to C the velocity is always in the same direction but during A to C the motion changes direction. A to C: υav-x =

*2.7. Set Up: The positions xt at time t are: x0 = 0, x1 = 1.0 m, x2 = 4.0 m, x3 = 9.0 m, x4 = 16.0 m. Solve: (a) The distance is x3 − x1 = 8.0 m.

x −x x −x x −x Δx . (i) υav, x = 1 0 = 1.0 m/s; (ii) υav, x = 2 1 = 3.0 m/s; (iii) υav, x = 3 2 = 5.0 m/s; 1.0 s 1.0 s 1.0 s Δt x4 − x3 x4 − x0 (iv) υav, x = = 7.0 m/s; (v) υav, x = = 4.0 m/s 1.0 s 4.0 s Reflect: In successive 1 s time intervals the boulder travels greater distances and the average velocity for the intervals increases from one interval to the next. (b) υav { x =

2.8. Set Up: υ x (t ) is the slope of the x versus t graph. In each case this slope is constant, so υ x is constant. Solve: The graphs of υ x versus t are sketched in the figure below.


Motion Along a Straight Line

2.9. Set Up: Let + x be the direction the runner travels. υav, x = Solve: (a) υav, x =

2-3

Δx . 1 mi/h = 1.466 ft/s = 0.4470 m/s Δt

1.00 mi = 15.0 mi/h (4.00 min)(1 h/60 min)

⎛ 1.466 ft/s ⎞ (b) (15.0 mi/h) ⎜ ⎟ = 22.0 ft/s ⎝ 1 mi/h ⎠ ⎛ 0.4470 m/s ⎞ (c) (15.0 mi/h) ⎜ ⎟ = 6.70 m/s ⎝ 1 mi/h ⎠

2.10. Set Up: Assume constant speed v, so d = vt. Solve: (a) t =

d 5.0 × 106 m ⎛ 1 min ⎞ = = (2158 s) ⎜ ⎟ = 36 min t 7(331 m/s) ⎝ 60 s ⎠

(b) d = υ t = 7(331 m/s)(11 s) = 2.5 × 104 m = 25 km

*2.11. Set Up: 1.0 century = 100 yr. 1 km = 105 cm. Solve: (a) d = υ t = (5.0 cm/yr)(100 yr) = 500 cm = 5.0 m (b) t =

d 550 × 105 cm = = 1.1 × 107 yr t 5.0 cm/yr

2.12. Set Up: The distance around the circular track is d = π (40.0 m) = 126 m. For a half-lap, d = 63 m. Use coordinates for which the origin is at her starting point and the x-axis is along a diameter, as shown in the figure below.

Solve: (a) After one lap she has returned to her starting point. Thus, Δx = 0 and υav, x = 0.

d 126 m = = 2.01 m/s t 62.5 s Δx 40.0 m d 63 m = 2.20 m/s = = = 1.39 m/s; average speed = = 28.7 s Δt t 28.7 s average speed =

(b) Δx = 40.0 m and υav, x

2.13. Set Up: Since sound travels at a constant speed, Δx = vx Δt ; also, from the appendix we find that 1 mile is 1.609 km. 1 mi ⎛ ⎞ Solve: Δx = (344 m/s)(7.5 s) ⎜ ⎟ = 1.6 mi ⎝ 1.609 × 103 m ⎠ 1 mi ⎛ ⎞ 1 Reflect: The speed of sound is (344 m/s) ⎜ ⎟ ≈ mi/s 3 ⎝ 1.609 × 10 m ⎠ 5 1.85 m d 1.85 m = 3.03 s 2.14. Solve: (a) t = . touch: t = = 0.0243 s; pain: t = 76.2 m/s 0.610 m/s υ (b) The difference between the two times in (a) is 3.01 s.


2-4

Chapter 2

*2.15. Set Up: The velocity of the truck relative to the road is equal to the velocity of the truck relative to me plus my velocity relative to the road: vT/R = vT/M + vM/R . Assume that all numbers are good to two significant figures. Solve: (a) As seen by me, the truck moves backwards 25 meters during 5.5 seconds. Thus, the truck’s velocity relative to me −25 m is vT/M = = − 4.55 m/s. My velocity relative to the road is vM/R = (110 km/h)(1000 m/km)(1 h/3600 s) = 30.6 m/s. 5.5 s Thus, the velocity of the truck relative to the road is vT/R = vT/M + vM/R = 24.55 m/s + 30.6 m/s = 26 m/s. (b) The distance the truck moves (relative to the road) is Δx = (26 m/s)(5.5 s) = 143 m = 1.4 × 102 m. Reflect: Note that the distance that I move is (30.6 m/s)(5.5 s) = 168 m, which is 25 meters further than the distance

that the truck travels.

2.16. Set Up: Since we know the position of the mouse as a function of time, we can compute its average velocity Δx . Δt Solve: Calculate the position of the mouse at t = 0 s; t = 1.0 s; and t = 4.0 s: x(0 s) = 0

from υav, x =

x(1.0 s) = (8.5 cm ⋅ s −1)(1.0 s) − (2.5 cm ⋅ s −2 )(1.0 s) 2 = 6.0 cm x(4.0 s) = (8.5 cm ⋅ s −1 )(4.0 s) − (2.5 cm ⋅ s −2 )(4.0 s) 2 = −6.0 cm The average velocities of the mouse from 0 to 1 second and from 0 to 4 seconds are (respectively) Δx 6.0 cm − 0 Δx −6.0 cm − 0 = = 6.0 cm/s; υav, x = = = −1.5 cm/s. υav, x = Δt 1.0 s Δt 4.0 s Reflect: Since the average velocity of the mouse changes sign, the mouse must have turned around. The x versus t graph for the mouse, which is an inverted parabola, also shows that the mouse reverses direction.

*2.17. Set Up: Use the normal driving time to find the distance. Use this distance to find the time on Friday. Solve: Δx = υav, x Δt = (105 km/h)(1.33 h) = 140 km. Then on Friday Δt =

Δx

υav, x

=

140 km = 2.00 h. The increase 70 km/h

in time is 2.00 h − 1.33 h = 0.67 h = 40 min. Reflect: A smaller average speed corresponds to a longer travel time when the distance is the same.

2.18. Set Up: Let d be the distance A runs in time t. Then B runs a distance 200.0 m − d in the same time t. Solve: d = υ At and 200.0 m − d = υ Bt . Combine these two equations to eliminate d. 200.0 m − υ A t = υ Bt and t=

200.0 m = 13.3 s. Then d = (8.0 m/s)(13.3 s) = 106 m; they will meet 106 m from where A starts. 8.0 m/s + 7.0 m/s

*2.19. Set Up: The instantaneous velocity is the slope of the tangent to the x versus t graph. Solve: (a) The velocity is zero where the graph is horizontal; point IV. (b) The velocity is constant and positive where the graph is a straight line with positive slope; point I. (c) The velocity is constant and negative where the graph is a straight line with negative slope; point V. (d) The slope is positive and increasing at point II. (e) The slope is positive and decreasing at point III.

2.20. Set Up: The instantaneous velocity at any point is the slope of the x versus t graph at that point. Estimate the slope from the graph. Solve: A: υ x = 6.7 m/s; B: υ x = 6.7 m/s; C: υ x = 0; D: υ x = − 40.0 m/s; E: υ x = − 40.0 m/s; F: υ x = − 40.0 m/s; G: υ x = 0. Reflect: The sign of υ x shows the direction the car is moving. υ x is constant when x versus t is a straight line.

2.21. Set Up: Values of xt at time t can be read from the graph: x0 = 0, x4 = 3.0 cm, x10 = 4.0 cm, and x18 = 4.0 cm. υ x is constant when x versus t is a straight line.



2-5

Solve: The motion consists of constant velocity segments. 3.0 cm − 0 = 0.75 cm/s; 4. 0 s 4.0 cm − 3.0 cm = 0.17 cm/s; t = 10.0 s to 18.0 s: υ x = 0. t = 4.0 s to 10.0 s: υ x = 6.0 s The graph of υ x versus t is shown in the figure below.

t = 0 to 4.0 s: υ x =

Reflect: υ x is the slope of x versus t.

2.22. Set Up: The instantaneous acceleration is the slope of the υ x versus t graph. Solve: t = 3 s: The graph is horizontal, so a x = 0.

44 m/s − 20 m/s = 6.0 m/s 2 ; a x = 6.0 m/s 2 . 4s 0 − 44 m/s t = 11 s: The graph is a straight line with slope = − 11 m/s 2 ; a x = − 11 m/s 2 . 4s t = 7 s: The graph is a straight line with slope

2.23. Set Up: aav, x =

Δυ x Δt

Solve: (a) 0 s to 2 s: aav, x = 0; 2 s to 4 s: aav, x = 1.0 m/s 2 ; 4 s to 6 s: aav, x = 1.5 m/s 2 ; 6 s to 8 s: aav, x = 2.5 m/s 2 ;

8 s to 10 s: aav, x = 2.5 m/s 2 ; 10 s to 12 s: aav, x = 2.5 m/s 2 ; 12 s to 14 s: aav, x = 1.0 m/s 2 ; 14 s to 16 s: aav, x = 0. The acceleration is not constant over the entire 16 s time interval. The acceleration is constant between 6 s and 12 s. (b) The graph of υ x versus t is given in the figure below. t = 9 s: a x = 2.5 m/s 2 ; t = 13 s: a x = 1.0 m/s 2 ; t = 15 s: a x = 0.


2-6

Chapter 2

Reflect: The acceleration is constant when the velocity changes at a constant rate. When the velocity is constant, the acceleration is zero.

2.24. Set Up: 1 ft = 0.3048 m. g = 9.8 m/s 2 . ⎛ 1 ft ⎞ 2 Solve: (a) 5 g = 49 m/s 2 and 5 g = (49 m/s 2 ) ⎜ ⎟ = 160 ft/s ⎝ 0.3048 m ⎠ ⎛ 1 ft ⎞ 2 (b) 60 g = 590 m/s 2 and 60 g = (590 m/s 2 ) ⎜ ⎟ = 1900 ft/s ⎝ 0.3048 m ⎠

⎛ 1g ⎞ (c) (1.67 m/s 2 ) ⎜ ⎟ = 0.17 g ⎝ 9.8 m/s 2 ⎠

(

)

⎛ 1g ⎞ (d) 24.3 m/s 2 ⎜ ⎟ = 2.5 g ⎝ 9.8 m/s 2 ⎠

*2.25. Set Up: The acceleration ax equals the slope of the υ x versus t curve. Solve: The qualitative graphs of acceleration as a function of time are given in the figure below.

The acceleration can be described as follows: (a) positive and constant, (b) positive and increasing, (c) negative and constant, (d) positive and decreasing. Reflect: When y x and a x have the same sign then the speed is increasing. In (c) the velocity and acceleration have opposite signs and the speed is decreasing.

2.26. Set Up: The acceleration ax is the slope of the graph of υ x versus t. Solve: (a) Reading from the graph, at t = 4.0 s, υ x = 2.7 cm/s, to the right and at t = 7.0 s, υ x = 1.3 cm/s, to the left. (b) υ x versus t is a straight line with slope −

8.0 cm/s = − 1.3 cm/s 2 . The acceleration is constant and equal to 6.0 s

1.3 cm/s 2 , to the left. (c) The graph of a x versus t is given in the figure below.

*2.27. Set Up: Assume constant acceleration. υ0 x = 88 ft/s, υ x = 110 ft/s, and t = 3.50 s. Let x0 = 0. υ − υ0 x 110 ft/s − 88 ft/s Solve: (a) υ x = υ0 x + a xt and a x = x = = 6.3 ft/s 2 . 3.50 s

t

(b) x = x0 + υ0 xt +

1 a t2 2 x

= (88 ft/s)(3.50 s) +

1 (6.3 2

ft/s 2 )(3.50 s) 2 = 347 ft



2-7

2.28. Set Up: 1 mph = 0.4470 m/s and1 m = 3.281 ft. Let x0 = 0. υ0 x = 0, t = 2.0 s, and υ x = 45 mph = 20.1 m/s. υ − υ0 x 20.1 m/s − 0 (a) υ x = υ0 x + a xt and a x = x = = 10 m/s 2 2.0 s

t

⎛ 3.281 ft ⎞ 2 ax = (10 m/s 2 ) ⎜ ⎟ = 33 ft/s ⎝ 1m ⎠ (b) x = x0 + υ0 xt + 12 axt 2 = 12 (10 m/s 2 )(2.0 s) 2 = 20 m, or x = 12 (33 ft/s 2 )(2.0 s)2 = 66 ft

*2.29. Set Up: Take the + y direction to be upward. For part (a) we assume that the cat is in freefall with a y = 2g. Since the cat falls a known distance, we can find its final velocity using v y 2 = v0 y 2 + 2a y ( y − y0 ). For parts (b) and (c) we assume that the cat has a constant (but unknown) acceleration due to its interaction with the floor. We may use the equations for constant acceleration. Solve: (a) Solving for v y we obtain v y = ± v0 y 2 + 2a y ( y − y0 ) Here we set Δy = ( −4.0 ft)(1 m/3.28 ft) = −1.22 m, a y = −g, and v y = 0. v y = − v0 y 2 + 2a y ( y − y0 ) = − −2g( − 1.22 m) = − 4.89 m/s = −4.9 m/s. Where we choose the negative root since the cat is falling. The speed of the cat just before impact is the magnitude of its velocity, which is 4.9 m/s. (b) During its impact with the floor, the cat is brought to rest over a distance of 12 cm. Thus, we have 1 v0 y = 24.89 m/s, v y = 0, and Δy = 20.12 m. Solving Δy = (v y + v0 y )t for time we obtain: 2 2Δy 2(−0.12 m) t= = = 0.049 s. (v y + v0 y ) (0 + 24.89 m/s) (c) Solving v y 2 = v0 y 2 + 2a y ( y − y0 ) for a y we obtain a y =

(v y 2 − v0 y 2 ) 2Δy

=

02 − ( − 4.89 m/s) 2 = 99.6 m/s 2 . Since this 2(−0.12 m)

answer is only accurate to two significant figures, we can write it as 1.0 × 102 m/s 2 or approximately 10 g's. Reflect: During freefall the cat has a negative velocity and a negative acceleration—so it is speeding up. In contrast, during impact with the ground the cat has a negative velocity and a positive acceleration—so it is slowing down.

2.30. Set Up: Let + x be in his direction of motion. Assume constant acceleration. (a) υ x = 3(331 m/s) = 993 m/s, υ0 x = 0, and ax = 5 g = 49.0 m/s 2 . (b) t = 5.0 s υ − υ0 x 993 m/s − 0 Solve: (a) υ x = υ0 x + a xt and t = x = = 20.3 s 2 ax

49.0 m/s

Yes, the time required is larger than 5.0 s. (b) υ x = υ0 x + a xt = 0 + (49.0 m/s 2 )(5.0 s) = 245 m/s

2.31. Set Up: Assume the ball starts from rest and moves in the + x -direction. We may use the equations for constant acceleration. Solve: (a) x − x0 = 1.50 m, vx = 45.0 m/s and v0 x = 0. Using vx 2 = v0 x 2 + 2a x ( x − x0 ) gives ax =

vx 2 − v0 x 2 (45.0 m/s) 2 = = 675 m/s 2 . 2( x − x0 ) 2(1.50 m)

2( x − x0 ) 2(1.50 m) ⎛v +v ⎞ (b) Using x − x0 = ⎜ 0 x x ⎟ t gives t = = = 0.0667 s 2 ⎠ 45.0 m/s v0 x + vx ⎝


2-8

Chapter 2

Reflect: We could also use vx = v0 x + a xt to find t =

vx 45.0 m/s = = 0.0667 s, which agrees with our previous a x 675 m/s 2

result. The acceleration of the ball is very large.

2.32. Set Up: Let + x be the direction the jet travels and take x0 = 0. ax = 4 g = 39.2 m/s 2 , υ x = 4(331 m/s) = 1324 m/s, and v0 x = 0. Solve: (a) υ x = υ0 x + a xt and t =

υ x − υ0 x ax

=

1324 m/s − 0 39.2 m/s 2

= 33.8 s

(b) x = x0 + υ0 xt + 12 axt 2 = 12 (39.2 m/s 2 )(33.8 s)2 = 2.24 × 104 m = 22.4 km

2.33. Set Up: Let +x be the direction the person travels. υ x = 0 (stops), t = 36 ms = 3.6 × 10−2 s, ax = −60 g = −588 m/s 2 . a x is negative since it is opposite to the direction of the motion. Solve: υ x = υ0 x + a xt so υ0 x = − a xt . Then x = x0 + υ0 xt + 12 a xt 2 gives x = − 12 a xt 2 .

x = − 12 (−588 m/s 2 )(3.6 × 1022 s)2 = 38 cm Reflect: We could also find the initial speed: υ0 x = − axt = − (−588 m/s 2 )(36 × 1023 s) = 21 m/s = 47 mph

2.34. Set Up: Take the + x direction to be the direction of motion of the boulder. Solve: (a) Use the motion during the first second to find the acceleration. υ0 x = 0, x0 = 0, x = 2.00 m, and t = 1.00 s. 2 x 2(2.00 m) x = x0 + υ0 xt + 12 a xt 2 and a x = 2 = = 4.00 m/s 2 2 t (1.00 s) υ x = υ0 x + axt = (4.00 m/s 2 )(1.00 s) = 4.00 m/s For the second second, υ0 x = 4.00 m/s, a x = 4.00 m/s 2 , and t = 1.00 s.

x = x0 + υ0 xt + 12 axt 2 = (4.00 m/s)(1.00 s) + 12 (4.00 m/s 2 )(1.00 s)2 = 6.00 m We can also solve for the location at t = 2.00 s, starting at t = 0 : x = x0 + υ0 xt + 12 a xt 2 = 12 (4.00 m/s 2 )(2.00 s) 2 = 8.00 m, which agrees with 2.00 m in the first second and 6.00 m in the second second. The boulder speeds up so it travels farther in each successive second. (b) We have already found υ x = 4.00 m/s after the first second. After the second second,

υ x = υ0 x + axt = 4.00 m/s + (4.00 m/s 2 )(1.00 s) = 8.00 m/s

*2.35. Set Up: Let + x be in the direction of motion of the bullet. υ0 x = 0, x0 = 0, υ x = 335 m/s, and x = 0.127 m. Solve: (a) υ x 2 = υ0 x 2 + 2a x ( x − x0 ) and

ax = (b) υ x = υ0 x + a xt so t =

υ x 2 − υ0 x 2 2( x − x0 )

υ x − υ0 x ax

=

=

(335 m/s) 2 − 0 = 4.42 × 105 m/s 2 = 4.51 × 104 g 2(0.127 m)

335 m/s − 0 4.42 × 105 m/s 2

= 0.758 ms

⎛ υ + υx ⎞ Reflect: The acceleration is very large compared to g. In (b) we could also use ( x − x0 ) = ⎜ 0 x ⎟ t to calculate 2 ⎝ ⎠ 2( x − x0 ) 2(0.127 m) = = 0.758 ms t= υx 335 m/s

2.36. Set Up: Take + x in the direction to be in the direction the airplane travels. υ0 x = 0. ( x − x0 ) = 280 m.



2-9

2( x − x0 ) 2(280 m) ⎛ υ + υx ⎞ Solve: ( x − x0 ) = ⎜ 0 x = = 70.0 m/s ⎟ t gives vx = t 8.00 s 2 ⎝ ⎠

*2.37. Set Up: Let + x be the direction the car is moving. We can use the equations for constant acceleration. Solve: (a) From Eq. (2.13), with v0 x = 0, a x =

vx 2 (20 m/s) 2 = = 1.67 m/s 2 . 2( x − x0 ) 2(120 m)

(b) Using Eq. (2.14), t = 2( x − x0 )/vx = 2(120 m)/(20 m/s) = 12 s. (c) (12 s)(20 m/s) = 240 m. Reflect: The average velocity of the car is half the constant speed of the traffic, so the traffic travels twice as far.

2.38. Set Up: 1 mi/h = 1.466 ft/s. The car travels at constant speed during the reaction time. Let + x be the direction the car is traveling, so a x = − 12.0 ft/s 2 after the brakes are applied. ⎛ 1.466 ft/s ⎞ Solve: (a) υ0 x = (15.0 mi/h) ⎜ ⎟ = 22.0 ft/s. During the reaction time the car travels a distance of (22.0 ft/s) ⎝ 1 mi/h ⎠ (0.7 s) = 15.4 ft.

For the motion after the brakes are applied, υ0 x = 22.0 ft/s, a x = − 12.0 ft/s 2 , and υ x = 0. υ x 2 = υ0 x 2 + 2a x ( x − x0 ) gives ( x − x0 ) =

υ x 2 − υ0 x 2

0 − (22.0 ft/s)2

= 20.2 ft. 2(−12.0 ft/s 2 ) The total distance is 15.4 ft + 20.2 ft = 35.6 ft. 2a x

=

⎛ 1.466 ft/s ⎞ (b) υ0 x = (55.0 mi/h) ⎜ ⎟ = 80.6 ft/s. A calculation similar to that of part (a) gives a total stopping distance ⎝ 1 mi/h ⎠ of ( x − x0 ) = 56.4 ft + 270.7 ft = 327 ft.

*2.39. Set Up: 0.250 mi = 1320 ft. 60.0 mph = 88.0 ft/s. Let + x be the direction the car is traveling. Solve: (a) braking: υ0 x = 88.0 ft/s, x − x0 = 146 ft, υ x = 0. υ x 2 = υ0 x 2 + 2a x ( x − x0 ) gives

ax =

υ x 2 − υ0 x 2 2( x − x0 )

=

0 − (88.0 ft/s) 2 = − 26.5 ft/s 2 2(146 ft)

Speeding up: υ0 x = 0, x − x0 = 1320 ft, t = 19.9 s. x − x0 = υ0 xt + 12 axt 2 gives ax =

2( x − x0 )

t2

=

2(1320 ft) (19.9 s) 2

= 6.67 ft/s 2

(b) υ x = υ0 x + a xt = 0 + (6.67 ft/s 2 )(19.9 s) = 133 ft/s = 90.5 mph (c) t =

υ x − υ0 x

0 − 88.0 ft/s

= 3.32 s 226.5 ft/s 2 Reflect: The magnitude of the acceleration while braking is much larger than when speeding up. That is why it takes much longer to go from 0 to 60 mph than to go from 60 mph to 0. ax

=

2.40. Set Up: Let + x be the direction the train is traveling. Find x − x0 for each segment of the motion. Solve: t = 0 to 14.0 s: x − x0 = υ0 xt + 12 axt 2 = 12 (1.60 m/s 2 )(14.0 s) 2 = 157 m. At t = 14.0 s, the speed is

υ x = υ0 x + a x = (1.60 m/s 2 )(14.0 s) = 22.4 m/s. In the next 70.0 s, ax = 0 and x − x0 = υ0 xt = (22.4 m/s)(70.0 s) = 1568 m. For the interval during which the train is slowing down, υ0 x = 22.4 m/s, ax = − 3.50 m/s 2 and υ x = 0. υ x 2 = υ0 x 2 + 2ax ( x − x0 ) gives x − x0 =

υ x 2 − υ0 x 2

=

0 − (22.4 m/s) 2

2(−3.50 m/s 2 ) The total distance traveled is 157 m + 1568 m + 72 m = 1800 m. 2a x

= 72 m.


2-10

Chapter 2

2.41. Set Up: A = π r 2 and C = 2π r , where r is the radius. Solve:

A1

=

r12

A2

r22

2

2

⎛r ⎞ ⎛ 2r ⎞ and A2 = ⎜ 2 ⎟ A1 = ⎜ 1 ⎟ A = 4 A r ⎝ 1⎠ ⎝ r1 ⎠

⎛r ⎞ ⎛ 2r ⎞ C1 C2 = and C2 = ⎜ 2 ⎟ C1 = ⎜ 1 ⎟ C = 2C r1 r2 ⎝ r1 ⎠ ⎝ r1 ⎠

2.42. Set Up: Let L be the length of each side of the cube. The cube has 6 faces of area L2 , so A = 6 L2 . V = L3 . Solve: V1

L13

=

A1

L12 V2

L23

=

A2

L22

2

2

⎛L ⎞ ⎛ 3L ⎞ and A2 = ⎜ 2 ⎟ A1 = ⎜ 1 ⎟ A1 = 9 A1; surface area increases by a factor of 9. L ⎝ 1⎠ ⎝ L1 ⎠ 3

3

⎛L ⎞ ⎛ 3L ⎞ and V2 = ⎜ 2 ⎟ V1 = ⎜ 1 ⎟ V1 = 27V1; volume increases by a factor of 27. ⎝ L1 ⎠ ⎝ L1 ⎠

*2.43. Set Up: The volume of a cylinder of radius R and height H is given by V = π R 2 H . We know the ratio of the heights of the two tanks and their volumes. From this information we can determine the ratio of their radii. Solve: We take the ratio of the volumes of the two tanks:

have used

H large H small

Vlarge Vsmall

2

=

= 1.20. Solving for the ratio of the radii we obtain

2 ⎛ Rlarge ⎞ 218 π Rlarge H large = = 1.20 ⎜ ⎟ , where we 2 150 π Rsmall H small ⎝ Rsmall ⎠

Rlarge Rsmall

⎛ 1 ⎞⎛ 218 ⎞ = ⎜ ⎟⎜ ⎟ = 1.10. Thus, the larger ⎝ 1.20 ⎠⎝ 150 ⎠

radius is 10% larger than the smaller radius. Reflect: All of the ratios used are dimensionless, and independent of the units used for measurement.

2.44. Set Up: Assuming that the point guard and center have the same density, the ratio of their masses will be the ratio of their volumes. The volume of a rectangular solid of length l, height h, and width w is given by V = lhw. If

each linear dimension of the solid is increased by a factor of k, its volume will be increased by a factor of k 3 since (kl )( kh)( kw) = k 3lhw. This result is general and applies to objects of any shape, provided that all of its linear dimensions are increased by the same factor k. Solve: First convert the height of each player to inches: the height of the point guard is 70 inches and the height of ⎛ 86 ⎞ the center is 86 inches. Assume that each dimension is increased by a factor of ⎜ ⎟ . Thus we have ⎝ 70 ⎠ mass of center volume of center ⎛ 86 ⎞ = =⎜ ⎟ 175 lbs volume of point guard ⎝ 70 ⎠ Solving for the mass of the center we obtain

3

3

⎛ 86 ⎞ mass of center = (175 lbs ) ⎜ ⎟ = 325 lbs. ⎝ 70 ⎠ Reflect: If we double all of the dimensions of an object, its volume is increased by a factor of 23 = 8 and its surface

area is increased by a factor of 22 = 4. If the density of the object remains fixed, its mass will also increase by a factor of 8.

*2.45. Set Up: a A = aB , x0 A = x0 B = 0, υ0 x, A = υ0 x, B = 0, and t A = 2t B . Solve: (a) x = x0 + υ0 xt + 12 axt 2 gives x A = 12 a At A2 and xB = 12 aBt B 2 . a A = aB gives 2

⎛t ⎞ xB = ⎜ B ⎟ x A = ⎝ tA ⎠

( 12 )

2

xA

t A2

=

xB

tB 2

and

(250 km) = 62.5 km


Motion Along a Straight Line (b) υ x = υ0 x + a xt gives a A =

υA tA

and aB =

υB tB

. Since a A = aB ,

⎛ tB ⎞ ⎟υ A = ⎝ tA ⎠

υB = ⎜

υA

=

tA

υB tB

2-11

and

( 12 ) (350 m/s) = 175 m/s.

Reflect: vx is proportional to t and for v0 x = 0, x is proportional to t 2 .

2.46. Set Up: a A = 3aB and υ0 A = υ0 B . Let x0 A = x0 B = 0. Since cars stop, υ A = υ B = 0. ⎛a ⎞ Solve: (a) υ x 2 = υ0 x 2 + 2a x ( x − x0 ) gives a A x A = aB xB , and xB = ⎜ A ⎟ x A = 3D ⎝ aB ⎠ ⎛a ⎞ (b) υ x = υ0 x + a xt gives a At A = aBt B , so t A = ⎜ B ⎟ t B = 13 T ⎝ aA ⎠

2.47. Set Up: υ0 A = υ0 B = 0. Let x0 A = x0 B = 0. a A = aB and υ B = 2υ A . Solve: (a) υ x 2 = υ0 x 2 + 2a x ( x − x0 ) gives a x =

υ x 2 − υ0 x 2 2( x − x0 )

and

υ A2 xA

=

υB2 xB

.

2

⎛υ ⎞ xB = ⎜ B ⎟ x A = 4(500 m) = 2000 m ⎝ υA ⎠ ⎛υ ⎞ υ υ υ (b) υ x = υ0 x + axt gives ax = x . A = B and t B = ⎜ B ⎟ t A = 2T . t t A tB ⎝ υA ⎠ Reflect: x is proportional to υ x 2 and t is proportional to υ x .

2.48. Set Up: Let + y be upward. a y = − 9.80 m/s 2 . v y = 0 at the maximum height. Solve: (a) y − y0 = 0.220 m, a y = − 9.80 m/s 2 , υ y = 0. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ0 y = 22a y ( y − y0 ) = 22(− 9.80 m/s 2 )(0.220 m) = 2.08 m/s. (b) When the flea returns to ground, υ y = − υ0 y . υ y = υ0 y + a yt gives

t=

υ y − υ0 y ay

=

22.08 m/s − 2.08 m/s 29.80 m/s 2

= 0.424 s

(c) a = 9.80 m/s 2 , downward, at all points in the motion.

2.49. Set Up: Let + y be downward. a y = 9.80 m/s 2 Solve: (a) υ0 y = 0, t = 2.50 s, a y = 9.80 m/s 2 .

y − y0 = υ0 yt + 12 a yt 2 = 12 (9.80 m/s 2 )(2.50 s) 2 = 30.6 m. The building is 30.6 m tall. (b) υ y = υ0 y + a yt = 0 + (9.80 m/s 2 )(2.50 s) = 24.5 m/s (c) The graphs of a y , υ y , and y versus t are given in the figure below. Take y = 0 at the ground.


2-12

Chapter 2

2.50. Set Up: Take + y to be downward and y0 = 0. y = 14,600 ft, υ0 y = 0, and a y = 32 ft/s 2 . Solve: (a) y = y0 + υ0 yt + 12 a yt 2 and t =

2y = ay

2(14,600 ft) 32 ft/s 2

= 30.2 s

(b) υ y = υ0 y + a yt = (32 ft/s 2 )(30.2 s) = 966 ft/s = 659 mph (c) It is a poor assumption to neglect air resistance and aerodynamic lift.

*2.51. Set Up: Take + y upward. υ y = 0 at the maximum height. a y = − 0.379 g = − 3.71 m/s2 . Solve: Consider the motion from the maximum height back to the initial level. For this motion υ0 y = 0 and

t = 4.25 s. y = y0 + υ0 yt + 12 a yt 2 = 12 (−3.71 m/s 2 )(4.25 s) 2 = − 33.5 m The ball went 33.5 m above its original position. (b) Consider the motion from just after it was hit to the maximum height. For this motion υ y = 0 and t = 4.25 s.

υ y = υ0 y + a yt gives υ0 y = − a yt = − (−3.71 m/s 2 )(4.25 s) = 15.8 m/s. (c) The graphs are sketched in the figure below.

Reflect: The answers can be checked several ways. For example, υ y = 0, υ0 y = 15.8 m/s, and a y = − 3.7 m/s 2 in

υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives y − y0 =

υ y 2 − υ0 y 2 2a y

=

0 − (15.8 m/s)2 2( −3.71 m/s 2 )

= 33.6 m,

which agrees with the height calculated in (a).

2.52. Set Up: Take + y to be downward. υ0 y = 0 and let y0 = 0. Solve: (a) y = y0 + υ0 yt + 12 a yt 2 gives a y =

2y t2

=

2(11.26 m) (3.17 s)2

= 2.24 m/s 2 = 0.229 g

(b) υ y = υ0 y + a yt = (2.24 m/s 2 )(3.17 s) = 7.10 m/s

*2.53. Set Up: Take + y upward. a y = − 9.80 m/s 2 . The initial velocity of the sandbag equals the velocity of the balloon, so υ0 y = + 5.00 m/s. When the balloon reaches the ground, y − y0 = − 40.0 m. At its maximum height the sandbag has υ y = 0. Solve: (a) t = 0.250 s:

y − y0 = υ0 yt + 12 a yt 2 = (5.00 m/s)(0.250 s) + 12 (−9.80 m/s2 )(0.250 s)2 = 0.94 m. The sandbag is 40.9 m above the ground. υ y = υ0 y + a yt = + 5.00 m/s + (−9.80 m/s 2 )(0.250 s) = 2.55 m/s. t = 1.00 s:

y − y0 = (5.00 m/s)(1.00 s) + 12 (−9.80 m/s 2 )(1.00 s)2 = 0.10 m. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


2-13

The sandbag is 40.1 m above the ground. υ y = υ0 y + a yt = + 5.00 m/s + (−9.80 m/s 2 )(1.00 s) = − 4.80 m/s. (b) y − y0 = − 40.0 m, υ0 y = 5.00 m/s, a y = − 9.80 m/s 2 . y − y0 = υ0 yt + 12 a yt 2 gives − 40.0 m = (5.00 m/s) t −

(4.90 m/s 2 )t 2 . (4.90 m/s 2 )t 2 − (5.00 m/s)t − 40.0 m = 0 and

(

)

1 5.00 ± (−5.00)2 − 4(4.90)(−40.0) s = (0.51 ± 2.90) s. 9.80 t must be positive, so t = 3.41 s. t=

(c) υ y = υ0 y + a yt = + 5.00 m/s + (−9.80 m/s 2 )(3.41 s) = − 28.4 m/s (d) υ0 y = 5.00 m/s, a y = − 9.80 m/s 2 , υ y = 0. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

y − y0 =

υ y 2 − υ0 y 2 2a y

=

0 − (5.00 m/s) 2 2(−9.80 m/s 2 )

= 1.28 m.

The maximum height is 41.3 m above the ground. (e) The graphs of a y , υ y , and y versus t are given in the figure below. Take y = 0 at the ground.

2.54. Set Up: Take + y upward. (a) Set Up: Consider the motion from when he applies the acceleration to when the shot leaves his hand: v0 y = 0,

v y = ?, a y = 45.0 m/s 2 , y − y0 = 0.640 m . Use v y 2 = v0 y 2 + 2a y ( y − y0 ). Solve: v y = 2a y ( y − y0 ) = 2(45.0 m/s 2 )(0.640 m) = 7.59 m/s (b) Set Up: Consider the motion of the shot from the point where he releases it to its maximum height, where v = 0.

Take y = 0 at the ground. y0 = 2.20 m, y = ?, a y = 29.80 m/s 2 (free fall), v0 y = 7.59 m/s [from part (a)], and v y = 0 at maximum height. Use v y 2 = v0 y 2 + 2a y ( y − y0 ) . Solve: y − y0 =

v y 2 − v0 y 2 2a y

=

0 − (7.59 m/s)2 2( −9.80 m/s 2 )

= 2.94 m

Thus, y = 2.20 m + 2.94 m = 5.14 m. (c) Set Up: Consider the motion of the shot from the point where he releases it to when it returns to the height of his head. Take y = 0 at the ground. y0 = 2.20 m, y = 1.83 m, a y = 29.80 m/s 2 , v0 y = 17.59 m/s, t = ?. Use y − y0 = v0 yt + 12 a yt 2 . Solve: 1.83 m − 2.20 m = (7.59 m/s)t + 12 (−9.80 m/s 2 )t 2 = (7.59 m/s)t − (4.90 m/s 2 )t 2

4.90t 2 − 7.59t − 0.37 = 0, with t in seconds. Use the quadratic formula to solve for t: 1 t= 7.59 ± (7.59) 2 − 4(4.90)(−0.37) = 0.774 ± 0.822 9.80 Note that t must be positive, so t = 0.774 s + 0.822 s = 1.60 s

(

)


2-14

Chapter 2

Reflect: Calculate the time to the maximum height: v y = v0 y + a yt , so t = (vy − v0 y )/ay = −(7.59 m/s)/( − 9.80 m/s2 ) =

0.77 s. It also takes 0.77 s to return to 2.2 m above the ground, for a total time of 1.54 s. His head is a little lower than 2.20 m, so it is reasonable for the shot to reach the level of his head a little later than 1.54 s after being thrown; the answer of 1.60 s in part (c) makes sense.

*2.55. Set Up: aM = 0.170aE . Take + y to be upward and y0 = 0. Solve: (a) υ0E = υ0M . υ y 2 = υ0 y 2 + 2a y ( y − y0 ) with υ y = 0 at the maximum height gives 2a y y = − υ0 y 2 , so

aM yM = aE yE . ⎛a ⎞ ⎛ 1 ⎞ y M = ⎜ E ⎟ yE = ⎜ ⎟ (12.0 m) = 70.6 m a ⎝ 0.170 ⎠ ⎝ M⎠ (b) Consider the time to the maximum height on the earth. The total travel time is twice this. First solve for υ0 y ,

with υ y = 0 and y = 12.0 m. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ0 y = 22(− aE ) y = 22(−9.8 m/s 2 )(12.0 m) = 15.3 m/s. Then υ y = υ0 y + a yt gives t=

υ y − υ0 y ay

=

0 − 15.3 m/s −9.8 m/s 2

= 1.56 s.

The total time is 2(1.56 s) = 3.12 s. Then, on the moon υ y = υ0 y + a yt with υ0 y = 15.3 m/s, υ y = 0, and a = − 1.666 m/s 2 gives

t=

υ y − υ0 y ay

=

0 − 15.3 m/s −1.666 m/s 2

= 9.18 s.

The total time is 18.4 s. It takes 15.3 s longer on the moon. Reflect: The maximum height is proportional to 1/a, so the height on the moon is greater. Since the acceleration is the rate of change of the speed, the wrench loses speed at a slower rate on the moon and it takes more time for its speed to reach υ = 0 at the maximum height. In fact, tM /tE = aE /aM = 1/0.170 = 5.9, which agrees with our calculated times. But to find the difference in the times we had to solve for the actual times, not just their ratios.

2.56. Set Up: Take + y downward. a y = + 9.80 m/s 2 . Solve: (a) υ y = υ0 y + a yt = 15.0 m/s + (9.80 m/s 2 )(2.00 s) = 34.6 m/s (b) y − y0 = υ0 yt + 12 a yt 2 = (15.0 m/s)(2.00 s) + 12 (9.80 m/s 2 )(2.00 s) 2 = 49.6 m (c) υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives υ y = (15.0 m/s) 2 + 2(9.80 m/s 2 )(10.0 m) = 20.5 m/s

2.57. Set Up: Take + y upward. a y = − 9.80 m/s 2 . When the rock reaches the ground, y − y0 = − 60.0 m. Solve: (a) y − y0 = υ0 yt + 12 a yt 2 gives −60.0 m = (12.0 m/s)t − (4.9 m/s 2 )t 2 . (4.90 m/s 2 )t 2 − (12.0 m/s)t −

60.0 m = 0 and

(

)

1 12.0 ± ( −12.0) 2 − 4(4.90)( −60.0) s = (1.22 ± 3.71) s. 9.80 t must be positive, so t = 4.93 s.

t=

(b) υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives υ y = − (12.0 m/s) 2 + 2(−9.80 m/s 2 )(−60.0 m) = − 36.3 m/s. Reflect: We could have taken downward to be + y. Then y − y0 , υ y , and a y are all positive, but υ0 y is negative.

The same results are obtained with this alternative choice of coordinates.



2-15

2.58. Set Up: Take + x to be in the direction the sled travels. 1610 km/h = 447 m/s. 1020 km/h = 283 m/s. Assume the acceleration is constant. υ − υ0 x 447 m/s − 0 = = 248 m/s 2 = 25.3 g Solve: (a) υ 0 x = 0. υ x = υ0 x + axt gives ax = x t 1.80 s ⎛ υ + υ x ⎞ ⎛ 447 m/s ⎞ (b) ( x − x0 ) = ⎜ 0 x ⎟t = ⎜ ⎟ (1.80 s) = 402 m 2 2 ⎝ ⎠ ⎝ ⎠ (c) Solve for ax and compare to 40g. υ x = 0. ax =

υ x − υ0 x t

=

0 − 283 m/s = − 202 m/s 2 = − 20.6 g . 1.40 s

The figures are inconsistent, if the acceleration while stopping is constant. The acceleration while stopping could reach 40g if the acceleration wasn’t constant.

*2.59. Set Up: Use subscripts f and s to refer to the faster and slower stones, respectively. Take + y to be upward and y0 = 0 for both stones. υ0f = 3υ0s . When a stone reaches the ground, y = 0. 2υ0 y υ υ Solve: (a) y = y0 + υ0 yt + 12 a yt 2 gives a y = − . Since both stones have the same a y , 0f = 0s and t

tf

⎛υ ts = tf ⎜ 0s ⎝ υ0f

⎞ ⎟= ⎠

( 13 )10 s = 3.3 s

(b) Since υ y = 0 at the maximum height, then υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives a y = −

υ0f 2

ts

υ0 y 2 2y

. Since both have the

2

υ0s 2

⎛υ ⎞ same a y , = and yf = ys ⎜ 0f ⎟ = 9 H . yf ys ⎝ υ0s ⎠ Reflect: The faster stone reaches a greater height so it travels a greater distance than the slower stone and takes more time to return to the ground.

2.60. Set Up: Take + y to be downward and y0 = 0. Both coconuts have the same acceleration, a y = g . Let A be the coconut that falls from the greater height and let B be the other coconut. y A = 2 yB . υ0 A = υ0 B = 0. Solve: (a) υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives a y = (b) y = y0 + υ0 yt + 12 a yt 2 gives a y =

G

2y

t2

and

υ y2 2y

yA

t A2

=

and

yB

tB 2

υ A2 yA

=

υB2

. t A = tB

yB

. υB = υ A

yB =V yA

1 2

=V/ 2

yA = 2T yB

G

*2.61. Set Up: υ T/E = 65 mph, north. υ VW/E = 42 mph, south. Let + y be north. G G G G G G Solve: υ T/E = υ T/VW + υ VW/E and υ T/VW = υ T/E − υ VW/E

(a) (υT/VW ) y = (υT/E ) y − (υVW/E ) y = 65 mph − ( −42 mph) = 107 mph. Relative to the VW, the Toyota is traveling G G north at 107 mph. υ VW/T = − υ T/VW . Relative to the Toyota the VW is traveling south at 107 mph. (b) The answers are the same as in (a).

G

2.62. Set Up: A = air, E = eagle, G = ground. υ A/G = 35 mph, east.

G G G Solve: υ E/G = υ E/A + υ A/G G G G G (a) υ E/A = 22 mph, east. υ E/A and υ A/G are both east and υE/G = υE/A + υ A/G = 57 mph. υ E/G is east. G G G G (b) υ E/A = 22 mph, west. υ E/A and υ A/G are in opposite directions and υE/G = υA/G − υE/A = 13 mph. υ E/G is east.


2-16

Chapter 2

2.63. Set Up: Use coordinates with + y downward. Relative to the earth the package has υ0 y = + 3.50 m/s and a y = 9.80 m/s 2 . Solve: The velocity of the package relative to the ground just before it hits is

υ y = υ0 y 2 + 2a y ( y − y0 ) = (3.50 m/s)2 + 2(9.80 m/s 2 )(8.50 m) = 13.4 m/s G G G G G G G G (a) υP/G = 13.4 m/s, downward. υH/G = 3.50 m/s, downward. υP/G = υ P/H + υH/G and υP/H = υP/G − υH/G . G υ P/H = 9. 9 m/s, downward. G G G (b) υH/P = − υ P/H , so υH/P = 9.9 m/s, upward. Reflect: Since the helicopter is traveling downward, the package is moving slower relative to the helicopter than its speed relative to the ground.

G

2.64. Set Up: υP/A = 600 mph and is east for the first 200 mi and west for the return 200 mi. The time is the distance relative to the ground divided by the speed relative to the ground. G G G Solve: υP/E = υP/A + υA/E 4000 mi (a) υA/E = 0 and υP/E = 600 mph. t = = 6.67 h 600 mi/h G G (b) San Francisco to Chicago: υA/E = 150 mph, east. υP/A = 600 mph, east. 2000 mi = 2.67 h 750 mi/h = 600 mph, west.

υP/E = υP/A + υA/E = 750 mph. t = G G Chicago to San Francisco: υA/E = 150 mph, east. υP/A

υP/E = υP/A − υA/E = 450 mph. t = The total time is 2.67 h + 4.44 h = 7.11 h.

2000 mi = 4.44 h 450 mi/h

*2.65. Set Up: 1 light year = (3.00 × 108 m/s)(3.156 × 107 s) = 9.47 × 1015 m Solve: t =

d

υ

=

(4.25 light years)(9.47 × 1015 m/light year) 1000 × 103 m/s

= 4.02 × 1010 s = 1300 yr

2.66. Set Up: At t = 0 the auto and truck are at the same position. The auto overtakes the truck when after time T they have both traveled a distance d. Solve: (a) Apply x − x0 = υ0 xt + 12 axt 2 to the motion of each vehicle. The auto has υ0 x = 0 and a x = 2.50 m/s 2 , so d = 12 (2.50 m/s 2 )T 2 . The truck has υ0 x = 15.0 m/s and ax = 0, so d = (15.0 m/s)T . Combining these two equations gives (1.25 m/s 2 )T 2 = (15.0 m/s)T and T = 12.0 s. Then d = (15.0 m/s)(12.0 s) = 180 m. (b) υ x = υ0 x + axt = 0 + (2.50 m/s 2 )(12.0 s) = 30.0 m/s

*2.67. Set Up: The average speed is the total distance traveled divided by the total time. The elapsed time is the distance traveled divided by the average speed. The total distance traveled is 20 mi. With an average speed of 8 mi/h 10 mi for 10 mi, the time for that first 10 miles is = 1.25 h. 8 mi/h 20 mi Solve: (a) An average speed of 4 mi/h for 20 mi gives a total time of = 5.0 h. The second 10 mi must be 4 mi/h 10 mi = 2.7 mi/h. covered in 5.0 h − 1.25 h = 3.75 h. This corresponds to an average speed of 3.75 h



2-17

20 mi = 1.67 h. The second 10 mi must be covered 12 mi/h 10 mi in 1.67 h − 1.25 h = 0.42 h. This corresponds to an average speed of = 24 mi/h. 0.42 h 20 mi (c) An average speed of 16 mi/h for 20 mi gives a total time of = 1.25 h. But 1.25 h was already spent during 16 mi/h the first 10 miles and the second 10 miles would have to be covered in zero time. This is not possible and an average speed of 16 mi/h for the 20-mile ride is not possible. Reflect: The average speed for the total trip is not the average of the average speeds for each 10-mile segment. The rider spends a different amount of time traveling at each of the two average speeds. (b) An average speed of 12 mi/h for 20 mi gives a total time of

2.68. Set Up: Both you and your friend run at a steady pace over the 10 km race so the equation Δx = vxt applies. You can calculate the time that it takes for your friend to complete the race and then subtract 15 minutes (900 seconds) to obtain the time that you need to achieve your goal. The race is 10 km, which is 1.0 × 104 m. Solve: The time your friend takes to complete the race is t =

Δx 1.0 × 104 m = = 4.0 × 103 s. Thus, you need to vx 2.5 m/s

Δx 1.0 × 104 m = = 3.2 m/s. t 3.1 × 103 s Reflect: If your friend had a 15-minute head start, this is also the speed that you would need to just catch up at the end of the race.

complete the race in 4.0 × 103 s − 900 s = 3.1 × 103 s. The required speed to do this is vx =

*2.69. Set Up: Let + y to be upward and y0 = 0. aM = aE /6. At the maximum height υ y = 0. Solve: (a) υ y 2 = υ0 y 2 + 2a y ( y − y0 ). Since υ y = 0 and υ0 y is the same for both rocks, aM yM = aE yE and

(b) υ y = υ0 y + a yt . aMtM = aEtE and tM

⎛a ⎞ yE = ⎜ M ⎟ yM = H /6 ⎝ aE ⎠ ⎛a ⎞ = ⎜ E ⎟ tE = 6(4.0 s) = 24.0 s ⎝ aM ⎠

Reflect: On the moon, where the acceleration is less, the rock reaches a greater height and takes more time to reach that maximum height.

2.70. Set Up: Let + y to be downward. υ0 y = 2.0 m/s, υ y = 1.3 m/s, and y − y0 = 0.020 m. Solve: (a) υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

ay =

υ y 2 − υ0 y 2 2( y − y0 )

=

(1.3 m/s) 2 − (2.0 m/s) 2 = − 58 m/s 2 = − 5.9 g 2(0.020 m)

⎛ υ0 y + υ y ⎞ 2( y − y0 ) 2(0.020 m) = = 12 ms (b) y − y0 = ⎜ ⎟ t gives t = + 2 . 0 m/s + 1.3 m/s 2 υ υ 0y y ⎝ ⎠

2.71. Set Up: υ0 x = 0, υ x = 5.0 × 103 m/s, and x − x0 = 4.0 m Solve: (a) υ x 2 = υ0 x 2 + 2ax ( x − x0 ) gives

ax = (b) υ x = υ0 x + axt gives t =

υ x 2 − υ0 x 2 2( x − x0 )

υ x − υ0 x

(5.0 × 103 m/s) 2 = 3.1 × 106 m/s 2 = 3.2 × 105 g 2(4.0 m)

5.0 × 103 m/s

= 1.6 ms 3.1 × 106 m/s 2 (c) The calculated a is less than 450,000g so the acceleration required doesn’t rule out this hypothesis.

ax

=

=


2-18

Chapter 2

2.72. Set Up: Assume an initial height of 200 m and a constant acceleration of 9.80 m/s 2 . Let + y be downward, 1 km/h = 0.2778 m/s and 1 mi/h = 0.4470 m/s. Solve: (a) y − y0 = 200 m, a y = 9.80 m/s 2 , v0 y = 0. v y 2 = v0 y 2 + 2a y ( y − y0 ) gives v y =

2(9.80 m/s 2 )(200 m) =

60 m/s = 200 km/h = 140 mi/h. (b) Raindrops actually have a speed of about 1 m/s as they strike the ground. (c) The actual speed at the ground is much less than the speed calculated assuming free-fall, so neglect of air resistance is a very poor approximation for falling raindrops. Reflect: In the absence of air resistance raindrops would land with speeds that would make them very dangerous.

*2.73. Set Up: Let + y be downward. The egg has υ0 y = 0 and a y = 9.80 m/s 2 . Find the distance the professor walks during the time t it takes the egg to fall to the height of his head. At this height, the egg has y − y0 = 44.2 m. Solve: y − y0 = υ0 yt + 12 a yt 2 gives

t=

2( y − y0 ) = ay

2(44.2 m) 9.80 m/s 2

= 3 .00 s.

The professor walks a distance x − x0 = υ0 xt = (1.20 m/s)(3.00 s) = 3.60 m. Release the egg when your professor is 3.60 m from the point directly below you. Reflect: Just before the egg lands its speed is (9.80 m/s 2 )(3.00 s) = 29.4 m/s. It is traveling much faster than the professor.

2.74. Set Up: Let + x be in the direction down the incline. The final velocity for the first 10.0 s is the initial speed for the second 10.0 s of motion. Solve: For the first 10.0 s of motion υ0 x = 0 and υ x = a x (10.0 s). For the second 10.0 s of motion, υ0 x = a x (10.0 s), x − x0 = 150 m and t = 10.0 s. x − x0 = υ0 xt + 12 axt 2 gives 150 m = (10.0 m/s) 2 a x +

1 a (10.0 2 x 2

m/s) 2 = 150.0ax and

a x = 1.0 m/s 2 . Then for the first 5.0 s, x − x0 = υ0 xt + 12 a xt 2 = 0 + 12 (1.00 m/s 2 )(5.0 s) = 12.5 m.

*2.75. Set Up: Since air resistance is ignored, the boulder is in free-fall and has a constant downward acceleration of magnitude 9.80 m/s 2 . Apply the constant acceleration equations to the motion of the boulder. Take + y to be upward. Solve: (a) v0 y = 140.0 m/s, v y = +20.0 m/s, a y = 29.80 m/s 2 . v y = v0 y + a yt gives t= (b) v y = 220.0 m/s. t =

v y − v0 y ay

=

v y − v0 y ay

=

20.0 m/s − 40.0 m/s −9.80 m/s 2

−20.0 m/s − 40.0 m/s −9.80 m/s 2

= +2.04 s.

= +6.12 s.

(c) y − y0 = 0, v0 y = +40.0 m/s, a y = −9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives t = 0 and

t=−

2v0 y ay

=2

2(40.0 m/s) −9.80 m/s 2

= +8.16 s.

(d) v y = 0, v0 y = +40.0 m/s, a y = −9.80 m/s 2 . v y = v0 y + a yt gives t =

v y − v0 y ay

=

0 − 40.0 m/s −9.80 m/s 2

= 4.08 s.

(e) The acceleration is 9.80 m/s 2 , downward, at all points in the motion. (f) The graphs are sketched in the following figure.



2-19

Reflect: We have v y = 0 at the maximum height. The time to reach the maximum height is half the total time in the

air, so the answer in part (d) is half the answer in part (c). Also note that 2.04 s < 4.08 s < 6.12 s . The boulder is going upward until it reaches its maximum height, and after the maximum height it is traveling downward.

2.76. Set Up: Take + x to be the direction the car moves when speeding up. Use the acceleration and stopping data to find a in each case. 1 mph = 0.4470 m/s. Design the on-ramp for the less powerful car and the off-ramp for the car with bald tires. Solve: on-ramp (speeding up): υ0 x = 0, υ x = 60 mph = 26.8 m/s, t = 10.0 s and υ x = υ0 x + a xt gives ax =

υ x − υ0 x t

= 2.68 m/s 2 .

Then υ0 x = 0, υ x = 70 mph = 31.3 m/s and a x = 2.68 m/s 2 in υ x 2 = υ0 x 2 + 2a x ( x − x0 ) gives x − x0 =

υ x 2 − υ0 x 2 2a x

= 180 m.

This is the required length of the on-ramp. off-ramp (slowing down): υ0 x = 60 mph = 26.8 m/s, υ x = 0, t = 20.0 s and υ x = υ0 x + a xt gives ax =

υ x − υ0 x t

= − 1.34 m/s 2 .

Then υ0 x = 70 mph = 31.3 m/s, υ x = 0 and a x = − 1.34 m/s 2 in υ x 2 = υ0 x 2 + 2a x ( x − x0 ) gives x − x0 =

υ x 2 − υ0 x 2 2a x

= 370 m.

This is the required length of the off-ramp.

2.77. Set Up: The average acceleration is given by aav, x =

Δυ x 1 60 . The time for each beat is minutes, or Δt 72 72

seconds.

Δυ x 425 cm/s − 0 = = 510 cm/s 2 . Δt 60/72 seconds 1 Reflect: This is about g. 2 Solve: aav, x =

2.78. Set Up: Let + x be to the right. (a) Δx is constant so υ x is constant and positive. (b) Δx increases so vx is positive and increasing. (c) Δx decreases so υ x is positive and decreasing. (d) Δx increases and then decreases so υ x increases and then decreases.


2-20

Chapter 2

Solve: The graphs are sketched qualitatively in the figures below.

*2.79. Set Up: Take + y to be upward. There are two periods of constant acceleration: a y = + 2.50 m/s 2 while the engines fire and a y = − 9.8 m/s 2 after they shut off. Constant acceleration equations can be applied within each period of constant acceleration. Solve: (a) Find the speed and height at the end of the first 20.0 s. a y = + 2.50 m/s 2 , υ0 y = 0, and y0 = 0.

υ y = υ0 y + a yt = (2.50 m/s 2 )(20.0 s) = 50.0 m/s

and

y = y0 + υ0 yt + 12 a yt 2 = 12 (2.50 m/s 2 )(20.0 s) 2 = 500 m.

Next consider the motion from this point to the maximum height. y0 = 500 m, υ y = 0, υ0 y = 50.0 m/s, and a y = − 9.8 m/s 2 . υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives y − y0 =

υ y 2 − υ0 y 2 2a y

=

0 − (50.0 m/s) 2 2(−9.8 m/s 2 )

= + 128 m,

so y = 628 m. The duration of this part of the motion is obtained from υ y = υ0 y + a yt : t=

υ y − υ0 y ay

=

−50 m/s −9.8 m/s 2

= 5.10 s

(b) At the highest point, υ y = 0 and a y = 9.8 m/s 2 , downward. (c) Consider the motion from the maximum height back to the ground. a y = − 9.8 m/s 2 , υ0 y = 0, y = 0, and

y0 = 628 m. y = y0 + υ0 yt + 12 a yt 2 gives t=

2( y − y0 ) = 11.3 s. ay

The total time the rocket is in the air is 20.0 s + 5.10 s + 11.3 s = 36.4 s. υ y = υ0 y + a yt = ( −9.8 m/s 2 )(11.3 s) = −111 m/s. Just before it hits the ground the rocket will have speed 111 m/s. Reflect: We could calculate the time of free fall directly by considering the motion from the point of engine shutoff to the ground: υ0 y = 50.0 m/s, y − y0 = 500 m and a y = −9.8 m/s 2 . y − y0 = υ0 yt + 12 a yt 2 gives t = 16.4 s, which agrees with a total time of 36.4 s.



2-21

2.80. Set Up: The velocity is the slope of the x versus t graph. The sign of the slope specifies the direction of the velocity. Solve: (a) The mouse is to the right of the origin when x is positive. This is for 0 < t < 6.0 s and for t > 8.5 s. The mouse is to the left of the origin when x is negative. This is for 6.0 s < t < 8.5 s. The mouse is at the origin when t = 0, 6.0 s, 8.5 s. Δx 40.0 cm = = 13.3 cm/s. Δt 3.0 s (c) The acceleration is not constant. For constant a the x-t graph would be a section of a parabola, and it is not. (d) The speed is greatest when the magnitude of the slope is greatest, and this is between 5.0 s and 6.0 s. The speed in this interval is 40.0 cm = 40.0 cm/s. 1.0 s During this part of the motion the velocity is negative and the mouse is moving to the left. (e) The mouse is moving to the right when x is increasing. This happens for 0 < t < 3.0 s and for t > 7.0 s. The mouse is moving to the left when x is decreasing. This happens for 5.0 s < t < 7.0 s. The mouse is instantaneously at (b) At t = 0, υ x is positive and has magnitude

rest when the slope of the tangent to the x versus t graph is zero, when the tangent is horizontal. This happens for 3.0 s < t < 5.0 s and at t = 7.0 s. (f) In the first 3 seconds the mouse travels from 0 to 40.0 cm so it travels 40.0 cm. In the first 10 seconds the mouse travels from 0 cm to 40.0 cm, from 40.0 cm to −10.0 cm and from −10.0 cm to 15.0 cm. The total distance traveled is 40.0 cm + 50.0 cm + 25.0 cm = 115 cm. (g) The mouse is speeding up when the slope of x versus t is increasing. This happens at 5.0 s and between 7.0 s and 8.5 s. The mouse is slowing down at 3.0 s, between 6.0 s and 7.0 s and between 8.5 s and 10.0 s. (h) No, the acceleration is not constant during this entire interval and these formulas apply only for constant acceleration.

2.81. Set Up: In time tS the S-waves travel a distance d = υStS and in time tP the P-waves travel a distance d = υPtP . Solve: (a) tS = tP + 33 s. (b) t =

d

=

d

υS

=

d

υP

1 1 ⎛ ⎞ − + 33 s. d ⎜ ⎟ = 33 s and d = 250 km. 3 5 km/s 6 5 km/s . . ⎝ ⎠

375 km d 375 km = 107 s. tP = = = 58 s. tS − tP = 49 s 3.5 km/s υP 6.5 km/s

υ 2.82. Set Up: Assume straight line motion along the + x-axis. For the starting phase we may use the kinematic equations for constant acceleration: v0 x = 0, vx = 8.00 m/s, and Δt = 1.40 s; vx = Solve: (a) vx =

Δvx v +v , and Δx = x 0 x Δt. Δt 2

Δvx 8.00 m/s − 0 = = 5.71 m/s 2 . Δt 1.40 s

(b) His acceleration to a top speed of 11.8 m/s occurs over a time of 7.02 s. Thus, aav, x =

Δvx 11.8 m/s − 0 = = 1.68 m/s 2 . Δt 7.02 s

⎛ 8.00 m/s + 0 ⎞ (c) The distance traveled during the starting phase is Δx = ⎜ ⎟ (1.40 s ) = 5.60 m. 2 ⎝ ⎠ Reflect: We cannot calculate the distance that the sprinter travels during the second phase of the race, since we do not know that his acceleration is constant during this phase.

*2.83. Set Up: Let tfall be the time for the rock to fall to the ground and let ts be the time it takes the sound to travel from the impact point back to you. tfall + ts = 10.0 s. Both the rock and sound travel a distance d that is equal to the height of the cliff. Take + y downward for the motion of the rock. The rock has υ0 y = 0 and a y = 9.80 m/s 2 . Solve: (a) For the rock, y − y0 = υ0 yt + 12 a yt 2 gives tfall =

2d 9.80 m/s 2

.


2-22

Chapter 2

For the sound, ts =

d = 10.0 s. Let α 2 = d . 0.00303α 2 + 0.4518α − 10.0 = 0. α = 19.6 and d = 384 m. 330 m/s

(b) You would have calculated d = 12 (9.80 m/s 2 )(10.0 s) 2 = 490 m. You would have overestimated the height of the

cliff. It actually takes the rock less time than 10.0 s to fall to the ground. Reflect: Once we know d we can calculate that tfall = 8.8 s and ts = 1.2 s. The time for the sound of impact to travel back to you is 12% of the total time and cannot be neglected. The rock has speed 86 m/s just before it strikes the ground.

Solutions to age Problems 2.84. Set Up: Since the blood momentarily comes to rest, v0 x = 0. Also, we know that vx = 1.0 m/s and Δt = 250 ms. Solve: aav, x =

Δvx 1.0 m/s − 0 = = 4.0 m/s. Thus, the correct answer is D. Δt 0.250 s

*2.85. Set Up: Assuming that the aorta and arteries are circular in cross-sectional area, we can use 2

⎛d ⎞ A = π r 2 = π ⎜ ⎟ . Let d a be the diameter of the aorta and d b be the diameter of each branch. ⎝2⎠ 2

2

2

⎛d ⎞ ⎛d ⎞ ⎛d ⎞ Solve: Since the combined area of the two arteries is equal to that of the aorta we have π ⎜ b ⎟ + π ⎜ b ⎟ = π ⎜ a ⎟ , ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ which reduces to 2d b 2 = d a 2 . Thus, we have d b = d a / 2. The correct answer is B.

2.86. Set Up: We are asked to find the average velocity (not the average speed): vav, x = Solve: For the round-trip, we have xf = x0 so Δx = 0 m. Thus, we find vav, x =

Δx . Δt

Δx = 0 m/s. Δt

The correct answer is C.


3

MOTION IN A PLANE

Answers to Multiple-Choice Problems 1. B 2. A 3. C 4. C 15. D 16. B 17. B

5. A

6. B

7. B

8. A

9. C

10. C

11. A

12. C

13. D

14. B

Solutions to Problems 3.1. Set Up: Since distances are given in km, use km/s as units for velocity. Δx 6.24 km − 5.00 km Δy 0.925 km − 1.20 km Solve: (a) (υ x )av = = = 1.09 km/s; (υ y )av = = = − 0.24 km/s Δt 1.14 s Δt 1.14 s G (b) υav and its components are shown in the figure below.

υav = (υ x )av 2 + (υ y )av 2 = 1.12 km/s; tan φ =

(υ y )av (υ x )av

and φ = 12.4°, below the +x axis.

3.2. Set Up: Since the velocity is expressed in units of km/h and the time interval is in seconds, it is convenient to express the acceleration in mixed units of km/h ⋅ s. Solve: (a) Δυ y 365 km/h − 415 km/h Δυ x 838 km/h − 625 km/h (a x )av = = = 79.8 km/h ⋅ s; (a y )av = = = − 18.7 km/h ⋅ s Δt 6.52 s − 3.85 s 6.52 s − 3.85 s Δt G (b) aav and its components are shown in the figure below.

aav = ( ax )av 2 + (a y )av 2 = 820 km/h ⋅ s; tan φ =

(υ y )av (υ x )av

and φ = 13.2°, below the + x { axis.


3-1

3-2

Chapter 3

*3.3. Set Up: Coordinates of point A are (2.0 m, 1.0 m) and for point B they are (10.0 m, 6.0 m). Solve: (a) At A, x = 2.0 m, y = 1.0 m.

G y (b) r and its components are shown in Figure (a) below. r = x 2 + y 2 = 2.2 m; tanθ = and θ = 26.6D , counterx clockwise from the +x-axis.

(c) (υ x )av =

Δx 10.0 m − 2.0 m Δy 6.0 m − 1.0 m = = 5.3 m/s; (υ y )av = = = 3.3 m/s Δt 1.50 s Δt 1.50 s

(υ y )av G (d) υav and its components are shown in Figure (b) above. υav = (υ x )av 2 + (υ y )av 2 = 6.2 m/s; tan φ = and (υ x )av

φ = 32D , counterclockwise from the +x-axis.

G Reflect: The displacement of the dragonfly is in the direction of υav .

3.4. Set Up: Let the x-axis point due east and the y-axis point due north. The x-velocity changes from +8.00 m/s to 0 m/s and the y-velocity changes from 0 m/s to −8.80 m/s in 4.00 seconds. Δυ y −8.80 m/s − 0 m/s Δυ x 0 m/s − 8.00 m/s Solve: (a) aav, x = = = − 2.00 m/s 2 ; aav, y = = = − 2.20 m/s 2 4.00 s 4.00 s Δt Δt aav, y −2.20 m/s 2 = = +1.10. (b) aav = aav, x 2 + aav, y 2 = ( − 2.00 m/s 2 ) 2 + (−2.20 m/s 2 )2 = 2.97 m/s 2 ; tan φ = aav, x −2.00 m/s 2 Thus, noting that the acceleration vector is in the third-quadrant (since both its components are negative), we have φ = 227.7D , counterclockwise from the +x-axis. The direction can also be written as θ = 270D − 227.7D = 42.3D west of due south. Reflect: Since the westward component of the acceleration vector is slightly less than its southward component, the angle between the acceleration vector and the southward direction is slightly less than 45D.

*3.5. Set Up: The coordinates of each point are: A, (−50 m, 0); B, (0, + 50 m); C, (+50 m, 0); D, (0, − 50 m). At each point the velocity is tangent to the circular path, as shown in the figure below. The components (υ x , υ y ) of the velocity at each point are: A, (0, + 6.0 m/s); B, (+6.0 m/s, 0); C, (0, − 6.0 m/s); D, (−6.0 m/s, 0).


Motion in a Plane

3-3

Solve: (a) A to B The time for one full lap is

t= A to B is one-quarter lap and takes

1 (52.4 4

2π r

υ

=

2π (50 m) = 52.4 s. 6.0 m/s

s) = 13.1 s.

Δx 0 − (−50 m) Δy +50 m − 0 = = 3.8 m/s; (υ y )av = = = 3.8 m/s Δt 13.1 s Δt 13.1 s Δυ y 0 − 6.0 m/s Δυ x 6.0 m/s − 0 = = = 0.46 m/s 2 ; (a y )av = = = − 0.46 m/s 2 13.1 s Δt 13.1 s Δt

(υ x )av = (a x )av

(b) A to C t = 12 (52.4 s) = 26.2 s

Δx +50 m − (−50 m) Δy = = 3.8 m/s; (υ y )av = =0 Δt 26.2 s Δt Δυ y −6.0 m/s − 6.0 m/s Δυ x = = 0; (a y )av = = = − 0.46 m/s 2 26.2 s Δt Δt

(υ x )av = (a x )av

(c) C to D t = 14 (52.4 s) = 13.1 s

Δx 0 − 50 m Δy −50 m − 0 = = − 3.8 m/s; (υ y )av = = = − 3.8 m/s Δt 13.1 s Δt 13.1 s Δυ y 0 − (−6.0 m/s) Δυ x −6.0 m/s − 0 = = = − 0.46 m/s 2 ; (a y )av = = = 0.46 m/s 2 Δt 13.1 s 13.1 s Δt (υ x )av =

(a x )av

(d) A to A Δx = Δy = 0 so (υ x )av = (υ y )av = 0, and Δυ x = Δυ y = 0 so (ax )av = ( a y )av = 0 (e) For A to B, υav = (υ x )av 2 + (υ y )av 2 = (3.8 m/s) 2 + (3.8 m/s) 2 = 5.4 m/s. The speed is constant so the average

speed is 6.0 m/s. The average speed is larger than the magnitude of the average velocity because the distance traveled is larger than the displacement. (f) Velocity is a vector, with both magnitude and direction. The magnitude of the velocity is constant but its direction is changing. Reflect: For this motion the acceleration describes the rate of change of the direction of the velocity.

3.6. Set Up: Use coordinates with the origin at the initial position of the stone and take +y downward. υ0 x = 30 m/s, υ0 y = 0, ax = 0, a y = g = 9.8 m/s 2 Solve: (a) x0 = y0 = 0. x = υ0 xt = (30 m/s)t and y = 12 gt 2 = (4.9 m/s 2 )t 2 . t = 2 s: x = 60 m, y = 19.6 m; t = 4 s:

x = 120 m, y = 490 m

y = 78.4 m; t = 6 s: x = 180 m,

y = 176 m; t = 8 s: x = 240 m,

y = 314 m; t = 10 s: x = 300 m,


3-4

Chapter 3

(b) The graph of y versus x is shown in the figure below.

3.7. Set Up: We can use the equations for projectile motion with v0 x = 42.0 m/s, v0 y = 0, and Δx = 60.5 ft. Note that 1 ft = 0.3048 m. Solve: x − x0 = υ0 xt gives

t=

x − x0

υ0 x

=

(60.5 ft)(0.3048 m/ft) = 0.4391 s 42.0 m/s

The vertical displacement during this time is y − y0 = υ0 yt − 12 gt 2 = − 12 (9.80 m/s 2 )(0.4391 s)2 = − 0.945 m. Thus, the ball drops 0.945 meters. Reflect: In the absence of air resistance, the motion of a baseball is simply two-dimensional parabolic motion. In reality, air resistance is very important in baseball—curve balls are deflected by slight differences in air pressure on opposite sides of the spinning baseball.

3.8. Set Up: Take +y downward, so ax = 0, a y = + 9.80 m/s 2 , υ0 x = 1.10 m/s and υ0 y = 0. Solve: (a) y − y0 = υ0 yt + 12 a yt 2 = 12 (9.80 m/s 2 )(0.350 s)2 = 0.600 m. This is the height of the tabletop. (b) x − x0 = υ0 xt + 12 axt 2 = (1.10 m/s)(0.350 s) = 0.385 m (c) υ x = υ0 x = 1.10 m/s, υ y = υ0 y + a yt = (9.80 m/s 2 )(0.350 s) = 3.43 m/s. υ = υ x 2 + υ y 2 = 3.60 m/s.

tan θ =

qυ y q qυ x q

=

3.43 m/s 1.10 m/s

and θ = 72.2°. The velocity of the book just before it hits the floor has magnitude 3.60 m/s and is directed at 72.2D below the horizontal.

*3.9. Set Up: Take +y downward, so ax = 0, a y = +9.80 m/s 2 and υ0 y = 0. When the ball reaches the floor, y − y0 = 0.750 m. Solve: (a) y − y0 = υ0 yt + 12 a yt 2 gives t = (b) x − x0 = υ0 xt + 12 axt 2 gives υ0 x =

2( y − y0 ) = ay

2(0.750 m) 9.80 m/s 2

= 0.391 s.

x − x0 1.40 m = = 3.58 m/s. Since υ0 y = 0, υ0 = υ0 x = 3.58 m/s. t 0.391 s

(c) υ x = υ0 x = 3.58 m/s. υ y = υ0 y + a yt = (9.80 m/s 2 )(0.391 s) = 3.83 m/s. υ = υ x 2 + υ y 2 = 5.24 m/s. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Motion in a Plane

tan θ =

| υ y| | υ x|

=

3-5

3.83 m/s 3.58 m/s

and θ = 46.9°. The final velocity of the ball has magnitude 5.24 m/s and is directed at 46.9° below the horizontal. Reflect: The time for the ball to reach the floor is the same as if it had been dropped from a height of 0.750 m; the horizontal component of velocity has no effect on the vertical motion.

3.10. Set Up: The initial velocity of the bomb is the same as that of the helicopter. Take +y downward, so ax = 0, a y = + 9.80 m/s 2 , υ0 x = 60.0 m/s and υ0 y = 0. Solve: (a) y − y0 = υ0 yt + 12 a yt 2 with y − y0 = 300 m gives t =

2( y − y0 ) 2(300 m) = = 7.82 s. ay 9.80 m/s 2

(b) The bomb travels a horizontal distance x − x0 = υ0 xt + 12 axt 2 = (60.0 m/s)(7.82 s) = 470 m. (c) υ x = υ0 x = 60.0 m/s. υ y = υ0 y + a yt = (9.80 m/s 2 )(7.82 s) = 76.6 m/s. (d) Graphs of x versus t and of y versus t are given in the figure below.

(e) Because the airplane and the bomb always have the same x-component of velocity and position, the plane will be 300 m directly above the bomb at impact.

*3.11. Set Up: The ball moves with projectile motion with an initial velocity that is horizontal and has magnitude v0 . In both cases, the height h of the table and v0 have the same values; but the acceleration due to gravity changes from g E = 9.80 m/s 2 on earth to g X on planet X. Let +x be horizontal and in the direction of the initial velocity of the marble and let +y be upward. v0 x = v0 , v0 y = 0, ax = 0, a y = − g , where g is either g E or g X . Solve: Use the vertical motion to find the time in the air: y − y0 = − h. y − y0 = v0 yt + 12 a yt 2 gives t =

x − x0 = v0 xt + 12 axt 2 gives x − x0 = v0 xt = v0

2h . Then g

2h . x − x0 = D on earth and 2.76D on Planet X. ( x − x0 ) g = v0 2h , g

which is constant, so D g E = 2.76 D g X . g X =

gE

(2.76) 2

= 0.131g E = 1.28 m/s 2 .

Reflect: On Planet X the acceleration due to gravity is less, it takes the ball longer to reach the floor, and it travels farther horizontally.

3.12. Set Up: Take +y downward. ax = 0, a y = + 9.80 m/s 2 . υ0 x = υ0 , υ0 y = 0. She must travel 1.75 m horizontally during the time she falls 9.00 m vertically. Solve: Time to fall 9.00 m: y − y0 = υ0 yt + 12 a yt 2 gives © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3-6

Chapter 3

t=

2( y − y0 ) = ay

2(9.00 m) 9.80 m/s 2

= 1.36 s.

Speed needed to travel 1.75 m horizontally during this time: x − x0 = υ0 xt + 12 axt 2 gives

υ0 = υ0 x =

x − x0 1.75 m = = 1.29 m/s. t 1.36 s

*3.13. Set Up: Take +y to be downward. ax = 0, a y = + 9.80 m/s 2 . υ0 x = υ0 , υ0 y = 0. The car travels 21.3 m − 1.80 m = 19.5 m downward during the time it travels 61.0 m horizontally. Solve: Use the vertical motion to find the time in the air: 2( y − y0 ) 2(19.5 m) y − y0 = υ0 yt + 12 a yt 2 gives t = = = 1.995 s ay 9.80 m/s 2 Then x − x0 = υ0 xt + 12 axt 2 gives υ0 = υ0 x =

x − x0 61.0 m = = 30.6 m/s. t 1.995 s

(b) υ x = 30.6 m/s since a x = 0. υ y = υ0 y + a yt = − 19.6 m/s. υ = υ x 2 + υ y 2 = 36.3 m/s. Reflect: We calculate the final velocity by calculating its x and y components.

3.14. Set Up: Take +y to be upward, so υ0 x = 18.0 m/s and υ0 y = 15.0 m/s. At the highest point in the trajectory, υ0 y = 0. ax = 0, a y = − 9.80 m/s 2 . When the object reaches the ground, y − y0 = 0. Solve: (a) υ y = υ0 y + a yt and t =

υ y − υ0 y ay

=

0 − 15.0 m/s 29.80 m/s 2

= 1.53 s

(b) The vertical displacement at t = 1.53 s is

y − y0 = υ0 yt + 12 a yt 2 = (15.0 m/s)(1.53 s) + 12 (−9.80 m/s 2 )(1.53 s)2 = 11.5 m (c) Use y − y0 = υ0 yt + 12 a yt 2 . y − y0 = 0 and

t=−

2υ0 y ay

=−

2(15.0 m/s) −9.80 m/s 2

= 3.06 s.

The time from when it is thrown to when it returns to its original height is twice the time it takes it to reach its maximum height. (d) x − x0 = υ0 xt + 12 axt 2 = (18.0 m/s)(3.06 s) = 55.1 m

3.15. Set Up: Use coordinates with the origin at the initial position of the ball and +y upward. ax = 0 and a y = − 9.80 m/s 2 . The trajectory of the ball is sketched in the figure below.

Solve: (a) From the figure above, υ0 x = υ0 cos57° = 13 m/s and υ0 y = υ0 sin 57° = 20 m/s. (b) At the maximum height υ y = 0, so υ y 2 = υ0 y 2 + 2a y ( y − y0 ) with y = h and y0 = 0 gives


Motion in a Plane

h= (c) υ y = υ0 y + a yt gives t =

υ y − υ0 y ay

=

υ y 2 − υ0 y 2

=

2a y

0 − 20 m/s 29.80 m/s 2

0 − (20 m/s) 2 2( −9.80 m/s 2 )

3-7

= 20 m

= 2.0 s

G G (d) a = g , downward at all points of the trajectory. At the maximum height υ y = 0 and υ x = υ0 x , so υ = 13 m/s, horizontal. (e) The total time in the air is twice the time to the maximum height, so is 4.0 s. (f) x − x0 = υ0 xt + 12 axt 2 with x = R and x0 = 0 gives R = υ0 xt = (13 m/s)(4.0 s) = 52 m. Reflect: Our results for h and R agree with the general expressions derived in Example 3.5.

3.16. Set Up: The flare moves with projectile motion. The equations derived in Example 3.5 can be used to find the v0 2 sin 2 θ0 v0 2 sin 2θ0

maximum height h and range R. From Example 3.5, h = Solve: (a) h =

(125 m/s) 2 (sin 55.0°)2 2(9.80 m/s 2 )

= 535 m. R =

and R =

2g

(125 m/s)2 (sin110.0°) 9.80 m/s 2

g

.

= 1500 m.

⎛ 9.80 m/s 2 ⎞ (535 m) = 3140 m and (b) h and R are proportional to 1/g , so on the Moon, h = ⎜ ⎜ 1.67 m/s 2 ⎟⎟ ⎝ ⎠ ⎛ 9.80 m/s 2 ⎞ R =⎜ (1500 m) = 8800 m. ⎜ 1.67 m/s 2 ⎟⎟ ⎝ ⎠ Reflect: The projectile travels on a parabolic trajectory. It is incorrect to say that h = ( R /2) tan θ0 .

3.17. Set Up: Take +y upward. ax = 0,

a y = + 9.80 m/s 2 . υ0 x = υ0 cosθ0 = (30.0 m/s)cos36.9D = 24.0 m/s.

υ0 y = υ0 sin θ0 = (30.0 m/s)sin 36.9° = 18.0 m/s Solve: (a) y − y0 = υ0 yt + 12 a yt 2 with y − y0 = 10.0 m gives t=

(18.0 m/s) ± (18.0 m/s)2 − 2(9.80 m/s 2 )(10.0 m)

(b) The x component of velocity will be υ0 x

= 0.68 s, 2.99 s 9.80 m/s 2 = 24.0 m/s at all times. The y-component, obtained from υ y = υ0 y + a yt ,

is 11.3 m/s at the earlier time and −11.3 m/s at the later. (c) The magnitude is the same, 30.0 m/s, but the direction is now 36.9° below the horizontal.

3.18. Set Up: Use coordinates with the origin at the initial position of the stone and +y downward. ax = 0, a y = − 9.80 m/s 2 . The stone has υ0 x = 15.0 m/s and υ0 y = 20.0 m/s. Solve: (a) Find the vertical distance the rock travels in 4.00 s. y = y0 + υ0 yt + 12 a yt 2 = (20.0 m/s)(4.00 s) + 12 (9.80 m/s 2 )(4.00 s)2 = 158 m (b) x − x0 = υ0 xt + 12 axt 2 = (15.0 m/s)(4.00 s) = 60.0 m (c) The basket descends (20.0 m/s)(4.00 s) = 80.0 m in 4.00 s so is 158 m − 80.0 m = 78.0 m above the ground when

the stone reaches the ground. The stone is a distance d = x 2 + y 2 = (60.0 m) 2 + (78.0 m)2 = 98.4 m from the basket when it hits the ground.

*3.19. Set Up: The horizontal displacement when the ball returns to its original height is υ0 2 sin (2θ0 ) R=

g

.


3-8

Chapter 3

At its maximum height υ y = 0. g = 32 ft/s 2 . ax = 0, a y = − g . Solve: (a) υ0 =

Rg = sin (2θ 0 )

(375 ft)(32 ft/s 2 ) = 118 ft/s sin (60.0°)

(b) υ0 y = υ0 sin θ 0 = (118 ft/s)sin 30.0° = 59 ft/s. υ y 2 = υ0 y 2 + 2a y ( y − y0 ).

y − y0 =

υ y 2 − υ0 y 2

0 − (59 ft/s) 2

=

2a y

2(−32 ft/s 2 )

= 54.4 ft

Reflect: At the maximum height υ y = 0. But υ ≠ 0 there because the ball still has its constant horizontal

component of velocity. The horizontal range equation,

υ0 2 sin (2θ0 )

R=

, g can be used only when the initial and final points of the motion are at the same elevation.

3.20. Set Up: Take +y upward. ax = 0, a y = − 9.80 m/s 2 . υ0 x = υ0 cosθ 0 = 25.2 m/s. υ0 y = υ0 sin θ0 = 16.3 m/s. Solve: (a) With υ y = 0 in υ y 2 = υ0 y 2 + 2a y ( y − y0 ),

y − y0 = −

υ0 y 2 2a y

=−

(16.3 m/s) 2 2(−9.80 m/s 2 )

= 13.6 m

(b) υ x = υ0 y = 25.2 m/s. With y − y0 = − 15.0 m, υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ y = − υ0 y 2 + 2a y ( y − y0 ) = − (16.3 m/s) 2 + 2(−9.80 m/s 2 )(−15.0 m) = − 23.7 m/s. Then υ = υ x 2 + υ y 2 = 34.6 m/s. (c) Use the vertical motion to find the time in the air:

υ y = υ0 y + a yt gives t =

υ y − υ0 y ay

=

−23.7 m/s − 16.3 m/s −9.80 m/s 2

= 4.08 s.

Then x − x0 = υ0 xt + 12 axt 2 = (25.2 m/s)(4.08 s) = 103 m.

*3.21. Set Up: Use coordinates with the origin at the ground and +y upward. ax = 0, a y = −9.80 m/s 2 . At the maximum height υ y = 0. Solve: (a) υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ y = −2a y ( y − y0 ) = −2( −9.80 m/s 2 )(0.587 m) = 3.39 m/s υ0 y = υ0 sin θ0 so υ0 =

υ0 y sin θ0

=

3.39 m/s = 4.00 m/s sin 58.0°

(b) Use the vertical motion to find the time in the air. When the froghopper has returned to the ground, y − y0 = 0.

y − y0 = υ0 yt + 12 a yt 2 gives t = −

2υ0 y ay

=−

2(3.39 m/s) −9.80 m/s 2

= 0.692 s

Then x − x0 = v0 xt + 12 axt 2 = (υ0 cosθ0 )t = (4.00 m/s)(cos58.0°)(0.692 s) = 1.47 m Reflect: υ y = 0 when t = −

υ0 y ay

=−

3.39 m/s −9.80 m/s 2

= 0.346 s. The total time in the air is twice this.

3.22. Set Up: Use coordinates with the origin at the ground and +y upward. ax = 0, a y = −9.80 m/s 2 . Solve: (a) υ y = 0 when y − y0 = 0.0674 m © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Motion in a Plane

3-9

υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives υ0 y = 22a y ( y − y0 ) = −2(−9.80 m/s 2 )(0.0674 m) = 1.15 m/s υ0 y = υ0 sin θ0 so υ0 =

υ0 y

=

sin θ0

1.15 m/s = 1.50 m/s sin 50.0°

(b) Use the horizontal motion to find the time in the air. The grasshopper travels horizontally x − x0 = 1.06 m.

x − x0 = υ0 xt + 12 axt 2 gives t=

x − x0

υ0 x

=

x − x0 = 1.10 s υ0 cos50.0°

Find the vertical displacement of the grasshopper at t = 1.10 s: y − y0 = υ0 yt + 12 a yt 2 = (1.15 m/s)(1.10 s) + 12 (−9.80 m/s 2 )(1.10 s) 2 = − 4.66 m The height of the cliff is 4.66 m.

3.23. Set Up: Let +y be upward. ax = 0, a y = − 9.80 m/s 2 . υ0 x = υ0 cosθ0 , υ0 y = υ0 sin θ 0 . Solve: (a) x − x0 = υ0 xt + 12 axt 2 gives x − x0 = υ0 (cosθ0 )t and

cosθ0 =

45.0 m = 0.600; θ 0 = 53.1D (25.0 m/s)(3.00 s)

(b) At the highest point υ x = υ0 x = (25.0 m/s)cos53.1° = 15.0 m/s, υ y = 0 and υ = υ x 2 + υ y 2 = 15.0 m/s. At all points

in the motion, a = 9.80 m/s 2 downward. (c) Find y − y0 when t = 3.00 s:

y − y0 = υ0 yt + 12 a yt 2 = (25.0 m/s)(sin 53.1°)(3.00 s) + 12 (−9.80 m/s 2 )(3.00 s)2 = 15.9 m

υ x = υ0 x = 15.0 m/s, υ y = υ0 y + a yt = (25.0 m/s)(sin 53.1°) − (9.80 m/s 2 )(3.00 s) = − 9.41 m/s, and υ = υ x 2 + υ y 2 = (15.0 m/s) 2 + ( −9.41 m/s) 2 = 17.7 m/s Reflect: The acceleration is the same at all points of the motion. It takes the water υ0 y 20.0 m/s =− = 2.04 s t=− ay −9.80 m/s 2

to reach its maximum height. When the water reaches the building it has ed its maximum height and its vertical component of velocity is downward.

3.24. Set Up: Example 3.5 derives R =

υ0 2 sin 2θ 0 g

.

Solve: The maximum range occurs when sin 2θ 0 = 1, 2θ 0 = 90° and θ 0 = 45D.

*3.25. Set Up: Example 3.5 gives R =

υ02 sin 2θ 0 g

and t =

2υ0 sin θ 0 . g

Solve: (a) The maximum range occurs when sin 2θ 0 = 1, 2θ 0 = 90° and θ 0 = 45D. (b) Rmax =

υ0 2

(25.0 m/s) 2

= 63.8 m 9.80 m/s 2 2(25.0 m/s)sin 45° = 3.61 s (c) t = g g

=

Reflect: The time that the balloon is in the air can also be calculated from t =

Δx

υ0 x

=

63.8 m (25.0 m/s)cos 45D

= 3.61 s.


3-10

Chapter 3

3.26. Set Up: Example 3.5 derives R = Solve: Rmax =

υ0 2 g

υ02 sin 2θ 0 g

.

.

Rmax is proportional to v0 2 so when v0 is tripled, Rmax is increased by a factor of 9. Rmax = 9(1500 m) =

1.35 × 104 m = 13.5 km.


υ02 sin 2θ 0 g

. The maximum allowed velocity occurs when R = Rmax = 3.50 m.

Rmax g (3.50 m)(9.80 m/s 2 ) = = 6.69 m/s sin 2θ sin130D Reflect: We can only use the range equation in the special case where the projectile lands at the same elevation from which it was fired (i.e., when Δy = 0 ). Fortunately this is true for the surface of the water.

Solve: Solving for υ0 we obtain: υ0,max =


υ0 2 sin 2θ 0 g

.

Solve: Use the 45° data:

υ0 2 g

=

R 225 m = = 225 m. sin 2θ 0 sin 90°

Then for θ 0 = 60.0°, R = (225 m)sin120.0° = 195 m. The distance between the two arrows when they land is 225 m − 195 m = 30 m

*3.29. Set Up: We can determine the initial velocity of the bottle rocket from its maximum vertical height by using υ0 2 sin 2θ 0 the equation h =

determined from R =

, which is derived in Example 3.5. The maximum range of the projectile can be

2g

υ02 sin 2θ 0 g

, which is also derived in Example 3.5.

Solve: When fired vertically we have θ 0 = 90D and so h =

υ0 2 sin 2 90D 2g

=

υ02 2g

. Solving for υ0 we obtain υ0 = 2 gh .

The maximum range occurs when sin 2θ0 = 1 so θ 0 = 45D.

υ0 2 sin 2θ 0

(2gh)sin 90D = 2h = 50.0 m. g g Reflect: When a projectile is fired at a fixed initial speed without air resistance, the maximum range that it can achieve is twice the maximum height that it achieves when fired vertically.

Substituting this value into R =

we obtain R =

3.30. Set Up: The suitcase moves in projectile motion. The initial velocity of the suitcase equals the velocity of the airplane. Take +y to be upward. a x = 0, a y = − g . Solve: Use the vertical motion to find the time it takes the suitcase to reach the ground: v0 y = v0 sin23°, a y = 29.80 m/s 2 , y − y0 = 2114 m, t = ? y − y0 = v0 yt + 12 a yt 2 gives t = 9.60 s.

The distance the suitcase travels horizontally is x − x0 = v0 x = (v0 cos23.0°)t = 795 m. Reflect: An object released from rest at a height of 114 m strikes the ground at t =

2( y − y0 ) = 4.82 s. The −g

suitcase is in the air much longer than this since it initially has an upward component of velocity.


Motion in a Plane

3-11

*3.31. Set Up: The acceleration is arad . R = 0.75 m. Solve: arad =

υ2

so υ = Rarad = (0.75 m)(9.8 m/s 2 ) = 2.7 m/s

R

3.32. Set Up: The acceleration is arad . Part (a): T = 24 h = 8.64 × 104 s and R = 6.38 × 106 m. Part (b): T = 0.41 day = 9.84 h = 3.54 × 104 s and R = 7.18 × 107 m. Solve: (a) arad = (b) arad =

4π 2 R T2

4π 2 R

=

T

2

=

4π 2 (6.38 × 106 m) (8.64 × 104 s) 2

4π 2 (7.18 × 107 m) (3.54 × 104 s)2

= 0.0337 m/s 2 = 3.44 × 1023 g

= 2.3 m/s 2 = 0.23 g

*3.33. Set Up: The radius of the earth’s orbit is r = 1.50 × 1011 m and its orbital period is T = 365.3 days = 3.16 × 107 s. For mercury, r = 5.79 × 1010 m and T = 88.0 days = 7.60 × 106 s. Solve: (a) υ = (b) arad =

υ2 r

2π r = 2.98 × 104 m/s T

= 5.91 × 10−3 m/s 2 .

(c) υ = 4.79 × 104 m/s, and arad = 3.96 × 10−2 m/s 2 .

3.34. Set Up: 550 rev/min = 9.17 rev/s, corresponding to a period of 1 = 0.109 s. 9.17 rev/s Each blade tip moves in a circle of radius R = 3.40 m. T=

Solve: (a) υ = (b) arad =

υ2 R

2π R = 196 m/s. T

= 1.13 × 104 m/s 2 = 1.15 × 103 g .

3.35. Set Up: R = 0.150 m. T = 1 min = 60.0 s. 4π 2 R

4π 2 (0.150 m)

= 1.64 × 1023 m/s 2 T2 (60.0 s) 2 Reflect: Different points in the hand travel in circles of different radii but have the same period so they have different arad . arad is largest for the tip, since it has the largest R. Solve: arad =

=

3.36. Set Up: The centripetal acceleration for circular motion at a constant speed is given by arad = appendix we have 1 mph = 0.4470 m/s. Solve: First convert speeds from mph into m/s: ⎛ 0.4470 m/s ⎞ ⎛ 0.4470 m/s ⎞ (35 mph) ⎜ ⎟ = 15.6 m/s and (50 mph) ⎜ ⎟ = 22.4 m/s. Thus, we have ⎝ 1 mph ⎠ ⎝ 1 mph ⎠ arad =

υ2 r

=

(15.6 m/s )2 50.0 m

= 4.9 m/s 2 and arad =

( 22.4 m/s )2 50.0 m

υ2 r

. From the

=10 m/s 2 .


3-12

Chapter 3

Reflect: Since arad ∝ υ 2 , we can also calculate the second acceleration from the first acceleration: 2

⎛ 50 mph ⎞ 2 2 ⎜ ⎟ (4.9 m/s ) = 10 m/s . 35 mph ⎝ ⎠

*3.37. Set Up: R = 0.070 m. For 3.0 rev/s, the period T (time for one revolution) is T = (1.0 s)/(3.0 rev) = 0.333 s. Solve: arad =

4π 2 R T2

=

4π 2 (0.070 m) (0.333 s)2

= 25 m/s 2 = 2.5 g

Reflect: The acceleration is large and the force on the fluid must be 2.5 times its weight.

3.38. Set Up: arad = 5.5 g = 53.9 m/s 2 . 1 mph = 0.4470 m/s. Solve: arad =

υ2 R

so υ = Rarad = (350 m)(53.9 m/s 2 ) = 140 m/s = 310 mph G

*3.39. Set Up: Apply the relative velocity relation. The relative velocities are vC/E , the canoe relative to the earth, G G v R/E , the velocity of the river relative to the earth and vC/R , the velocity of the canoe relative to the river. G G G G G G G Solve: vC/E = vC/R + v R/E and therefore vC/R = vC/E − v R/E . The velocity components of vC/R are −0.50 m/s +

(0.40 m/s)/ 2, east and (0.40 m/s)/ 2, south, for a velocity relative to the river of 0.36 m/s, at 52.5D south of west. Reflect: The velocity of the canoe relative to the river has a smaller magnitude than the velocity of the canoe relative to the earth.

G

G

3.40. Set Up: The relative velocities are the water relative to the earth, υ W/E , the boat relative to the water, υB/W , G G G and the boat relative to the earth, υB/E . υB/W is due east and has magnitude 4.2 m/s. υ W/E is due south and has G G G magnitude 2.0 m/s. υB/E = υB/W + υ W/E . The relative velocity addition diagram is given in the figure below.

Solve: (a) υB/E = υ W/E 2 + υB/W 2 = (2.0 m/s) 2 + (4.2 m/s) 2 = 4.65 m/s. tan θ =

υ W/E 2.0 m/s = . θ = 25.5°, south of east. υB/W 4.2 m/s

G (b) To cross the river the boat must travel 800 m east, relative to the earth. The eastward component of y B/E is

υB/W = 4.2 m/s. t=

800 m = 190 s. 4.2 m/s

G (c) The component of y B/E in the southward direction is υW/E = 2.0 m/s. In the 190 s it takes him to cross the river he travels south a distance of (2.0 m/s)(190 s) = 380 m.


Motion in a Plane

G

3-13

G

*3.41. Set Up: The relative velocities are the water relative to the earth, υ W/E , the boat relative to the water, υB/W , G G G and the boat relative to the earth, υB/E . υB/E is due east, υ W/E is due south and has magnitude G G G G 2.0 m/s. υB/W = 4.2 m/s. υB/E = υB/W + υ W/E . The velocity addition diagram is given in the figure below.

G υ 2.0 m/s Solve: (a) Find the direction of ν B/W . sin θ = W/E = . θ = 28.4°, north of east. υB/W 4.2 m/s (b) υB/E = υB/W 2 − υ W/E 2 = (4.2 m/s) 2 − (2.0 m/s) 2 = 3.7 m/s (c) t =

800 m

υB/E

=

800 m = 216 s. 3.7 m/s

Reflect: It takes longer to cross the river in this problem than it did in Problem 3.40. In the direction straight across the river (east) the component of his velocity relative to the earth is less than 4.2 m/s.

G

3.42. Set Up: The relative velocities are the raindrops relative to the earth, υ R/E , the raindrops relative to you, G G G G G G υ R/Y , and you relative to the earth, υ Y/E : υ R/E = υ R/Y + υ Y/E . Suppose that you are running due east so that υ Y/E is

G G due east and has magnitude 5.0 m/s. Thus, υ R/Y is 30D west of vertical and υ R/E is vertical (since there is no wind).

The relative velocity addition diagram is shown below. Solve: υR/E =

υY/E

tan 30.0°

=

5.0 m/s = 8.7 m/s. tan 30°

Reflect: We can also find the magnitude of the speed of the raindrops relative to you:

υR/Y =

υY/E

sin 30°

=

5.0 m/s = 10 m/s. sin30°

G

3.43. Set Up: υB/A = 100 km/h. y A/G = 40 km/h, east.


3-14

Chapter 3

G G G G Solve: y B/G = y B/A + y A/G . We want υ B/G to be due south. The relative velocity addition diagram is shown in the

figure below.

(a) sin φ =

υA/G 40 km/h = , φ = 24°, west of south. υB/A 100 km/h

(b) υB/G = υB/A 2 − υA/G 2 = 91.7 km/h. t =

d

υB/G

=

500 km = 5.5 h 91.7 km/h

Reflect: The speed of the bird relative to the ground is less than its speed relative to the air. Part of its velocity relative to the air is directed to oppose the effect of the wind.

3.44. Set Up: Once the rocket leaves the incline it moves in projectile motion. The acceleration along the incline determines the initial velocity and initial position for the projectile motion. For motion along the incline let +x be

directed up the incline. vx 2 = v0 x 2 + 2a x ( x − x0 ) gives vx = 2(1.25 m/s 2 )(200 m) = 22.36 m/s. When the projectile motion begins the rocket has v0 = 22.36 m/s at 35.0D above the horizontal and is at a vertical height of (200.0 m)sin 35.0° = 114.7 m. For the projectile motion let +x be horizontal to the right and let +y be upward. Let

y=0

at the ground. Then

y0 = 114.7 m, v0 x = v0 cos35.0D = 18.32 m/s, v0 y = v0 sin 35.0D = 12.83 m/s, ax = 0,

a y = −9.80 m/s 2 . Let x = 0 at point A, so x0 = (200.0 m)cos35.0° = 163.8 m. Solve: (a) At the maximum height we have v y = 0. v y 2 = v0 y 2 + 2a y ( y − y0 ) gives

y − y0 =

v y 2 − v0 y 2 2a y

=

0 − (12.83 m/s) 2 2(−9.80 m/s 2 )

= 8.40 m and y = 114.7 m + 8.40 m = 123 m. The maximum height above

ground is 123 m. (b) The time in the air can be calculated from the vertical component of the projectile motion: y − y0 = 2114.7 m, v0 y = 12.83 m/s, a y = 29.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives (4.90 m/s 2 )t 2 − (12.83 m/s)t − 114.7 m. The quadratic formula gives t =

(

)

1 12.83 ± (12.83) 2 + 4(4.90)(114.7) s. The positive root is t = 6.32 s. Then x − x0 = 9.80

v0 xt + 12 axt 2 = (18.32 m/s)(6.32 s) = 115.8 m and x = 163.8 m + 115.8 m = 280 m. The horizontal range of the rocket is 280 m. Reflect: The expressions for h and R derived in Example 3.5 do not apply here. They are only for a projectile fired on level ground.

*3.45. Set Up: For the motion of the football take +y upward. υ0 x = υ0 cos 40.0° = 9.19 m/s, υ0 y = υ0 sin 40.0° = 7.71 m/s, ax = 0, and a y = − 9.80 m/s 2 . Use the y component motion to find the time the football is in the air. This same time applies to the horizontal motion. Solve: When the ball returns to the ground, y − y0 = 0. y − y0 = υ0 yt + 12 a yt 2 gives 0 = υ0 yt + 12 a yt 2 and


Motion in a Plane

t=−

2υ0 y ay

=−

2(7.71 m/s) 29.80 m/s 2

3-15

= 1.57 s.

Then x − x0 = υ0 xt + 12 axt 2 = (9.19 m/s)(1.57 s) = 14.4 m. In the 1.57 s that the ball is in the air the second player must travel 30.0 m − 14.4 m = 15.6 m. Therefore, his (constant) velocity must be

υx =

x − x0 15.6 m = = 9.94 m/s. t 1.57 s

3.46. Set Up: The range for a projectile that lands at the same height from which it was launched is R =

υ02 sin 2θ g

.

The maximum range is for θ = 45°. Solve: Assuming θ = 45°, and R = 50 m, υ0 = gR = 22 m/s. Reflect: We have assumed that debris was launched at all angles, including the angle of 45° that gives maximum range.

*3.47. Set Up: Take +y downward. υ0 x = 64.0 m/s , υ0 y = 0. Solve: Use the vertical motion to find the time in the air: y − y0 = υ0 yt + 12 a yt 2 with y − y0 = 90.0 m gives

t=

2( y − y0 ) = ay

2(90.0 m) 9.80 m/s 2

= 4.29 s.

Then x − x0 = (64.0 m/s)(4.29 s) = 275 m.

3.48. Set Up: Use coordinates with +y upward and +x horizontal. The bale has υ0 x = υ0 cosθ 0 = (75 m/s)cos55° = 43.0 m/s and υ0 y = υ0 sin θ 0 = (75 m/s)sin 55° = 61.4 m/s. y0 = 150 m and y = 0. Solve: Use the vertical motion to find t: y − y0 = υ0 yt + 12 a yt 2 gives −150 m = (61.4 m/s)t − (4.90 m/s 2 )t 2 . The

quadratic formula gives t = 6.27 ± 8.36 s. The physical value is the positive one, and t = 14.6 s. Then

x − x0 = υ0 xt + 12 axt 2 = (43.0 m/s)(14.6 s) = 630 m. Reflect: If the airplane maintains constant velocity after it releases the bales, it will also travel horizontally 630 m during the time it takes the bales to fall to the ground, so the airplane will be directly over the impact spot when the bales land.

3.49. Set Up: Let +y be upward and +x be to the right. The missile has υ0 x = 30.0 m/s, υ0 y = 40.0 m/s, ax = 0 and a y = − 9.80 m/s 2 . The cart has ax = 0 and υ0 x = 30.0 m/s. Solve: (a) At the maximum height, υ y = 0.


υ y 2 − υ0 y 2 2a y

=

0 − (40.0 m/s) 2 2( −9.80 m/s 2 )

= 81.6 m

(b) Find t for y − y0 = 0 (returns to initial level).

y − y0 = υ0 yt + 12 a yt 2 gives t = −

2υ0 y ay

=−

2(40.0 m/s) −9.80 m/s 2

= 8.16 s

Then x − x0 = υ0 xt + 12 axt 2 = (30.0 m/s)(8.16 s) = 245 m (c) The missile also travels horizontally 245 m so the missile lands in the cart.


3-16

Chapter 3

3.50. Set Up: Let +y be upward. y − y0 = − 0.9 m when x − x0 = 188 m. With x0 = y0 = 0, x and y are related by y = (tan θ0 ) x −

g 2

2υ0 cos 2 θ0

x2 .

Solve: (a) θ 0 = 45° so tan θ0 = 1 and cosθ0 = 12 . y = x −

gx 2 = x−y

υ0 =

gx 2

υ0 2

.

(9.80 m/s 2 )(188 m) 2 = 42.8 m/s 188 m + 0.9 m

(b) For x = 116 m,

y=x−

gx 2

(9.80 m/s 2 )(116 m) 2

= 44.0 m. (42.8 m/s) 2 The ball will be 44.9 m above the ground and therefore 41.9 m above the top of the fence.

υ0 2

*3.51. Set Up: Example 3.5 gives R =

= 116 m −

υ0 2 sin 2θ 0 g

.

Solve: We have the maximum range when θ 0 = 45° so Rmax =

υ0 2 g mars

=

(70.0 m/s) 2 3.71 m/s 2

= 1.32 × 103 m.

Reflect: The maximum range is inversely proportional to g. Thus, the range on the earth would be ⎛g ⎞ (1320 m) ⎜ mars ⎟ = 500 m, ignoring air resistance. g ⎝ earth ⎠

3.52. Set Up: Let +y be upward. ax = 0, a y = − 9.80 m/s 2 . With x0 = y0 = 0, x and y are related by y = (tan θ 0 ) x −

g 2υ02 cos 2 θ 0

x2.

θ 0 = 60.0°. y = 8.00 m when x = 18.0 m. Solve: (a) υ0 =

gx 2 2

2(cos θ 0 )( x tan θ 0 − y )

= 16.6 m/s

(b) υ x = υ0 x = υ0 cosθ 0 = 8.3 m/s. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ y = − (υ0 sin θ0 ) 2 + 2a y ( y − y0 ) = − (14.4 m/s) 2 + 2( −9.80 m/s 2 )(8.00 m) = − 7.1 m/s υ = υ x 2 + υ y 2 = 10.9 m/s. tan θ =

| υ y| | υ x|

=

7.1 and θ = 40.5°, below the horizontal. 8.3

Reflect: We can check our calculated v0 . t=

Then y − y0 = υ0 yt +

1 a t2 2 y

x − x0

υ0 x

=

18.0 m = 2.17 s. 8.3 m/s

= (14.4 m/s)(2.17 s) − (4.9 m/s 2 )(2.17 s) 2 = 8 m, which checks.

*3.53. Set Up: Use coordinates with the origin at the boy and with +y downward. The ball has υ0 y = 0, υ0 y = 0, υ0 x = 8.50 m/s, ax = 0 and a y = 9.80 m/s 2 . Solve: (a) The dog must travel horizontally the same distance the ball travels horizontally, so the dog must have speed 8.50 m/s. (b) Use the vertical motion of the ball to find its time in the air. y − y0 = υ0 yt + 12 a yt 2 gives


Motion in a Plane

t=

3-17

2( y − y0 ) 2(12.0 m) = = 1.56 s ay 9.80 m/s 2

Then x − x0 = υ0 xt + 12 axt 2 = (8.50 m/s)(1.56 s) = 13.3 m

3.54. Set Up: Use coordinates with the origin at the boy and with +y upward. The ball has υ0 x = υ0 cosθ 0 = (8.50 m/s)cos 60.0D = 4.25 m/s, υ0 y = υ0 sin θ0 = (8.50 m/s)sin 60.0D = 7.36 m/s, ax = 0 and a y = − 9.80 m/s 2 .

Solve: (a) The dog must travel horizontally the same distance the ball travels horizontally, so the dog must have speed 4.25 m/s. (b) Use the vertical motion of the ball to find its time in the air. y − y0 = υ0 yt + 12 a yt 2 gives

−12.0 m = (7.36 m/s)t − (4.90 m/s 2 )t 2 . The quadratic formula gives t = 0.751 ± 1.74 s. The negative value is unphysical, so t = 2.49 s. Then x − x0 =

υ0 xt + 12 axt 2 = (4.25 m/s)(2.49 s) = 10.6 m Reflect: The ball is in the air longer than when it is thrown horizontally (Problem 3.56), but it doesn’t travel as far horizontally. The dog doesn’t have to run as far or as fast as in Problem 3.53.

3.55. Set Up: Use coordinates with +y upward. Once the water leaves the cannon it is in free-fall and has ax = 0 and a y = −9.80 m/s 2 . The water has υ0 x = υ0 cosθ 0 = 15.0 m/s and υ0 y = υ0 sin θ0 = 20.0 m/s. Solve: Use the vertical motion to find t that gives y − y0 = 10.0 m: y − y0 = υ0 yt + 12 a yt 2 gives

10.0 m = (20.0 m/s)t − (4.90 m/s 2 )t 2 . The quadratic formula gives t = 2.04 ± 1.45 s, and t = 0.59 s or t = 3.49 s. Both answers are physical. For t = 0.59 s, x − x0 = υ0 xt = (15.0 m/s)(0.59 s) = 8.8 m. For t = 3.49 s, x − x0 = υ0 xt = (15.0 m/s)(3.49 s) = 52.4 m. When the cannon is 8.8 m from the building the water hits this spot on the wall on its way up to its maximum height. When is it 52.4 m from the building it hits this spot after it has ed through its maximum height.

3.56. Set Up: Use coordinates with +y upward. υ0 x = υ0 cosθ0 = (100.0 m/s)cos60.0° = 50.0 m/s; υ0 y = υ0 sin θ0 = (100.0 m/s)sin 60.0D = 86.7 m/s, ax = 0 and a y = −9.80 m/s 2 . Solve: υ x = υ0 x since ax = 0. υ y = υ0 y + a yt = 86.7 m/s − (9.80 m/s 2 )t . The graphs of υ x versus t and υ y versus t

are given in the figure below. The graph of υ x is a straight horizontal line and the graph of υ y is a straight line with slope −9.80 m/s 2 . υ y = 0 at the maximum height.

*3.57. Set Up: Let +y be downward. ax = 0, a y = + 9.80 m/s 2 . υ0 x = υ0 cosθ0 = 5.36 m/s, υ0 y = υ0 sin θ0 = 4.50 m/s.


3-18

Chapter 3

Solve: Use the vertical motion to find the time in the air: y − y0 = υ0 yt + 12 a yt 2 with y − y0 = 14.0 m gives 14.0 m = (4.50 m/s)t + (4.9 m/s 2 )t 2 . The quadratic formula gives

t=

(

)

1 −4.50 ± (4.50) 2 − 4(4.9)(−14.0) s. 2(4.9)

The positive root is t = 1.29 s. Then x − x0 = υ0 xt + 12 axt 2 = (5.36 m/s)(1.29 s) = 6.91 m.

3.58. Set Up: The velocity has a horizontal tangential component and a vertical component. The vertical component υx2 of acceleration is zero and the horizontal component is arad =

R

. Let +y be upward and +x be in the direction of the

tangential velocity at the instant we are considering. Solve: (a) The bird’s tangential velocity can be found from circumference 2π (8.00 m) 50.27 m υx = = = = 10.05 m/s time of rotation 5.00 s 5.00 s Thus its velocity consists of the components υ x = 10.05 m/s and υ y = 3.00 m/s. The speed relative to the ground is then υ = υ x2 + υ y2 = 10.5 m/s. (b) The bird’s speed is constant, so its acceleration is strictly centripetal—entirely in the horizontal direction, toward υ 2 (10.05 m/s)2 = 12.6 m/s 2 . the center of its spiral path—and has magnitude arad = x = r 8.00 m 3.00 m/s (c) Using the vertical and horizontal velocity components θ = tan21 = 16.6°. 10.05 m/s Reflect: The angle between the bird’s velocity and the horizontal remains constant as the bird rises.

*3.59. Set Up: The water moves in projectile motion. Let x0 = y0 = 0 and take +y to be positive. ax = 0, a y = − g . Solve: The equations of motions are y = (v0sinθ0 )t − 12 gt 2 and x = (v0cos θ 0 )t. When the water goes in the tank for

the minimum velocity, y = 2 D and x = 6 D. When the water goes in the tank for the maximum velocity, y = 2 D and x = 7 D. In both cases we have θ 0 = 45D , so sin θ 0 = cosθ 0 = 2 /2. To reach the minimum distance: 6 D =

2 2 v0t , and 2 D = v0t − 12 gt 2 . Solving the first equation for t gives 2 2 2

t=

⎛ 6D 2 ⎞ 6D 2 . Substituting this into the second equation gives 2 D = 6 D − 12 g ⎜⎜ ⎟⎟ . Solving this for v0 gives v0 ⎝ v0 ⎠

v0 = 3 gD . To reach the maximum distance: 7 D =

2 2 v0t , and 2 D = v0t − 12 gt 2 . Solving the first equation for t gives 2 2 2

t=

⎛ 7D 2 ⎞ 7D 2 . Substituting this into the second equation gives 2 D = 7 D − 12 g ⎜⎜ ⎟⎟ . Solving this for v0 gives v0 ⎝ v0 ⎠

v0 = 49 gD/ 5 = 3.13 gD , which, as expected, is larger than the previous result.

Reflect: A launch speed of v0 = 6 gD = 2.45 gD is required for a horizontal range of 6D. The minimum speed

required is greater than this, because the water must be at a height of at least 2D when it reaches the front of the tank.

3.60. Set Up: Use coordinates with the origin at the ground and +y upward. The shot put has y0 = 2.00 m, υ0 x = υ0 cos θ 0 , υ0 y = υ0 sin θ 0 , ax = 0 and a y = − 9.80 m/s 2 . 1 mph = 0.4470 m/s Solve: x − x0 = υ0 xt + 12 axt 2 gives 23.11 m = (υ0 cos 40.0°)t and υ0t = 30.17 m


Motion in a Plane

3-19

y − y0 = υ0 yt + 12 a yt 2 gives 0 = 2.00 m + (υ0 sin 40.0°)t − (4.90 m/s 2 )t 2 . Use υ0t = 30.17 m and solve for t. This gives t = 2.09 s. Then υ0 =

30.17 m = 14.4 m/s = 32.2 mph 2.09 s

Reflect: Since the initial and final heights are not the same, the expression R =

υ0 2 sin (2θ0 ) g

does not apply.

*3.61. Set Up: For circular motion the acceleration has magnitude υ2 arad =

R and is directed toward the center of the circle. The period T (time for one revolution) is 2π R T= .

υ

Solve: (a) arad

υ2

(7.00 m/s) 2 = = = 3.50 m/s 2 , upward. R 14.0 m

(b) arad = 3.50 m/s 2 , downward. (c) T =

2π R

υ

=

2π (14.0 m) = 12.6 s. 7.00 m/s

3.62. Set Up: Let +y be upward. ax = 0, a y = −9.80 m/s 2 . With x0 = y0 = 0, x and y are related by y = (tan θ 0 ) x −

g 2

2

2υ0 cos θ 0

x2.

Solve: (a) x = 40.0 m when y = − 15.0 m.

υ0 =

gx 2

=

2

2(cos θ0 )( x tan θ 0 − y )

(9.80 m/s 2 )(40.0 m) 2 2(cos53.0°) 2 ([40.0 m]tan 53.0° + 15.0 m)

= 17.8 m/s

(b) Find x for v0 = 8.9 m/s and y = − 100 m.

y = (tan θ0 ) x −

g 2

2

x 2 gives −100 = 1.33 x − 0.171x 2 .

2v0 cos θ0 The quadratic formula gives x = 28.4 m. He lands in the river 28.4 m from the near bank. Reflect: Before doing the calculation for part (b) we don’t know that he landed in the river. If we had found x > 40 m for y = − 100 m, we would know he hit the vertical face of the far bank before he reached the water.

3.63. Set Up: Take +y downward. υ0 x = υ0 , υ0 y = 0. ax = 0, a y = +9.80 m/s 2 . Solve: (a) Use the vertical motion to find the time for the boulder to reach the level of the lake: y − y0 = υ0 yt + 12 a yt 2 with

y − y0 = +20 m gives t=

2( y − y0 ) 2(20 m) = = 2.02 s. ay 9.80 m/s 2

The rock must travel horizontally 100 m during this time. x − x0 = υ0 xt + 12 axt 2 gives x − x0 100 m = = 49 m/s t 2.02 s (b) In going from the edge of the cliff to the plain, the boulder travels downward a distance of y − y0 = 45 m.

υ0 = υ0 x =

t=

2( y − y0 ) 2(45 m) = = 3.03 s ay 9.80 m/s 2

and x − x0 = υ0 xt = (49 m/s)(3.03 s) = 148 m. The rock lands 148 m − 100 m = 48 m beyond the foot of the dam. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3-20

Chapter 3

Reflect: The boulder es over the dam 2.02 s after it leaves the cliff and then travels an additional 1.01 s before landing on the plain. If the boulder has an initial speed that is less than 49 m/s, then it lands in the lake.

3.64. Set Up: Since Δy = 0 we may use the equations derived in Example 3.5 for the time of flight and range of the 2υ0 sin θ0 υ 2 sin 2θ0 , R= 0 . The speed that the outfielder needs can be calculated from this time and the g g initial distance of the outfielder from the location where the ball lands. υ 2 sin 2θ 0 (35.0 m/s) 2 sin130D Solve: The range of the ball is R = 0 = = 95.76 m. The time of flight for the ball is g 9.80 m/s 2 ball: t =

t=

2υ0 sin θ0 2(35.0 m/s)sin 65D = = 6.474 s. The distance that the outfielder must travel in this time is g (9.80 m/s 2 )

distance 25.76 m = = 3.98 m/s. time 6.474 s Reflect: The time of flight can also be calculated directly from the range and the x-component of velocity: 95.76 m R = = 6.47 s. t= υ0 cosθ (35.0 m/s)cos65D 95.76 − 70.0 = 25.76 meters. Thus, the speed required is: speed =

*3.65. Set Up: Use coordinates with +y upward. Both the shell and the projectile have ax = 0 and a y = − 9.80 m/s 2 as they move through the air. Solve: (a) Find the maximum height of the shell: υ0 y = υ0 sin θ 0 = 120 m/s and v y = 0


υ y 2 − υ0 y 2 2a y

=

0 − (120 m/s) 2 2(−9.80 m/s 2 )

= 735 m

At its maximum height the shell has υ x = υ0 x = υ0 cos53° = 90.3 m/s and υ y = 0. Relative to the ground the projectile has velocity components υ0 x = 90.3 m/s + (100.0 m/s)cos30.0° = 176.9 m/s and υ0 y = (100.0 m/s)sin 30.0° = 50.0 m/s Find the maximum height of the projectile: y0 = 735 m. υ y = 0. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives y − y0 =

υ y 2 − υ0 y 2 2a y

=

0 − (50.0 m/s) 2 2(−9.80 m/s 2 )

= 128 m

and the maximum height is y = 735 m + 128 m = 863 m. (b) Find the time for the shell to reach its maximum height: υ y = υ0 y + a yt gives

t=

υ y − υ0 y ay

=

0 − 120 m/s

29.80 m/s 2

= 12.2 s

During this time the shell has traveled horizontally a distance x − x0 = υ0 xt = (υ 0 cosθ 0 )t = (150 m/s)(cos53D )(12.2 s) = 1100 m Find the time for the projectile to reach the ground from the point where it is launched. It starts at y0 = 735 m and ends up at y = 0, so y − y0 = − 735 m.

y − y0 = υ0 yt + 12 a yt 2 gives −735 m = (50.0 m/s)t − (4.90 m/s 2 )t 2 and t = 18.4 s.

During this time the projectile has a horizontal displacement x − x0 = υ0 xt = (176.9 m/s)(18.4 s) = 3255 m. It is at x = x0 + 3255 m = 1100 m + 3255 m = 4360 m from the original launch point when it reaches the ground. Reflect: If the shell didn’t fire the projectile its horizontal range would be υ 2 sin (2θ0 ) (150 m/s)2 (sin106°) R= 0 = = 2210 m. g 9.80 m/s 2 The projectile gets an extra boost when it is fired and travels farther than this.


Motion in a Plane

3-21

Solutions to age Problems 3.66. Set Up: We may use the equation for the maximum height of a projectile derived in Example 3.5. Setting υ 2 sin 2 θ 0 υ0 2 θ 0 = 0D we obtain h = 0 = . Alternatively, we can use the equations for constant acceleration, which 2g

2g

were derived in Chapter 2, to obtain the same result. Assuming υ0 is the same on both the earth and the moon, the maximum height is inversely proportional to the acceleration of gravity at each location. Solve: We compare the maximum height on the moon to that on the earth:

hmoon υ0 2 /2g moon g earth = = = 6. Thus, hearth υ02 /2g earth g moon

we have hmoon = 6hearth = 30 meters. The correct answer is C.

*3.67. Set Up: We may use the equation for the range of a projectile derived in Example 3.5: R =

υ02 sin 2θ0 g

.

Assume that υ0 is the same on both the earth and the moon. The actual value of θ 0 is irrelevant as long as it is the same on both the earth and the moon. Rmoon υ0 2 sin 2θ 0 /g moon g earth = = = 6. Thus, the Rearth υ02 sin 2θ0 /gearth g moon range on the moon is greater than that on the earth by a factor of 6. The correct answer is B.

Solve: We compare the range on the moon to that on the earth:

3.68. Set Up: We may use the equation for the time of flight of a projectile derived in Example 3.5: t =

2υ0 sin θ 0 . g

Assume that υ0 is the same on both the earth and the moon. The actual value of θ 0 is irrelevant as long as it is the same on both the earth and the moon. tmoon 2υ0 sin θ 0 /g moon gearth = = = 6. Thus, tearth 2υ0 sin θ 0 /gearth g moon the time of flight on the moon is greater than that on the earth by a factor of 6. The correct answer is C.

Solve: We compare the time of flight on the moon to that on the earth:

*3.69. Set Up: We may use the equation for the maximum height of a projectile derived in Example 3.5: υ 2 sin 2 θ0 h= 0 . Assume that υ0 is the same on both the earth and the moon. The actual value of θ 0 is irrelevant as 2g long as it is the same on both the earth and the moon.

Solve: We compare the maximum height of the projectile on the moon to that on the earth: hmoon υ02 sin 2 θ0 /2 g moon g earth = = = 6. Thus, the maximum height of the projectile on the moon is greater than that on hearth υ02 sin 2 θ0 /2 gearth g moon the earth by a factor of 6. The correct answer is B.

3.70. Set Up: From Example 3.5, the vertical velocity will be zero when the ball reaches its maximum height, and υ0 sin θ 0 2υ0 sin θ 0 this occurs at tmaximum height =

g

. The time required for the entire flight is tflight =

Solve: We compare these two times to obtain:

tmaximum height tflight

=

g

.

υ0 sin θ 0 /g 1 = . Thus, the time of flight is twice as 2υ0 sin θ 0 /g 2

large as the time to reach the maximum height and zero vertical velocity. The correct answer is C.


4

NEWTON’S LAWS OF MOTION

Answers to Multiple-Choice Problems 1. C 2. C 3. B 15. B 16. C

4. A

5. D

6. C

7. C

8. D

9. A

10. C

11. A, D

12. A

13. B

14. E

Solutions to Problems 4.1. Set Up and Solve: The horizontal component of the force is (10 N)cos 45° = 7.1 N, to the right, and the vertical component is (10 N)sin 45° = 7.1 N, downward. 4.2. Set Up: Take + x to be along the direction of the pull of A. The coordinates and forces are shown in the figure below. Rx = ∑ Fx and R y = ∑ Fy .

Solve: FAx = FA = 270 N. FAy = 0. FBx = FB cos60.0° = 150 N. FBy = FB sin 60.0° = 260 N.

Rx = FAx + FBx = 270 N + 150 N = 420 N. R y = FAy + FBy = 260 N. R = Rx 2 + R y 2 = 494 N. tan θ =

G R makes an angle of 31.8° with the rope of dog A.

Ry Rx

=

260 N . 420 N

*4.3. Set Up: The force and its horizontal and vertical components are shown in the figure below. The force is directed at an angle of 20.0° + 30.0° = 50.0° above the horizontal. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

4-1

4-2

Chapter 4

Solve: Fh = F cos50.0° = 241 N; Fv = F sin 50.0° = 287 N. G Reflect: We could also find the components of F in the directions parallel and perpendicular to the incline but the problem asks for horizontal and vertical components.

4.4. Set Up: Take + y to be upward, so ∑ Fy = 5.00 N. The strap on each side of the jaw exerts a force F directed at an angle of 52.5° above the horizontal, as shown in the figure below.

Solve: ∑ Fy = 2 F sin 52.5° = 5.00 N, so F = 3.15 N.

*4.5. Set Up: Let F1 = 985 N, F2 = 788 N, and F3 = 411 N. The angles θ that each force makes with the + x { axis are θ1 = 31°, θ 2 = 122°, and θ3 = 233°. Solve: (a) F1x = F1 cosθ1 = 844 N; F1 y = F1 sin θ1 = 507 N F2 x = F2 cosθ 2 = − 418 N; F2 y = F2 sin θ 2 = 668 N F3 x = F3 cosθ3 = − 247 N; F3 y = F3 sin θ3 = − 328 N (b) Rx = F1x + F2 x + F3 x = 179 N; R y = F1 y + F2 y + F3 y = 847 N

R = Rx 2 + Ry 2 = 886 N; tan θ =

G R and its components are shown in the figure below.

Ry Rx

so θ = 78.1°


Newton’s Laws of Motion

4-3

Reflect: Adding the forces as vectors gives a very different result from adding their magnitudes.

4.6. Set Up: Let + x be the direction of the force and acceleration. ∑ Fx = 48.0 N. Solve: ∑ Fx = ma x gives m =

∑ Fx 48.0 N = = 16.0 kg. ax 3.00 m/s 2

4.7. Set Up: Let + x be the direction of the force. ∑ Fx = 140 N. Use a constant acceleration equation to calculate the displacement x − x0 of the probe. Solve: (a) a x =

∑ Fx 140 N = = 4.31 m/s 2 m 32.5 kg

(b) υ0 x = 0, a x = 4.31 m/s 2 , t = 10.0 s. x − x0 = υ0 xt + 12 axt 2 = 12 (4.31 m/s 2 )(10.0 s) 2 = 215 m

4.8. Set Up: Take + x to be the direction in which the skater is moving initially. υ x = 0 (comes to rest). υ − υ0 x 0 − 2.40 m/s Solve: υ x = υ0 x + a xt so a x = x = = − 0.682 m/s 2 t 3.52 s The only horizontal force on the skater is the friction force, so f x = max = (68.5 kg)(−0.682 m/s 2 ) = − 46.7 N.

The force is 46.7 N, directed opposite to the motion of the skater.

*4.9. Set Up: Take + x to be in the direction in which the cheetah moves. υ0 x = 0. υ − υ0 x 20.1 m/s − 0 Solve: (a) υ x = υ0 x + a xt so a x = x = = 10.05 m/s 2 2.0 s

t

Fx = ma x = (68 kg)(10.05 m/s 2 ) = 680 N

(b) The force is exerted on the cheetah by the ground. Reflect: The net force on the cheetah is in the same direction as the acceleration of the cheetah.

4.10. Set Up: Let + x be the direction of the force. ∑ Fx = 0.250 N. Use ∑ Fx = max to calculate the acceleration of the puck. Then use constant acceleration equations to find the speed and displacement of the puck for t = 2.00 s. ∑ Fx 0.250 N Solve: ax = = = 1.56 m/s 2 . v0 x = 0. m 0.160 kg x − x0 = υ0 xt + 12 axt 2 = 12 (1.56 m/s 2 )(2.00 s)2 = 3.12 m

υ x = υ0 x + axt = (1.56 m/s 2 )(2.00 s) = 3.12 m/s

*4.11. Set Up: Let + x be the direction of the force. ∑ Fx = 80.0 N. Use the information about the motion to find the acceleration and then use ∑ Fx = max to calculate m. Solve: x − x0 = 11.0 m, t = 5.00 s, υ0 x = 0. x − x0 = υ0 xt + 12 axt 2 gives

ax =

2( x − x0 ) t

2

=

2(11.0 m) (5.00 s) 2

= 0.880 m/s 2 .

∑ Fx 80.0 N = = 90.9 kg. ax 0.880 m/s 2 Reflect: The mass determines the amount of acceleration produced by a given force. m=

4.12. Set Up: w = mg . g = 9.80 m/s 2 Solve: (a) m =

w 3.20 N = = 0.327 kg g 9.80 m/s 2

(b) w = mg = (14.0 kg)(9.80 m/s 2 ) = 137 N © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

4-4

Chapter 4

4.13. Set Up: F = ma. We must use w = mg to find the mass of the boulder. Solve: m =

w 2400 N = = 244.9 kg g 9.80 m/s 2

Then F = ma = (244.9 kg)(12.0 m/s 2 ) = 2940 N. Reflect: We must use mass in Newton’s second law. Mass and weight are proportional.

4.14. Set Up: 1 N = 0.2248 lb. An object of mass 1 kg weighs 2.205 lb. ⎛ 1N ⎞ Solve: (a) (150 lbs) ⎜ ⎟ = 667 N. ⎝ 0.2248 lb ⎠ ⎛ 0.2248 lb ⎞ (b) (40 N) ⎜ ⎟ = 9.0 lb. Yes, an adult collie should weigh much more than this, perhaps 40 lbs. ⎝ 1N ⎠ ⎛ 2.205 lb ⎞ (c) (200 kg) ⎜ ⎟ = 441 lbs. Yes, this is over twice the weight of an average person. ⎝ 1 kg ⎠

*4.15. Set Up: 1 μ g = 1026 g = 1029 kg. 1 mg = 1023 g = 1026 kg. 1 N = 0.2248 lb. g = 9.80 m/s 2 = 32.2 ft/s 2 . Solve: (a) m = 210 μ g = 2.10 × 1027 kg. w = mg = (2.10 × 1027 kg)(9.80 m/s 2 ) = 2.06 × 1026 N (b) m = 12.3 mg = 1.23 × 1025 kg. w = mg = (1.23 × 1025 kg)(9.80 m/s 2 ) = 1.21 × 1024 N

w 45 N ⎛ 0.2248 lb ⎞ (c) (45 N) ⎜ = 4.6 kg ⎟ = 10.1 lb. m = = g 9.80 m/s 2 ⎝ 1N ⎠

4.16. Set Up: We find the following conversions in the appendix: 1 kg = 103 g, 1 kg has a weight of 2.205 lb when g = 9.80 m/s 2 . ⎛ 1 kg ⎞ Solve: (a) (160 lb) ⎜ ⎟ = 73 kg. ⎝ 2.205 lb ⎠ ⎛ 1 kg ⎞ (b) (1.9 lb) ⎜ ⎟ = 0.86 kg. ⎝ 2.205 lb ⎠

⎛ 2.205 lb ⎞ (c) (2300 kg) ⎜ ⎟ = 5100 lb. ⎝ 1 kg ⎠ ⎛ 1 kg ⎞ ⎛ 2.205 lb ⎞ (d) We may write our answer in either pounds or ounces: (22 g) ⎜⎜ 3 ⎟⎟ ⎜ ⎟ = 0.049 lb or (0.49 lb) (16 oz/lb) = ⎝ 10 g ⎠ ⎝ 1 kg ⎠ 0.78 oz. Reflect: The pound is a unit of force and the kilogram is a unit of mass. However, on the earth, mass and weight are proportional, so it is possible to convert between pounds and kilograms.

4.17. Set Up: Weight and mass are related by w = mg . The mass is constant but g and w depend on the location. On the earth, we have g = 9.80 m/s 2 . Solve: (a) We have

w w w = m, which is constant, so E = A . We know that wE = 17.5 N, g E = 9.80 m/s 2 , and g gE gA

⎛w ⎞ ⎛ 3.24 N ⎞ 2 2 wA = 3.24 N. Thus, we have g A = ⎜ A ⎟ g E = ⎜ ⎟ (9.80 m/s ) = 1.81 m/s . ⎝ 17.5 N ⎠ ⎝ wE ⎠ 17.5 N w = 1.79 kg. (b) m = E = g E 9.80 m/s 2 Reflect: The weight at a location and the acceleration due to gravity at that location are directly proportional. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


4-5

4.18. Set Up: 1 N = 0.2248 lb. g = 9.80 m/s2 = 32.2 ft/s 2 . ⎛ 0.2248 lb ⎞ Solve: (a) w = mg = (150 kg)(9.80 m/s 2 ) = (1470 N) ⎜ ⎟ = 330 lb; too heavy for a normal adult. ⎝ 1N ⎠ (b) w = mg = (4.25 slugs)(32.2 ft/s 2 ) = 137 lbs; this is reasonable. (c) 65,000 grams = 65 kg. w = mg = 637 N = 143 lb; this is reasonable.

*4.19. Set Up: The weight of an object depends on its mass and the value of g at its location and is independent of the motion of the object. Solve: (a) 138 N (b) 138 N (c) w = mg = (138 kg)(9.80 m/s 2 ) = 1350 N, for both (a) and (b). (d) They would be the same, 138 N. Reflect: The weight of an object is the gravitational force exerted on it by the earth. 4.20. Set Up: The weight of an object depends on its location. The mass of an object is an intrinsic property of the object and doesn’t depend on the location. w = mg . On earth, g = 9.80 m/s 2 . Solve: (a) m =

w 44.0 N = = 4.49 kg g 9.80 m/s 2

(b) m = 4.49 kg, the same as on earth. wIo = mg Io = (4.49 kg)(1.81 m/s 2 ) = 8.13 N

*4.21. Set Up: On Earth, g E = 9.80 m/s 2 . wE 85.2 N = = 8.69 kg. The mass m is independent of the location of the object. g E 9.80 m/s 2 w 32.2 N wM = mg M so g M = M = = 3.71 m/s 2 . g 8.69 kg (b) m = 8.69 kg, the same at either location. Solve: (a) wE = mg E so m =

Reflect: The object weighs more on earth because the acceleration due to gravity is greater there.

4.22. Set Up: Use coordinates with + y upward and the origin at the initial level of the rock. When the rock returns to the initial level, y = 0. Solve: y = y0 + υ0 yt + 12 a yt 2 gives a y = −

2υ0 y t

=

22(13.0 m/s) = − 17.2 m/s 2 . 1.51 s

Therefore, g X = 17.2 m/s 2 . wX = mg X = (5.24 kg)(17.2 m/s 2 ) = 90.2 N.

*4.23. Set Up: Use coordinates for which the +x-direction is the direction the car is traveling initially. m = 1750 kg. When the car has stopped, υ x = 0. Solve: (a) f = 0.25mg and is in the − x { direction. ∑ Fx = ma x gives − f = ma x and a x =

2 f 20.25mg = = − 0.25 g = − 2.45 m/s 2 m m

The acceleration is 2.45 m/s 2 , directed opposite to the motion. (b) υ0 x = 110 km/h = 30.6 m/s, υ x = 0, a x = − 2.45 m/s 2 . υ x 2 = υ0 x 2 + 2a x ( x − x0 ) so

x − x0 =

υ x 2 − υ0 x 2

0 − (30.6 m/s)2

= 191 m 2( −2.45 m/s 2 ) Reflect: The stopping distance is proportional to the square of the initial speed. 2a x

=

4.24. Set Up: Each force is applied to the box by another object.


4-6

Chapter 4

Solve: (a) The box pulls on the rope. (b) The friction force is a horizontal force applied to the box by the floor. The reaction force is a horizontal force in the opposite direction applied by the box to the floor. (c) The normal force is an upward force applied by the floor to the box. The reaction force is a downward force applied by the box to the floor. (d) The weight of the box is a downward force applied to the box by the earth. The reaction force is an upward force applied to the earth by the box.

G

G

4.25. Set Up: ∑ F = ma refers to forces that all act on one object. The third law refers to forces that a pair of objects exert on each other. Solve: (a) the earth (gravity) (b) 4 N; the book (c) no, these two forces are exerted on the same object (d) 4 N; the earth; the book; upward (e) 4 N, the hand; the book; downward (f) second (The two forces are exerted on the same object and this object has zero acceleration.) (g) third (The forces are between a pair of objects.) G

G

4.26. Set Up: ∑ F = ma refers to forces that are all on one object. An object is in equilibrium if the vector sum of all the forces on it is zero. A third law pair of forces have the same magnitude regardless of the motion of either object. Solve: (a) No. There is a net upward force on the book equal to 1 N. (b) No. The force exerted on the book by your hand is 5 N, upward. The force exerted on the book by the earth is 4 N, downward. (c) Yes. These forces form a third-law pair and are equal in magnitude and opposite in direction. (d) Yes. These forces form a third-law pair and are equal in magnitude and opposite in direction. (e) One, only the gravity force. (f) No. There is a net downward force of 5 N exerted on the book.

*4.27. Set Up: The reaction forces in Newton’s third law are always between a pair of objects. In Newton’s second

law all the forces act on a single object. Let + y be downward. m = w /g . Solve: The reaction to the upward normal force on the enger is the downward normal force, also of magnitude 620 N, that the enger exerts on the floor. The reaction to the enger’s weight is the gravitational force that the enger ∑ Fy 650 N − 620 N exerts on the earth, upward and also of magnitude 650 N. = a y gives a y = = 0.452 m/s 2 . The m (650 N)/(9.80 m/s 2 )

enger’s acceleration is 0.452 m/s 2 , downward. Reflect: There is a net downward force on the enger and the enger has a downward acceleration.

4.28. Set Up: Neglect air resistance. No object is in with the stone. Solve: The free-body diagram is the same as for (a), (b), and (c) and is shown in the figure below. The only force is the downward force of gravity.

*4.29. Set Up: The brakes cause a friction force from the road that is directed opposite to the motion. Solve: (a) The free-body diagram is shown in Figure (a) below. w is the gravity force, n is the normal force, and f is the friction force. The friction force is directed opposite to the direction the car is traveling.



4-7

(b) The free-body diagram is shown in Figure (b) above. The force P is applied to the enger by the car seat and is in the direction the car is moving. Reflect: In each case there is a net horizontal force on the object and this force produces an acceleration in the direction of the net force.

4.30. Set Up: Neglect air resistance. The gravity force w on the ball is independent of its motion. Solve: The free-body diagrams for (a) and (b) are identical and are shown in Figure (a) below. (c) The wall exerts a horizontal force F on the ball. The free-body diagram is shown in Figure (b) below.

*4.31. Set Up: Identify the forces on each object. In each case the forces are the non force of gravity (the weight) and the forces applied by objects that are in with each crate. Each crate touches the floor and the G other crate, and some object applies F to crate A. Solve: (a) The free-body diagrams for each crate are given in the figure below. FAB (the force on m A due to mB ) and FBA (the force on mB due to m A ) form an action-reaction pair. (b) Since there is no horizontal force opposing F, any value of F, no matter how small, will cause the crates to accelerate to the right. The weight of the two crates acts at a right angle to the horizontal and is in any case balanced by the upward force of the surface on them. Reflect: Crate B is accelerated by FBA and crate A is accelerated by the net force F − FAB . The greater the total weight of the two crates, the greater their total mass and the smaller will be their acceleration.

4.32. Set Up: Since the observer in the train sees the ball hang motionless, the ball must have the same acceleration as the train car. By Newton’s second law, there must be a net force on the ball in the same direction as its G acceleration. The forces on the ball are gravity, which is w, downward, and the tension T in the string, which is directed along the string. Solve: (a) The acceleration of the train is zero, so the acceleration of the ball is zero. There is no net horizontal force on the ball and the string must hang vertically. The free-body diagram is sketched in Figure (a) below. (b) The train has a constant acceleration directed east so the ball must have a constant eastward acceleration. There must be a net horizontal force on the ball, directed to the east. This net force must come from an eastward component of G T and the ball hangs with the string displaced west of vertical. The free-body diagram is sketched in Figure (b) below. Reflect: When the motion of an object is described in an inertial frame, there must be a net force in the direction of the acceleration. \


4-8

Chapter 4

G

4.33. Set Up: The suitcase is in with the floor and this surface exerts an upward force n and a horizontal

G force f that is directed opposite to the motion of the suitcase.

Solve: The free-body diagram for the suitcase is shown in the figure below.

4.34. Set Up: There is a friction force between the floor and the bottom crate and a different friction force between the two crates. Call the bottom crate A and the top crate B. wA = 250 N and wB = 135 N. Assume that the crates accelerate to the right; the crates will accelerate since the total friction force from the floor is less than the worker’s push. Solve: The free-body diagram for the top crate is given in Figure (a) below and for the bottom crate in Figure (b) below. f A and f A= are the friction forces between the crates and n A and n=A are the normal forces the crates exert on each other. f A and f A= are a Newton’s third law pair and are equal in magnitude and opposite in direction. The same is true about n A and n=A . f f and nf are the friction and normal forces exerted on block A by the floor.

*4.35. Set Up: Call the blocks A and B, with wA = 850 N and wB = 750 N. Since the worker pulls to the right, the

G friction force on each block is to the left. Let F be the force the worker applies and let T be the tension in the rope. Solve: The free-body diagrams are shown in the figure below. The rope pulls to the right on block A and to the left on block B.

G Reflect: The force F acts only on block B. G

4.36. Set Up: Call the blocks A and B, with wA = 50 lb and wB = 85 lb. Let F be the force the worker applies to block A. Since the worker pushes to the right, the friction force on each block is to the left. Let n=AB be the normal force that A exerts on B and let n=BA be the normal force that B exerts on A. Solve: The free-body diagrams are shown in the figure below. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


4-9

4.37. Set Up: Use coordinates with + y upward. The objects are at rest so they have a = 0. Solve: (a) The free-body diagram for the chandelier is shown in Figure (a) below. Tb , the tension in the bottom link of the chain, equals the magnitude of the upward force that the bottom of the chain exerts on the chandelier. ∑ Fy = ma y gives Tb − w = 0 so Tb = w = 490 N.

(b) Apply ∑ Fy = ma y to the chain. The free-body diagram for the chain is shown in Figure (b) above. The chandelier

pulls downward on the chain with force Tb . Tt is the tension in the top link of the chain. w′ = 245 N is the weight of the chain. We know from (a) that Tb = 490 N. ∑ Fy = ma y gives Tt − Tb − w′ = 0 and Tt = Tb + w′ = 735 N. (c) Apply oFy = ma y to the bottom half of the chain. The free-body diagram is shown in Figure (c) above.

w′h = 122 N is the weight of half the chain and Tm is the tension in the middle link of the chain. ∑ Fy = ma y gives Tm − Tb − w′h = 0 so Tm = Tb + wh′ = 612 N. Reflect: The tension increases from the top to the bottom of the chain.

4.38. Set Up: For simplicity, represent each acrobat as a point mass. The upper acrobat experiences the following forces: FU, T , FU, L , and w U , which are the force the trapeze exerts on the upper acrobat, the force that the lower acrobat exerts on the upper acrobat, and the weight of the upper acrobat (respectively). Similarly, the lower acrobat experiences the following forces: FL, U and w L , which are the force that the upper acrobat exerts on the lower acrobat and the weight of the lower acrobat (respectively). Note that FU, L and FL, U are equal in magnitude by Newton’s third law. Solve: (a) The free-body diagram for the upper acrobat is shown below in Figure (a). (b) The free-body diagram for the lower acrobat is shown below in Figure (b). Reflect: A more realistic free-body diagram would show each acrobat as an extended object. The force on each leg or arm would be shown as a separate vector, which need not be perfectly vertical.


4-10

Chapter 4

*4.39. Set Up: Take + y to be upward. The mass of the bucket is mB = wB /g = 28.1 kg and the mass of the chain is mC = wC /g = 12.8 kg. The chain also has an upward acceleration of 2.50 m/s 2 . Solve: (a) Consider the chain and bucket as a combined object of mass mtot = 28.1 kg + 12.8 kg = 40.9 kg and

weight wtot = 400 N. The free-body diagram is shown in Figure (a) below. Tt is the tension in the top link. ∑ Fy = ma y gives Tt − wtot = mtot a and Tt = wtot + mtot a = 400 N + (40.9 kg)(2.50 m/s 2 ) = 502 N

(b) Apply ∑ Fy = ma y to the bucket. The free-body diagram (Figure (b) above) gives Tb − wB = mBa so Tb = wB + m Ba =

275 N + (28.1 kg)(2.50 m/s 2 ) = 345 N (c) Apply ∑ Fy = ma y to the bottom half of the chain. The free-body diagram is shown in Figure (c) above. The

bottom half of the chain has mass mh = 6.4 kg and weight wh = 62.5 N. Tm is the tension in the middle link of the chain. ∑ Fy = ma y gives Tm − wh − Tb = mh a. Tm = wh + Tb + mh a = 62.5 N + 345 N + (6.4 kg)(2.5 m/s 2 ) = 424 N

Reflect: Part (b) can also be done by applying ∑ Fy = ma y to the chain. The free-body diagram is given in Figure (d) above.

∑ Fy = ma y gives Tt − wc − Tb = mc a and Tb = Tt − wc − mc a = 502 N − 125 N − (12.8 kg)(2.50 m/s 2 ) = 345 N, which agrees with the value obtained previously. The tension increases from the bottom to the top of the chain.

4.40. Set Up: Let the acceleration be in the + x direction. Apply ∑ Fx = max to find the resultant horizontal force. Solve: (a) ∑ Fx = ma x = (55 kg)(15 m/s 2 ) = 825 N (b) The force is exerted by the blocks. The blocks push on the sprinter because the sprinter pushes on the blocks.

*4.41. Set Up: Identify the forces on the chair. The floor exerts a normal force and a friction force. Let + y be upward and let + x be in the direction of the motion of the chair. Solve: (a) The free-body diagram for the chair is given in the figure below. (b) For the chair, a y = 0 so ∑ Fy = ma y gives n − mg − Fsin37° = 0 and n = 142 N . G Reflect: n is larger than the weight because F has a downward component.

4.42. Set Up: Use coordinates where + x is in the direction the ball is thrown.



4-11

Solve: (a) Solve for a x : x − x0 = 1.0 m, υ0 x = 0, υ x = 46 m/s. υ x 2 = υ0 x 2 + 2ax ( x − x0 ) gives

ax =

υ x 2 − υ0 x 2 2( x − x)

=

(46 m/s) 2 − 0 = 1058 m/s2 2(1.0 m)

G The free-body diagram for the ball during the pitch is shown in Figure (a) below. The force F is applied to the ball by the pitcher’s hand. ∑ Fx = ma x gives F = (0.145 kg)(1058 m/s 2 ) = 153 N

(b) The free-body diagram after the ball leaves the hand is given in Figure (b) above. The only force on the ball is the downward force of gravity.

*4.43. Set Up: Your mass is m = w/g = 63.8 kg. Both you and the package have the same acceleration as the elevator. Take + y to be upward, in the direction of the acceleration of the elevator. Solve: (a) Your free-body diagram is shown in Figure (a) below. n is the scale reading. ∑ Fy = ma y gives n − w = ma

and n = w + ma = 625 N + (63.8 kg)(2.50 m/s 2 ) = 784 N

(b) The free-body diagram for the package is given in Figure (b) above. ∑ Fy = ma y gives T − w = ma so T = w + ma =

(3.85 kg)(9.80 m/s 2 + 2.50 m/s 2 ) = 47.4 N Reflect: The objects accelerate upward so for each object the upward force is greater than the downward force.

G

G

4.44. Set Up: We know that ∑ F = ma must be satisfied for each object. Newton’s third law says that the force

G G FC on T that the car exerts on the truck is equal in magnitude and opposite in direction to the force FT on C that the G truck exerts on the car. The only horizontal force on the car is the force FT on C exerted by the truck. The car exerts a G G force FC on T on the truck. There is also a horizontal friction force f that the highway surface exerts on the truck.

.

Assume the system is accelerating to the right in the free-body diagrams. Solve: (a) The free-body diagram for the car is sketched in Figure (a) below. (b) The free-body diagram for the truck is sketched in Figure (b) below. G (c) The friction force f accelerates the system forward. The tires of the truck push backwards on the highway surface as they rotate, so by Newton’s third law the roadway pushes forward on the tires. G Reflect: FT on C and FC on T each are equal to the tension T in the rope. Both objects have the same acceleration a T = mC a and f − T = mT a, so f = ( mC + mT ) a. The acceleration of the two objects is proportional to f.


4-12

Chapter 4

4.45. Set Up: Take + y to be upward. 4973 km/h = 1381 m/s. 45 km = 4.5 × 104 m. The astronaut and the shuttle will have the same acceleration. Solve: (a) υ0 y = 0, υ y = 1381 m/s, y − y0 = 4.5 × 104 m. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives ay =

υ y 2 − υ0 y 2 2( y − y0 )

=

(1381 m/s)2 − 0 2(4.5 × 104 m)

= 21.2 m/s 2

(b) The free-body diagram for the astronaut is shown in the figure below. ∑ Fy = ma y gives n − mg = ma and

n = m( g + a ) = (55.0 kg)(9.80 m/s 2 + 21.2 m/s 2 ) = 1700 N

(c) She would think the scale reading is her weight, 1700 N. And she would therefore think her mass is m = w/g = 173 kg. Reflect: Since she has an upward acceleration the upward force n is greater than the downward force of gravity.

4.46. Set Up: Let P be the magnitude of the upward force that the woman’s hand exerts on the briefcase. Take the + y direction upward so that Py = P and wy = − mg . Also, since the elevator is accelerating downward, we have a y = −1.50 m/s 2 . Solve: The free-body diagram for the briefcase is shown in the figure below. Apply Newton’s Second Law to the briefcase: ∑ Fy = ma y gives P + (− mg ) = ma y . Solving for P we obtain: P = m( g + a y ) = (2.5 kg)(9.80 m/s 2 − 1.50 m/s2 ) = 20.8 N = 21 N By Newton’s third law, this is also the magnitude of the downward force that the briefcase exerts on the woman’s hand. Reflect: Since the acceleration is downward, P is less than the weight of the briefcase, which is mg = (2.5 kg) (9.80 m/s 2 ) = 25 N. If the acceleration had been upward, P would be larger than the weight of the briefcase.

*4.47. Set Up: Use constant acceleration equations to calculate the acceleration ax that would be required. Then use ∑ Fx = ma x to find the necessary force. Let + x be the direction of the initial motion of the auto.



Solve: vx 2 = v0 x 2 + 2a x ( x − x0 ) with vx = 0 gives a x = −

and a x = 2

4-13

v0 x 2 . The force F is directed opposite to the motion 2( x − x0 )

F . Equating these two expressions for a x gives m F =m

v0 x 2 (12.5 m/s) 2 = (850 kg) = 3.7 × 106 N. 2( x − x0 ) 2(1.8 × 10−2 m)

Reflect: A very large force is required to stop such a massive object in such a short distance.

4.48. Set Up: Use coordinates where + x is along the barrel, in the direction of the motion of the bullet. The expanding gases in the barrel produce a horizontal force F on the bullet. Solve: (a) υ0 x = 0, υ x = 965 m/s, x − x0 = 0.450 m. υ x 2 = υ0 x 2 + 2ax ( x − x0 ) gives ax =

υ x 2 − υ0 x 2 2( x − x0 )

=

(965 m/s)2 − 0 = 1.03 × 106 m/s 2 2(0.450 m)

∑ Fx = ma x gives F = ma x = (4.20 × 1023 kg)(1.03 × 106 m/s 2 ) = 4330 N (b) (i) The free-body diagram for the bullet while it is in the barrel is given in Figure (a) below. (ii) The free-body diagram for the bullet after it leaves the barrel is given in Figure (b) below; the forces F and n from the rifle no longer act on the bullet.

1g ⎛ ⎞ (c) a = (1.03 × 106 m/s 2 ) ⎜ = 1.05 × 105 g 2⎟ ⎝ 9.80 m/s ⎠ υ − υ0 x 965 m/s − 0 (d) υ x = υ0 x + a xt gives t = x = = 0.937 ms ax 1.03 × 106 m/s 2 G

G

G

*4.49. Set Up: Apply ∑ F = ma to the parachutist. Let + y be upward. Fair is the force of air resistance. Solve: (a) w = mg = (55.0 kg)(9.80 m/s2 ) = 539 N (b) The free-body diagram is given in the figure below. ∑ Fy = Fair − w = 620 N − 539 N = 81 N. The net force is upward.

∑ Fy

81 N = = 1.5 m/s 2 . The acceleration is upward. m 55.0 kg Reflect: Both the net force and the acceleration are upward. Since her velocity is downward and her acceleration is upward, her speed decreases. (c) a y =


4-14

Chapter 4

G

G

4.50. Set Up: Apply ∑ F = ma and solve for the mass m of the spacecraft. Let + y be upward. The weight of the spacecraft near the surface of Planet X is w. Solve: (a) The velocity of the spacecraft is downward. When it is slowing down, the acceleration is upward. When it is speeding up, the acceleration is downward. (b) In each case the net force is in the direction of the acceleration. Speeding up: w > F and the net force is downward. Slowing down: w < F and the net force is upward. (c) Denote the y-component of the acceleration when the thrust is F1 by a1 and the y-component of the acceleration

when the thrust is F2 by a2 . a1 = +1.20 m/s 2 and a2 = −0.80 m/s 2 . The forces and accelerations are then related by F1 − w = ma1, F2 − w = ma2 . Dividing the first of these by the second to eliminate the mass gives

solving for the weight w gives w = w=

F1 − w a1 = , and F2 − w a2

a1F2 − a2 F1 . Substituting the given numbers, with + y upward, gives a1 − a2

(1.20 m/s 2 )(10.0 × 103 N) − (−0.80 m/s 2 )(25.0 × 103 N) 1.20 m/s 2 − (−0.80 m/s 2 )

= 16.0 × 103 N.

Reflect: The acceleration due to gravity at the surface of Planet X did not need to be found.

*4.51. Set Up: Let + y be upward. At his maximum height, υ y = 0. While he is in the air, a y = − 9.80 m/s 2 . m = w/g = 90.8 kg Solve: (a) υ0 y = 0, a y = − 9.80 m/s 2 and y − y0 = 1.2 m. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ0 y = 22a y ( y − y0 ) = 22(−9.80 m/s 2 )(1.2 m) = 4.85 m/s (b) For the motion while he is pushing against the floor, v0 y = 0, v y = 4.85 m/s, and t = 0.300 s. υ y = υ0 y + a y gives

ay =

υ y − υ0 y t

=

4.85 m/s − 0 = 16.2 m/s 2 . The acceleration is upward. 0.300 s

G (c) The free-body diagram while he is pushing against the floor is given in the figure below. F is the vertical force the floor applies to him.

(d) ∑ Fy = ma y gives F − mg = ma and F = m( g + a ) = (90.8 kg)(9.80 m/s 2 + 16.2 m/s 2 ) = 2.36 × 103 N. The force

he applied to the ground has this same magnitude and is downward. Reflect: The ground must push upward on him with a force greater than his weight in order to give him an upward acceleration.

4.52. Set Up: w = 683 N is the force of gravity exerted on you, independent of your motion. Your mass is m = w/g = 69.7 kg. Use coordinates with + y upward. Your free-body diagram is given in the figure below. n is the

scale reading, the force the scale exerts on you. You and the elevator have the same acceleration.



4-15

n−w . m

Solve: ∑ Fy = ma y gives n − w = ma y so a y = (a) n = 725 N, so a y =

725 N − 683 N = 0.603 m/s 2 . a y is positive so the acceleration is upward. 69.7 kg

(b) n = 595 N, so a y =

595 N − 683 N = − 1.26 m/s 2 . a y is negative so the acceleration is downward. 69.7 kg

4.53. Set Up: Use coordinates with + x horizontal and in the direction of the motion of the ball and with + y upward. Solve: (a) υ0 x = 0, υ x = 73.14 m/s, t = 3.00 × 1022 s. υ x = υ0 x + a xt gives ax =

υ x − υ0 x t

=

73.14 m/s − 0 3.00 × 1022 s

= 2.44 × 103 m/s 2 .

∑ Fx = ma x gives F = ma x = (57 × 1023 kg)(2.44 × 103 m/s 2 ) = 140 N

G (b) The free-body diagram while the ball is in with the racket is given in Figure (a) below. F is the force G exerted on the ball by the racket. After the ball leaves the racket, F ceases to act, as shown in Figure (b) below.

4.54. Set Up: Let m be the mass of the tool. There is no friction. Use coordinates where + x is horizontal, in the direction of the 12.0 N force, and let + y be downward. Solve: First find the mass m: x − x0 = 16.0 m, t = 2.00 s, υ0 x = 0. x − x0 = υ0 xt + 12 a xt 2 gives

ax = ∑ Fx = ma x gives F = ma x and m =

2( x − x0 ) t

2

=

2(16.0 m) (2.00 s)2

= 8.00 m/s 2 .

F 12.0 N = = 1.50 kg a x 8.00 m/s 2

Find g N , the acceleration due to gravity on Newtonia. y − y0 = 10.0 m, υ0 y = 0, t = 2.58 s.

y − y0 = υ0 yt + 12 a yt 2 gives a y =

2( y − y0 ) t2

=

2(10.0 m) (2.58 s) 2

= 3.00 m/s 2 ; g N = 3.00 m/s 2 .

The weight on Newtonia is wN = mg N = (1.50 kg)(3.00 m/s 2 ) = 4.50 N. The weight on Earth is wE = mg E = (1.50 kg)(9.80 m/s 2 ) = 14.7 N. G

G

*4.55. Set Up: Apply ∑ F = ma to the barbell and to the athlete. Use the motion of the barbell to calculate its acceleration. Let + y be upward.


4-16

Chapter 4

Solve: (a) The free-body diagrams for the barbell and for the athlete are sketched in the figure below. (b) The athlete’s weight is mg = (90.0 kg)(9.80 m/s2 ) = 882 N. The upward acceleration of the barbell is found from

y − y0 = v0 yt + 12 a yt 2 . a y =

2( y − y0 ) t2

=

2(0.600 m) (1.6 s) 2

= 0.469 m/s 2 . The force needed to lift the barbell is given by

Flift − wbarbell = ma y . The barbell’s mass is (490 N)/(9.80 m/s 2 ) = 50.0 kg, so Flift = wbarbell + ma = 490 N + (50.0 kg)

(0.469 m/s 2 ) = 490 N + 23 N = 513 N. The athlete is not accelerating, so Ffloor − Flift − wathlete = 0. Thus, Ffloor = Flift + wathlete = 513 N + 882 N = 1395 N, which is the total force that the ground exerts on the athlete’s feet as he lifts the barbell. Reflect: Since the athlete pushes upward on the barbell with a force greater than its weight, the barbell pushes down on him with a force that is greater than its weight, and so the normal force on the athlete is greater than the total weight, 1372 N, of the athlete plus barbell.

4.56. Set Up: Take + y downward. While he is in the air, before he touches the ground, his acceleration is a y = 9.80 m/s 2 . Solve: (a) υ0 y = 0, y − y0 = 3.10 m, and a y = 9.80 m/s 2 . υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ y = 2a y ( y − y0 ) = 2(9.80 m/s 2 )(3.10 m) = 7.79 m/s (b) υ0 y = 7.79 m/s, υ y = 0, y − y0 = 0.60 m. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

ay =

υ y 2 − υ0 y 2 2( y − y0 )

=

0 − (7.79 m/s) 2 = − 50.6 m/s 2 . 2(0.60 m)

The acceleration is upward. G (c) The free-body diagram is given in the figure below. F is the force the ground exerts on him.

(d) ∑ Fy = ma y gives mg − F = − ma. F = m( g + a ) = (75.0 kg)(9.80 m/s 2 + 50.6 m/s 2 ) = 4.53 × 103 N, upward.

F 4.53 × 103 N = = 6.16, so F = 6.16w. w (75.0 kg)(9.80 m/s 2 ) (e) The reaction force is the man pushing downward on the ground with a force of 4.53 × 103 N. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


4-17

4.57. Set Up: Let + x be in the direction of motion of the electron. Solve: (a) υ0 x = 0, ( x − x0 ) = 1.80 × 1022 m, υ x = 3.00 × 106 m/s. υ x 2 = υ0 x 2 + 2ax ( x − x0 ) gives ax = (b) υ x = υ0 x + a xt gives t =

υ x 2 − υ0 x 2 2( x − x0 )

υ x − υ0 x ax

=

=

(3.00 × 106 m/s) 2 − 0 2(1.80 × 1022 m)

3.00 × 106 m/s − 0 2.50 × 1014 m/s 2

= 2.50 × 1014 m/s 2

= 1.2 × 1028 s

(c) ∑ Fx = ma x = (9.11 × 10231 kg)(2.50 × 1014 m/s 2 ) = 2.28 × 10216 N.

Solutions to age Problems 4.58. Set Up: Assume that the bacterium is moving along the + x axis and slowing with a constant acceleration so that we can use the kinematic equations for constant acceleration: υ x 2 = υ02x + 2a x Δx. Note that all values are only good to one significant figure and 1 nm = 1029 m. Solve: a x =

υ x 2 − υ0 x 2 2Δx

=

02 − (3 × 10−3 m/s) 2 2(10−11 m)

= 24.5 × 105 m/s 2 . Thus, the magnitude of the acceleration (rounded

to one significant figure) is 5 × 105 m/s 2 . The correct answer is B.

*4.59. Set Up: We can find the magnitude of the viscous force from Newton’s second law: F = ma. The mass of the bacterium is 10212 g = 10215 kg. Solve: The viscous force needed to stop the bacterium in the required distance is F = ma = (1 × 10215 kg)

(5 × 105 m/s 2 ) = 5 × 10210 N. The correct answer is D.

4.60. Set Up: According to Newton’s second law ∑ Fx = max . Solve: If the bacterium is moving at a constant velocity, its acceleration must be zero. Thus, the net force on the bacterium must be zero. This means that the force exerted by the flagellum must be equal to the viscous force, which, from the previous problem, is 5 × 10210 N. The correct answer is B.

*4.61. Set Up: According to Newton’s Second Law ∑ Fx = max . Solve: To change from a constant speed (where the net force is zero) to an acceleration of 0.001 m/s 2 , the flagellum

must increase the net force to F = ma = (10215 kg)(1023 m/s 2 ) = 10218 N. The correct answer is A.


5

APPLICATIONS OF NEWTON’S LAWS

Answers to Multiple-Choice Problems 1. A

2. A

3. B

4. D

5. A

6. A

7. B

8. B

9. A

10. C

11. C

12. B

13. D

Solutions to Problems *5.1. Set Up: Constant speed means a = 0. Use coordinates where +y is upward. The breaking strength for the rope in each part is (a) 50 N, (b) 4000 N, and (c) 6 × 104 N. The free-body diagram for the bucket plus its contents is given in the figure below.

Solve: oFy = ma y gives T − w = 0 so T = w. (a) w = 50 N, so 35 N of cement. The mass of cement is 3.6 kg. (b) w = 4000 N, so 3985 N of cement. The mass of cement is 410 kg. (c) w = 6 × 104 N, so 6 × 104 N of cement. The mass of cement is 6 × 103 kg. Reflect: Since a = 0, the upward pull of the rope equals the total weight of the object.

5.2. Set Up: Use coordinates where +y is upward. a y = 0 for each object. Call the tension in the wires Tt (top), Tm (middle), and Tb (bottom). The free-body diagram for each object is given in the figure below.


5-1

5-2

Chapter 5

Solve: (a) and (b) oFy = ma y for each object gives Tb − W = 0, Tm − W − Tb = 0, and Tt − Tm − W = 0. Tb = W ,

Tm = Tb + W = 2W , and Tt = Tm + W = 3W . The top wire has the greatest tension. If Tt = 75.0 N

then

W = Tt /3 = 25.0 N and Tm = 2W = 50.0 N and Tb = W = 25.0 N.

5.3. Set Up: a = 0 for each object. Apply oFy = ma y to each weight and to the pulley. Take +y upward. The pulley has negligible mass. Let Tr be the tension in the rope and let Tc be the tension in the chain. Solve: (a) The free-body diagram for each weight is the same and is given in Figure (a) below. oFy = ma y gives

Tr = w = 25.0 N.

(b) The free-body diagram for the pulley is given in Figure (b) above. Tc = 2Tr = 50.0 N. Reflect: The tension is the same at all points along the rope.

5.4. Set Up: For each object use coordinates where +y is upward. Each object has a = 0. Call the objects 1 and 2, with w1 = 72 N and w2 = 150 N. Solve: (a) The free-body diagrams for each object are shown in the figure below. TB and TC are replaced by their

x and y components.

(b) For object 2, oFy = ma y gives TA − w2 = 0 and TA = 150 N. (c) For object 2, oFx = max gives TC cos60D − TB cos60D = 0 so TB = TC

oFy = ma y gives TB sin 60D + TC sin 60D − TA − w1 = 0. TB =

TA + w1

2sin 60D

G

=

150 N + 72 N = 128 N; TC = 128 N 2sin 60°

G

*5.5. Set Up: Take +y upward. Apply oF = ma to the person. a = 0. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Applications of Newton’s Laws

5-3

Solve: (a) The free-body diagram for the person is given in the figure below. The two sides of the rope each exert a force with vertical component T sin θ . oFy = ma y gives 2T sin θ − w = 0.

T=

(90.0 kg)(9.80 m/s 2 ) mg = = 2.54 × 103 N 2sin θ 2sin10.0D

(b) Set T = 2.50 × 104 N and solve for θ :

sin θ =

mg (90.0 kg)(9.80 m/s 2 ) = and θ = 1.01D. 2T 2(2.50 × 104 N)

Reflect: Only the vertical components of the tension on each side of the person act to hold him up. The tension in the rope is much greater than his weight.

5.6. Set Up: Apply Newton’s first law to the car. Use x and y coordinates that are parallel and perpendicular to the ramp, respectively. Solve: (a) The free-body diagram for the car is given in the figure below. The vertical weight w and the tension T in the cable have each been replaced by their x and y components. sin 25.0D sin 25.0D = (1130 kg)(9.80 m/s 2 ) = 5460 N. (b) aFx = 0 gives T cos31.0D − w sin 25.0D = 0 and T = w D cos31.0 cos31.0D (c) aFy = 0 gives n + T sin 31.0D − w cos 25.0D = 0 and

n = w cos 25.0D − T sin 31.0D = (1130 kg)(9.80 m/s 2 )cos 25.0D − (5460 N)sin 31.0D = 7220 N Reflect: We could also use coordinates that are horizontal and vertical and would obtain the same values of n and T.


5-4

Chapter 5

*5.7. Set Up: The SI units for force and area are N and m 2 . 1 N = 1 kg ⋅ m/s 2 . Fmax so σ has units of N/m 2 or kg/m ⋅ s 2 A 755 N = = 2.5 × 10−3 m 2 = 25 cm 2 3.0 × 105 N/m 2

Solve: (a) σ = (b) A =

Fmax

σ

5.8. Set Up: For each muscle Fmax = σ A, where σ = 3.0 × 105 N/m 2 . 30 cm 2 = 3.0 × 1023 m 2 . Use coordinates

G where +y is along the Achilles tendon. The force Fmax exerted by each muscle then makes an angle of 22.5D with the +y axis. Apply the equilibrium conditions to the point where the muscles attach to the tendon. The free-body diagram is given in the figure below. T is the tension in the tendon.

Solve: oFy = 0 gives T = 2 Fmax cos 22.5D = 0

Fmax = σ A = (3.0 × 105 N/m 2 )(3.0 × 10−3 m 2 ) = 900 N. T = 2 Fmax cos 22.5D = 1700 N.

5.9. Set Up: Use coordinates where +y is upward and +x is to the right. The force at the foot is horizontal. The tension T in the cable that has one end attached to the leg and the other end attached to the weight is equal to W and is the same everywhere along the cable. Solve: (a) The free-body diagram for the leg is given in Figure (a) below. FT is the traction force on the leg. oFy = 0 gives T − w = 0 and T = w = 47.0 N so W = 47.0 N.



5-5

(b) The free-body diagram for the pulley that is attached to the foot is given in Figure (b) above. oFx = 0 gives

2T cos 25D − FT = 0 so FT = 2T cos 25D = 85.2 N. Reflect: The tractive force on the leg is greater than W.

5.10. Set Up: Use coordinates where the +x is along the dashed line in the figure in the problem. The free-body diagram for the pulley attached to the foot is given in the figure below. The tension in the cable is the same all along the cable and is equal to W. The tension forces have been replaced by their x and y components.

Solve: oFx = 0 gives 2T cos35D − FT = 0 and T =

FT

2cos35D

=

25 N 2cos35D

= 15 N. W = 15 N.

*5.11. Set Up: Use coordinates with +y upward and +x to the right. Call the objects A and B, with wA = 175 N and mB = 32 kg. Label the tensions in the three wires T1, T2 , and T3 . Solve: (a) The free-body diagrams for each object are given in the figure below. T3 has been replaced by its x and

y components.

(b) oFy = 0 for B gives T2 − wB = 0 and T2 = mB g = 314 N. oFy = 0 for A gives T3 cos53°− T2 − wA = 0 so

T3 =

T2 + wA = 813 N. oFx = 0 for A gives T3 sin 53° − T1 = 0 and T1 = 649 N. cos53°

(c) The tensions are unaffected by the length of the wires, so long as the third wire still makes a 53D angle with one wall. Reflect: T3 cos53° equals the combined weight of both objects, 489 N.

5.12. Set Up: Use coordinates with +y upward and +x to the right. Label the tensions T1 and T2 . Solve: (a) The free-body diagram for the officer is given in the figure below. Each tension is replaced by its x and y components.


5-6

Chapter 5

⎛ cos35D ⎞ T (b) oFx = 0 gives T2 cos 48D − T1 cos35D = 0; T2 = ⎜ ⎜ cos 48D ⎟⎟ 1 ⎝ ⎠ oFy = 0 gives T1 sin 35D + T2 sin 48D − w = 0

T1 (sin 35D + cos35D tan 48D ) = w and T1 = 482 N. Then T2 = 590 N.

G

G

5.13. Set Up: Apply oF = ma to the ball. Since the post is frictionless, the force it exerts on the ball is a horizontal normal force. Use coordinates with +y upward and +x to the right. Solve: The angle θ between the string and post can be calculated from Figure (a) below. l is the length of the string and r is the radius of the ball. r 0.110 m sin θ = = and θ = 4.18D. r + l 1.40 m + 0.110 m

(a) The free-body diagram of the ball is given in Figure (b) above. mg (0.270 kg)(9.80 m/s 2 ) = = 2.65 N (b) oFy = ma y gives T cosθ − mg = 0 and T = cosθ cos 4.18D (c) oFx = ma x gives T sin θ − n = 0 and n = T sin θ = (2.65 N)sin 4.18D = 0.193 N.



G

5-7

G

5.14. Set Up: The tension in C in each case is TC = 250 N. Apply oF = ma to the knot where the three cords are ed. Use coordinates with +y upward and +x to the right. There is no motion and a = 0. Solve: (a) The free-body diagram for the knot is given in Figure (a) below. Tensions TA and TB are replaced by

their x and y components.

oFx = 0 gives TB cos 45D − TA cos30D = 0 and TB = 1.225TA oFy = 0 gives TA sin 30D + TB sin 45D − TC = 0. Using TB = 1.225TA gives 0.500TA + 0.866TA = 250 N and TA = 183 N.

TB = 1.225TA = 224 N. (b) The free-body diagram for the knot is given in Figure (b) above. Tensions TA and TB are replaced by their x and

y components. oFx = 0 gives TB cos 45D − TA sin 60D = 0 and TB = 1.225TA oFy = 0 gives TB sin 45D − TA cos60D − TC = 0. Using TB = 1.225TA gives 0.866TA − 0.500TA = 250 N and TA = 683 N.

Then TB = 837 N.

G

G

*5.15. Set Up: Apply oF = ma to each block. a = 0. Take +y perpendicular to the incline and +x parallel to the incline. Solve: The free-body diagrams for each block, A and B, are given in the figure below.

(a) For B, oFx = ma x gives T1 − w sin α = 0 and T1 = w sin α . (b) For block A, oFx = ma x gives T1 − T2 − w sin α = 0 and T2 = 2 w sin α. (c) oFy = ma y for each block gives n A = nB = w cos α . (d) For α → 0, T1 = T2 → 0 and n A = nB → w. For α → 90D , T1 = w, T2 = 2 w and n A = nB = 0. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

5-8

Chapter 5

5.16. Set Up: For constant speed in either direction, a = 0. Take +y perpendicular to the incline and +x parallel to the incline. Solve: (a) The free-body diagram for the piano is given in Figure (a) below. Since a = 0, oFx = 0. There is a component of w down the incline, so the push P exerted by the man must be directed up the incline.

oFx = ma x gives P − mg sin α = 0 and

P = mg sin α = (180 kg)(9.80 m/s 2 )sin11.0D = 337 N. (b) The piano has a = 0 just as in part (a), so P = 337 N. G (c) The free-body diagram is shown in Figure (b) above. Now P has a component P cos θ directed up the incline. oFx = ma x gives P cosθ − mg sin θ = 0.

P = mg tan θ = (180 kg)(9.80 m/s 2 ) tan11.0D = 343 N.

*5.17. Set Up: 60 g = 588 m/s 2 . Use coordinates where the +x direction is in the direction of the acceleration, opposite to the initial velocity of the car. Solve: (a) The free-body diagram for the person is given in the figure below. F is the force the air bag exerts and n is the normal force from the car seat. oFx = ma x gives F = ma = (75 kg)(588 m/s 2 ) = 4.41 × 104 N

(b)

4.41 × 104 N F F = = = 60, so F = 60 w. w mg (75 kg)(9.80 m/s 2 )

5.18. Set Up: Let the person’s weight be W, so W = 680 N. Assume constant acceleration during the speeding up motion and assume that the body moves upward 15 cm in 0.50 s while speeding up. The free-body diagram for the person is given in the figure below. F is the force exerted on him by his arms.



5-9

Solve: υ0 y = 0, y − y0 = 0.15 m, t = 0.50 s. y − y0 = υ0 yt + 12 a yt 2 gives

ay =

2( y − y0 )

t

2

=

oFy = ma y gives F − W = ma. m =

2(0.15 m) (0.50 s)

2

= 1.2 m/s 2

⎛ W a⎞ , so F = W ⎜1 + ⎟ = 1.12W = 762 N. g g ⎝ ⎠

5.19. Set Up: Solve for the net force Fnet on the ball. Solve: υ0 = 0, υ = 73.14 m/s, t = 30 × 1023 s. υ = υ0 + at gives a=

υ − υ0 t

=

73.14 m/s 23

30 × 10

s

= 2.44 × 103 m/s 2 .

Fnet = ma = (W /g )a = 250W . Reflect: The acceleration is much larger than g and the net force is much larger than the weight of the ball.

5.20. Set Up: The mass of the spaceship is w/g = 7714 kg. Take +y to be upward, away from the ground. Solve: (a) The free-body diagram is given in the figure below. F is the braking force.

(b) υ0 y = − 1.00 × 103 m/s, υ y = 0, t = 120 s. υ y = υ0 y + a yt gives

ay =

υ y − υ0 y t

=

0 − (−1.00 × 103 m/s) = 8.33 m/s 2 . 120 s

oFy = ma y gives F − w = ma y and F = w + ma y = 75,600 N + (7714 kg)(8.33 m/s 2 ) = 1.40 × 105 N

*5.21. Set Up: Take +y to be upward. After he leaves the ground the person travels upward 60 cm and his acceleration is g = 9.80 m/s 2 , downward. His weight is W so his mass is W /g. Solve: (a) υ y = 0 (at the maximum height), y − y0 = 0.60 m, a y = − 9.80 m/s 2 . υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ0 y = 22a y ( y − y0 ) = 22(−9.80 m/s 2 )(0.60 m) = 3.4 m/s.


5-10

Chapter 5

(b) The free-body diagram for the person while he is pushing up against the ground is given in the figure below.

(c) For the jump, υ0 y = 0, υ y = 3.4 m/s (from part (a)), y − y0 = 0.50 m. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

ay =

υ y 2 − υ0 y 2 2( y − y0 )

=

(3.4 m/s) 2 − 0 = 11.6 m/s 2 2(0.50 m)

⎛ a⎞ oFy = ma y gives n − W = ma. n = W + ma = W ⎜1 + ⎟ = 2.2W g ⎝ ⎠ Reflect: To accelerate the person upward during the jump, the upward force from the ground must exceed the downward pull of gravity. The ground pushes up on him because he pushes down on the ground.

5.22. Set Up: Let the acceleration be in the +x direction.

Solve: (a) Apply oFx = ma x to the four cars taken as a single object. The free-body diagram for this object is given

in Figure (a) below. m is the mass of one car and F1 is the force that the engine exerts on the first car.

oFx = ma x gives F1 = (4m)a and F1 = 4(38,000 kg)(1.10 m/s 2 ) = 1.67 × 105 N. (b) Apply oFx = ma x to the last three cars taken as a single object. The free-body diagram for this object is given in

Figure (b) above. F2 is the force that the first car exerts on the second car. oFx = ma x gives F2 = (3m) a = 3(38,000 kg)(1.10 m/s 2 ) = 1.25 × 105 N. We could also consider the forces on the first car: F1 − F2 = ma and

F2 = F1 − ma = 1.25 × 105 N. (c) Apply oFx = ma x to the last two cars taken as a single object. The free-body diagram for this object is given in

Figure (c) above. F3 is the force that the second car exerts on the third car. oFx = max gives F3 = (2m) a = 2(38,000 kg)(1.10 m/s 2 ) = 8.36 × 104 N. We could also consider the forces on the second car: F2 − F3 = ma and

F3 = F2 − ma = 8.36 × 104 N.



5-11

(d) Apply oFx = ma x to the fourth car. The free-body diagram for this object is given in Figure (d) above. F4 is the

force that the third car exerts on the fourth car. oFx = max gives F4 = ma = (38,000 kg)(1.10 m/s 2 ) = 4.18 × 104 N. We could also consider the forces on the third car: F3 − F4 = ma and F4 = F3 − ma = 4.18 × 104 N.

G

*5.23. Set Up: Take +y to be upward. The fish has the same upward acceleration as the elevator. Let F be the upward force exerted on the fish by the spring balance; F is what the balance reads. Solve: (a) The free-body diagram for the fish is sketched in the figure below.

oFy = ma y gives F − mg = ma and

F 60.0 N = = 4.90 kg. g + a 9.80 m/s 2 + 2.45 m/s 2 The weight of the fish is w = mg = 48.0 N. m=

(b) F − mg = ma y and m = 4.90 kg from part (a).

a=

F − mg 35.0 N − (4.90 kg)(9.80 m/s 2 ) = = − 2.65 m/s 2 m 4.90 kg

The elevator is accelerating downward with a = 2.65 m/s 2 . (d) If the cable breaks, a y = − 9.80 m/s 2 . F − mg = m( −9.80 m/s 2 ) and F = 0. The elevator is in free-fall and the

balance reads zero.

G

G

5.24. Set Up: The maximum tension in the chain is at the top of the chain. Apply ∑ F = ma to the composite object of chain and boulder. Use the constant acceleration kinematic equations to relate the acceleration to time. Let +y be upward. The free-body diagram for the composite object is given in the figure below. T = 2.50wchain .

mtot = mchain + mboulder = 1325 kg. Solve: (a) ∑ Fy = ma y gives T − mtot g = mtot a.

a=

T − mtot g 2.50mchain g − mtot g ⎛ 2.50mchain ⎞ = =⎜ − 1⎟ g mtot mtot mtot ⎝ ⎠

⎛ 2.50 [ 575 kg ] ⎞ a=⎜ − 1⎟ (9.80 m/s 2 ) = 0.832 m/s 2 . ⎝ 1325 kg ⎠ (b) Assume the acceleration has its maximum value: a y = 0.832 m/s 2 , y − y0 = 125 m and v0 y = 0.

y − y0 = v0 yt + 12 a yt 2 gives t =

2( y − y0 ) 2(125 m) = = 17.3 s ay 0.832 m/s 2

Reflect: The tension in the chain is T = 1.41 × 104 N and the total weight is 1.30 × 104 N. The upward force exceeds the downward force and the acceleration is upward.


5-12

Chapter 5

*5.25. Set Up: The camera is at rest relative to the train car and has the same horizontal acceleration as the car. Use coordinates with +y upward and +x in the direction of motion of the train car. Solve: (a) The free-body-diagram is given in the figure below. The tension T in the strap has been replaced by its x and y components. φ = 12.0D

(b) oFx = ma x gives T sin φ = ma. oFy = ma y gives T cos φ − mg = 0. Combining these two equations to eliminate

T gives a = g tan φ = 2.08 m/s 2 . And the y component equation gives mg 3.00 N = = 3.07 N cos φ cos12.0° Reflect: As a increases there must be a larger horizontal component of T and the angle φ increases. T=

5.26. Set Up: The hanging mass is at rest with respect to the cart, so its acceleration equals that of the cart. Use coordinates with +x in the direction of the acceleration. Solve: (a) Since the speed is increasing the acceleration is in the direction of the motion and is to the right. The free-body diagram of the mass is given in the figure below. +x is to the right and T has been replaced by its x and y components.



oFy = ma y gives T cos β − mg = 0 and T =

5-13

mg cos β

oFx = ma x gives T sin β = ma.

⎛ mg ⎜ ⎝ cos β

⎞ a ⎟ sin β = ma and tan β = . g ⎠

For a = 3.0 m/s 2 , β = 17°. The accelerometer deflects 17D to the left of the vertical. (b) Since the speed is decreasing, the acceleration is opposite to the direction of the motion and is to the right, as in part (a). The same equation applies as in part (a) and a 4.5 m/s 2 tan β = = . g 9.80 m/s 2

β = 25D. The accelerometer deflects 25D to the left of vertical. (c) Constant speed means a = 0. The hanging mass has no acceleration and hence no horizontal force on it. The accelerometer hangs vertically.

5.27. Set Up: Use coordinates where +x is parallel to the slope and directed up the slope. Let the skier have mass m. The normal force n exerted by the ground is perpendicular to the slope. Solve: (a) The free-body diagram for the skier is given in the figure below. The weight mg of the skier has been replaced by its x and y components. oFx = ma x gives − mg sin 24D = ma x and a x = − g sin 24D = −4.0 m/s 2 . Her acceleration is 4.0 m/s 2 , directed down the slope.

(b) and (c) Nothing changes in the free-body diagram and her acceleration is the same as in (a).

5.28. Set Up: In each free-body diagram take a positive coordinate direction to be the direction of the acceleration of the object. The box accelerates to the right and the bucket accelerates downward. Their accelerations have the same magnitude. The box has mass mbox = (375 N)/(9.80 m/s 2 ) = 38.3 kg. Solve: (a) The free-body diagram for the box and bucket are shown in the figure below.


5-14

Chapter 5

(b) oFx = ma x applied to the box gives T = mbox a. oFy = ma y applied to the bucket gives mbucket g − T = mbucket a.

Combining these two equations to eliminate T gives

⎛ ⎞ mbucket 2 a=⎜ ⎟ g = 3.58 m/s m m + bucket ⎠ ⎝ box (c) υ 0 y = 0, a y = 3.58 m/s 2 , y − y0 = 1.50 m. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υy =

2a y ( y − y0 ) =

2(3.58 m/s 2 )(1.50 m) = 3.28 m/s

*5.29. Set Up: Let m1 = 50.0 kg and m2 = 30.0 kg. The two boxes have the same magnitude of acceleration. For the 30.0 kg box let +y be downward and for the 50.0 kg box use coordinates parallel and perpendicular to the ramp, with + x up the ramp. Solve: (a) The free-body diagrams are given in the figure below. φ = 30.0D

(b) The downward force on the 30.0 kg box is m2 g = 294 N. The force pulling the system in the opposite direction is

the component of m1g directed down the ramp, m1g sin φ = 245 N. m2 g > m1g sin o and the 30.0 kg box moves downward. Therefore, the 50.0 kg box moves up the ramp. (c) oFx = ma x applied to m1 gives T − m1g sin o = m1a. oFy = ma y applied to m2 gives m2 g − T = m2a.

Combining these two equations to eliminate T gives a=

m2 g − m1g sin φ 294 N − 245 N = = 0.612 m/s 2 m1 + m2 50.0 kg + 30.0 kg

The 50.0 kg box accelerates up the ramp at 0.612 m/s 2 and the 30.0 kg box accelerates downward at 0.612 m/s 2 . Reflect: Only the component of the weight of the 50 kg box that is parallel to the ramp acts to oppose the weight of the 30 kg box. Therefore, the 30 kg box pulls the 50 kg box up the ramp even though its weight is less than the weight of the 50 kg box.

5.30. Set Up: Assume the box is initially at rest. The box remains at rest and friction is static if the pull is less than the maximum possible static friction force. If the pull is larger than this, the box moves and the friction is kinetic. Solve: n = w = 80 N. The maximum possible static friction force is μs n = 40 N. The kinetic friction force is

μk n = 20 N. The actual static friction force is only as large as necessary to prevent motion. (a) fs = 0 and the box remains at rest. (b) fs = 25 N and the box remains at rest. (c) fs = 39 N and the box remains at rest. (d) The box slides and f k = 20 N. (e) The box slides and f k = 20 N.



5-15

*5.31. Set Up: Since the only vertical forces are n and w, the normal force on the box equals its weight. Static friction is as large as it needs to be to prevent relative motion between the box and the surface, up to its maximum possible value of fs max = μs n. If the box is sliding then the friction force is f k = μk n. Solve: (a) If there is no applied force, no friction force is needed to keep the box at rest. (b) fs max = μs n = (0.40)(40.0 N) = 16.0 N. If a horizontal force of 6.0 N is applied to the box, then fs = 6.0 N in the

opposite direction. (c) The monkey must apply a force equal to fs max , 16.0 N. (d) Once the box has started moving, a force equal to f k = μk n = 8.0 N is required to keep it moving at constant

velocity. Reflect: μk < μs and less force must be applied to the box to maintain its motion than to start it moving.

5.32. Set Up: Assume that the normal force n at each hip is vertical. Solve: (a) The free-body diagram for the upper body of the person is shown in the figure below. oFy = ma y gives

2n − wu = 0 and n = wu /2.

(b) f k = μk n = (0.0050)(2.5)(0.65) mg = 5.2 N (c) f k = μk n = (0.30)(2.5)(0.65) mg = (0.30/0.0050)(5.2 N) = 310 N

*5.33. Set Up: The free-body diagram for the two crates treated as a single object, weight wC , is shown in Figure (a) below. The system doesn’t move so the friction force exerted by the roof is static friction. For the heaviest pallet of bricks this force has its maximum possible, fs = μs n. The free-body diagram for the pallet of bricks is given in Figure (b) below.

Solve: (a) For the crates, oFy = ma y gives n − wC = 0 and n = 400 lb. Then

fs = μs n = (0.666)(400 lb) = 266 lb. oFx = ma x gives T − fs = 0 and T = fs = 266 lb


5-16

Chapter 5

For the bricks, oFy = ma y gives T − wb = 0 and wb = T = 266 lb (b) For the upper crate the only horizontal force on the crate would be friction. This crate has a x = 0 so oFx = 0 and

the friction force is zero. Reflect: If some bricks are removed, so the weight of the pallet is reduced, the system remains at rest. The friction force on the crate is equal to the new weight of the pallet and is less than μs n.

G

G

5.34. Set Up: Constant velocity means a = 0 for each crate. Apply oF = ma to each crate, with +x to the right. Solve: The free-body diagrams for each crate are shown in the figure below.

(a) oFy = ma y for each crate gives n A = m A g and nB = mB g . f A = μk n A = μk m A g and f B = μk nB = μk mB g .

oFx = ma x for A gives T − f A = 0. oFx = ma x for B gives F − T − f B = 0. Adding these two equations gives

F − f A − f B = 0. F = f A + f B = μk g (m A + mB ). (b) T = f A = μs m A g

G

G

5.35. Set Up: Use the information about the motion to find the acceleration of the puck and then use oF = ma to relate a to the friction force. Take +x to be in the direction the puck is moving. Solve: υ 0 x = 9.9 m/s, x = 32.0 m and υ x = 0. υ x 2 = υ0 x 2 + 2a x x gives ax =

υ x 2 − υ0 x 2 2x

=

0 − (9.9 m/s) 2 = − 1.53 m/s 2 . 2(32.0 m)

The free-body diagram for the puck is given in the figure below. oFy = ma y gives n − mg = 0 and n = mg . oFx = ma x gives − μk n = max .



μk = − G

5-17

ma x a 21.53 m/s 2 =− x =− = 0.16. mg g 9.80 m/s 2

G

5.36. Set Up: Use oF = ma to find the acceleration that can be given to the car by the kinetic friction force. Then use a constant acceleration equation. Take +x in the direction the car is moving. Solve: (a) The free-body diagram for the car is shown in the figure below.

oFy = ma y gives n = mg . oFx = ma x gives − μk n = max . − μk mg = ma x and a x = − μk g . Then υ x = 0 and

υ x 2 = υ0 x 2 + 2a x ( x − x0 ) gives ( x − x0 ) = −

υ0 x 2 2a x

=+

υ0 x 2 (29.1 m/s) 2 = = 54.0 m. 2 μk g 2(0.80)(9.80 m/s 2 )

(b) υ0 x = 2 μ k g ( x − x0 ) = 2(0.25)(9.80 m/s 2 )(54.0 m) = 16.3 m/s

G

G

*5.37. Set Up: Apply ∑ F = ma to the box and calculate the normal and friction forces. The coefficient of kinetic fk . Let + x be in the direction of motion. ax = −0.90 m/s 2 . The box has mass 8.67 kg. n Solve: The normal force has magnitude 85 N + 25 N = 110 N. The friction force, from FH − f k = ma is friction is the ratio

28 N = 0.25. 110 N Reflect: The normal force is greater than the weight of the box, because of the downward component of the push force.

f k = FH − ma = 20 N − (8.67 kg)(−0.90 m/s 2 ) = 28 N. μ k =

G

G

5.38. Set Up: Apply aF = ma to the tires. The friction slows the tires so f = μr n = ma, where n = mg. Solve: ax =

v 2 − v0 2 , where L is the distance covered before the wheel’s speed is reduced to half its original speed 2L

and v = v0 /2. μ r =

2 1 2 3v 2 a v0 2 − v 2 v0 − 4 v0 = = = 0 . g 2 Lg 2 Lg 8 Lg

Low pressure, L = 18.1 m and High pressure, L = 92.9 m and

3 (3.50 m/s) 2 = 0.0259. 8 (18.1 m)(9.80 m/s 2 ) 3 (3.50 m/s) 2 = 0.00505. 8 (92.9 m)(9.80m/s) 2

Reflect: μ r is inversely proportional to the distance L, so

μr1 L2 = . μr2 L1


5-18

Chapter 5

*5.39. Set Up: n = mg . For constant speed, ax = 0. Apply oFx = max with + x in the direction of motion of the box. Solve: (a) The free-body diagram for the box is given in the figure below. oFx = max gives P − f k = 0 and

P = f k = μk mg = (0.200)(11.2 kg)(9.80 m/s 2 ) = 22.0 N.

fk = − μk g = − 1.96 m/s 2 . m G G Reflect: In (b) the friction force and the acceleration are directed opposite to the motion. Since a and y are in opposite directions, the box slows down. (b) P = 0 and − f k = max . ax = −

5.40. Set Up: Once the crate has started to move, friction is kinetic. The free-body diagram for the crate is given in G the figure below. F is the force the worker applies. The crate has mass m = w/g = 48.6 kg.

Solve: oFy = ma y so n = w. The force F to get the box moving is F = μs n = (0.615)(476 N) = 293 N. After the

box is moving, friction is f k = μ k n = (0.420)(476 N) = 200 N. oFx = max gives F − f k = ma and

a=

293 N − 200 N F − fk = = 1.91 m/s 2 m 48.6 kg

*5.41. Set Up: The free-body diagram for the box is given in the figure below. When the box is either at rest or sliding at constant speed, the acceleration of the box is zero. Use coordinates as shown, with the x axis parallel to the incline. The weight mg has been resolved into its x and y components.



5-19

Solve: oFy = ma y so n − mg cosθ = 0 and n = mg cosθ . oFx = max so f − mg sin θ = 0

f = μ n = μ mg cosθ so μ mg cosθ = mg sin θ and μ = tan θ . At θ = θ1, f = fs and μs = tan θ1. At θ = θ 2 , f = f k and μ k = tan θ 2 . Reflect: Usually μ k < μs , so θ 2 < θ1. If the angle is kept at θ1 the box will accelerate down the incline once it has

been set into motion.

5.42. Set Up: Take +x to be parallel to the incline and directed down the incline.

Solve: (a) In the absence of friction the free-body diagram for the van is as given in Figure (a) below. oFx = max

gives mg sin θ = ma and a = g sin θ = (9.80 m/s 2 )sin 40.0° = 6.30 m/s 2

(b) The free-body diagram with friction is given in Figure (b) above. oFy = ma y gives n = mg cosθ . oFx = max

gives mg sin θ − μk n = ma. mg sin θ − μ k mg cosθ = ma and a = g (sin θ − μ k cosθ ) = 4.80 m/s 2 .

5.43. Set Up: Take +x to be down the incline. At the maximum angle, the static friction force has its maximum value, fs = μs n. Solve: The free-body diagram for the patient is given in the figure below. oFy = ma y gives n = mg cosθ .

oFx = max gives mg sin θ − μs n = 0. mg sin θ − μs mg cosθ = 0. tan θ = μs and θ = 50°.


5-20

Chapter 5

5.44. Set Up: (a) The free-body diagram for the person is given in Figure (a) below. F is the traction force along the spinal column and w = mg is the person’s weight. (b) The free-body diagram for the collar where the cables are attached is given in Figure (b) below. The tension in each cable has been resolved into its x and y components.

Solve: (a) n = w and F = fs = μs n = 0.75w = 0.75(9.80 m/s2 )(78.5 kg) = 577 N (b) 2T sin 65° − F = 0 so

T=

F 0.75w = = 0.41w = (0.41)(9.80 m/s 2 )(78.5 kg) = 315 N ° 2sin 65 2sin 65°

*5.45. Set Up: The free-body diagram for the crate is given in the figure below. Use coordinates with axes parallel and perpendicular to the ramp. Constant speed means a = 0. The mass of the crate is m = w/g = 38.3 kg. The weight

mg has been resolved in to its x and y components.

Solve: (a) oFy = ma y gives n = mg cos φ . f k = μ k n = μ k mg cos φ . oFx = max with ax = 0 gives T − mg sin φ −

f k = 0, so T = mg sin φ + f k = (375 N)sin 33° + (0.250)(375 N)cos33° = 283 N (b) T = 0 so − f k − mg sin φ = max

− μ k mg cos φ − mg sin φ = max and ax = − g ( μ k cos φ + sin φ ) = − 0.754 g The acceleration would be 0.754 g = 7.39 m/s 2 , directed down the ramp. Reflect: In part (a) the force up the incline is 283 N and the force down the incline is 283 N. When the rope breaks T → 0 and the force down the incline remains 283 N. The acceleration down the incline is (283 N)/(38.3 kg) =

7.39 m/s 2 . © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


5-21

5.46. Set Up: The free-body diagram is given in Figure (a) (for going up the hill) and in Figure (b) below (for going down the hill). In each case the kinetic friction force is directed opposite to the velocity. Use coordinates where the axes are parallel and perpendicular to the ground. Let m be the mass of the toboggan.

Solve: In either case oFy = ma y gives n = mg cos φ and f k = μ k mg cos φ . (a) oFx = max gives − f k − mg sin φ = max

− μ k mg cos φ − mg sin φ = ma x and ax = − g ( μ k cos φ + sin φ ) = − (9.80 m/s2 )(0.30cos 40.0° + sin 40.0°) = −8.55 m/s2 The acceleration is 8.55 m/s 2 , directed down the hill. (b) oFx = max gives − mg sin φ + f k = max and ax = − g (sin φ − μ k cos φ ) = −4.05 m/s 2 . The acceleration is

4.05 m/s 2 , directed down the hill. G

G

*5.47. Set Up: Apply aF = ma to the box. When the box is ready to slip, the static friction force has its maximum possible value, fs = μs n. Use coordinates parallel and perpendicular to the ramp. Solve: (a) The normal force will be wcosα and the component of the gravitational force along the ramp is w sin α . The box begins to slip when w sin α > μs w cos α , or tan α > μs = 0.35, so slipping occurs at α = arctan(0.35) = 19.3D. (b) When moving, the friction force along the ramp is μ k w cos α , the component of the gravitational force along the

ramp is w sin α , so the acceleration is ( w sin α − wμ k cos α ) m = g (sin α − μ k cos α ) = 0.92 m s 2 . (c) Since v0 x = 0, 2ax = v 2 , so v = (2ax)1/ 2 , or v = [(2)(0.92m/s 2 )(5 m)]1/ 2 = 3 m/s. Reflect: When the box starts to move, friction changes from static to kinetic and the friction force becomes smaller.

5.48. Set Up: Use coordinates where + y is upward and + x is horizontal to the right. The applied force pushes to the left so the friction force is to the right. G Solve: (a) The free-body diagram is given in the figure below. The applied force F has been replaced by its x and y components.


5-22

Chapter 5

(b) oFy = ma y gives n − F sin 30° − w = 0 and n = w + F sin 30° = 125 N + (75 N)(sin 30°) = 163 N

(c) oFx = max gives f − F cos30° = 0 and f = F cos30° = 65 N (d) The maximum possible static friction force is fs = μs n = (0.80)(163 N) = 130 N (e) The vertical component of F is now upward. oFy = ma y now gives n + F sin 30° − w = 0.

n = w − F sin 30° = 88 N. G

G

5.49. Set Up: Apply aF = ma to the crate. fs ≤ μs n and f k = μk n. Let + y be upward and let + x be in the direction of the push. Since the floor is horizontal and the push is horizontal, the normal force equals the weight of the crate: n = mg = 441 N . The force it takes to start the crate moving equals max fs and the force required to keep it moving equals f k Solve: (a) max fs = 313 N, so μs =

313 N 208 N = 0.710. f k = 208 N, so μ k = = 0.472. 441 N 441 N

(b) The friction is kinetic. aFx = max gives F − f k = ma and

F = f k + ma = 208 N + (45.0 kg)(1.10 m/s 2 ) = 258 N. (c) (i) The normal force now is mg = 72.9 N. To cause it to move, F = max fs = μs n = (0.710)(72.9 N) = 51.8 N.

(ii) F = f k + ma and a =

F − f k 258 N − (0.472)(72.9 N) = = 4.97 m/s 2 m 45.0 kg

Reflect: The kinetic friction force is independent of the speed of the object. On the moon, the mass of the crate is the same as on earth, but the weight and normal force are less. G G 5.50. Set Up: Apply ∑ F = ma to the package. Calculate a and then use a constant acceleration equation to describe the motion. Let + x be directed up the ramp. Solve: (a) Fnet = 2mg sin 37° − f k = 2mg sin 37° − μ k mg cos37° = ma and a = −(9.8 m/s 2 )(0.602 + (0.30)(0.799)) = 28.25m/s 2

Since we know the length of the slope, we can use vx2 = v0 x 2 + 2ax ( x − x0 ) with x0 = 0 and vx = 0 at the top. v0 2 = −2ax = −2( −8.25 m/s 2 )(8.0 m) = 132 m 2 /s 2 and v0 = 132 m 2 /s 2 = 11.5 m/s

(b) For the trip back down the slope, gravity and the friction force operate in opposite directions to each other. Fnet = 2mg sin 37° + μ k mg cos37° = ma and a = g (− sin 37° + 0.30 cos37°) = (9.8 m/s 2 )(( −0.602) + (0.30)(0.799)) = −3.55 m/s 2 .

Now we have v0 = 0, x0 = 28.0 m, x = 0 and v 2 = v0 2 + 2a( x − x0 ) = 0 + 2(−3.55 m/s 2 )(−8.0 m) = 56.8 m 2 /s 2 , so

v=

56.8 m 2 /s 2 = 7.54 m/s.

Reflect: In both cases, moving up the incline and moving down the incline, the acceleration is directed down the incline. The magnitude of a is greater when the package is going up the incline, because mg sin 37° and f k are in the

same direction whereas when the package is going down these two forces are in opposite directions.

mg . We know D, so we must D 4 find the mass of the raindrop. The density of water is 103 kg/m3 and the volume of a sphere is V = π r 3. 3 4 3 4 −3 3 −7 3 Solve: The volume of the raindrop is V = π r = π (4.15 × 10 m) = 2.994 × 10 m . The mass of the raindrop 3 3

*5.51. Set Up: From Example 5.13 the terminal velocity of the raindrop is υ T =

is

103 kg 1 m3

⋅ (2.994 × 10−7 m3 ) = 2.994 × 10−4 kg. Thus, the terminal velocity of the raindrop is



υT =

(2.994 × 10−4 kg)(9.80 m/s 2 ) 2.43 × 10−5 kg/m

5-23

= 11.0 m/s.

Reflect: At terminal velocity the upward drag force on the raindrop equals the downward gravitational force it experiences, and the acceleration of the raindrop is zero. A radius of 4 mm represents a very large raindrop, but a speed of 11 m/s (about 25 mph) is reasonable for a raindrop or hailstone of this size. mg . The acceleration of the raindrop can be determined from D D Newton’s second law: mg − Dυ y2 = ma y , which gives a y = g − υ y 2 . m

5.52. Set Up: According to Example 5.13 υ T =

2

D⎛1 ⎞ D 2 1 3 1 ⎛ D ⎞ ⎛ mg ⎞ Solve: When υ y = υT , the acceleration is a y = g − ⎜ υT ⎟ = g − υT = g − ⎜ ⎟ ⎜ ⎟ = g − g = g . ⎝ 4m ⎠ ⎝ D ⎠ 4m 4 4 m⎝2 ⎠ 2

Thus, the acceleration of the raindrop is

3 g when it reaches half its terminal velocity. 4

Reflect: At terminal velocity the force of air resistance is equal to the weight of the object. At one-half terminal velocity, the force of air resistance is reduced to one-quarter the weight of the object.

*5.53. Set Up: Take + y to be downward. Solve: (a) FD = Dυ 2 . The FD versus υ graph is sketched in Figure (a) below. (b) Fnet = − FD + mg = − Dυ 2 + mg . See Figure (b) below. (c) υ = 0 at t = 0 and υ → υ t as t increases. When υ = υ t , Fnet = 0. See Figure (c) below. (d) See Figure (d) below. (e) a = Fnet /m; graph has same shape as Fnet versus t. See Figure (e) below.

5.54. Set Up: qFspr q = mg so mg = kx.


5-24

Chapter 5

mg (1.25 kg)(9.80 m/s 2 ) = = 327 N/m 0.0375 m x kx (327 N/m)(0.0813 m) = = 2.71 kg (b) m = g 9.80 m/s 2 Solve: (a) k =

*5.55. Set Up: qFspr q = mg so mg = kx. x is the change in length of the spring. mg (0.875 kg)(9.80 m/s 2 ) = = 357 N/m 0.0240 m x kx (357 N/m)(0.0572 m) = = 2.08 kg (b) m = g 9.80 m/s 2 Reflect: The elongation of the vertical spring is proportional to the mass hung from it. So, the mass in (b) is Solve: (a) k =

⎛ 5.72 cm ⎞ ⎜ ⎟ (0.875 kg) = 2.08 kg. ⎝ 2.40 cm ⎠

5.56. Set Up: qFspr q = F , where F is the pull on the strip or the force the strip exerts. F 1.50 N = = 40.0 N/m x 0.0375 m (b) F = kx = (40.0 N/m)(0.0114 m) = 0.456 N Solve: (a) F = kx so k =

5.57. Set Up: According to Hooke’s law, the force exerted by a spring is given by Fspr = 2kx, where x is the amount that the spring is stretched ( x > 0) or compressed ( x < 0) from its original length. The negative sign indicates that the force is opposite the displacement of the spring. Solve: (a) When the 60 kg person stands on the pogo stick, the spring compresses by x = (5.0 cm − 46 cm) = −41 cm = −0.41 m, where the negative sign indicates a compression, which happens to be in the downward direction in this instance. Thus, the maximum force exerted by the spring is Fspr = − kx = − (1.4 × 104 N/m)(−0.41 m) = 5740 N, which is in the upward direction. The net force on the 60 kg person is the sum of the upward spring force and her downward weight ( wy = 2mg ): aFy = 5740 N − (60 kg)(9.80 m/s2 ) = 5152 N = 5200 N. (b) Apply Newton’s second law: aFy = ma y and solve for a y to obtain a y

aFy

m

=

5152 N = 85.9 m/s 2 = 86 m/s 2 . 60 kg

m ⎞⎛ 1g ⎛ ⎞ Finally convert this into g’s to obtain ⎜ 85.9 2 ⎟⎜ ⎟ = 8.8 g . s ⎠⎝ 9.80 m/s 2 ⎠ ⎝ Reflect: This is a large acceleration and this pogo stick is definitely “extreme”! Note, to avoid rounding errors, it is best to keep one or two extra digits in your intermediate calculations—you can underline the significant digits for clarity.

5.58. Set Up: F = − Fspr so F = kx. Solve: (a) Yes, Hooke’s law says F = kx and this is true here since the graph of F versus x is a straight line. (b) k is the slope of the graph of F versus x: k = (c) F = kx = (150 N/m)(0.17 m) = 25 N.

15.0 N = 150 N/m 0.100 m

5.59. Set Up: mg = qFspr q = kx ⎛k⎞ Solve: (a) Yes. Hooke’s law says m = ⎜ ⎟ x and this is true here since the graph of m versus x is a straight line. ⎝g⎠ k 0.500 kg = 2.50 kg/m and k = (2.50 kg/m) g = 24.5 N/m (b) The slope of the graph of m versus x is k /g , so = g 0.200 m



(c) x =

5-25

mg (2.35 kg)(9.80 m/s 2 ) = = 0.94 m = 94 cm k 24.5 N/m

⎛ 2.35 kg ⎞ Reflect: A 0.50 kg mass stretches the spring 20 cm, so a 2.35 kg mass stretches it ⎜ ⎟ (20 cm). ⎝ 0.50 kg ⎠

5.60. Set Up: Use coordinates with +y upward. Label the masses 1, 2, and 3 and call the amounts the springs are stretched x1, x2 , and x3. Each spring force is kx. Solve: (a) The three free-body diagrams are shown in the figure below.

(b) kx3 = mg so x3 =

mg (6.40 kg)(9.80 m/s 2 ) = = 0.800 cm k 7.80 × 103 N/m

⎛ mg ⎞ kx2 = mg + kx3 = 2mg so x2 = 2 ⎜ ⎟ = 1.60 cm ⎝ k ⎠ ⎛ mg ⎞ kx1 = mg + kx2 = 3mg so x3 = 3 ⎜ ⎟ = 2.40 cm ⎝ k ⎠ The length of the springs, starting from the top one, are 14.4 cm, 13.6 cm, and 12.8 cm.

*5.61. Set Up: The magnitude of the spring force is kx. The force diagrams are given in the figure below.

ma (9.50 kg)(2.00 m/s 2 ) = = 0.152 m = 15.2 cm k 125 N/m ma 0.152 m = = 0.176 m = 17.6 cm (b) kx cos30.0° = ma and x = k cos30.0° cos30.0° Reflect: Only the horizontal component of the spring force accelerates the sled, so a greater spring force is needed in part (b). Solve: (a) kx = ma and x =


5-26

Chapter 5

5.62. Set Up: The free-body diagrams are given in the figure below. There is a kinetic friction force f k = μk n.

Solve: (a) n = mg and f k = μk mg . kx − f k = ma so kx − μk mg = ma

m( μk g + a) (9.50 kg)(0.200 [9.80 m/s 2 ] + 2.00 m/s 2 ) = = 30.1 cm k 125 N/m (b) oFy = ma y gives n + kx sin φ = mg so n = mg − kx sin o and f k = μk mg − μk kx sin φ x=

oFx = max gives kx cos φ − f k = ma and kx cos φ − μk mg + μk kx sin φ = ma

x=

m( μ k g + a ) (9.50 kg)(0.200[9.80 m/s 2 ] + 2.00 m/s 2 ) = = 32.1 cm k (cos φ + μk sin φ ) (125 N/m)(cos30.0° + [0.200][sin 30.0°])

*5.63. Set Up: Take + y downward. (a) Assume the hip is in free-fall. (b) The free-body diagram for the person is given in the figure below. It is assumed that the whole mass of the person has the same acceleration as her hip.

Solve: (a) υ0 y = 0, y − y0 = 1.0 m, a y = +9.80 m/s 2 . υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ y = 2a y ( y − y0 ) = 2(9.80 m/s 2 )(1.0 m) = 4.4 m/s (b) υ0 y = 4.4 m/s, y − y0 = 0.020 m, υ y = 1.3 m/s. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

ay =

υ y 2 − υ0 y 2 2( y − y0 )

=

(1.3 m/s) 2 − (4.4 m/s)2 = −440 m/s 2 2(0.020 m)

The acceleration is 440 m/s 2 , upward. oFy = ma y gives w − n = − ma and n = w + ma = m( a + g ) = (55 kg)(440 m/s 2 + 9.80 m/s 2 ) = 25,000 N

(c) υ y = υ0 y + a yt gives t =

υ y − υ0 y ay

=

1.3 m/s − 4.4 m/s

2440 m/s 2

= 7.0 ms

Reflect: When the velocity change occurs over a small distance the acceleration is large.



5-27

5.64. Set Up: Hooke’s law says the magnitude F of the force from the pad is given by F = kx, where x is the distance the pad is compressed. From Problem 5.63, set F = 25,000 N when x = 0.020 m. Solve: k =

F 25,000 N = = 1.3 × 106 N/m x 0.020 m

5.65. Set Up: Assume that both friction and the mass of the bungee cord are negligible. You can determine the Δυ x

acceleration of the wagon by using ax = Newton’s second law, Fyou

. The force that you apply to the wagon can then be determined from Δt = max . Finally (by Newton’s third law), this force must be equal in magnitude—but

opposite in direction—to the force that the bungee cord applies to you. Thus, we have Fyou = 1kx or k =

Fyou x

.

Δυ x 1.5 m/s − 0 = = 0.75 m/s 2 . The total weight of the system (wagon Δt 2.0 s w 295 N = 30.1 kg. plus sister plus bungee cord) is 220 N + 75 N = 295 N, so the mass of the system is m = = g 9.80 m/s 2

Solve: The acceleration of the wagon is ax =

The net force you apply to the system through the bungee cord must be Fx = max = (30.1 kg)(0.75 m/s 2 ) = 22.6 N. Thus, we find k =

Fyou x

=

22.6 N = 30 N/m, where the answer has been rounded to one significant figure. (2.0 m − 1.3 m)

Reflect: The force that the spring applies to you is Fspring = − kx, but the force that you apply to the spring is

Fyou = − Fspring = + kx. Note that x is the elongation of the spring—not its position on the x-axis. G

G

5.66. Set Up: Apply aF = ma to the load of bricks and to the counterweight. The tension is the same at each end of the rope. The rope pulls up with the same force (T ) on the bricks and on the counterweight. The counterweight accelerates downward and the bricks accelerate upward; these accelerations have the same magnitude. Solve: (a) The free-body diagrams for the bricks and counterweight are given in the figure below.

(b) Apply aFy = ma y to each object. The acceleration magnitude is the same for the two objects. For the bricks G G take + y to be upward since a for the bricks is upward. For the counterweight take + y to be downward since a is

downward. bricks: aFy = ma y T − m1g = m1a counterweight: aFy = ma y m2 g − T = m2 a Add these two equations to eliminate T: (m2 − m1 ) g = (m1 + m2 ) a


5-28

Chapter 5

⎛ m − m1 ⎞ ⎛ 28.0 kg − 15.0 kg ⎞ 2 2 a =⎜ 2 ⎟g =⎜ ⎟ (9.80 m/s ) = 2.96 m/s ⎝ 15.0 kg + 28.0 kg ⎠ ⎝ m1 + m2 ⎠ (c) T − m1g = m1a gives T = m1 (a + g ) = (15.0 kg)(2.96 m/s 2 + 9.80 m/s 2 ) = 191 N

As a check, calculate T using the other equation. m2 g − T = m2 a gives T = m2 ( g − a ) = 28.0 kg(9.80 m/s 2 − 2.96 m/s 2 ) = 191 N, which checks. Reflect: The tension is 1.30 times the weight of the bricks; this causes the bricks to accelerate upward. The tension is 0.696 times the weight of the counterweight; this causes the counterweight to accelerate downward. If m1 = m2 , a = 0

and T = m1g = m2 g . In this special case the objects don’t move. If m1 = 0, a = g and T = 0; in this special case the counterweight is in free-fall. Our general result is correct in these two special cases.

5.67. Set Up: For the climber, a = 0. The rock face exerts a horizontal force n on the climber and no friction force. Let +y be upward. Solve: The free-body diagram for the climber is given in the figure below.

oFy = ma y gives T cos15° − mg = 0.

T=

mg 600.0 N = = 621 N. cos15° cos15°

(b) oFx = max gives n − T sin15° = 0. n = T sin15° = 161 N. Reflect: The tension in the rope is greater than his weight because only the vertical component of the tension holds him up.

5.68. Set Up: The free-body diagram for one bowling ball is given in the figure below. The tension T has been resolved into its x and y components. n is the force the other ball exerts on this one.

Solve: (a) oFy = ma y gives T cos 25° − w = 0 and T =

w 71.2 N = = 78.6 N cos 25° cos 25°

(b) oFx = max gives T sin 25° − n = 0 and n = T sin 25° = 33.2 N



5-29

*5.69. Set Up: The free-body diagrams for the rocket (weight wr ) and astronaut (weight W ) are given in Figures (a) and (b) below. FT is the thrust and n is the normal force the rocket exerts on the astronaut. The speed of sound is 331 m/s.

(a) Apply oFy = ma y to the rocket: FT − wr = ma. a = 4 g and wr = mg , so

F = m(5 g ) = (2.25 × 106 kg)(5)(9.80 m/s 2 ) = 1.10 × 108 N (b) Apply oFy = ma y to the astronaut: n − W = ma. a = 4 g and m =

⎛W n =W + ⎜ ⎝ g

W , so g

⎞ ⎟ (4 g ) = 5W ⎠

(c) υ 0 = 0, υ = 331 m/s and a = 4 g = 39.2 m/s 2 . υ = υ0 + at gives

t=

υ − υ0 a

=

331 m/s 39.2 m/s 2

= 8.4 s

5.70. Set Up: Construct a standard coordinate system with a horizontal x-axis and a vertical y-axis. Forces TA , TB , and TC are directed along strands A, B, and C, respectively. Note that TC = mg = (15 × 10−6 kg)(9.8 m/s 2 ) = 1.47 × 10−4 N = 0.15 mN is the weight of the spider.

Solve: (a) Apply oFx = 0 to obtain TA cos120D + TB cos 20D = 0. Solving for TB we obtain (i) TB = 2TA

cos120D cos 20D

.

Next apply oFy = 0 to obtain TA sin120D + TB sin 20D + mg sin 270D = 0, which can be written as (ii) TA sin120D + TB sin 20D − mg = 0. Substitute equation (i) into equation (ii) to obtain

TA sin120D − TA

cos120D cos 20D

sin 20D − mg = 0. Finally, solve this for TA to obtain TA =

(15 × 10−6 kg)(9.80 m/s 2 ) sin120D − tan 20D cos120D

Substituting this value into equation (i) gives TB = − (0.14 mN)

= 0.14 mN.

cos120D

= 0.075 mN. cos 20D Thus, the three forces are TA = 0.14 mN, TB = 0.075 mN, and TC = 0.15 mN.

(b) Using Hooke’s law, we have the magnitude of the elongation of each thread is given by x =

elongations of A, B, and C are

F . Thus, the k

0.14 mN 0.075 mN 0.147 mN = 0.13 cm, = 0.068 cm, and = 0.13 cm, 1.10 mN/cm 1.10 mN/cm 1.10 mN/cm

respectively.


5-30

Chapter 5

Reflect: Because strand A is nearly vertical and strand B is nearly horizontal, the tension in strand A is nearly equal to the weight of the spider and the tension in strand B is small.

G

G

*5.71. Set Up: a = 0 for both objects. The tension in the vertical wire is w. Apply oF = ma to the knot where the wires are ed and then to block A. The static friction force on block A will be whatever is required to keep block A at rest. Solve: The free-body diagram for the knot is given in Figure (a) below. oFy = ma y gives T1 sin 45° = w. oFx = max gives T2 cos 45.0° = T3 . sin 45.0° = cos 45.0° and T3 = w. The free-body diagram for block A is given in

Figure (b) below. fs = T3 = w = 12.0 N

(b) fs max = μs n = μs mg = (0.25)(60.0 N) = 15.0 N. Then w = fs max = 15.0 N. Reflect: The friction force found in part (a) is less than the maximum possible friction force.

G

G

5.72. Set Up: Apply aF = ma to the box. The box has an upward acceleration of a = 1.90 m/s 2 .

Solve: The floor exerts an upward force n on the box, obtained from n − mg = ma, or n = m( a + g ). The friction

force that needs to be balanced is μ k n = μk m(a + g ) = (0.32)(28.0 kg)(1.90 m/s 2 + 9.80 m/s 2 ) = 105 N. Reflect: If the elevator wasn’t accelerating the normal force would be n = mg and the friction force that would have to be overcome would be 87.8 N. The upward acceleration increases the normal force and that increases the friction force.

*5.73. Set Up: The unstretched length L0 of the tendon is 20.0 cm. The weight W suspended from the end of the tendon equals the force applied by the tendon. Solve: (a) x = L − L0 so Hooke’s law says W = kx = kL − kL0 . The graph of W versus L is of this form. 15.0 N = 50.0 N/m 0.500 m − 0.200 m (c) W = kx = (50.0 N/m)(0.080 m) = 4.0 N (b) k is the slope of W versus x, so k =

5.74. Set Up: The net force acting on the parachutist is given by Fy ,net = ma y . The acceleration of the parachutist can be calculated from the equation υ y 2 = υ0 y 2 + 2a y Δy. Choose the +y direction to be upward. Solve: During impact, the velocity goes from υ0 y = −6.30 m/s to υ y = 0 over a displacement of Δy = −0.92 m and

the acceleration is a y =

02 − (−6.30 m/s)2 = 21.6 m/s 2 . Thus, the net force acting on the parachutist is Fy ,net = ma y = 2(−0.92 m)



5-31

(65.0 kg)(21.6 m/s 2 ) = 1400 N (rounding to two significant figures). The net force on the parachutist is the sum of two forces: the upward force of the ground, Fground and the downward force of gravity acting on the parachutist. Thus, Fground − mg = 1400 N so that Fground = 1400 N + (65.0 kg)(9.80 m/s 2 ) = 2000 N (rounding to two significant figures). Reflect: The acceleration of the parachutists gives the net force acting on him. Since the net force is the sum of two opposing forces, it is smaller than the force that the ground exerts on the parachutist during impact.

5.75. Set Up: On Mars, g M = 0.379 g E = 3.71 m/s 2 . The free-body diagrams in each stage are shown in figures below. FD is the air drag force in Stage I, FP is the force from the parachute in Stage II and FR is the force from the retrorockets in Stage III.

Solve: (a) υ0 y = 19,300 km/h = 5361 m/s, υ y = 1600 km/h = 444 m/s, t = 240 s. υ y = υ0 y + a yt gives

ay =

υ y − υ0 y t

=

444 m/s − 5361 m/s = − 20.5 m/s 2 . 240 s

oFy = ma y gives mg − FD = ma y .

FD = m( g − a y ) = (827 kg)(3.71 m/s 2 − [−20.5 m/s 2 ]) = 2.0 × 104 N (b) υ0 y = 1600 km/h = 444 m/s, υ y = 321 km/h = 89.2 m/s, t = 94 s. υ y = υ0 y + a yt gives

ay =

υ y − υ0 y t

=

89.2 m/s − 444 m/s = − 3.77 m/s 2 . 94 s

oFy = ma y gives mg − FP = ma y .

FP = m( g − a y ) = (827 kg)(3.71 m/s 2 − [ −3.77 m/s 2 ]) = 6.19 × 103 N (c) υ0 y = 321 km/h = 89.2 m/s, υ y = 0, t = 2.5 s. υ y = υ0 y + a yt gives

ay =

υ y − υ0 y t

=

0 − 89.2 m/s = − 35.7 m/s 2 . 2.5 s

oFy = ma y gives mg − FR = ma y . FR = m( g − a y ) = (827 kg)(3.71 m/s2 − [−35.7 m/s 2 ]) = 3.26 × 104 N Reflect: In each stage the upward force is greater than the downward force, the acceleration is upward, opposite to G υ , and the speed decreases.

5.76. Set Up: Constant speed means a = 0. Use Newton’s third law to relate forces on A and on B. Solve: (a) Treat A and B as a single object of weight w = wA + wB = 4.80 N. The free-body diagram for this combined object is given in Figure (a) below. oFy = ma y gives n = w = 4.80 N. f k = μk n = 1.44 N. oFx = max gives F = f k = 1.44 N


5-32

Chapter 5

(b) The free-body force diagrams for blocks A and B are given in Figures (b) and (c) above. n and f k are the normal

and friction forces applied to block B by the tabletop and are the same as in part (a). f kB is the friction force that A applies to B. It is to the right because the force from A opposes the motion of B. nB is the downward force that A exerts on B. f kA is the friction force that B applies to A. It is to the left because block B wants A to move with it. n A is the normal force that block B exerts on A. By Newton’s third law, f kB = f kA and these forces are in opposite directions. Also, n A = nB and these forces are in opposite directions. oFy = ma y n A = wA = 1.20 N, so nB = 1.20 N. f kA = μk n A = (0.300)(1.20 N) = 0.36 N, and

for block A gives

f kB = 0.36 N.

oFx = max for

block A gives T = f kA = 0.36 N. oFx = max for block B gives F = f kB + f k = 0.36 N + 1.44 N = 1.80 N (c) In part (a) block A is at rest with respect to B and it has zero acceleration. There is no horizontal force on A besides friction, and the friction force on A is zero.

*5.77. Set Up: When the spring is stretched an amount x the spring force has magnitude kx and is directed to the right in the figure in the problem. The free-body diagram for the weight is given in Figure (a) below. When the spring is compressed a distance q x q the spring force has magnitude k q x q and is directed to the left. The free-body diagram for this case is given in Figure (b) below.

ma (1.10 kg)(2.50 m/s 2 ) = = 250 N/m x 0.0110 m k q xq (250 N/m)(0.0230 m) =− = −5.23 m/s 2 . The acceleration is and ax = − m 1.10 kg

Solve: (a) oFx = max gives kx = ma and k = (b) oFx = max gives − k q x q = max

5.23 m/s 2 , in the backward direction. G

G

5.78. Set Up: Apply aF = ma to each block. Constant speed means a = 0. When the blocks are moving, the friction force is f k and when they are at rest, the friction force is fs .



5-33

Solve: (a) The tension in the cord must be m2 g in order that the hanging block move at constant speed. This tension

must overcome friction and the component of the gravitational force along the incline, so m2 g = (m1g sin α + μk m1g cos α ) and m2 = m1 (sin α + μk cos α ). (b) In this case, the friction force acts in the same direction as the tension on the block of mass m1, so

m2 g = (m1g sin α − μk m1g cos α ), or m2 = m1 (sin α − μk cos α ). (c) Similar to the analysis of parts (a) and (b), the largest m2 could be is m1 (sinα + μscosα ) and the smallest m2

could be is m1 (sinα − μscosα ). Reflect: In parts (a) and (b) the friction force changes direction when the direction of the motion of m1 changes. In

part (c), for the largest m2 the static friction force on m1 is directed down the incline and for the smallest m2 the static friction force on m1 is directed up the incline.

5.79. Set Up: If the truck accelerates to the right the friction force on the box is to the right, to try to prevent the box from sliding relative to the truck. The free-body diagram for the box is given in the figure below. The maximum acceleration of the box occurs when fs has its maximum value, fs = μs n. If the box doesn’t slide its acceleration equals the acceleration of the truck.

Solve: n = mg . oFx = max gives fs = ma so μs mg = ma and a = μs g = 6.37 m/s 2 . υ0 x = 0, υ x = 30.0 m/s.

υ x = υ0 x + axt gives t=

υ x − υ0 x ax

=

30.0 m/s − 0 6.37 m/s 2

= 4.71 s

Reflect: If the truck has a smaller acceleration it is still true that fs = ma, but now fs < μs n.

5.80. Set Up: The free-body for the car is given in the figure below.

Solve: oFx = max gives − f k = ma x . − μk mg = ma x so ax = − μk g = − 3.92 m/s 2

υ x = 0 (stops), x − x0 = 219.75 ft = 67.0 m. υ x 2 = υ0 x 2 + 2ax ( x − x0 ) gives υ0 x = 22ax ( x − x0 ) = 22( −3.92 m/s 2 )(67.0 m) = 22.9 m/s = 51 mph He was speeding.


5-34

Chapter 5

G

G

*5.81. Set Up: Apply aF = ma to the brush. Constant speed means a = 0. Target variables are two of the forces on the brush. Note that the normal force exerted by the wall is horizontal, since it is perpendicular to the wall. The kinetic friction force exerted by the wall is parallel to the wall and opposes the motion, so it is vertically downward. The free-body diagram is given in the figure below.

Solve: Using aFx = max , we obtain n − F cos53.1° = 0, so n = F cos53.1°; thus, f k = μk n = μk F cos53.1°.

Using aFy = ma y we obtain F sin 53.1° − w − f k = 0, so F sin 53.1° − w − μk F cos53.1° = 0, which factors to give F (sin 53.1° − μk cos53.1°) = w . Solving for F we obtain F = (a) F =

w sin 53.1° − μk cos53.1°

=

w . sin 53.1° − μk cos53.1°

12.0 N = 16.9 N sin 53.1° − (0.150)cos53.1°

(b) n = F cos53.1° = (16.9 N)cos53.1° = 10.1 N Reflect: In the absence of friction, we have w = F sin 53.1°, which agrees with our expression.

5.82. Set Up: The tension throughout the cable connected to W is the weight W of the hanging mass. The free-body diagram for the pulley connected to the feet is given in the figure below.

Solve: (a) Ty = 51.5 N, to the leg. oFy = 0 gives W sin 46° − W sin 37° − Ty = 0. W (sin 46° − sin 37°) = 51.5 N

and W = 438 N. The mass of W is 44.7 kg. (b) The tractive force along the leg is Tx . oFx = 0 gives W cos 46° + W cos37° − Tx = 0

Tx = W (cos 46° + cos37°) = 654 N

*5.83. Set Up: The normal force is horizontal, perpendicular to the wall and the friction force is upward, parallel to the wall.



5-35

Solve: (a) The free-body diagram is shown in the figure below.

(b) f = W = 20 N. Note: n = F = 50 N. The maximum possible static friction force is fs = μs n = (0.50)(50 N) = 25 N

and the actual friction force is less than this. (c) Constant speed means a = 0. Moving means f = f k = μk n. f k = W so μk n = W and n =

W

μk

=

20 N = 100 N. 0.20

5.84. Set Up: Use coordinates with axes parallel and perpendicular to the incline, with + x in the direction of the acceleration. Solve: With the rock is sliding up the hill the friction force is down the hill. The free-body diagram is given in Figure (a) below.

oFy = ma y gives n = mg cos φ and f k = μk n = μk mg cos φ oFx = max gives mg sin φ + μk mg cos φ = ma

a = g (sin φ + μk cos φ ) = (9.80 m/s 2 )(sin 36° + [0.45]cos36°) a = 9.33 m/s 2 , down the incline. (b) The component of gravity down the incline is mg sin φ = 0.588mg . The maximum possible static friction force is fs = μs n = μs mg cos φ = 0.526mg . fs can’t be as large as mg sin φ and the rock slides back down. As the rock slides down, f k is up the incline. The free-body diagram is given in Figure (b) above.

oFy = ma y gives n = mg cos φ and f k = μk n = μk mg cos φ oFx = max gives mg sin φ − μk mg cos φ = ma a = g (sin φ − μ k cos φ ) = 2.19 m/s 2 , down the incline.

G

G

5.85. Set Up: Apply aF = ma to the enger to find the maximum allowed acceleration. Then use a constant acceleration equation to find the maximum speed. The free-body diagram for the enger is given in the following figure.


5-36

Chapter 5

Solve: aFy = ma y gives n − mg = ma. Set n = 1.6mg to obtain a = 0.60 g = 5.88 m/s 2 . Also, y − y0 = 3.0 m,

a y = 5.88 m/s 2 , v0 y = 0, so v y 2 = v0 y 2 + 2a y ( y − y0 ) gives v y = 5.9 m/s. Reflect: A larger final speed would require a larger value of a y , which would mean a larger normal force on the person.

5.86. Set Up: The free-body diagram for the car is given in the figure below. Use coordinates with axes parallel and perpendicular to the bridge surface.

Solve: (a) fs = μs n. n = mg cos φ so fs = μs mg cos φ

oFx = max gives mg sin φ − fs = 0. mg sin φ − μs mg cos φ = 0 so μs = tan φ tan φ = 0.750 and φ = 36.9° (b) Now friction is f k = μk n = μk mg cos φ . oFx = max gives mg sin φ − μk mg cos φ = ma.

a = g (sin φ − μk cos φ ) = (9.80 m/s 2 )(sin 36.9° − [0.550]cos36.9°) = 1.57 m/s 2

*5.87. Set Up: The block has the same horizontal acceleration a as the cart. Let +x be to the right and +y be upward. To find the minimum acceleration required, set the static friction force equal to its maximum value, μs n. Solve: The free-body diagram for the block is given in the figure below.



oFx = max gives n = ma. oFy = ma y gives fs − mg = 0. μs n = mg and μs ma = mg . a =

g

μs

5-37

.

Reflect: The smaller μs is the greater a must be to prevent slipping. Increasing a increases the normal force n and

that increases the maximum fs for a given μs .

5.88. Set Up: The monkey and bananas have the same mass and the tension in the rope has the same upward value at the bananas and at the monkey. Therefore, the monkey and bananas will have the same net force and hence the same acceleration, in both magnitude and direction. Solve: (a) For the monkey to move up, T > mg . The bananas also move up. (b) The bananas and monkey move with the same acceleration and the distance between them remains constant. (c) Both the monkey and bananas are in free fall. They have the same initial velocity and as they fall the distance between them doesn’t change. (d) The bananas will slow down at the same rate as the monkey. If the monkey comes to a stop, so will the bananas.

Solutions to age Problems *5.89. Set Up: Consider the analysis of Example 5.6 and use an x-axis that is parallel to the hill: this example shows that the net force on the toboggan (without the sail) is equal to w sin α and is directed downhill. Solve: Once the toboggan is moving at a constant speed, the net force on it must be zero. We know that the sum of the gravitational force and normal force is w sin α (directed downhill), so the sail must provide an equal force directed uphill. The correct answer is B. 5.90. Set Up: Consider the analysis of Example 5.6 and use an x-axis that is parallel to the hill: this example shows that the net force on the toboggan (without the sail) is equal to w sin α and is directed downhill. Solve: Assuming that the force of the wind remains the same magnitude as in the previous example, the net force on the toboggan would be 2w sin α (and directed downhill). Thus, the acceleration of the toboggan would be F 2mg sin α a= = = 2 g sin α (directed downhill). The correct answer is C. m m *5.91. Set Up: Consider the analysis of Example 5.11 and use an x-axis that is parallel to the hill. Solve: Since the wind blows parallel to the hill, the force of the wind has no y-component and we have

aFy = n + (− w cos α ) = 0 so N = w cos α as in Example 5.11. In the x-direction we have two changes: (i) since the

sled is moving uphill, friction is directed downhill (in the + x direction), and (ii) the magnitude of the wind force (which acts in the negative x-direction) must be subtracted. Thus, we have (for constant speed motion) aFx = w sin α + μk w cos α − f wind = ma x = 0. We can solve this for the force of the wind to obtain f wind =

w sin α + μk w cos α. The correct answer is D.


6

CIRCULAR MOTION AND GRAVITATION

Answers to Multiple-Choice Problems 1. D 2. B 15. C

3. D

4. A

5. C

6. C

7. C

8. A

9. D

10. B

11. B

12. C

13. A

14. A

Solutions to Problems *6.1. Set Up: The path shows the relative size of the radius of curvature R of each part of the track. RB > RC . A straight line is an arc of a circle of infinite radius, so RA → ∞.

υ2

G . At A, arad = 0. arad, B < arad, C . In each case, arad is directed toward the center of the circular R G path. The net force is proportional to arad and is shown in the figure below.

Solve: arad =

Reflect: The road must exert a greater friction force on the car when it makes tighter turns that have smaller radii of curvature.

6.2. Set Up: Since the stone travels in a circular path, its acceleration is arad = υ 2 /R, directed toward the center of the circle. The only horizontal force on the stone is the tension of the string. Set the tension in the string equal to its maximum value. Solve: (a) The free-body diagram for the stone is given in the figure below. In the diagram the stone is at a point to the right of the center of the path. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6-1

6-2

Chapter 6

(b) ∑ Fx = ma x gives T = m

υ2 R

.

υ=

TR = m

(60.0 N)(0.90 m) = 8.22 m/s. 0.80 kg

*6.3. Set Up: Each hand travels in a circle of radius 0.750 m and has mass (0.0125)(52 kg) = 0.65 kg and weight 6.4 N. The period for each hand is T = (1.0 s)/(2.0) = 0.50 s. Let + x be toward the center of the circular path. G (a) The free-body diagram for one hand is given in the figure below. F is the force exerted on the hand by the wrist. This force has both horizontal and vertical components.

(b) arad =

4π 2 R T

2

=

4π 2 (0.750 m) (0.50 s)

2

= 118 m/s 2

∑ Fx = max gives Fx = marad = (0.65 kg)(118 m/s 2 ) = 77 N F 77 N = = 12. The horizontal force from the wrist is 12 times the weight of the hand. w 6.4 N Reflect: The wrist must also exert a vertical force on the hand equal to its weight. (c)

6.4. Set Up: Since the car travels in an arc of a circle, it has acceleration arad = υ 2 /R, directed toward the center of the arc. The only horizontal force on the car is the static friction force exerted by the roadway. To calculate the minimum coefficient of friction that is required, set the static friction force equal to its maximum value, fs = μs n. Friction is static friction because the car is not sliding in the radial direction. Solve: (a) The free-body diagram for the car is given in the figure below. The diagram assumes the center of the curve is to the left of the car.


Circular Motion and Gravitation

(b) ∑ Fy = ma y gives n = mg . ∑ Fx = max gives μs n = m

μs =

υ2 gR

=

υ2 R

. μs mg = m

(25.0 m/s)2 (9.80 m/s 2 )(220 m)

υ2

6-3

and

R

= 0.290

*6.5. Set Up: The person moves in a circle of radius R = 3.00 m + (5.00 m)sin 30.0° = 5.50 m. The acceleration of the person is arad = υ 2 /R, directed horizontally to the left in the figure in the problem. The time for one revolution is the period T =

2π R

υ

.

G Solve: (a) The free-body diagram is given in the figure below. F is the force applied to the seat by the rod.

(b) ∑ Fy = ma y gives F cos30.0° = mg and F =

υ2 mg . ∑ Fx = max gives F sin 30.0° = m . Combining cos30.0° R

these two equations gives

υ = Rg tan θ = (5.50 m)(9.80 m/s 2 ) tan 30.0° = 5.58 m/s. 2π R 2π (5.50 m) = = 6.19 s. Then the period is T = 5.58 m/s υ

G G (c) The net force is proportional to m so in ∑ F = ma the mass divides out and the angle for a given rate of rotation is independent of the mass of the engers. Reflect: The person moves in a horizontal circle so the acceleration is horizontal. The net inward force required for circular motion is produced by a component of the force exerted on the seat by the rod.

G

G

6.6. Set Up: Apply ∑ F = ma to the button. The button moves in a circle, so it has acceleration arad . The situation is equivalent to that of Example 6.3. 2 2π R v2 Solve: (a) μs = . Expressing v in of the period T, v = so μs = 4π2 R . A platform speed of 40.0 rev/min T Rg T g corresponds to a period of 1.50 s, so μs =

4π 2 (0.150 m)

= 0.269. (1.50 s ) 2 (9.80 m / s 2 ) (b) For the same coefficient of static friction, the maximum radius is proportional to the square of the period (longer periods mean slower speeds, so the button may be moved farther out) and so is inversely proportional to the square of 2

⎛ 40.0 ⎞ = 0.067 m. the speed. Thus, at the higher speed, the maximum radius is (0.150 m) ⎜ ⎝ 60.0 ⎟⎠ 4π 2 R

. The maximum radial acceleration that friction can give is μs mg . At the faster rotation rate T T2 is smaller so R must be smaller to keep arad the same. Reflect: arad =


6-4

Chapter 6

6.7. Set Up: For Venus, R = 1.08 × 1011 m, T = 224.7 d = 1.94 × 107 s, and m = 4.87 × 1024 kg. Use coordinates where + x is toward the sun. 2π R 2π (1.08 × 1011 m) = = 3.50 × 104 m/s = 35.0 km/s T 1.94 × 107 s (b) The free-body diagram is shown in the figure below. Solve: (a) υ =

arad =

υ2 R

=

(3.50 × 104 m/s) 2 1.08 × 1011 m

= 0.0113 m/s 2

F = ma = (4.87 × 1024 kg)(0.0113 m/s 2 ) = 5.50 × 1022 N

6.8. Set Up: An object moving with speed v in an arc of a circle with radius R is arad =

υ2

, in toward the center R of the circular path. The normal force is perpendicular to the roadway and the gravity force is downward. The car G travels in a horizontal circle, so the acceleration is horizontal. Take + x to be horizontal, in the direction of a . 55.0 mph = 80.7 ft/s. Solve: (a) The free-body diagram for the car is given in the figure below.

∑ Fy = ma y gives n cos β = mg . ∑ Fx = max gives n sin β = m

n= tan β =

υ2 gR

=

υ2 R

.

⎛ mg ⎞ mg υ2 and ⎜ ⎟ sin β = m . cos β R ⎝ cos β ⎠ (80.7 m/s)2 (32 ft/s 2 )(900.0 ft)

= 0.226 and β = 12.7°.

*6.9. Set Up: At the safe speed no friction force is required. Consider a car of mass m. The free-body diagram for the car is given in the figure below. Use coordinates where + x is toward the center of the horizontal turn. Each turn is one-quarter of a circle so the radius R is given by 0.25 mi = 14 (2π R ). 9°12′ is 9.2°.



6-5

Solve: R = 4(0.25 mi)/2π = 0.1592 mi = 256 m

∑ Fy = ma y gives n y = w so n =

mg cos φ

∑ Fx = max gives nx = marad so n sin φ = m

υ2 R

2

⎛ mg ⎞ υ and υ = Rg tan φ = (256 m)(9.80 m/s 2 ) tan 9.2° = 20.0 m/s = 44.7 mph ⎜ ⎟ sin φ = m φ cos R ⎝ ⎠ Reflect: Race cars travel much faster than this. The track must exert a large friction force on the cars to maintain their circular motion.

6.10. Set Up: The ball has acceleration arad = υ 2 /R, directed toward the center of the circular path. When the ball is at the bottom of the swing, its acceleration is upward. Take + y upward, in the direction of the acceleration. The bowling ball has mass m = w/g = 7.27 kg. (4.20 m/s) 2 = 4.64 m/s, upward. R 3.80 m (b) The free-body diagram is given in the figure below. ∑ Fy = ma y gives T − mg = marad . Solve: (a) arad =

υ2

=

T = m( g + arad ) = (7.27 kg)(9.80 m/s 2 + 4.64 m/s 2 ) = 105 N

*6.11. Set Up: The period is T = 60.0 s and T =

2π R

υ

. The apparent weight of a person is the normal force

exerted on him by the seat he is sitting on. His acceleration is arad = υ 2 /R, directed toward the center of the circle. The enger has mass m = w/g = 90.0 kg. Solve: (a) υ =

υ 2 (5.24 m/s) 2 2π R 2π (50 m) = = 5.24 m/s. Note that arad = = = 0.549 m/s 2 . T R 50.0 m 60.0 s

(b) The free-body diagram for the person at the top of his path is given in Figure (a) below. The acceleration is downward, so take + y downward. ∑ Fy = ma y gives mg − n = marad .


6-6

Chapter 6

n = m( g − arad ) = (90.0 kg)(9.80 m/s2 − 0.549 m/s2 ) = 833 N.

The free-body diagram for the person at the bottom of his path is given in Figure (b) above. The acceleration is upward, so take + y upward. ∑ Fy = ma y gives n − mg = marad and n = m( g + arad ) = 931 N. (c) Apparent weight = 0 means n = 0 and mg = marad .

g= The time for one revolution would be T=

υ2 R

and υ = gR = 22.1 m/s.

2π R

υ

=

2π (50.0 m) = 14.2 s. 22.1 m/s

Note that arad = g . (d) n = m( g + arad ) = 2mg = 2(882 N) = 1760 N, twice his true weight. Reflect: At the top of his path his apparent weight is less than his true weight and at the bottom of his path his apparent weight is greater than his true weight.

G

6.12. Set Up: arad = υ 2 /R, directed toward the center of the circular path. At the bottom of the dive, arad is upward. The apparent weight of the pilot is the normal force exerted on her by the seat on which she is sitting. υ2 υ2 (95.0 m/s) 2 = = 230 m. Solve: (a) arad = gives R = R arad 4.00(9.80 m/s 2 ) (b) The free-body diagram for the pilot is given in the figure below. ∑ Fy = ma y gives n − mg = marad .

n = m( g + arad ) = m( g + 4.00 g ) = 5.00mg = (5.00)(50.0 kg)(9.80 m/s 2 ) = 2450 N of a full rotation, so in 12 s a hand travels through a distance of G 1 (2π R). In (c) use coordinates where + y is upward, in the direction of arad at the bottom of the swing. 8

6.13. Set Up: R = 0.700 m. A 45° angle is

1 8

1 ⎛ 2π R ⎞ υ 2 (1.10 m/s) 2 = = 1.73 m/s 2 Solve: (a) υ = ⎜ = 1.10 m/s and arad = ⎟ R 8 ⎝ 0.50 s ⎠ 0.700 m



6-7

(b) The free-body diagram is shown in the figure below. F is the force exerted by the blood vessel.

(c) ∑ Fy = ma y gives F − w = marad and F = m( g + arad ) = (1.00 × 1023 kg)(9.80 m/s 2 + 1.73 m/s 2 ) = 1.15 × 1022 N, upward

(d) When the arm hangs vertically and at rest, arad = 0 so F = w = mg = 9.8 × 1023 N. Reflect: The acceleration of the hand is only about 20% of g, so the increase in the force on the blood drop when the arm swings is about 20%.

6.14. Set Up: The forces on the water are the downward force of gravity and the downward normal force exerted by the bottom of the pail on the water. The acceleration of the water is arad = υ 2 /R, downward. At the minimum speed, the water starts to fall from the bucket and n → 0. Solve: The free-body diagram for the water is given in the figure below. ∑ Fy = ma y , with n = 0, gives

mg = m

υ2 R

.

υ = gR = (9.80 m/s2 )(0.600 m) = 2.42 m/s

*6.15. Set Up: According to the hint, “red out” can occur when the non-gravitational part of the centripetal acceleration exceeds 2.5 g’s. Thus, the maximum allowed downward centripetal acceleration is g + 2.5 g = 3.5 g .

v2 . Also, from the appendix we know that 1 km/h = 0.2778 m/s. r m/s ⎞ ⎛ Solve: First convert the speed of the plane into m/s: (320 km/h) ⎜ 0.2778 ⎟ = 88.9 m/s. The minimum allowed ⎝ km/h ⎠ We know that arad =

radius occurs when the centripetal acceleration has its maximum allowed value of 3.5 g = obtain r =

v2 . Solving for r we r

v2 (88.9 m/s) 2 = = 230 m. 3.5 g 3.5(9.80 m/s 2 )


6-8

Chapter 6

Reflect: The force of gravity tends to cause all objects to accelerate together and does not contribute to the redistribution of blood necessary for “red out” to occur—it is the non-gravitational part of the acceleration that must be limited below 2.5g.

6.16. Set Up: The spheres exert a gravitational force on each other. Each sphere has mass m. G = 6.673 × 10211 N ⋅ m 2 /kg 2

Solve: Fg = G

m2 r2

. m=r

Fg G

= (0.25 m)

3.0 × 1029 N 6.673 × 10211 N ⋅ m 2 /kg 2

= 1.7 kg

*6.17. Set Up: A proton has mass mp = 1.673 × 10227 kg. 1.0 fm = 1.0 × 10215 m Solve: Fg = G

m2 r

2

= (6.673 × 10211 N ⋅ m 2 /kg 2 )

(1.673 × 10227 kg) 2 215

(1.0 × 10

m)

2

= 1.87 × 10234 N

6.18. Set Up: The net force on each astronaut is the gravity force exerted by the other astronaut. Call the astronauts A and B, where m A = 65 kg and mB = 72 kg. Solve: (a) The free-body diagram for astronaut A is given in Figure (a) and for astronaut B in Figure (b) below. F F ∑ Fx = max for A gives FA = m Aa A and a A = A . And for B, aB = B . mA mB

FA = FB = G

m AmB r2

= (6.673 × 10211 N ⋅ m 2 /kg 2 )

(65 kg)(72 kg) (20.0 m)2

aA =

7.807 × 10210 N = 1.2 × 10211 m/s 2 ; 65 kg

aB =

7.807 × 10210 N = 1.1 × 10211 m/s 2 72 kg

= 7.807 × 10210 N

(b) x = x0 + υ0 xt + 12 axt 2 gives x A = 12 a At 2 and xB = 12 aBt 2

x A + xB = 20.0 m so 20.0 m = 12 (a A + aB )t 2 and t=

2(20.0 m) 211

1.2 × 10

m/s 2 + 1.1 × 10211 m/s 2

= 1.32 × 106 s = 15 days

(c) Their accelerations would increase as they moved closer and the gravitational attraction between them increased.

*6.19. Set Up: The gravitational force between two objects is Fg = G

m1m2 r2

. The mass of the earth is

mE = 5.97 × 1024 kg and the mass of the sun is mS = 1.99 × 1030 kg. The distance from the earth to the sun is rSE = 1.50 × 1011 m and the distance from the earth to the moon is rEM = 3.84 × 108 m. Solve: FSM = G

mSmM rSM 2

=G

mSmM rSE 2

. FEM = G

mE mM rEM 2

.



6-9

2

2

FSM ⎛ mS ⎞ ⎛ rEM ⎞ ⎛ 1.99 × 1030 kg ⎞⎛ 3.84 × 108 m ⎞ =⎜ = 2.18 ⎟⎜ ⎟ =⎜ ⎟⎜ ⎟⎜ 1.50 × 1011 m ⎟⎟ FEM ⎝ mE ⎠⎝ rSE ⎠ ⎝⎜ 5.97 × 1024 kg ⎠⎝ ⎠ It is more accurate to say the moon orbits the sun. Reflect: The sun is farther away but has a much greater mass than the earth.

6.20. Set Up: The gravitational force is Fg = G

mE m r2

, where r is the distance of the satellite from the center of the

earth. RE = 6.38 × 106 m, so r = RE + h = 6.38 × 106 m + 7.80 × 105 m = 7.16 × 106 m. mE = 5.97 × 1024 kg. Solve: Fg = (6.673 × 10211 N ⋅ m 2 /kg 2 )

(5.97 × 1024 kg)(2150 kg) (7.16 × 106 m)2

= 1.67 × 104 N

The weight of the satellite at the earth’s surface is w = mg = 2.11 × 104 N.

Fg w

= 0.791, so Fg = 79.1, of w.

6.21. Set Up: Use Eq. (6.5) to calculate Fg exerted by the earth and by the sun and add these forces as vectors. Let G G FE and FS be the gravitational forces exerted on the spaceship by the earth and by the sun. The forces and distances

are shown in figure below.

Solve: (a) The distance from the earth to the sun is r = 1.50 × 1011 m. Let the ship be a distance x from the earth; it is then a distance r − x from the sun.

FE = FS gives GmmE /x 2 = GmmS /(r − x) 2 mE /x 2 = ms /(r − x) 2 and (r − x) 2 = x 2 (mS /mE )

(

r − x = x mS /mE and r = x 1 + mS /mE x=

)

11

r 1.50 × 10 m = = 2.59 × 108 m (from center of earth) 1 + mS /mE 1 + 1.99 × 1030 kg/5.97 × 1024 kg

(b) Reflect: At the instant when the spaceship es through this point its acceleration is zero. But if its speed is not zero when it reaches this point, it will through it. Since mS W mE this equal-force point is much closer to the

earth than to the sun.

6.22. Set Up: Use coordinates where + x is to the right. Each gravitational force is attractive, so is toward the mass exerting it. Treat the masses as uniform spheres, so the gravitational force is the same as for point masses with the same center-to-center distances. The free-body diagrams for (a) and (b) are given in Figures (a) and (b) below.


6-10

Chapter 6

Solve: (a) FB = G

m AmB rAB 2

= (6.673 × 10211 N ⋅ m 2 /kg 2 )

FC = G

m AmC rAC

2

(2.00 kg) 2 (0.50 m) 2

= (6.673 × 10211 N ⋅ m 2 /kg 2 )

= 1.069 × 1029 N (2.00 kg) 2 (0.10 m)2

= 2.669 × 1028 N

Fnet, x = FBx + FCx = 1.069 × 10−9 N + 2.669 × 10−8 N = 2.8 × 10−8 N The net force on A is 2.8 × 1028 N, to the right. (b) FB = G

m AmB rAB

2

= (6.673 × 10−11 N ⋅ m 2 /kg 2 ) FC = G

m AmC rAC 2

(2.00 kg) 2 (0.40 m) 2

= 1.668 × 10−9 N

= (6.673 × 10−11 N ⋅ m 2 /kg 2 )

(2.00 kg) 2 (0.10 m) 2

= 2.669 × 10−8 N

Fnet, x = FBx + FCx = 1.668 × 10−9 N − 2.669 × 10−8 N = − 2.5 × 10−8 N The net force on A is 2.5 × 1028 N, to the left.

*6.23. Set Up: The force exerted on the particle by the earth is w = mg , where m is the mass of the particle. The force exerted by the 100 kg ball is Fg =

Gm1m2 r2

, where r is the distance of the particle from the center of the ball.

G = 6.67 × 10211 N ⋅ m 2 /kg 2 , g = 9.80 m/s 2 .

Solve: Fg = w gives

r=

Gmmball r2

= mg and

Gmball (6.67 × 10211 N ⋅ m 2 /kg 2 )(100 kg) = = 2.61 × 1025 m = 0.0261 mm. It is not feasible to do this; a 100 kg g 9.80 m/s 2

ball would have a radius much larger than 0.0261 mm. Reflect: The gravitational force between ordinary objects is very small. The gravitational force exerted by the earth on objects near its surface is large enough to be important because the mass of the earth is very large.

6.24. Set Up: The forces on A are shown in Figures (a) and (b) below. Use the coordinates as shown. For (b) take the + y axis to be along the direction from A to D.

Solve: (a) FB = FC = G

m AmB rAB

2

= (6.673 × 10211 N ⋅ m 2 /kg 2 )

(3.00 kg)2 (0.100 m) 2

= 6.00 × 1028 N. Fx = FBx + FCx = 0, since

FCx = − FBx Fy = FBy + FCy = 2 FB cos30° = 1.04 × 1027 N The net force is 1.04 × 1027 N, in the direction from A toward the center of the line connecting B and C.



6-11

(b) The distance between A and D is rAD = 0.141 m.

FD = G

m AmD rAD

FB = FC = G

2

= (6.673 × 10211 N ⋅ m 2 /kg 2 )

m AmB rAB

(3.00 kg) 2

= (6.673 × 10211 N ⋅ m 2 /kg 2 )

2

= 3.02 × 1028 N

(0.141 m) 2

(3.00 kg)2 (0.100 m) 2

= 6.00 × 1028 N

Fx = FBx + FCx + FDx = 0, since FCx = − FBx and FDx = 0

Fy = FBy + FCy + FDy = FD + 2 FB cos 45° Fy = 3.02 × 1028 N + 2(6.00 × 1028 N)cos45° = 1.15 × 1027 N The net force is 1.15 × 1027 N, toward D.

6.25. Set Up: Use Eq. (6.5) to find the force each point mass exerts on the particle, find the net force, and use Newton’s second law to calculate the acceleration. Each force is attractive. The particle (mass m) is a distance r1 = 0.200 m from m1 = 8.00 kg and therefore a distance r2 = 0.300 m from m2 = 15.0 kg. Let + x be toward the 15.0 kg mass. Solve: F1 = F2 =

Gm2m r22

Gm1m r12

= (6.67 × 10211 N ⋅ m 2 /kg 2 )

= (6.67 × 10211 N ⋅ m 2 /kg 2 )

(8.00 kg) m (0.200 m) 2

(15.0 kg) m (0.300 m) 2

= (1.334 × 1028 N/kg)m, in the −x -direction.

= (1.112 × 1028 N/kg) m, in the + x -direction. The net force is

Fx = F1x + F2 x = (−1.334 × 1028 N/kg + 1.112 × 1028 N/kg) m = (−2.2 × 1029 N/kg)m. ax =

Fx = 22.2 × 1029 m/s 2 . m

The acceleration is 2.2 × 1029 m/s 2 , toward the 8.00 kg mass. Reflect: The smaller mass exerts the greater force, because the particle is closer to the smaller mass.

6.26. Set Up: The earth has radius RE = 6.38 × 106 m. Let the earth’s mass be mE and your mass be m. Solve: At the surface of the earth your weight is w1 = G

w2 = G

mmE ( h + RE )

2

mmE RE 2

. At height h above the earth’s surface,

, w2 = w1 /2 so

1 2 RE

2

=

1 (h + RE ) 2

.

2 RE = h + RE and h = ( 2 − 1) RE = 0.414 RE = 2.64 × 103 km

*6.27. Set Up: The circumference c is related to the radius Rp of the planet by c = 2π Rp . Take + y downward and use the measured motion to find a y = g , the acceleration due to gravity at the surface of the planet.

mE = 5.97 × 1024 kg Solve: (a) υ0 y = 0, t = 0.811 s, y − y0 = 5.00 m

y = y0 + υ0 yt + 12 a yt 2 gives a y =

2( y − y0 )

t

2

Rp = g=

Gmp Rp 2

so mp =

=

2(5.00 m) (0.811 s)2

= 15.2 m/s 2

c 6.24 × 107 m = = 9.93 × 106 m 2π 2π

gRp 2 G

=

(15.2 m/s 2 )(9.93 × 106 m) 2 6.673 × 10211 N ⋅ m 2 /kg 2

= 2.25 × 1025 kg

(b) mp = 3.76mE © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6-12

Chapter 6

Reflect: Rp = 1.56 RE and mp = 3.76mE so g p =

3.76 (1.56)2

g E = 15.2 m/s 2 , which checks.

6.28. Set Up: Let m′ and R′ be the revised values of the mass and radius of the earth and let w′ be the revised weight of the object. W = G

mmE RE

Solve: (a) m′ = 2mE . w′ = G (b) R ′ = RE /2. w′ = G

mmE

2

; w′ = G

m(2mE )

( RE /2) 2

RE 2

mm′ ( R′ )2

= 2W

= 4W

(c) R ′ = RE /2, m′ = mE /2. w′ = G (d) R ′ = 2 RE , m′ = 2mE . w′ = G

m( mE /2) ( RE /2) 2

m(2mE ) (2 RE ) 2

= 2W

= W /2

*6.29. Set Up: The mass of the probe on earth is m =

wE 3120 N = = 318.4 kg. The radius of Titan is g E 9.80 m/s 2

RT = 12 (5150 km) = 2.575 × 106 m. Solve: w = G

mmT

=

(6.673 × 10211 N ⋅ m 2 /kg 2 )(318.4 kg)(1.35 × 1023 kg)

RT 2 (2.575 × 106 m) 2 Reflect: The mass of the probe is independent of its location.

6.30. Set Up: g m =

Gmm Rm 2

= 433 N

, where the subscript m refers to the moon. For the earth, g E =

GmE RE 2

. mm = mE /81 and

Rm = RE /4. Solve: g m =

G ( mE /81) ( RE /4) 4

=

16 GmE 16 16 = g E = (9.80 m/s 2 ) = 1.9 m/s 2 81 RE 2 81 81

6.31. Set Up: Rn = 10.0 × 103 m. mn = 1.99 × 1030 kg. g n = G

mn

Rn 2

, where the subscript n refers to the neutron

star. Your mass is

m= Solve: g n = (6.673 × 10211 N ⋅ m 2 /kg 2 )

w 675 N = = 68.9 kg. g 9.80 m/s 2

1.99 × 1030 kg (10.0 × 103 m) 2

= 1.33 × 1012 m/s 2

Your weight on the neutron star would be wn = mg n = (68.9 kg)(1.33 × 1012 m/s 2 ) = 9.16 × 1013 N.

6.32. Set Up: (a) Although equation 6.7 was derived for the earth it can be applied to the asteroid, which we assume to be spherically symmetric; thus, we have ga =

Gma Ra 2

. Also, we know that G = 6.67 × 10211 N ⋅ m 2 / kg 2 ,

ma = 4.0 × 1016 kg, and ra = 16 km = 1.6 × 104 m.

(b) The astronaut’s weight on the asteroid can be calculated using Wa = mga . (c) We note that g a is nearly constant for a fall of 1.0 meter so we can use the equations of motion for constant acceleration:

1 use equation 2.12 (Δx = v0 xt + axt 2 ) with the positive x-axis upward, v0 x = 0, Δx = 21.0 m, and ax = 2ga . 2 © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


6-13

(d) Assuming that g a is nearly constant over the height of the jump, we can use the equations of motion for constant

acceleration: use equation 2.13 ( vx 2 = v0 x 2 + 2ax Δx ). At the top of the astronaut’s jump we have vx = 0 and we will assume that the astronaut jumps with the same initial velocity, v0 x , on both the earth and the asteroid.

Gma

Solve: (a) ga = (b) Wa = mga =

Ra

2

=

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(4.0 × 1016 kg) (1.6 × 104 m) 2

= 0.01042 m/s 2 = 1.0 cm/s 2 .

WE 650 N ga = (0.01042 m/s 2 ) = 0.69 N gE 9.80 m/s 2

(c) We can solve equation 2.12 for time: t =

2Δx 2(1.0 m) = = 14 seconds. ga 0.01042 m/s 2

(d) We can solve equation 2.13 for Δx and use it to compare what happens on the asteroid to what happens on

Δxa (02 − v0 x 2 )/( − 2 ga ) g E gE 9.80 m/s 2 = 2 = . (0.60 m) = 5.6 × 102 m. Δ x = Δ x = Thus, we have a E ΔxE (0 − v0 x 2 )/( − 2 g E ) ga ga 0.01042 m/s 2 Reflect: Contrary to our assumption, the acceleration of the astronaut does not remain perfectly constant during this incredible 560 meter jump. Using the inverse-square law, we can show that the astronaut’s acceleration decreases by only about 6%—and so we have probably underestimated the height of the astronaut’s jump slightly.

earth:

*6.33. Set Up: T =

2π r

υ

G G . r and υ are also related by applying ∑ F = ma to the motion of the satellite. The satellite

has arad = υ 2 /R, The only force on the satellite is the gravitational force, Fg = G

mE m r2

. mE = 5.97 × 1024 kg

Solve: (a) and (c) The free-body diagram is given in the figure below. Fg = marad gives G

G

(b) arad =

υ

r

=

r

2

=m

υ2 r

.

Gm (6.673 × 10211 N ⋅ m 2 /kg 2 )(5.97 × 1024 kg) mE = 1.04 × 107 m. = υ 2 and r = 2E = 2 r (6200 m/s) υ T=

2

mE m

(6200 m/s)

2

1.04 × 107 m

6.34. Set Up: T =

2π r

υ

υ

=

2π (1.04 × 107 m) = 1.05 × 104 s = 176 min 6200 m/s

= 3.70 m/s 2

G G . r and υ are related by applying ∑ F = ma to the satellite.

Solve: Fg = marad and G

υ=

2π r

mE m r2

=

GmE = r

mυ2 . r (6.673 × 10211 N ⋅ m 2 /kg 2 )(5.97 × 1024 kg) 7880 × 103 m

= 7.11 × 103 m/s


6-14

Chapter 6

T=

2π r

υ

=

2π (7880 × 103 m) 7.11 × 103 m/s

G

= 6.96 × 103 s = 116 min

G

6.35. Set Up: r and υ are related by applying ∑ F = ma to the satellite. r = h + RE = 780 × 103 m + 6.38 × 106 m = 7.16 × 106 m Solve: Fg = marad gives G

υ=

mE m r2

=m

υ2 r

and

(6.673 × 10211 N ⋅ m 2 /kg 2 )(5.97 × 1024 kg)

GmE = r

7.16 × 106 m

= 7.46 × 103 m/s = 7.46 km/s

6.36. Set Up: r = 10.5 × 109 m. T = 6.3 days = 5.443 × 105 s. The mass of the sun is mS = 1.99 × 1030 kg. Solve: T =

2π r 3/2 so GmHD

mHD =

4π 2 r 3

T 2G

=

4π 2 (10.5 × 109 m)3 (5.443 × 105 s) 2 (6.673 × 10211 N ⋅ m 2 /kg 2 )

= 2.3 × 1030 kg; mHD = 1.2mS

*6.37. Set Up: The rotational period of the earth is 1 day = 8.64 × 104 s. The radius of the satellite’s path is r = h + RE , where h is its height above the surface of the earth and RE = 6.38 × 106 m. υ and r are related by G G applying ∑ F = ma to the motion of the satellite. mE = 5.97 × 1024 kg Solve: (a) 24 h = 8.64 × 104 s (b)-(c) The free-body diagram for the satellite is given in Figure (a) below. Fg = marad gives G

υ2 =

mE m r

2

=m

υ2 r

and

2π r 2π r 4π 2 r 2 GmE GmE . T= so υ = and = . r υ T r T2

1/ 3

⎛ GmET 2 ⎞ r =⎜ ⎜ 4π 2 ⎟⎟ ⎝ ⎠

1/3

⎡ (6.673 × 10211 N ⋅ m 2 /kg 2 )(5.97 × 1024 kg)(8.64 × 104 s) 2 ⎤ =⎢ ⎥ 4π 2 ⎣⎢ ⎦⎥

= 4.22 × 107 m

h = r − RE = 3.58 × 107 m = 3.58 × 104 km (d)

r = 5.6. The sketch of the earth and orbit of the satellite is given in Figure (b) above. RE



6-15

Reflect: The orbital period is proportional to r 3/2 . To achieve a period as large as 24 h, the satellite is at a large altitude above the earth’s surface.

6.38. Set Up: According to problem 6.32 the mass of the asteroid is ma = 4.0 × 1016 kg and its radius is Ra = 1.6 × 104 m. If we assume that the rock is in a circular orbit just above the surface of a spherically symmetric asteroid we can use equations 6.8 and 6.10 (after replacing the mass of the earth with the mass of the asteroid): thus, 2π Ra3/ 2 Gma . υ= and T = Ra Gma

Gma (6.67 × 10211 N ⋅ m 2 /kg 2 )( 4.0 × 1016 kg) = = 13 m/s Ra 1.6 × 104 m

Solve: (a) υ = (b) T =

2π Ra 3/2 2π (1.6 × 104 m)3/2 = = 7.8 × 103 s = 2.2 h Gma (6.67 × 10−11 N ⋅ m 2 /kg 2 )( 4.0 × 1016 kg)

Reflect: It would be easy to put a rock into orbit on 234 Ida since13 m/s is just under 30 mph. Since the orbit period is 2.2 hours, you have a long time to duck!

*6.39. Set Up: The period of each satellite is given by equation (6.10). Set up a ratio involving T and r: 2π r 3/2 2π T T1 T2 T=

Gmp

gives

r 3/2

⎛r ⎞ Solve: T2 = T1 ⎜ 2 ⎟ ⎝ r1 ⎠

3/2

=

= constant, so

Gmp

r13/2

⎛ 48,000 km ⎞ = (6.39 days) ⎜ ⎟ ⎝ 19,600 km ⎠

⎛ 64,000 km ⎞ T2 = (6.39 days) ⎜ ⎝ 19,600 km ⎟⎠

=

r23/2

.

3/2

= 24.5 days. For the other satellite we have

3/2

= 37.7 days.

Reflect: T increases when r increases.

6.40. Set Up: RE = 6.38 × 106 m. mE = 5.97 × 1024 kg. For h = 400 km , r = h + RE = 6.78 × 106 m Solve: (a) m =

w 750 N = = 76.53 kg g 9.80 m/s 2 w = Fg = G

mmE 2

= (6.673 × 10211 N ⋅ m 2 /kg 2 )

(76.53 kg)(5.97 × 1024 kg)

r (b) The orbit of the satellite is drawn to scale in the figure below.

(6.78 × 106 m) 2

= 664 N

(c) No. The astronauts appear weightless because they are in free-fall.

G

G

6.41. Set Up: Apply ∑ F = ma to the motion of the baseball. rD = 6 × 103 m. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6-16

Chapter 6

Solve: Fg = marad gives G

mD m rD 2

=m

GmD = rD

υ=

υ2 rD

.

(6.673 × 10211 N ⋅ m 2 /kg 2 )(2.0 × 1015 kg) 6 × 103 m

= 4.7 m/s

4.7 m/s = 11 mph, which is easy to achieve.

6.42. Set Up: mE = 5.97 × 1024 kg, RE = 6.38 × 106 m. The satellite makes 15.65 revolutions in 1 day = 8.64 × 104 s Solve: (a) The free-body diagram for the satellite is given in the figure below.

(b) T =

Fg = m T=

8.64 × 104 s = 5.52 × 103 s 15.65

υ2 r

2π r

υ

so G

mmE r

and υ 2 =

2

=m

4π 2 r 2 T2

υ2 r

and

GmE = υ2 r 1/3

.

⎛ GmET 2 ⎞ GmE 4π 2 r 2 = . r =⎜ 2 ⎜ 4π 2 ⎟⎟ r T ⎝ ⎠

, so 1/3

⎡ (6.673 × 10211 N ⋅ m 2 /kg 2 )(5.97 × 1024 kg)(5.52 × 103 s) 2 ⎤ r=⎢ ⎥ 4π 2 ⎢⎣ ⎥⎦

= 6.75 × 106 m

h = r − RE = 3.7 × 105 m = 370 km

*6.43. Set Up: r = 137.5 m. The distance a point on the rim travels in 1 revolution is 2π r. υ2

so υ = rarad = (137.5 m)(9.80 m/s 2 ) = 36.7 m/s R In one minute a point on the rim travels a distance (36.7 m/s)(60 s) = 2202 m. The number of revolutions in one Solve: arad =

minute is

2202 m = 2.55. The station must turn at 2.55 rpm. 2π r

6.44. Set Up: mM = 7.35 × 1022 kg, RM = 1.74 × 106 m. The mass ed by the hip is (0.65)(65 kg) + 43 kg = 85.25 kg Solve: (a) g M =

GmM RM 2

=

(6.673 × 10211 N ⋅ m 2 /kg 2 )(7.35 × 1022 kg) (1.74 × 106 m) 2

= 1.62 m/s 2

(b) n = (85.25 kg) g M = 138 N and f k = μ k n = (0.0050)(138 N) = 0.69 N (c) n = (85.25 kg) g E = 835 N and f k = μ k n = 4.2 N.



6-17

6.45. Set Up: R = 1.815 × 106 m. Use coordinates where +y is upward. At the maximum height, υ0 y = 0, a y = − g. Solve: At the surface of Io, g =

Gm R

2

=

(6.673 × 10211 N ⋅ m 2 /kg 2 )(8.94 × 1022 kg) (1.815 × 106 m)2

= 1.81 m/s 2

υ y = υ0 y + 2a y ( y − y0 ) so υ0 y = 22a y ( y − y0 ) =

− 2( −1.81 m/s 2 )(5.00 × 105 m) = 1.345 × 103 m/s

Now solve for y − y0 when υ0 y = 1.345 × 103 m/s and a y = − 9.80 m/s 2 . υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

y − y0 =

υ y 2 − υ0 y 2 2a y

=

2(1.345 × 103 m/s) 2 2(−9.80 m/s 2 )

= 92 km

6.46. Set Up: Your friend slides toward you when the friction force on her isn’t sufficient to make her turn as you G υ2 G . turn. Set fs = μs n and apply ∑ F = ma to your friend. Set her acceleration equal to that of the car, arad = r

Solve: (a) In the absence of friction your friend moves in a straight line, so turn to the right, toward her. (b) The free-body diagram for your friend is given in the figure below.

oFy = ma y gives n = mg . oFx = max gives μs n = marad . μs mg = m

R=

υ2 R

and

υ2 (15 m/s) 2 = = 42 m. μs g (0.55)(9.80 m/s 2 )

Reflect: The larger the coefficient of friction, the smaller must be the radius of the turn.

*6.47. Set Up: The person moves in a horizontal circle of radius r = 2.5 m. Set the static friction force equal to its maximum value, fs = μs n. The person has an acceleration

arad =

υ2 r

,

directed toward the center of the circle. The period is

T=

1 = 1.67 s 0.60 rev/s

and the person has speed

υ=

2π r = 9.41 m/s. T


6-18

Chapter 6

Solve: (a) The free-body diagram is given in the figure below. The diagram is for when the wall is to the right of her, so the center of the cylinder is to the left.

(b) oFx = max gives n = m

μs m

υ2 r

υ2 r

. oFy = ma y gives fs = mg , so μs n = mg . Combining these two equations gives

= mg and

μs =

gr

υ2

=

(9.80 m/s 2 )(2.5 m) (9.41 m/s)2

= 0.28.

(c) The mass m of the person divides out of the equation for μs ; the answer to (b) does not depend on the mass of the

person. Reflect: The greater the rotation rate the larger the normal force exerted by the wall and the larger the friction force. Therefore, for smaller μs the rotation rate must be larger.

6.48. Set Up: Use coordinates where + y is upward. A 30° arc is weights travel a distance of

1 12

of a revolution and in one 30° swing the

1 (2π r ). 12

Solve: (a) The free-body diagram is given in the figure below. F is the force exerted by the hand.

(b) υ =

1 (2π r ) 12 1 s 3

=

1 (2π )(0.72 12 1 s 3

m)

= 1.13 m/s and arad =

υ2 R

= 1.77 m/s 2

oFy = ma y gives F − w = marad and F = m( g + arad ) = (8.0 kg)(9.80 m/s 2 + 1.77 m/s 2 ) = 93 N



*6.49. Set Up: density = mass/volume. The volume of a sphere is

4 π R3 . 3

Toro is the same as that of earth to calculate the mass of Toro. Then g T = G

6-19

Use the assumption that the density of mT RT

2

G G . Apply oF = ma to the object to

find its speed when it is in a circular orbit around Toro. mE mT Solve: (a) = gives 4πR 3 4πR 3 E T 3 3 3

3

⎛ 5.0 × 103 m ⎞ ⎛R ⎞ mT = mE ⎜ T ⎟ = (5.97 × 1024 kg) ⎜ = 2.9 × 1015 kg. ⎜ 6.38 × 106 m ⎟⎟ R ⎝ E⎠ ⎝ ⎠

gT = G

mT RT

2

=

(6.673 × 10211 N ⋅ m 2 /kg 2 )(2.9 × 1015 kg) (5.0 × 103 m) 2

= 7.7 × 1023 m/s 2

(b) The gravity force on the object is mgT . In a circular orbit just above the surface of Toro, its acceleration is

υ2 RT

.

G υ2 G and Then oF = ma gives mgT = m RT

υ = gT RT = (7.7 × 1023 m/s 2 )(5.0 × 103 m) = 6.2 m/s. A speed of 6.2 m/s corresponds to running 100 m in 16.1 s, which is barely possible for the average person.

6.50. Set Up: We can use the analysis done in Example 6.4. As in that example, we assume friction is negligible. From Example 6.4, the banking angle β is given by tan β =

v2 . Also, we have n = mg / cos β . 65.0 mi/h = 29.1 m/s. gR

(29.1 m/s) 2

and β = 21.0°. The expression for tan β does not involve the mass of the (9.80 m/s 2 )(225 m) vehicle, so the truck and car should travel at the same speed. (1125 kg)(9.80 m/s 2 ) (b) For the car, ncar = = 1.18 × 104 N and ntruck = 2ncar = 2.36 × 104 N, since mtruck = 2mcar . cos21.0° Reflect: The vertical component of the normal force must equal the weight of the vehicle, so the normal force is proportional to m. Solve: (a) tan β =

6.51. Set Up: REuropa = 1.565 × 106 m. Equating the weight to marad gives υmax = gR , where R = 1.0 m is the length of her leg. Solve: At the surface of Europa, mEuropa (6.673 × 10211 N ⋅ m 2 /kg 2 )(4.78 × 1022 kg) g =G = = 1.30 m/s 2 2 REuropa (1.565 × 106 m) 2

υmax = (1.30 m/s 2 )(1.0 m) = 1.1 m/s

6.52. Set Up: The orbital speed is given by v = Gm /r , where m is the mass of the star. The orbital period is given 2π r 30 11

by T =

v

. The sun has mass mS = 1.99 × 10

kg. The orbit radius of the earth is 1.50 × 10

m.

Solve: (a) v = Gm /r . v = (6.673 × 10211 N ⋅ m 2 /kg 2 )(0.85 × 1.99 × 1030 kg)/((1.50 × 1011 m)(0.11)) = 8.27 × 104 m/s. (b) 2π r /v = 1.25 × 106 s = 14.5 days (about two weeks). Reflect: The orbital period is less than the 88-day orbital period of Mercury; this planet is orbiting very close to its star, compared to the orbital radius of Mercury. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6-20

Chapter 6

*6.53. Set Up: The comet has radius R = 2.5 × 103 m, volume V = 43 π R3 and density ρ = 2.1 × 103 kg/m3 . Solve: (a) The comet has mass

mC = ρCVC = ρC 43 π R3 = (2.1 × 103 kg/m3 )( 43 π )(2.5 × 103 m)3 = 1.374 × 1014 kg. The gravitational force is Fg = G

mmC r2

=

(6.673 × 10211 N ⋅ m 2 /kg 2 )(385 kg)(1.374 × 1014 kg) (237 × 103 m) 2

= 6.28 × 1025 N

(b) At the surface of the earth, the earth’s gravity force is w = mg = (385 kg)(9.80 m/s 2 ) = 3.77 × 103 N. The force

exerted by the comet is much less than this.

6.54. Set Up: The block moves in a horizontal circle of radius r = (1.25 m)2 − (1.00 m)2 = 0.75 m. Each string makes an angle θ with the vertical. cos θ =

1.00 m , so θ = 36.9°. 1.25 m

The block has acceleration arad = υ 2 /r , directed to the left in the figure in the problem. Solve: (a) The free-body diagram for the block is given in the figure below.

oFy = ma y

gives

Tu cos θ − Tl cos θ − mg = 0.

Tl = Tu −

mg (4.00 kg)(9.80 m/s 2 ) = 80.0 N − = 31.0 N cos θ cos36.9°

(b) oFx = ma x gives (Tu + Tl )sin θ = m

υ=

v2 . r

r (Tu + Tl )sin θ = m

(0.75 m)(80.0 N + 31.0 N)sin 36.9° = 3.53 m/s 4.00 kg

Reflect: The tension in the upper string must be greater than the tension in the lower string so that together they produce an upward component of force that balances the weight of the block.

*6.55. Set Up: The package moves in the arc of a circle so it has acceleration arad =

υ2 r

, directed toward the

center of the curve. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


6-21

Solve: The free-body diagram for the package is given in the figure below. oFy = ma y gives T cos30.0° = mg .

oFx = ma x gives T sin 30.0° = m υ=

υ2 r

. Therefore, T =

υ2 mg and mg tan 30.0° = m . r cos30.0°

gr tan 30.0° = (9.80 m/s 2 )(50.0 m) tan 30.0° = 16.8 m/s

6.56. Set Up: On earth, g E = g = 9.80 m/s 2 . On Mars, gM = G

mM RM

2

= (6.673 × 10211 N ⋅ m 2 /kg 2 )

6.42 × 1023 kg 6

(3.40 × 10 m)

2

= 3.71 m/s 2 .

2

The artificial gravity acceleration is arad = υ /r . r = 400.0 m Solve: (a) υ = rarad = rg = (400.0 m)(9.80 m/s 2 ) = 62.6 m/s (b) υ = rg M = (400.0 m)(3.71 m/s 2 ) = 38.5 m/s (c) The free-body diagram is given in the figure below. The diagram is drawn for when the rim of the station is to the right in the figure.

Solutions to age Problems *6.57. Set Up: Assume that the diameter of the space station is 1000 meters so that its radius is 500 meters. For circular motion at a constant speed, the acceleration of the astronaut is ar =

v2 . r

Solve: v = rar = (500 m)(5 m/s 2 ) = 50 m/s. The correct answer is D.

6.58. Set Up: The period for motion at a constant speed v on a circle of circumference C is T = Solve: T =

C . v

C 3000 m = = 60 s. The correct answer is D. v 50 m/s

*6.59. Set Up: The radial force required to keep the helmet moving on its circular path is F = m

v2 = mar . r

Solve: F = mar = (2 kg)(5 m/s 2 ) = 10 N. The correct answer is B. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7

WORK AND ENERGY

Answers to Multiple Choice Problems 1. B 2. D 3. A 4. C 13. C 14. A 15. C

5. C

6. C

7. A, D

8. B, C, E

9. A, D, E

10. B, D

11. B, D

12. B, D

Solutions to Problems 7.1. Set Up: Assume the fisherman is holding the pole straight out in front of him so that the pole and fishing line are roughly parallel to the water. Thus φ = 0° may be used in the relation W = F|| s = ( F cos φ ) s. Solve: W = F|| s = ( F cos φ ) s = [(25.0 N)(cos0°)](12.0 m) = 300 J Reflect: Because φ = 0°, all of the force that the fisherman applies through the line is used to perform work on the fish.

7.2. Set Up: Use W = F|| s = ( F cos φ ) s. On the way up, the displacement is upward and the gravity force is downward, so φ = 180D. On the way down, both the displacement and force are downward, so φ = 0°. Solve: On the way up: W = ( mg cos180°) s = (5.80 × 10−2 kg)(9.80 m/s 2 )(cos180°)(6.17 m) = − 3.51 J

On the way down: W = (mg cos0°) s = (5.80 × 10−2 kg)(9.80 m/s 2 )(cos0°)(6.17 m) = 3.51 J Reflect: When the force and displacement are in opposite directions, the work done is negative.

*7.3. Set Up: Use W = F|| s = ( F cos φ ) s with φ = 15.0°. Solve: W = ( F cos φ ) s = (180 N)(cos15.0°)(300.0 m) = 5.22 × 104 J Reflect: Since cos15.0° ≈ 0.97, the relatively small angle of 15.0° allows the boat to apply approximately 97% of

the 180 N force to pulling the skier.

7.4. Set Up: Use W = F|| s = ( F cos φ ) s. In part (a), φ = 0°. In part (b), solve for φ. Solve: (a) W = ( F cos φ ) s = (8.50 N)(cos0°)(17.4 m) = 148 J (b) cos φ = W/(Fs ) = 65.0 J/[(8.50 N)(17.4 m)] = 0.439 and φ = 63.9° Reflect: More work is done in part (a) where F|| = F . In part (b), because the box is pulled at an angle other than

zero, F|| < F and less work is done even though the force and displacement are the same.


7-1

7-2

Chapter 7

*7.5. Set Up: For parts (a) through (d), identify the appropriate value of φ and use the relation W = F|| s = ( F cos φ ) s. In part (e), apply the relation Wnet = Wstudent + Wgrav + Wn + W f . Solve: (a) Since you are applying a horizontal force, φ = 0°. Thus,

Wstudent = (2.40 N)(cos0°)(1.50 m) = 3.60 J (b) The friction force acts in the horizontal direction, opposite to the motion, so φ = 180°.

W f = ( F f cos φ ) s = (0.600 N)(cos180°)(1.50 m) = − 0.900 J (c) Since the normal force acts upward and perpendicular to the tabletop, φ = 90°.

Wn = (n cos φ ) s = (ns )(cos90°) = 0.0 J (d) Since gravity acts downward and perpendicular to the tabletop, φ = 270°.

Wgrav = (mg cos φ ) s = (mgs )(cos 270°) = 0.0 J (e) Wnet = Wstudent + Wgrav + Wn + W f = 3.60 J + 0.0 J + 0.0 J − 0.900 J = 2.70 J Reflect: Whenever a force acts perpendicular to the direction of motion, its contribution to the net work is zero.

7.6. Set Up: Use W = F|| s = ( F cos φ )s. Calculate the work done by each force. In each case, identify the angle φ. In part (d), the net work is the algebraic sum of the work done by each force. Solve: (a) Since the force exerted by the rope and the displacement are in the same direction, φ = 0D and Wrope = (72.0 N)(cos0°)(5.20 m) = + 374 J (b) Gravity is downward and the displacement is at 30.0° above the horizontal, so φ = 90.0° + 30.0° = 120.0°.

Wgrav = (128.0 N)(cos120°)(5.20 m) = − 333 J (c) The normal force n is perpendicular to the surface of the ramp while the displacement is parallel to the surface of the ramp, so φ = 90° and Wn = 0. (d) Wnet = Wrope + Wgrav + Wn = + 374 J − 333 J + 0 = + 41 J (e) Now φ = 50.0° − 30.0° = 20.0° and Wrope = (72.0 N)(cos 20.0°)(5.20 m) = + 352 J Reflect: In part (b), gravity does negative work since the gravity force acts downward and the carton moves upward. Less work is done by the rope in part (e), but the net work is still positive.

*7.7. Set Up: In order to move the crate at constant velocity, the worker must apply a force that equals the force of friction, Fworker = f k = μk n. Each force can be used in the relation W = F|| s = ( F cos φ ) s for parts (b) through (d). For part (e), apply the net work relation as Wnet = Wworker + Wgrav + Wn + W f . Solve: (a) The magnitude of the force the worker must apply is:

⎛π ⎞ Fworker = f k = μk n = μk mg = (0.25)(30.0 kg)(9.80 m/s 2 ) = 74 N ⎜ − θ ⎟ ⎝2 ⎠ (b) Since the force applied by the worker is horizontal and in the direction of the displacement, φ = 0° and the work is:

Wworker = ( Fworker cos φ ) s = [(74 N)(cos0°)](4.5 m) = + 333 J (c) Friction acts in the direction opposite of motion, thus φ = 180° and the work of friction is:

W f = ( f k cos φ ) s = [(74 N)(cos180°)](4.5 m) = − 333 J (d) Both gravity and the normal force act perpendicular to the direction of displacement. Thus, neither force does any work on the crate and Wgrav = Wn = 0.0 J. (e) Substituting into the net work relation, the net work done on the crate is: Wnet = Wworker + Wgrav + Wn + W f = + 333 J + 0.0 J + 0.0 J − 333 J = 0.0 J Reflect: The net work done on the crate is zero because the two contributing forces, Fworker and F f , are equal in

magnitude and opposite in direction.


Work and Energy

7-3

7.8. Set Up: Use W = F|| s = ( F cos φ ) s to calculate the work done in each of parts (a) through (c). In part (d), the net work consists of the contributions due to all three forces, or Wnet = Wgrav + Wn + W f

Solve: (a) As the package slides, work is done by the frictional force which acts at φ = 180° to the displacement.

The normal force is mg cos53.0°. Thus for μ k = 0.40,

W f = F|| s = ( f k cos φ ) s = (μk n cos φ ) s = [μk (mg cos53.0°)](cos180°) s = (0.40)[(8.00 kg)(9.80 m/s 2 )(cos53.0°)](cos180°)(2.00 m) = − 38 J (b) Work is done by the component of the gravitational force parallel to the displacement. As seen in the figure above φ = 90° − 53° = 37° and the work of gravity is: Wgrav = (mg cos φ ) s = [(8.00 kg)(9.80 m/s 2 )(cos37.0°)](2.00 m) = +125 J (c) Wn = 0 since the normal force is perpendicular to the displacement. (d) The net work done on the package is:

Wnet = Wgrav + Wn + W f = 125 J + 0.0 J − 38 J = 87 J

*7.9. Set Up: WF = ( F cos φ ) s, since the forces are constant. We can calculate the total work by summing the work done by each force. The forces are sketched in the figure below.

Solve: Using W1 = F1s cos φ1 we obtain W1 = (1.80 × 106 N)(0.75 × 103 m)cos14° so W1 = 1.31 × 109 J. Also, note

that W2 = F2 s cos φ2 = W1 . Thus, Wtot = W1 + W2 = 2(1.31 × 109 J) = 2.62 × 109 J Reflect: Only the component F cos φ of force in the direction of the displacement does work. These components are G in the direction of s so the forces do positive work.


7-4

Chapter 7

7.10. Set Up: In each case the forces are constant and the displacement is along a straight line, so W = Fs cos φ. In part (a), when the cable pulls horizontally φ = 0° and when it pulls at 35.0° above the horizontal φ = 35.0D. In part (b), if the cable pulls horizontally φ = 180D. If the cable pulls on the car at 35.0D above the horizontal it pulls on the truck at 35.0D below the horizontal and φ = 145.0D. For the gravity force φ = 90D , since the force is vertical and the displacement is horizontal. Solve: (a) When the cable is horizontal, W = (850 N)(5.00 × 103 m)cos 0D = 4.25 × 106 J. When the cable is 35.0° above the horizontal, W = (850 N)(5.00 × 103 m)cos35.0D = 3.48 × 106 J. (b) Since cos180D = − cos0D and cos145.0D = − cos35.0D , the answers are −4.25 × 106 J and −3.48 × 106 J. (c) Since cos φ = cos90D = 0, we have W = 0 in both cases. Reflect: If the car and truck are taken together as the system, the tension in the cable does no net work.

7.11. Set Up: Since the speed is constant, the acceleration and the net force on the monitor are zero. Use this fact to develop expressions for the friction force, f k , and the normal force, n (which are shown in the figure below). Then use W = F|| s = ( F cos φ ) s to calculate W.

Solve: (a) Summing forces along the incline, oF = ma = 0 = f k − mg sin θ

giving f k = mg cosθ , directed up the incline. Substituting, W f = ( f k cos φ ) s = [(mg sin θ )cos φ ]s = [(10.0 kg)(9.80 m/s 2 )(sin 36.9°)](cos0°)(5.50 m) = + 324 J (b) The gravity force is downward and the displacement is directed up the incline so φ = 126.9°. Wgrav = (10.0 kg)(9.80 m/s 2 )(cos126.9°)(5.50 m) = − 324 J (c) The normal force, n, is perpendicular to the displacement and thus does zero work. Reflect: Friction does positive work and gravity does negative work. The net work done is zero.

7.12. Set Up: Use K = 12 mυ 2 to calculate υ , where K is equal to the change in thermal energy. Solve: (a) υ = 2 K/m = 2(4.186 J)/(2.0 × 1023 kg) = 65 m/s (b) υ = 2(4.186 J)/(4.0 × 1023 kg) = 46 m/s Reflect: When the mass increases by a factor of two, the speed required decreases by a factor of 1/ 2.


Work and Energy

7-5

*7.13. Set Up: Use K = 12 mυ 2 and solve for m. Solve: m = 2 K/υ 2 = 2(1960 J)/(965 m/s) 2 = 4.21 × 1023 kg = 4.21 g Reflect: The kinetic energy of an object is proportional to the mass of the object.

7.14. Set Up: Use K = 12 mυ 2 to relate υ and K. Solve: (a) K = 12 (70 kg)(32 m/s) 2 = 3.6 × 104 J (b) K is proportional to υ 2 , so K increases by a factor of 4 when υ doubles.

*7.15. Set Up: Use K = 12 mυ 2 to relate υ and K. Let υ1 and K1 be speed and kinetic energy when υ1 = 10.0 m/s. Solve: (a) K 2 = K1/2 and K1 /K 2 = υ12 /υ 22 so υ2 = υ1 K 2 /K1 = (10.0 m/s) 1/2 = 7.07 m/s. (b) Since υ2 = υ1 /2, we obtain K 2 = K1(υ 22 /υ12 ) = K1(1/4). This means that the decrease in K is 3K1 /4.

7.16. Set Up: Applying K = 12 mυ 2 to relate υ and K results in K1/K 2 = υ12 /υ2 2 . Solve: (a) υ2 = υ1 K 2 /K1 , where υ1 = υ , K1 = 10.0 J and K 2 = 20.0 J. Substituting,

υ2 = υ (20.0 J)/(10.0 J) = 2υ = 1.41υ (b) Now let K1 = 20.0 J, υ1 = 1.41υ , and υ2 = υ 1/ 2 = 1.41υ / 2. Substituting, 2

⎡ (1.41υ / 2) ⎤ K 2 = K1(υ2 /υ1) 2 = (20.0 J) ⎢ ⎥ = (20.0 J) /4 = 5.00 J ⎣ 1.41υ ⎦ Reflect: When K changes by a factor of two, the speed υ changes by a factor of

2. When υ changes by a factor

2

of two, the kinetic energy changes by a factor of 2 = 4.

7.17. Set Up: The work done by the pitcher equals the change in kinetic energy, W = Kf − Ki . The pitcher imparts a velocity, υf = 32.0 m/s, to the ball which was initially at rest in his hand, υ i = 0.00 m/s. Solve: W = K f − Ki = 12 m(υf 2 − υi 2 ) = 12 (0.145 kg)[(32.0 m/s)2 − (0.00 m/s)2 ] = 74.2 J

7.18. Set Up: Use the work-kinetic energy theorem Wnet = Kf − Ki , with K = 12 mυ 2 . In part (a) use Ki and K f to calculate W. In parts (b) and (c) use Ki and W to calculate K f . Solve: (a) Substituting the notation i = A and f = B, Wnet = K B − K A = 12 (1.50 kg)[(1.25 m/s) 2 − (3.21 m/s) 2 ] = − 6.56 J (b) Noting i = B and f = C , KC = K B + Wnet = 12 (1.50 kg)(1.25 m/s) 2 − 0.750 J = + 0.422 J

KC = 12 mυC 2 so υC = 2 KC /m = 0.750 m/s (c) Similarly, KC = 12 (1.50 kg)(1.25 m/s) 2 + 0.750 J = 1.922 J and υC = 1.60 m/s Reflect: Negative Wnet corresponds to a decrease in kinetic energy (slowing down) and positive Wnet corresponds

to an increase in kinetic energy (speeding up).

*7.19. Set Up: Use the work-kinetic energy theorem: Wnet = K f − Ki = Fnet s. Since the net force is due to friction, Fnet s = − f k s = − μk mgs. Also, since the car stops, K f = 0. Solve: (a) Wnet = K f − Ki = Fnet s gives − 12 mυ i 2 = − μk mgs. Solving for the distance,

s=

υ i2 2μk g

=

(23.0 m/s) 2 2(0.700)(9.80 m/s 2 )

= 38.6 m


7-6

Chapter 7

(b) Since s is proportional to υ i 2 , doubling υ i increases s by a factor of 4; s therefore becomes 154 m. (c) The original kinetic energy was converted into thermal energy by the negative work of friction. Reflect: To stop the car friction must do negative work equal in magnitude to the initial kinetic energy of the car.

7.20. Set Up: Since the meteor comes to rest the energy it delivers to the ground equals its original kinetic energy: K = 12 mv 2 . We know that v = 12 km/s = 1.2 × 104 m/s. Solve: (a) K = 12 (1.4 × 108 kg)(1.2 × 104 m/s) 2 = 1.0 × 1016 J. (b) It would take a typical coal-fired power plant

1.0 × 1016 J 109 J/s

⎛ 1 h ⎞⎛ 1 day ⎞ = 107 s, which is equal to (107 s) ⎜ ⎟⎜ ⎟= ⎝ 3600 s ⎠⎝ 24 h ⎠

102 days, to produce this much energy. Reflect: Part of the energy transferred to the ground lifts soil and rocks into the air and creates a large crater.

*7.21. Set Up: From the work-energy relation, W = Wgrav = ΔK rock or F|| s = Kf − Ki . As the rock rises, the gravitational force, F = mg , does work on the rock. Since this force acts in the direction opposite to the motion and displacement, s, the work is negative. Solve: (a) Applying F|| s = K f − Ki we obtain:

− mgh = 12 mυ f 2 − 12 mυ i 2 Dividing by m and solving for υ i , υ i = υf 2 + 2gh . Substituting h = 15.0 m and υ f = 25.0 m/s,

υ i = (25.0 m/s) 2 + 2(9.80 m/s 2 )(15.0 m) = 30.3 m/s (b) Solve the same work-energy relation for h. At the maximum height υf = 0.

− mgh = 12 mυ f 2 − 12 mυ i 2 h=

υ i2 − υ f 2 2g

=

(30.3 m/s) 2 − (0.0 m/s) 2 2(9.80 m/s 2 )

= 46.8 m

Reflect: Note that the weight of 20 N was never used in the calculations because both gravitational potential and kinetic energy are proportional to mass, m. Thus any object, that attains 25.0 m/s at a height of 15.0 m, must have an initial velocity of 30.3 m/s. As the rock moves upward gravity does negative work and this reduces the kinetic energy of the rock.

7.22. Set Up: Apply the work-energy principle Wnet = K f − Ki . Since the ball’s motion continues along the ground, Wgrav = mgh = 0. Assume negligible friction between the ball and the ground so W f = 0. Consequently, the net work is simply Wnet = F|| s = Fkick s. Solve: Applying Wnet = F|| s = Fkick s = 12 m(υ f 2 − υ i 2 ) and solving for the distance, s

s=

1 m(υ 2 f 2

− υ i2 )

Fkick

=

1 (0.420 2

kg)[(6.00 m/s) 2 − (2.00 m/s) 2 ] ⎛ 100 cm ⎞ ⎜ ⎟ = 16.8 cm 40.0 N ⎝ m ⎠

Reflect: Check the answer by converting it to inches: s = 16.8 cm[(1 in.)/(2.54 cm)] ≈ 6.6 in. Since a person’s foot

is roughly 6 to 12 inches in length, this result is very reasonable.


Work and Energy

7-7

*7.23. Set Up: Use Wnet = W f = − f k s = − μk mgs and Wnet = K f − Ki . The skier stops, so K f = 0. Solve: (a) Setting the two expression for net work equal, Wnet = − μk mgs = − 12 mυ i 2 . Solving for the coefficient of

kinetic friction,

μk =

υ i2 2 gs

=

(12.0 m/s)2 2(9.80 m/s 2 )(184 m)

= 3.99 × 1022

(b) The mass m of the skier divides out and μk is independent of m. If υ i is doubled while s is constant then μk

increases by a factor of 4; μk = 0.160. Reflect: To stop the skier friction does negative work that is equal in magnitude to the initial kinetic energy of the skier.

7.24. Set Up: Apply Wtot = K 2 − K1, where K1 = 0. The normal force does no work. The work W done by gravity is W = mgh, where h = Lsin θ is the vertical distance the block has dropped when it has traveled a distance L down the incline and θ is the angle the plane makes with the horizontal.

Solve: The work-energy theorem gives v =

2K 2W = = 2 gh = 2 gL sin θ . Using the given numbers, m m

v = 2(9.80 m/s 2 )(0.75 m)sin 36.9° = 2.97 m/s. Reflect: The final speed of the block is the same as if it had been dropped from a height h.

*7.25. Set Up: Use Won spring = + 12 kx 2 in part (a) and Fon spring = kx in part (b). Solve: (a) k =

2(Won spring ) x2

=

2(8.0 J) (0.025 m) 2

= 2.6 × 104 N/m.

(b) Fon spring = (2.6 × 104 N/m)(0.025 m) = 650 N.

7.26. Set Up: Use Fon spring = kx in parts (a) and (c) and Won spring = + 12 kx 2 in part (b). Solve: (a) k =

Fon spring x

=

25.0 N = 1.14 × 103 N/m 0.192 m − 0.170 m

(b) x = 0.022 m so Won spring = 12 (1.14 × 103 N/m)(0.022 m) 2 = 0.276 J (c) Since Fon spring = kx, doubling the force results in doubling the length of stretch giving x = 0.044 m. The new

length will therefore be 17.0 cm + 4.4 cm = 21.4 cm.

7.27. Set Up: Solve Fon spring = kx for x to determine the length of stretch and use Won spring = + 12 kx 2 to assess the corresponding work. Solve: x =

Fon spring k

=

15.0 N = 0.0500 m 300.0 N/m

The new length will be 0.240 m + 0.0500 m = 0.290 m. The corresponding work done is, Won spring =

1 (300.0 N/m)(0.0500 m)2 = 0.375 J 2

Reflect: Fon spring is always the force applied to one end of the spring, thus we did not need to double the 15.0 N

force. Consider a free-body diagram of a spring at rest; forces of equal magnitude and opposite direction are always applied to both ends of every section of the spring examined.


7-8

Chapter 7

7.28. Set Up: Use Fon spring = kx and Won spring = + 12 kx 2 for parts (a) and (b). Solve: (a) Fon spring = (1200 N/m)(1.0 m) = 1200 N. The corresponding work on the spring is:

1 (1200 N/m)(1.0 m)2 = 600 J 2 = (1200 N/m)(2.0 m) = 2400 N. To stretch the spring from x = 1.0 m to

Won spring = (b) Now x = 2.0 m resulting in Fon spring

x = 2.0 m requires Won spring = 12 (1200 N/m)[(2.0 m) 2 − (1.0 m) 2 ] = 1800 J of additional work. Reflect: Doubling x, doubles Fon spring . Also, since the total work to stretch the spring from 0.0 m to 2.0 m is 600 J + 1800 J = 2400 J, the total work increases by a factor of four.

*7.29. Set Up: The work done is the area under the graph of Fon spring versus x between the initial and final positions. Solve: (a) The area between x = 0 and x = 5.0 cm is a right triangle. The work is thus calculated as:

1 (5.0 × 10−2 m)(250 N) = 6.2 J 2 (b) Since the area between x = 2.0 cm and x = 7.0 cm is a trapezoid, the work is calculated as: Won spring =

1 (100 N + 350 N)(0.070 m − 0.020 m) = 11.2 J 2 Reflect: The force is larger for x = 7.0 cm than for 5.0 cm and the work done in (b) is greater than that done in (a), Won spring =

even though the displacement of the end of the spring is 5.0 cm in each case.

7.30. Set Up: To answer part (a), convert her weight to pounds. Use ΔU grav = mg ( yf − yi ) to calculate her increase in gravitational potential energy in part (b). For part (c), use ΔU grav = − Wby gravity .

⎛ 0.2248 lb ⎞ Solve: (a) w = (575 N) ⎜ ⎟ = 129 lb ⎝ 1N ⎠ (b) For yf − yi = (4.75 m)(sin 53°) and mg = 575 N, ΔU grav = (575 N)(4.75 m)(sin 53°) = 2.18 × 103 J. (c) Wby gravity = − ΔU grav = − 2.18 × 103 J Reflect: (a) This is a reasonable weight for an adult woman. (b) Note that when y increases, U grav increases and

Wby gravity is negative.

*7.31. Set Up: Use ΔU grav = mg ( yf − yi ). Solve: ΔU grav = (72 kg)(9.80 m/s 2 )(0.60 m) = 420 J. Reflect: This gravitational potential energy comes from elastic potential stored in his tensed muscles.

7.32. Set Up: Only gravity does work on the swimmer from the point where he has just left the board until just before he enters the water, so Eq. (7.11) applies. Let point 1 be his initial position just after he leaves the board and let point 2 be his final position just before he enters the water. Take + y in the upward direction and y = 0 at the water level. Solve: (a) We have v1 = 0, y2 = 0, and y1 = 3.25 m. From equation 7.11 we have

1 mv1 2 2

+ mgy1 = 12 mv2 2 + mgy2 ,

which gives us mgy1 = 12 mv2 2 . Solving for the final velocity we obtain v2 = 2 gy1 = 2(9.80 m/s 2 )(3.25 m) = 7.98 m/s. (b) We have v1 = 2.50 m/s, y2 = 0, y1 = 3.25 m. Using equation 7.11 we obtain

1 mv1 2 2

+ mgy1 = 12 mv2 2 , which

gives us v2 = v12 + 2 gy1 = (2.50 m/s) 2 + 2(9.80 m/s 2 )(3.25 m) = 8.36 m/s. (c) We have v1 = 2.5 m/s and v2 = 8.36 m/s. The solution is the same as in part (b). Reflect: Kinetic energy depends only on the speed, not on the direction of the velocity.


Work and Energy

7-9

F − kx = , where F is the force m m the spring exerts on the mass. Let point 1 be the initial position of the mass against the compressed spring, so K1 = 0 and U1 = 11.5 J. Let point 2 be the point where the mass leaves the spring, so U el, 2 = 0.

*7.33. Set Up: Only the spring does work and Eq. (7.15) applies. Also, we have a =

Solve: (a) K1 + U el,1 = K 2 + U el,2 gives U el,1 = K 2 . Thus,

1 mv2 2 2

= U el,1 and v2 =

2U el,1 m

=

2(11.5 J) = 3.03 m/s. 2.50 kg

K is largest when U el is least and this is when the mass leaves the spring. The mass achieves its maximum speed of 3.03 m/s as it leaves the spring and then slides along the surface with constant speed. (b) The acceleration is greatest when the force on the mass is the greatest, and this is when the spring has its 2U el 2(11.5 J) maximum compression. Since U el = 12 kx 2 we have x = − =− = −0.0959 m. The minus sign k 2500 N/m kx (2500 N/m)(−0.0959 m) indicates compression. Finally, we have F = − kx = ma x and a x = − = − = 95.9 m/s 2 . m 2.50 kg Reflect: If the end of the spring is displaced to the left when the spring is compressed, then a x in part (b) is to the

right, and vice versa.

7.34. Set Up: Solve Fon spring = kx for k in part (a). Use U el = 12 kx 2 for part (b), and Won spring = + 12 kx 2 for part (c). Solve: (a) The force constant of the spring is: Fon spring 800.0 N = = 4.00 × 103 N/m k= x 0.200 m (b) (i) and (ii) The elastic potential energy is the same whether the spring is stretched or compressed: U el = 12 kx 2 = 12 (4.00 × 103 N/m)(0.300 m) 2 = 180 J. (c) The work done on the spring is equal to the spring’s gain in potential energy: Won spring = U el = + 12 kx 2 = + 12 (4.00 × 103 N/m)(0.200 m) 2 = 80.0 J

*7.35. Set Up: Use Fon tendon = kx. In part (a), Fon tendon equals mg, the weight of the object suspended from it. In part (b), also apply U el = 12 kx 2 to calculate the stored energy. Solve: (a) k = (b) x =

Fon tendon (0.250 kg)(9.80 m/s 2 ) = = 199 N/m x 0.0123 m

Fon tendon 138 N = = 0.693 m = 69.3 cm; U el = 12 (199 N/m)(0.693 m) 2 = 47.8 J k 199 N/m

Reflect: The 250 g object has a weight of 2.45 N. The 138 N force is much larger than this and stretches the tendon a much greater distance.

7.36. Set Up: Use U el = 12 kx 2 for the potential energy. U1 for x1 and U 2 for x2 are related by U 2 /U1 = ( x2 /x1)2 . 2

2

⎛ 4.00 × 1022 m ⎞ ⎛x ⎞ = (10.0 J)(4.00) = 40.0 J Solve: (a) U 2 = U1 ⎜ 2 ⎟ = (10.0 J) ⎜ ⎜ 2.00 × 1022 m ⎟⎟ ⎝ x1 ⎠ ⎝ ⎠ (b) U el depends only on the magnitude of x, not on its sign; therefore, the elastic potential energy is U el = 10.0 J. (c) x2 = x1 (d) k =

2U x2

U2 20.0 J = (2.00 × 10−2 m) = 2.00(2.00 × 10−2 m)(100 cm/m) = 2.83 cm U1 10.0 J

=

2(10.0 J) (2.00 × 10−2 m) 2

= 5.00 × 104 N/m


7-10

Chapter 7

7.37. Set Up: Use U el = 12 kx 2 for part (a) to assess the spring constant and for (b) to determine the spring length required to achieve 850 J of stored energy. Solve: (a) k =

2U el x2

=

2(15.0 J)

= 2.08 × 105 N/m

(1.20 × 10−2 m)2

(b) U el = 12 (2.08 × 105 N/m) x 2 = 850 J 2(850 J)

x=

2.08 × 105 N/m

= 9.04 × 10−2 m = 9.04 cm

The spring is only 8.50 cm long and thus cannot be compressed 9.04 cm to achieve the desired 850 J of stored energy.

7.38. Set Up: The force applied to the springs is Fx = kx. The work done on a spring to move its end from x1 to x2 is W = 12 kx2 2 − 12 kx1 2 . Use the information that is given to calculate k. When the springs are compressed 0.200 m from their

uncompressed length, x1 = 0 and x2 = −0.200 m. When the platform is moved 0.200 m farther, x2 becomes −0.400 m. Solve: (a) k =

2W 2

x2 − x1

2

=

2(80.0 J) (0.200 m) 2 − 0

= 4000 N/m.

Fx = kx = (4000 N/m)(−0.200 m) = −800 N. The magnitude of force that is required is 800 N. (b) To compress the springs from x1 = 0 to x2 = −0.400 m, the work required is W = 12 kx2 2 − 12 kx1 2 = 12 (4000 N/m)

(−0.400 m)2 = 320 J. The additional work required is 320 J − 80 J = 240 J. For x = −0.400 m, Fx = kx = −1600 N. The magnitude of force required is 1600 N. Reflect: More work is required to move the end of the spring from x = −0.200 m to x = −0.400 m than to move it from x = 0 to x = −0.200 m, even though the displacement of the platform is the same in each case. The magnitude of the force increases as the compression of the spring increases.

*7.39. Set Up: The change in gravitational potential energy is ΔU grav = mg ( yf − yi ), while the increase in kinetic energy, for a zero initial velocity, is ΔK = 12 mυf 2 . Set the food energy, expressed in joules, equal to the mechanical energies developed. Solve: (a) The food energy equals mg ( yf − yi ), so

yf − yi =

(140 food calories)(4186 J/1 food calorie)

= 920 m (65 kg)(9.80 m/s 2 ) (b) The mechanical energy would be 20% of the results of part (a): Δy = (0.20)(920 m) = 180 m.

7.40. Set Up: Set 20% of the food energy (in joules) equal to the total work you do in raising the weight N times. Solve: For each arm raise, you do work equal to W = mgh where h = 35 cm. ⎛ 4186 J ⎞ = N (mgh) 0.20(350 food cal) ⎜ ⎝ 1 food cal ⎟⎠

N=

(0.20)(350)(4186 J) (5.0 kg)(9.80 m/s 2 )(0.35 m)

= 1.7 × 104

Reflect: It is not reasonable to do this many repetitions in a single exercise session.

7.41. Set Up: Set 20% of the food energy (in joules) equal to the total work you do in jumping upward a distance of h = 50 cm, N times.


Work and Energy

7-11

Solve: ⎛ 4186 J ⎞ 0.20(500 food cal) ⎜ ⎟ = N (mgh) ⎝ 1 food cal ⎠ (0.20)(500)(4186 J) N= = 1.1 × 103 jumps (75 kg)(9.80 m/s 2 )(0.50 m)

The total time it would take is thus N (2.0 s) = 2.2 × 103 s = 37 minutes. Reflect: This would be a very strenuous workout even for a very fit person.

7.42. Set Up: Use K f + U f = Ki + U i . Setting yi = h, yf = 0, and Ki = 0, gives 12 mυf 2 = mgh. Solve: υf = 2 gh = 2(9.80 m/s 2 )(100 m) = 44.3 m/s

*7.43. Set Up: For part (a), U f − U i = mg ( yf − yi ). Take yi = 2425 ft = 739.1 m and yf = 0. For part (b), use K f + U f = Ki + U i with Ki = U f = 0 and K f = 12 mυ f 2 ; this gives

1 mυ 2 f 2

= Ui.

2

Solve: (a) U i = − mg ( yf − yi ) = − (1.00 kg)(9.80 m/s )(0.00 m − 739.1 m) = 7.24 × 103 J (b) The final kinetic energy of each kilogram is equal to the initial potential energy, K f = U i = 7.24 × 103 J while the

corresponding speed is 2(7.24 × 103 J) = 120 m/s. 1.00 kg

υf =

2U i = m

υf =

2(7.24 × 103 J) = 14 m/s = 32 mph. 70 kg

By comparison, for a 70 kg person,

(c) To double K f , double U i ; since mg is constant, this means h must be doubled. You therefore need h = 4850 ft.

To double vf , you must increase h by a factor of 4 (2υf = 2 2U i /m = 8 gh ), resulting in h = 9700 ft.

7.44. Set Up: Use K f + U f = Ki + U i . Setting yi = h, yf = 0, Ki = 0 and K f = 12 mυf 2 gives

1 mυf 2 2

= mgh, or

υf = 2 gh . Solve: (a) υf = 2(9.80 m/s 2 )(500 m) = 99 m/s = 220 mph (b) Reflect: Hailstones actually have a much smaller speed when they reach the ground. Most of their initial potential energy is converted to thermal energy by the negative work done on them by the air drag force.

*7.45. Set Up: Let θ be the angle of the initial speed above the horizontal. In parts (a)-(c), θ has values of 90°, −90°, and 0°, respectively. Use K f + U f = Ki + U i with K f = 12 ( w/g )υf 2 and Ki = 12 ( w/g )υ0 2 . Let yi = h and yf = 0; this gives

1 ( w/g )υ 2 f 2

= 12 ( w/g )υ0 2 + wh.

Solve: (a)-(d) υf = υ0 2 + 2 gh . (e) Since K and U are both proportional to the weight w = mg , w divides out of the

expression and υf is unaffected by a change in weight. Reflect: The initial kinetic energy depends only on the initial speed and is independent of the direction of the initial velocity.

7.46. Set Up: For part (a) use K f + U f = Ki + U i . At the maximum height, υf = 0 so K f = 0. Let yf = h and yi = 0 and note that Ki = 12 mυ i 2 ; this gives mgh = 12 mυ i 2 . In part (b) use U grav = mgh. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7-12

Chapter 7

Solve: (a) υ i = 2 gh = 2(1.81 m/s 2 )(5.00 × 105 m) = 1350 m/s. (b) On Io, U grav = mgh = (25 kg)(1.81 m/s 2 )(5.00 × 105 m) = 2.3 × 107 J. On earth, g, and thus the potential energy, is

roughly five times greater: ⎛ 9.80 m/s 2 ⎞ U grav = (2.3 × 107 J) ⎜ = 1.2 × 108 J ⎜ 1.81 m/s 2 ⎟⎟ ⎝ ⎠

7.47. Set Up: Use K f + U f = Ki + U i . Let yi = 0 and yf = h and note that U i = 0 while K f = 0 at the maximum height. Consequently, conservation of energy becomes mgh = 12 mυi 2 . Solve: (a) υ i = 2 gh = 2(9.80 m/s 2 )(0.20 m) = 2.0 m/s. (b) Ki = mgh = (0.50 × 10−6 kg)(9.80 m/s 2 )(0.20 m) = 9.8 × 1027 J

Ki 9.8 × 1027 J = = 2.0 J/kg m 0.50 × 1026 kg (c) The human can jump to a height of

⎛l h h = hf ⎜ h ⎝ lf

⎛ ⎞ ⎞ 2.0 m ⎟⎟ = 200 m. ⎟ = (0.20 m) ⎜⎜ 23 ⎠ ⎝ 2.0 × 10 m ⎠

To attain this height, he would require a takeoff speed of: υi = 2 gh = 2(9.80 m/s 2 )(200 m) = 63 m/s. (d) The human’s kinetic energy per kilogram, Ki = gh = (9.80 m/s 2 )(0.60 m) = 5.9 J/kg. m (e) Reflect: The flea stores the energy in its tensed legs.

7.48. Set Up: Let yf = 0. For part (a), U i = mgyi . For part (b) use K f + U f = Ki + U i with U f = Ki = 0 and K f = 12 mυ f 2 ; the result is

1 mυ 2 f 2

= mgyi .

Solve: (a) The figure above shows that the difference in potential energy at the top of the swing is proportional to the height difference, yi = (2.20 m)(1 − cos 42°) = 0.56 m. The difference in potential energy is thus U i = mgyi = (25 kg)(9.80 m/s 2 )(0.56 m) = 140 J.

(b) υf = 2 gyi = 2(9.80 m/s 2 )(0.56 m) = 3.3 m/s (c) Reflect: The tension is radial while the displacement is tangent to the circular path; thus there is no component of the tension along the direction of the displacement and the tension in the ropes does no work on the child. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Work and Energy

7-13

*7.49. Set Up: Only gravity does work, so apply Eq. (7.11). We have vi = 0, so 12 mvf 2 = mg ( yi − yf ). Solve: Tarzan is lower than his original height by a distance yi − yf = l (cos30D − cos 45D ) so his speed is

v = 2 gl (cos30°− cos 45°) = 7.9 m/s, a bit quick for conversation. Reflect: The result is independent of Tarzan’s mass.

7.50. Set Up: Use K f + U f = Ki + U i with yf = h and yi = 0. Also apply U f = U grav = mgh and U i = U el = 12 kx 2 with Ki = K f = 0 to obtain mgh = 12 kx 2 .

h

Solve: (a)

x2

=

k h h , which is constant, so 22 = 12 and, 2mg x2 x1 2

2 ⎛x ⎞ ⎛ 40.0 cm ⎞ h2 = h1 ⎜ 2 ⎟ = (6.0 m) ⎜ ⎟ = 4(6.0 m) = 24 m ⎝ 20.0 cm ⎠ ⎝ x1 ⎠

(b) x2 = x1

h2 h1

= (20.0 cm)

12 m = 2(20.0 cm) = 28.3 cm 6.0 m

(c) Now mh = kx 2 /2 g is constant, so m1h1 = m2h2 . Solving for h2 ,

⎛m ⎞ ⎛1⎞ h2 = h1 ⎜ 1 ⎟ = (6.0 m) ⎜ ⎟ = 3.0 m m ⎝2⎠ ⎝ 2⎠ Reflect: The height that the pebble reaches is inversely proportional to its mass and directly proportional to the energy that is receives from the elastic band.

*7.51. Set Up: Apply K f + U f = Ki + U i with Ki = U f = 0, K f = K , and U i = 12 kx 2 ; the result is K = 12 kx 2 . Since

K x

2

is constant,

K1

x12

=

K2

x2 2

.

2 ⎛x 2⎞ ⎛ 10.0 cm ⎞ Solve: K 2 = K1 ⎜ 22 ⎟ = K ⎜ = 4K ⎟ ⎜x ⎟ ⎝ 5.0 cm ⎠ ⎝ 1 ⎠

7.52. Set Up: In part (a), substitute Ki = K f = 0 into K f + U f = Ki + U i to obtain U el,f = U el,i . Also let subscript 1 refer to the stronger spring and 2 to the weaker spring. Then U el,i = K f which results in

1 mυ max 2 2

Solve: (a) x2 = x1 k1 /k2 = (4.0 cm) (b) υmax = x1

1k x 2 2 1 1

= 12 k2 x2 2 and x2 = x1 k1/k2 . In part (b), apply

= 12 k1x12 . 32 N/cm = (4.0 cm)( 2) = 5.7 cm 16 N/cm

k1 3200 N/m = (4.0 × 10−2 m) = 1.8 m/s m 1.5 kg

7.53. Set Up: Use K f + U f = Ki + U i with Ki = Kf = 0. If we let yf = 0 then yi = x, the amount the spring will be compressed (as shown in the figure below). This results in

1 kx 2 2

= mgx.


7-14

Chapter 7

Solve: x =

2mg 2(12.0 N) = = 7.38 × 10−2 m = 7.38 cm k 325 N/m

Reflect: At the maximum compression the magnitude of the spring force is twice the weight of the object and the system is not in equilibrium.

7.54. Set Up: Use energy methods. There are changes in both elastic and gravitational potential energy; elastic; U = 12 kx 2 , gravitational: U = mgy. Solve: (a) U = 12 kx 2 so x =

2U 2(3.20 J) = = 0.0632 m = 6.32 cm k 1600 N/m

(b) Points 1 and 2 in the motion are sketched in the figure below.

We have K1 + U1 + Wother = K 2 + U 2 , where Wother = 0 (Only work is that done by gravity and spring force), K1 = 0, K 2 = 0, and y = 0 at final position of book. Using U1 = mg (h + d ) and U 2 = 12 kd 2 we obtain 0 + mg (h + d ) + 0 = 12 kd 2 . The original gravitational potential energy of the system is converted into potential energy of the compressed spring. Finally, we use the quadratic formula to solve for d: 12 kd 2 − mgd − mgh = 0, which gives ⎞ 1⎛ ⎛1 ⎞ d = ⎜ mg ± (mg ) 2 + 4 ⎜ k ⎟ ( mgh) ⎟ ⎟ k ⎜⎝ ⎝2 ⎠ ⎠

In our analysis we have assumed that d is positive, so our solution is 1 d= ((1.20 kg)(9.80 m/s 2 ) + ((1.20 kg)(9.80 m/s 2 ))2 + 2(1600 N/m)(1.20 kg)(9.80 m/s 2 )(0.80 m). 1600 N/m d = 0.0074 m + 0.1087 m = 0.12 m = 12 cm Thus, we have d = 0.0074 m + 0.1087 m = 0.12 m = 12 cm. Reflect: It was important to recognize that the total displacement was h + d; gravity continues to do work as the book moves against the spring. Also note that with the spring compressed 0.12 m it exerts an upward force (192 N) greater than the weight of the book (11.8 N). The book will be accelerated upward from this position.

*7.55. Set Up: Apply Emech = K + U with U i = U f and K f = 0; therefore, the lost mechanical energy is Ki = 12 mυ i 2 . Solve: Ki = 12 (1.50 kg)(13.0 m/s)2 = 127 J Reflect: Part of the initial mechanical energy is converted to thermal energy by the negative work done by friction. We did not use the stopping time in our solution.

7.56. Set Up: Apply Emech = K + U Emech,i − Emech,f = Ki − K f =

1 m(υi 2 2

with U i = U f

since the otter returns to the same height. Thus,

2

− υf ).

Solve: Emech,i − Emech,f = 12 (11.4 kg)[(5.75 m/s) 2 − (3.75 m/s) 2 ] = 108 J


Work and Energy

7-15

Reflect: Part of the initial mechanical energy was converted to thermal energy by the negative work done by friction.

*7.57. Set Up: Use Emech = K + U with yf = 0, yi = h, and Ki = U f = 0. Conservation of energy thus becomes: Emech,i − Emech,f = mgy i − 12 mυ f 2 . Solve: Emech,i − Emech,f = (12.0 × 10−3 kg)[(9.80 m/s 2 )(2.50 m) − 12 (3.20 m/s)2 ] = 0.233 J Reflect: In the absence of air resistance an object released from rest has a speed of 7.00 m/s after it has fallen 2.50 m. The speed of the ball is much less than this because of the negative work done by friction.

7.58. Set Up: The friction force is constant during each displacement and Eq. (7.1) can be used to calculate work, but the direction of the friction force can be different for different displacements. G We have f = μk mg = (0.25)(1.5 kg)(9.80 m/s 2 ) = 3.675 N and the direction of f is opposite to the motion. Solve: (a) The path of the book is sketched in the figure below:

G For the motion from you to Beth the friction force is directed opposite to the displacement s and W1 = − fs = −(3.675 N)(8.0 m) = −29.4 J. For the motion from Beth to Carlos the friction force is again directed

opposite to the displacement and W2 = −29.4 J. Wtot = W1 + W2 = −29.4 J − 29.4 J = −59 J (b) The path of the book is sketched in the figure below:

G G For the diagonal path we have s = 2(8.0 m) 2 = 11.3 m. We know that f is opposite to s, so W = − fs = − (3.675 N)(11.3 m) = −42 J. (c)

For the motion from you to Kim we have W = − fs , so W = −(3.675 N)(8.0 m) = −29.4 J.

For the motion from Kim to you we have W = − fs = −29.4 J. The total work for the round = trip is −29.4 J − 29.4 J = 259 J. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7-16

Chapter 7

(d) Reflect: Parts (a) and (b) show that for two different paths between you and Carlos, the work done by friction is different. Part (c) shows that when the starting and ending points are the same, the total work is not zero. Both these results show that the friction force is nonconservative.

7.59. Set Up: The energy of the box at the edge of the roof is given by: Emech,f = Emech,i − f k s. Setting yf = 0 at this point, yi = (4.25 m)sin36° = 2.50 m. Furthermore, by substituting Ki = 0 and K f = 12 mυf 2 into the conservation equation,

1 mυ 2 f 2

= mgyi − f k s or

υf = 2 gyi − 2 f k sg/w = 2 g ( yi − f k s/w) Solve: υf = 2(9.80 m/s 2 )[(2.50 m) − (22.0 N)(4.25 m)/(85.0 N)] = 5.24 m/s Reflect: Friction does negative work and removes mechanical energy from the system. In the absence of friction the final speed of the toolbox would be 7.00 m/s.

7.60. Set Up: Use the work-energy theorem, Eq. (7.16). The target variable μk will be a factor in the work done by friction. Let point 1 be where the block is initially released and let point 2 be where the block finally stops, as shown in the figure below. Thus, we have K1 + U1 + Wother = K 2 + U 2 .

Work is done on the block by the spring and by friction, so Wother = W f and U = U el . Solve: We have K1 = K 2 = 0, U1 = U1,el = 12 kx1 2 = 12 (100 N/m)(0.200 m)2 = 2.00 J, and U 2 = U 2,el = 0 (since after

the block leaves the spring has given up all its stored energy). Finally, we have Wother = W f = ( f k cos φ ) s =

μk mg (cos φ ) s = − μk mgs, since φ = 180° (The friction force is directed opposite to the displacement and does negative work). Putting all this into K1 + U1 + Wother = K 2 + U 2 gives U1,el + W f = 0, so μk mgs = U1,el . Solving for μk we obtain μk =

U1,el mgs

=

2.00 J (0.50 kg)(9.80 m/s 2 )(1.00 m)

= 0.41.

Reflect: U1,el + W f = 0 says that the potential energy originally stored in the spring is taken out of the system by the

negative work done by friction.

*7.61. Set Up: Use Emech,f = Emech,i − f k s to solve for f k . There is no change in height so U i = U f and the energy relation becomes K f = K i − f k s. Solve: (a) First solve for the kinetic energy values: Ki = 12 mυ i 2 = 12 (375 kg)(4.5 m/s) 2 = 3800 J K f = 12 mυ f 2 = 12 (375 kg)(3.0 m/s)2 = 1690 J Substituting, fk =

Ki − K f 3800 J − 1690 J = = 300 N s 7. 0 m

(b) (i) ( Ki − K f )/Ki = [(3800 J − 1690 J)/3800 J] × 100, = 56, (b) (ii) (υ f − υ i )/υ i = [(1.5 m/s)/(4.5 m/s)] × 100, = 33, Reflect: Friction does negative work and reduces the kinetic energy of the object.


Work and Energy

7-17

7.62. Set Up: For part (a) use Emech, f = Emech,i − f k s where f k = μk mg . Let yf = 0 at the bottom of the hill; then yi = 2.50 m along the rough patch. The energy equation is thus

1 mυ 2 f 2

= 12 mυ i 2 + mgyi − μk mgs and solving for her

final speed, υf = υi 2 + 2 gyi − 2μ k gs . For part (b), the internal energy is calculated as the negative of the work done by friction: −W f = + f k s = + μ k mgs. Solve: (a) υf = (6.50 m/s) 2 + 2(9.80 m/s 2 )(2.50 m) − 2(0.300)(9.80 m/s 2 )(3.50 m) = 8.41 m/s (b) Internal energy = μ k mgs = (0.300)(62.0 kg)(9.80 m/s 2 )(3.50 m) = 638 J

*7.63. Set Up: For part (a) use P = 1 mυ 2 2

W to solve for W, the energy the bulb uses. Then set this value equal to Δt

for part (b), and solve for the speed. In part (c), equate the W from part (a) to U grav = mgh and solve for the

height. Solve: (a) W = P Δt = (100 W)(3600 s) = 3.6 × 105 J (b) K = 3.6 × 105 J so υ =

2K = m

(c) U grav = 3.6 × 105 J so h =

U grav

2(3.6 × 105 J) = 100 m/s 70 kg 3.6 × 105 J

= 520 m (70 kg)(9.80 m/s 2 ) Reflect: (b) Olympic runners achieve speeds up to approximately 36 m/s, or roughly one = third the result calculated. (c) The tallest tree on record, a redwood, stands 364 ft or 110 m, or 4.7 times smaller than the result. mg

=

7.64. Set Up: Use the relation P = Fυ to relate the given power and velocity to the force required. Recall that a watt represents the rate of energy—a joule per second. Solve: The force required is thus F = (30.0 × 103 J/s)/(15.0 m/s) = 2.00 × 103 N.

*7.65. Set Up: Since the electricity cost rate is given in cents per kWh, each power must be multiplied by the corresponding time usage in determining the cost: cost = Pt × (cost/Pt ) = (kWh)[cents/(kWh)]. ⎛ 746 W ⎞ ⎛ kW ⎞ ⎤ ⎛ 7.35 . ⎞ ⎛ cos t ⎞ ⎡ Solve: (a) cos t = Pt × ⎜ ⎟⎜ ⎟ = ⎢(10.0 hp)(8.00 hr) × ⎜ ⎟⎥ ⎜ ⎟ = 439. = $4.39 Pt ⎝ ⎠ ⎣ ⎝ hp ⎠ ⎝ 1000 W ⎠ ⎦ ⎝ kWh ⎠ ⎛ 24.0 hr ⎞ ⎛ kW ⎞ ⎤ ⎛ 7.35 . ⎞ . ⎛ cos t ⎞ ⎡ (b) cos t = Pt × ⎜ = $0.13 per day ⎟⎜ ⎟ = ⎢(75 W) ⎜ ⎟⎥ ⎜ ⎟ = 13 Pt day 1000 W kWh day ⎝ ⎠ ⎣ ⎠⎦ ⎝ ⎠ ⎝ ⎠⎝ Reflect: Note that the cost of running the motor for an entire day is 3($4.39) = $13.2; thus, the cost of running the motor is ($13.2)/($0.13) = 102 ≈ 100 times the cost of running the light.

7.66. Set Up: Use the relation P = Fυ to relate the given force and velocity to the total power developed. Solve: The total power is P = Fυ = (165 N)(9.00 m/s) = 1.49 × 103 W. Each rider therefore contributes Peach rider = (1.49 × 103 W)/2 = 745 W ≈ 1 hp.

Reflect: The result of one horsepower is very large; a rider could not sustain this output for long periods of time.

7.67. Set Up: Use Eq. (7.17) to relate the power provided and the amount of work done against gravity in 16.0 s. The work done against gravity depends on the total weight, which depends on the number of engers. Solve: Find the total mass that can be lifted:

Pav =

P t ΔW mgh , so m = av = gh t Δt


7-18

Chapter 7

⎛ 746 W ⎞ 4 Pav = (40 hp) ⎜ ⎟ = 2.984 × 10 W ⎝ 1 hp ⎠ m=

Pavt (2.984 × 104 W)(16.0 s) = = 2.436 × 103 kg gh (9.80 m/s 2 )(20.0 m)

This is the total mass of elevator plus engers. The mass of the engers is 2.436 × 103 kg − 600 kg = 1.836 × 103 kg = 28.2. Thus, 28 engers can ride. 65.0 kg Reflect: Typical elevator capacities are about half this, in order to have a margin of safety. 1.836 × 103 kg. The maximum allowed number of engers is

7.68. Set Up: The watt represents power as joules per second; so convert the unit of year to seconds for part (a). For part (b), the rate of consumption is simply the power divided by the estimated number of consumers. 1.0 × 1019 J/y Solve: (a) P = = 3.2 × 1011 W = 3.2 × 108 kW (365 days/y)(24 h/day)(3600 s/h) (b) Power per person =

3.2 × 108 kW (310 × 106 persons)

= 1.0 kW/person

*7.69. Set Up: From Problem 7.68, the U.S. consumes 3.2 × 1011 W of power per year. The area required in part (a) may then be calculated as A = ( PU.S. )/(P radiation /m 2 ). However, for an array efficiency of 25%, the radiative power rate must be multiplied by 0.25 to for power losses. For part (b), since the array is square, A = L2 and the length of a side is simply L =

A.

Solve: (a) A = ( PU.S. )/(Pradiation /m 2 ) = (b) Lmetric =

⎛ km 2 ⎞ 3 2 ⎜ ⎟ = 1.3 × 10 km 0.25(1.0 × 103 W/m 2 ) ⎜⎝ 106 m 2 ⎟⎠ 3.2 × 1011 W

A = 1.3 × 103 km 2 = 36 km ⎛ 1 mi ⎞ LEnglish = (36 km) ⎜ ⎟ = 22 mi ⎝ 1.609 km ⎠

Reflect: An array of this size is approximately equivalent to the state of Rhode Island and thus feasible.

7.70. Set Up: The thermal energy is produced as a result of the force of friction, F = μk mg . The average thermal power is thus the average rate of work done by friction or P = Fυave where υave is based on the initial and final velocities. ⎛ 8.00 m/s − 0.00 m/s ⎞ Solve: P = Fυave = [(0.200)(20.0 kg)(9.80 m/s 2 )]⎜ ⎟⎠ = 157 W ⎝ 2 Reflect: The power could also be determined as the rate of change of kinetic energy, ΔK/t , where the time is calculated from υf = υ i + at and a is calculated from a force balance, oF = ma = μ k mg .

*7.71. Set Up: Use1.00 hp = 746 W to convert the given units. In part (c), also apply the energy relation W = P Δt. ⎛ 1.00 hp ⎞ Solve: (a) (100 W) ⎜ ⎟ = 0.134 hp ⎝ 746 W ⎠

⎛ 746 W ⎞ 4 (b) (75 hp) ⎜ ⎟ = 5.6 × 10 W ⎝ 1.00 hp ⎠


Work and Energy

7-19

⎛ 1.00 hp ⎞ −3 (c) (2.00 W) ⎜ ⎟ = 2.68 × 10 hp 746 W ⎝ ⎠ ⎛ 1.00 hp ⎞ 23 (d) (3.92 × 1026 W) ⎜ ⎟ = 5.25 × 10 hp ⎝ 746 W ⎠ ⎛ 746 W ⎞ 4 4 (e) (25 hp) ⎜ ⎟ = 1.865 × 10 W; W = (1.865 × 10 W)(1.5 h) = 28 kWh . 1 00 hp ⎝ ⎠

7.72. Set Up: Convert the power from hp to W and then use the energy relation W = P Δt to calculate the energy expended. Solve: The power in watts is:

⎛1 ⎞ ⎛ 746 W ⎞ P = ⎜ hp ⎟ ⎜ ⎟ = 249 W. 3 ⎝ ⎠ ⎝ 1.00 hp ⎠ The energy is thus,

⎛ 1 food cal ⎞ W = (249 W)(3600 s) = 8.96 × 105 J ⎜ ⎟ = 214 food calories. ⎝ 4186 J ⎠ Since only 20% of these food calories goes to mechanical energy, the number of food calories burned is: 214 food cal = 1100 food calories 0.20 Reflect: This would be an impossibly strenuous exercise session; the average calorie intake per day for a human is about 2700 food calories.

7.73. Set Up: Relate power, work, and time. Work done in each stroke is W = Fs and Pav = W/t. Solve: 100 strokes per second means Pav = 100 Fs/t with t = 1.00 s, F = 2mg and s = 0.010 m. Pav = 0.20 W. Reflect: For a 70 kg person to apply a force of twice his weight through a distance of 0.50 m for 100 times per second, the average power output would be 7.0 × 104 W. This power output is very far beyond the capability of a person.

7.74. Set Up: Calculate the power used to make the plane climb against gravity. Consider the vertical motion since gravity is vertical. Solve: The rate at which work is being done against gravity is P = Fv = mgv = (700 kg)(9.80 m/s 2 )(2.5 m/s) = 17.15 kW. This is the part of the engine power that is being used to make the airplane climb. The fraction this is of the total is 17.15 kW/75 kW = 0.23. Reflect: The power we calculate for making the airplane climb is considerably less than the power output of the engine.

*7.75. Set Up: For part (a), the work performed each day is W = Wgrav = mgh where h = 1.63 m. The mass of the blood is calculated using the density, ρ = m /V = 1050 kg/m3 , and the known volume of V = 7500 L. The power output is then found from P = Wgrav /t . Solve: (a) The work performed in a single day is:

kg ⎞ ⎛ 1 m3 ⎞ ⎛ m⎞ ⎛ 5 W = Wgrav = mgh = ρVgh = ⎜ 1050 3 ⎟ ⎜ 7500 L × ⎟⎟ ⎜ 9.80 2 ⎟ (1.63 m) = 1.26 × 10 J ⎜ 1000 L ⎠ ⎝ m ⎠⎝ s ⎠ ⎝ (1.26 × 105 J/day) = 1.46 J/s = 1.46 W t (24 hr/day)(3600 s/hr) (c) Reflect: The heart actually puts out more power than the value calculated because two forms of energy were not considered; the heart also provides kinetic energy to the blood and generates thermal energy by working against the friction of the vein and vessel walls. (b) P =

Wgrav

=


7-20

Chapter 7

7.76. Set Up: The mass of a given volume of air can be determined from its density, ρ , by using m = ρ V . The 1 kinetic energy of a mass m moving with a velocity υ is K = mυ 2 . The amount of wind ing through the blades 2 of the wind turbine in a given time is given by the volume of a cylinder that has a radius equal to the radius of its blades and a height equal to the distance the wind travels in the given amount of time. Recall that the volume of a cylinder of radius r and height h is given by V = π r 2 h.

Solve: (a) The mass of 1.00 m3 of air is m = ρ V = (1.2 kg/m3 )(1.00 m3 ) = 1.2 kg. For wind moving at 9.3 m/s the

1 1 kinetic energy associated with this volume is K = mυ 2 = (1.2 kg)(9.3 m/s)2 = 51.9 J. Thus, the wind energy per 2 2 cubic meter is 52 J/m3. (b) In one second, a cylindrical column of air with a radius of 41 m and a height of (9.3 m/s)(1 s) = 9.3 m es through the wind turbine. This column has a volume of π r 2 h = π (41 m) 2 (9.3 m) = 4.91 × 104 m3. Thus, the amount of wind energy available to the turbines in 1 second is (4.91 × 104 m3 )(51.9 J/m3 ) = 2.55 × 106 J. Since the wind turbine is 35% efficient, the energy output of the turbine would be (2.55 × 106 J)(0.35) = 8.9 × 105 J per second, which is 8.9 × 105 W. Reflect: The kinetic energy per cubic meter is proportional to υ 2 and the volume of wind ing through the blades in a given time is proportional to υ ; thus, assuming a fixed efficiency, the power output of the wind turbine is

proportional to υ 3!

7.77. Set Up: In part (a), the conservation of energy K f + U f = Ki + U i results in

1 kx 2 2 f

+ 0 = 0 + 12 mυ i 2 or

k = (myi 2 )/x 2 . Similarly, the maximum speed may be calculated in part (b) as υ = x k/m . Solve: (a) k = (mυ i 2 )/x 2 = [(1200 kg)(0.65 m/s) 2 ]/[(7.0 cm)(m/100 cm)]2 = 1.0 × 105 N/m

U grav = K = 12 mυ 2 = 12 (1200 kg)(0.65 m/s) 2 = 250 J (b) υ = x

5 k ⎛ m ⎞ 1.0 × 10 N/m = (5.0 cm) ⎜ = 0.46 m/s ⎟ m 1200 kg ⎝ 100 cm ⎠

Reflect: The maximum permissible speed calculated in (b) is equivalent to roughly 1 mi/h. A practical design for a parking garage should be sufficient for speeds up to 5 mi/h.

7.78. Set Up: Applying K f + U f = Ki + U i in part (a) gives mgh = 12 mυ 2 or υ = 2 gh . In part (b), use: Emech = K + U , where Emech,i = mgh and Emech,f = 12 mυ 2 Solve: (a) υ =

2(9.80 m/s 2 )(1000 m) = 140 m/s

(b) Emech,i − Emech,f = mgh − 12 mυ 2 = (70kg)(9.80m/s 2 )(1000m) − 12 (70 kg)(62 m/s) 2 = 5.5 × 105 J Reflect: Mechanical energy was converted to thermal energy by the negative work done by the air drag force.

*7.79. Set Up: Conservation of energy says the decrease in potential energy equals the gain in kinetic energy. Since the two animals are equidistant from the axis, they each have the same speed v. Solve: One mass rises while the other falls, so the net loss of potential energy is

(0.500 kg −

0.200 kg)(9.80 m/s 2 )(0.400 m) = 1.176 J. This is the sum of the kinetic energies of the animals and is equal to 1 m v2 , 2 tot

and v =

2(1.176 J) = 1.83 m/s. (0.700 kg)


Work and Energy

7-21

Reflect: The mouse gains both gravitational potential energy and kinetic energy. The rat’s gain in kinetic energy is less than its decrease of potential energy, and the energy difference is transferred to the mouse.

7.80. Set Up: Apply K f + U f = Ki + U i to the system of the two climbers for which Ki = 0. For the 60.0 kg person, who experiences a gravitational potential energy change, let yf = 0 so yi = 1.50 m. Both climbers move with the same speed υ. Substituting into the conservation equation,

1 mυ 2 2 1 f

+ 12 m2υf 2 = m2 gyi , where m1 = 75.0 kg

and m2 = 60.0 kg. Solve: υf =

2m2 gyi = m1 + m2

2(60.0 kg)(9.80 m/s 2 )(1.50 m) = 3.61 m/s 75.0 kg + 60.0 kg

7.81. Set Up: Use K f + U f = Ki + U i − f k s with f k = μk m1g , s = 1.50 m, and Ki = 0. For the 60.0 kg person, who experiences a gravitational potential energy change, let yf = 0 so yi = 1.50 m. Both climbers move with the same speed v. This results in Solve: υf =

1 mυ 2 2 1 f

+ 12 m2υf 2 = m2 gyi − μ k m1gd , where m1 = 75.0 kg and m2 = 60.0 kg.

2 gd (m2 − μ k m1 ) = m1 + m2

2(9.80 m/s 2 )(1.50 m)[60.0 kg − 0.250(75.0 kg)] = 3.00 m/s 75.0 kg + 60.0 kg

Reflect: Friction does negative work and removes mechanical energy from the system. The tension does positive work on one person and the same magnitude of negative work on the other person and therefore does no net work on the system.

7.82. Set Up: Apply Eq. (7.16) to the motion of the skier from the gate to the bottom of the ramp. We are given that Wother = 24000 J. Let y = 0 at the bottom of the ramp. Solve: For the skier to be moving at no more than 30.0 m/s, his kinetic energy at the bottom of the ramp can be no mv 2 (85.0 kg)(30.0 m/s) 2 = = 38,250 J. Friction does −4000 J of work on him during his run, which 2 2 means his combined U and K at the top of the ramp must be no more than 38,250 J + 4000 J = 42,250 J. His K at the

bigger than

mv 2 (85.0 kg)(2.0 m/s) 2 = = 170 J. His U at the top should thus be no more than 42,250 J − 170 J = 42,080 J, 2 2 42,080 J 42,080 J which gives a height above the bottom of the ramp of h = = = 50.5 m. mg (85.0 kg)(9.80 m/s 2 ) top is

Reflect: In the absence of air resistance, for this h his speed at the bottom of the ramp would be 31.5 m/s. The work done by air resistance is small compared to the kinetic and potential energies that enter into the calculation.

*7.83. Set Up: We know that Pav = F||vav . We can find the average velocity, vav = F = ma to calculate the force. Solve: Your friend’s average acceleration is a =

0 + 6.00 m/s = 3.00 m/s, and use 2

v − v0 6.00 m/s = = 2.00 m/s 2 . Since there are no other horizontal t 3.00 s

forces acting, the force you exert on her is given by Fnet = ma = (65.0 kg)(2.00 m/s 2 ) = 130 N. Pav = (130 N) × (3.00 m/s) = 390 W. Reflect: We could also use the work-energy theorem: W = K 2 − K1 = 12 (65.0 kg)(6.00 m/s)2 = 1170 J. Pav =

W = t

1170 J = 390 W, the same as obtained by our other approach. 3.00 s


7-22

Chapter 7

7.84. Set Up: From the work-energy principle, W f = K f − Ki where W f = − f k s = − (0.25mg ) s and subscripts “f” and “i” refer to the surface areas with and without friction, respectively. Solve: W f = K f − Ki

−(0.25mg ) s = 12 mυf 2 − 12 mυ i 2 −0.25 gs = 12 [(0.55υ i ) 2 − υ i 2 ] −0.25(9.80 m/s 2 ) s =

1 2

{[(0.55)(3.0 m/s)]2 − (3.0 m/s)2}

s = 1.3 m

7.85. Set Up: At any point, the potential energy is given by mgl (1 − cosθ ) where l is the string length and θ is the angle the string makes with the vertical. At the top of the swing, the kinetic energy is zero while the potential energy is a maximum since θ = θ max = 45°; whereas, at the bottom of travel, the kinetic energy becomes a maximum equal to the U grav of the top of the swing. In of conservation of energy, K f = U grav,i or

1 mυ 2 f 2

= mgl (1 − cos 45°)

where “f” refers to the vertical position of the pendulum and “i” refers to top of travel. Solve: Solving for the velocity,

υf = 2 gl (1 − cos 45°) = 2(9.80 m/s 2 )(0.80 m)(1 − cos 45°) = 2.1 m/s The tension in the string at this instant is found using a force balance of the weight and radial acceleration, T = mg +

mυf 2 2mgl (1 − cos 45°) = mg + = mg[1 + 2(1 − cos 45°)] l l

T = (0.12 kg)(9.80 m/s 2 )(1.5858) = 1.9 N Reflect: As the string es through the vertical the ball has an upward acceleration, due to its circular motion. Therefore, there is a net upward force and the tension is greater than the weight of the ball.

7.86. Set Up: The water is both raised and given speed, so the mechanical energy relation may be applied: Emech = mgy + 12 mυ 2 . One liter of water has a mass of 1 kg. Solve: W = Emech, f − Emech, i = mg ( yf − yi ) + 12 m(υf 2 − υ i 2 )

= (750 kg)(9.80 m/s 2 )(0 − ( −14.0 m)) + 12 (750 kg)[(18.0 m/s) 2 − (0.0 m/s) 2 ] = 2.24 × 105 J Reflect: Most of the work done on the water by the pump goes to increasing the kinetic energy of the water.

*7.87. Set Up: For part (a), apply conservation of energy to the motion from point A to point B: K B + U grav, B = K A + U grav, A with K A = 0. Defining yB = 0 and y A = 13.0 m, conservation of energy becomes

1 mυ B 2 2

= mgy A or

υ B = 2 gy A . In part (b), the free-body diagram for the roller coaster car at point B is shown in the figure below. oFy = ma y gives mg + n = marad , where arad = υ 2 /r. Solving for the normal force results in: ⎛υ2 ⎞ n = m⎜ − g⎟ ⎜ r ⎟ ⎝ ⎠


Work and Energy

7-23

Solve: (a) υ B = 2(9.80 m/s 2 )(13.0 m) = 16.0 m/s

⎡ (16.0 m/s) 2 ⎤ − 9.80 m/s2 ⎥ = 1.15 × 104 N (b) n = (350 kg) ⎢ 6 0 m . ⎣⎢ ⎦⎥ Reflect: The normal force n is the force that the tracks exert on the roller coaster car. The car exerts a force of equal magnitude and opposite direction on the tracks.

7.88. Set Up: For part (a), apply K f + U f = Ki + U i with K f = U i = 0 which results in

2 1 kx 2 max

= 12 mυ02 . For

parts (b) through (d), the maximum force F exerted by the air bag is expressed as Fmax = kxmax . Solve: (a) Solving

2 1 kx 2 max

= 12 mυ02 for the spring constant gives k=

mυ0 2

xmax 2

.

(b) Use the expression in part (a) to replace k in the relation Fmax = kxmax :

Fmax =

mυ0 2 . xmax

2

⎛ 1m ⎞ 5 (c) Fmax = (5.0 × 105 N/m 2 )(2500 cm 2 ) ⎜ 2 ⎟ = 1.25 × 10 N ⎝ 10 cm ⎠ (d) k =

Fmax 1.25 × 105 N = = 4.2 × 105 N/m. The expression in part (b) gives xmax 0.30 m

υ0 =

Fmax xmax (1.25 × 105 N)(0.30 m) = = 24 m/s = 54 mph m 65 kg

(e) Reflect: A safer design requires a larger distance xmax . This could be accomplished by extending the gas, brake,

and clutch pedals so that the driver could sit farther away from the steering wheel or by decreasing the stiffness k of the airbag. A stiffer air bag would increase k and Fmax and thus make the system less safe.

*7.89. Set Up: We can calculate the average speed of the horse from the number of trips that the horse makes around a circle of known circumference each hour. The force exerted by the horse can be calculated from its assumed power output by using Pav = F||υav . According to the appendix, 1 hp = 746 W = 550 ft ⋅ lb/s. ⎛ 144 trips ⎞ ⎛ 24π feet ⎞ ⎛ 1 hour ⎞ Solve: (a) The average speed of the horse is ⎜ ⎟⎜ ⎟⎜ ⎟ = 3.0 ft/s. To produce 1 hp, we ⎝ hour ⎠ ⎝ 1 trip ⎠ ⎝ 3600 s ⎠ P 550 ft ⋅ lb/s require a force that satisfies F|| = av = = 180 lb. 3.0 ft/s υav (b) The work output of the person must be at least her gain in gravitational potential energy, which is ΔU = mg Δ y. If this work occurs over a time t, the average power output is Pav =

mgh (70 kg)(9.80 m/s 2 )(3.0 m) = = 410 W. This t 5.0 s

⎛ 1 hp ⎞ can also be written as (410 W) ⎜ = 0.55 hp. ⎝ 746 W ⎟⎠ Reflect: The power output of a person can exceed 1 hp for a brief time.

7.90. Set Up: W = Pt

Solve: (a) The hummingbird produces energy at a rate of 0.7 J/s to 1.75 J/s. At 10 beats/s, the bird must expend

between 0.07 J/beat and 0.175 J/beat.


7-24

Chapter 7

(b) The steady output of the athlete is (500 W)/(70 kg) = 7 W/kg, which is below the 10 W/kg necessary to stay aloft. Though the athlete can expend 1400 W/70 kg = 20 W/kg for short periods of time, no human-powered aircraft could

stay aloft for very long. Reflect: Movies of early attempts at human-powered flight bear out our results.

7.91. Set Up: First set the final kinetic energy of the object to one half of the initial elastic potential energy: K f = 12 U el,i or 12 mυf 2 = 12 ( 12 kxi 2 ). Solving for the speed, υf = xi k/2m . Solve: υf = (0.120 m)

(35 N/cm)(100 cm/m) = 10 m/s 2(0.250 kg)

Reflect: U el,f = 12 U el,i . This gives xf 2 = 12 xi 2 and xf = xi / 2 = 0.71xi . The object has moved 3.5 cm, less than half

the total distance of 12.0 cm.

7.92. Set Up: Apply the adapted version of the conservation of energy for nonconservative forces: Wnc = Emech,f − Emech,i = mg ( yf − yi ) + 12 m(υf 2 − υi 2 ) where yf − yf = 0.0, υ f = 0.0 and Wnc = − μk mgs. Solve: (a) Solve for the initial velocity to determine the driver’s speed at impact: Wnc = − μk mgs = − 12 mυ i 2

⎛ 60 mph ⎞ ⎟ = 50 mph ⎝ 88 ft/s ⎠ The driver was thus 15 mph over the posted speed limit at the instant he struck the pedestrian. (b) The driver must pay a ticket fine of ($10.00/mph)(15 mph) = $150.

υ i = 2μk gs = 2(0.30)(32.18 ft/s 2 )(280 ft) = 73.5 ft/s ⎜

*7.93. Set Up: Apply K1 + U1 + Wother = K 2 + U 2 to the motion of the person. Point 1 is where he steps off the platform and point 2 is where he is stopped by the cord. Let y = 0 at point 2. y1 = 41.0 m. Wother = − 12 kx 2 , where x = 11.0 m is the amount the cord is stretched at point 2. The cord does negative work. Solve: K1 = K 2 = U 2 = 0, so mgy1 − 12 kx 2 = 0 and k = 631 N/m.

Now apply F = kx to the test pulls: F = kx so x = F /k = 0.602 m. Reflect: All his initial gravitational potential energy is taken away by the negative work done by the force exerted by the cord, and this amount of energy is stored as elastic potential energy in the stretched cord.

7.94. Set Up: At the top of the loop, the net force needed to produce the centripetal acceleration, Fnet = mυ 2 /R, must be equal to or greater than the gravitational force for the car to remain on the track. Thus, a force balance results in Fnet = mυ 2 /R = mg or υ = gR . Next calculate the height, h, necessary to produce this speed by applying conservation of energy, U A + K A = U B + K B . Solve: U A + K A = U B + K B

⇒ mghA + 0 = mghB + 12 mυ B 2 mg (2 R) + 12 mgR

5 = R 2 mg The height must be at least 2.5R = 2.5(20.0 m) = 50.0 m. Reflect: If h < 2.5 R the car is moving too slowly at B to stay on the track. In that case,

υ B = gR ⇒ hA =

υ2 R

< g,


Work and Energy

7-25

gravity provides more downward force than is needed to maintain the circular path, and the car falls out of the circle. If h > 2.5 R, then at B more downward force than gravity alone is needed and this additional downward force is supplied by the normal force that the track exerts on the car.

*7.95. Set Up: First apply conservation of energy to find the kinetic energy of the wood as it enters the rough section: U1 = K 2 , or K 2 = mgy1. Then determine the distance traveled using the work-energy relation, Wnc = K3 + U 3 − K 2 − U 2 where K3 = U 3 = U 2 = 0 and Wnc = W f = − μ k mgs. For part (b), use the same relation and the calculated distance to find the frictional work. Solve: (a) The work-energy relation reduces to Wnc = W f = − K 2 , which gives − μ k mgs = − K 2 . Substituting K 2 = mgy1 and solving for the distance, s: s=

mgy1 y 4.0 m = 1 = = 20 m 0.20 μk mg μk

(b) Wnc = W f = − μ k mgs = − (0.20)(2.0 kg)(9.80 m/s 2 )(20 m) = − 78 J

7.96. Set Up: For part (a), use the conservation of energy relation for nonconservative forces: K foot + U foot + Wnc = K top + U top . For part (b), compare the maximum static friction force, fs = μs mg cosθ , to the skier’s weight component down the slope. Solve: (a) Conservation of energy results in

2 1 mυ foot 2

− μ k (mg cosθ ) s = mgh, where s = h/sin θ . Substituting and

solving for the height, 2 1 mυ foot 2

⎡ h ⎤ − μ k (mg cosθ ) ⎢ ⎥ = mgh ⎣ sin θ ⎦

υfoot 2 2

⎡ μ gh(cosθ ) ⎤ = gh + ⎢ k ⎥ sin θ ⎣ ⎦

h=

υfoot 2 2 g {1 + [( μ k cosθ )/sin θ ]}

h=

(15 m/s) 2 = 8.8 m 2(9.80 m/s){1 + [(0.25cos 40.0°) / sin 40.0°]}

(b) The skier will begin to slide down the hill if the downhill component of her weight is greater than the maximum value of the static friction acting on her. So compare these two quantities. Maximum static friction: fs = μs mg cosθ = (0.75)(68 kg)(9.80 m/s 2 )cos(40.0°) = 380 N

Downhill weight component: wg,down = mg sin θ = (68 kg)(9.80 m/s 2 )(sin 40.0°) = 430 N Since 430 N > 380 N, the skier will slide down the hill. Reflect: As the skier moves up the hill part of her initial kinetic energy is converted to gravitational potential energy by the negative work done by gravity and the rest is removed by the negative work done by friction.

*7.97. Set Up: For parts (a) and (c), use the energy relation W = P Δt to find the energy in joules and then convert the respective results to food calorie units. Solve: (a) Energy = W = (80 J/s)(3600 s/h)(24 h/day) = 6.9 × 106 J/day

⎛ 1 food cal ⎞ W = (6.9 × 106 J/day) ⎜ = 1.6 × 103 food calories/day ⎝ 4186 J ⎟⎠ (b) The body’s metabolic processes convert food energy to thermal energy. (c) Energy = W = [(80 J/s)(3600 s/h)(16 hr) + (200 J/s)(3600 s/h)(8 h)]

= 1.04 × 107 J = 2.48 × 103 food calories


7-26

Chapter 7

7.98. Set Up: Apply P = F|| v, where F|| is the force F of water resistance. 1 hp = 746 W. 1 km/h = 0.2778 m/s Solve: F =

(0.70) P (0.70)(260,000 hp)(746 W/hp) = = 7.5 × 106 N. v (65 km/h)((0.2778 m/s)/(1 km/h))

Reflect: The power required depends on speed, because of the factor of v in P = F|| v and also because the resistive

force increases with speed.

7.99. Set Up: Ignore any friction losses in the rope. The 4 kg bucket will rise 2 m above the ground as the 12 kg bucket falls. Note also that the velocity of the buckets must be identical. Conservation of energy, K f + U f = U i , thus becomes:

1 (m 12kg 2

+ m4kg )υf 2 + m4kg gh = m12kg gh.

Solve: (a) 1 (m 12kg 2 1 (12.0 2

+ m4kg )υf 2 + m4kg gh = m12kg gh

kg + 4.0 kg)υf 2 + (4.0 kg)(9.80 m/s 2 )(2.00 m) = (12.0 kg)(9.80 m/s 2 )(2.00 m)

υf = 156.8/8.0 = 4.43 m/s (b) Reflect: The speed of the buckets would be less than 4.43 m/s because the potential energy of the buckets would be converted into kinetic energy shared by the mass of the pulley as well as the buckets.

7.100. Set Up: Apply conservation of energy: K f + U f = Ki + U i . Let yi = 0, so yf = h, the maximum height. At this maximum height, υ f, y = 0 and υ f, x = υ i, x thus υ f = υ i, x = (15 m/s)(cos60.0°) = 7.5 m/s. Substituting into conservation of energy gives Solve: h =

υ i 2 − (7.5 m/s) 2 2g

1 mυ 2 i 2

=

= mgh + 12 m(7.5 m/s) 2 .

(15 m/s) 2 − (7.5 m/s) 2 2(9.80 m/s 2 )

= 8.6 m

*7.101. Set Up: P = F|| v and 1 m/s = 3.6 km/h. Solve: (a) F =

P 28.0 × 103 W = = 1.68 × 103 N. v (60.0 km/h)((1 m/s)/(3.6 km/h))

(b) The speed is lowered by a factor of one-half, and the resisting force is lowered by a factor of (0.65 + 0.35/4), and so the power at the lower speed is (28.0 kW)(0.50)(0.65 + 0.35/4) = 10.3 kW = 13.8 hp. (c) Similarly, at the higher speed, (28.0 kW)(2.0)(0.65 + 0.35 × 4) = 114.8 kW = 154 hp. Reflect: At low speeds rolling friction dominates the power requirement but at high speeds air resistance dominates.

7.102. Set Up: The mass is the density times the volume V. For a sphere, V = 43 π r 3 . Use K = 12 mυ 2 and convert all quantities to SI units. Solve: (a) m = (3.5 × 103 kg/m3 ) 43 π r 3 = (3.5 × 103 kg/m3 )

( 43 )π (5 × 103 m)3 = 1.8 × 1015 kg

(b) K = 12 mυ 2 = 12 (1.8 × 1015 kg)(11 × 103 m/s)2 = 1.1 × 1023 J (c) The time required for the human race to use 1.1 × 1023 J is

t=

1.1 × 1023 J 500 × 1018 J/y

= 220 years.

7.103. Set Up: Use the constant acceleration equation x = [(υ i + υf ) t/2] to calculate υ f . Then calculate the energy K = 12 mυ 2 . From Problem 7.106, the mass is m = 1.8 × 1015 kg.


Work and Energy

Solve:

υf =

7-27

2 x 2(5.0 × 10−2 m) = = 0.20 m/s t 0.50 s

K = 12 (1.8 × 1015 kg)(0.20 m/s) 2 = 3.6 × 1013 J The number N of 1.0 megaton bombs is the total energy divided by the energy of one bomb: N =

3.6 × 1013 J 4.184 × 1015 J

=

8.6 × 10−3. A 1.0 megaton bomb provides more than enough energy.

7.104. Set Up: Apply Wtot = K 2 − K1 . Let x0 be the initial distance the spring is compressed. The work done by the spring is

1 kx 2 2 0

− 12 kx 2 , where x is the final distance the spring is compressed.

Solve: (a) Equating the work done by the spring to the gain in kinetic energy,

1 kx 2 2 0

= 12 mv 2 , so v =

k x0 = m

400 N / m (0.060 m) = 6.93 m/s. 0.0300 kg (b) Wtot must now include friction, so

1 mv 2 2

= Wtot = 12 kx0 2 − fx0 , where f is the magnitude of the friction force.

Then, v=

k 2 2f 400 N/m 2(6.00 N) x0 − x0 = (0.060 m) 2 − (0.060 m) = 4.90 m/s. m m 0.0300 kg (0.0300 kg)

(c) The greatest speed occurs when the acceleration (and the net force) is zero. Let x be the amount the spring is still f 6.00 N compressed, so the distance the ball has moved is x0 − x. kx = f , x = = = 0.0150 m. To find the speed, k 400 N/m

the net work is Wtot = 12 k ( x0 2 − x 2 ) − f ( x0 − x), so the maximum speed is vmax = vmax =

k 2f ( x0 2 − x 2 ) − ( x0 − x). m m

400 N/m 2(6.00 N) ((0.060 m) 2 − (0.0150 m) 2 ) − (0.060 m − 0.0150 m) = 5.20 m/s (0.0300 kg) (0.0300 kg)

Reflect: The maximum speed with friction present (part (c)) is larger than the result of part (b) but smaller than the result of part (a). *7.105. Set Up: Hooke’s law states U el = 12 kx 2 and F = kx. Solve: (a) Yes, since U is a linear function of x 2 . (b) The graph shows U el = 5.0 J when x 2 = 4.0 × 10−2 m 2 ; thus k = (c) x =

2U el x

2

=

2(5.0 J) 4.0 × 1022 m 2

= 250 N/m.

2U el 2(10.0 J) = = 0.283 m = 28.3 cm k 250 N/m

(d) F = kx = (250 N/m)(0.283 m) = 70.8 N (e) U el = 12 kx 2 ; the graph is a parabola as shown below. 7.106. Set Up: Use conservation of energy to find the speed of the sled at the edge of the cliff. Let yi = 0 so

yf = h = 11.0 m. K f + U f = Ki + U i gives

1 mυ 2 f 2

+ mgh = 12 mυ i 2 or υf = υ i 2 − 2gh . Then analyze the projectile

motion of the sled: use the vertical component of motion to find the time t that the sled is in the air; then use the horizontal component of the motion with a x = 0 to find the horizontal displacement. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7-28

Chapter 7

Solve: υf = (22.5 m/s) 2 − 2(9.80 m/s 2 )(11.0 m) = 17.1 m/s

2 yf 2( −11.0 m) = = 1.50 s ay 29.80 m/s 2

yf = υi, yt + 12 a yt 2:

t=

xf = υi, xt + 12 axt 2:

xf = υi, xt = (17.1 m/s)(1.50 s) = 25.6 m

Reflect: Conservation of energy can be used to find the speed of the sled at any point of the motion but does not specify how far the sled travels while it is in the air.


8

MOMENTUM

Answers to Multiple-Choice Problems 1. A, D 2. D 3. A, B, C 4. C, D 13. A, C, D 14. A, B 15. D

5. C

6. B

7. A

8. A, D 9. B

10. C 11. B, D

12. A, D, E

Solutions to Problems 8.1. Set Up: Use coordinates where + x is to the right. In (a) – (c) denote the two objects as A and B. In (d) call the third object C. Solve: (a) Px = p Ax + pBx = (5.0 kg)(4.0 m/s) + (3.0 kg)(8.0 m/s) = 44 kg ⋅ m/s (b) Px = (6.0 kg)(2.0 m/s) + (3.0 kg)( −4.0 m/s) = 0 (c) Px = (9.0 kg)(4.0 m/s) + (6.0 kg)( −8.0 m/s) = − 12 kg ⋅ m/s (d) Px = p Ax + pBx + pCx = (7.0 kg)(2.0 m/s) + (8.0 kg)(−5.0 m/s) + (4.0 kg)(6.0 m/s) = − 2.0 kg ⋅ m/s

8.2. Set Up: The signs of the velocity components indicate their directions. Call the objects A and B. Solve: (a) Px = p Ax = (8.0 kg)(45 m/s cos60°) = 180 kg ⋅ m/s

Py = p Ay = (8.0 kg)(45 m/s sin 60°) = 312 kg ⋅ m/s (b) Px = p Ax + pBx = 0 + (6.0 kg)(−4.0 m/s) = − 24 kg ⋅ m/s

Py = p Ay + pBy = (9.0 kg)(3.0 m/s) + 0 = 27 kg ⋅ m/s (c) Px = p Ax + pBx = 0 + (8.0 kg)(12 m/s cos72°) = 29.7 kg ⋅ m/s

Py = p Ay + pBy = (6.0 kg)( −5.0 m/s) + (8.0 kg)(12 m/s sin 72°) = 61.3 kg ⋅ m/s (d) Px = p Ax + pBx = (8.0 kg)(−5.0 m/s sin 35°) + (10.0 kg)(5.0 m/s sin 35°) = 5.7 kg ⋅ m/s

Py = p Ay + pBy = (8.0 kg)( −5.0 m/s cos35°) + (10.0 kg)( −5.0 m/s cos35°) = −73.7 kg ⋅ m/s

*8.3. Set Up: The signs of the velocity components indicate their directions. Solve: (a) Px = p Ax + pCx = 0 + (10.0 kg)(−3.0 m/s) = − 30 kg ⋅ m/s

Py = p Ay + pCy = (5.0 kg)( −11.0 m/s) + 0 = −55 kg ⋅ m/s (b) Px = pBx + pCx = (6.0 kg)(10.0 m/s cos60°) + (10.0 kg)(−3.0 m/s) = 0

Py = pBy + pCy = (6.0 kg)(10.0 m/s sin 60°) + 0 = 52 kg ⋅ m/s (c) Px = p Ax + pBx + pCx = 0 + (6.0 kg)(10.0 m/s cos60°) + (10.0 kg)( −3.0 m/s) = 0

Py = p Ay + pBy + pCy = (5.0 kg)( −11.0 m/s) + (6.0 kg)(10.0 m/s sin 60°) + 0 = − 3.0 kg ⋅ m/s


8-1

8-2

Chapter 8

Reflect: A has no x-component of momentum so Px is the same in (b) and (c). C has no y-component of momentum

so Py in (c) is the sum of Py in (a) and (b).

8.4. Set Up: υ = 65 mph = 29 m/s Solve: (a) p = mυ = (1200 kg)(29 m/s) = 3.5 × 104 kg ⋅ m/s

K = 12 mυ 2 = 12 (1200 kg)(29 m/s) 2 = 5.0 × 105 J (b) The SUV has twice the mass and therefore twice the momentum and kinetic energy: p = 7.0 × 104 kg ⋅ m/s, K = 1.0 × 106 J

*8.5. Set Up: 145 g = 0.145 kg; 57 g = 0.057 kg Solve: (a) p = mυ = (0.145 kg)(45 m/s) = 6.5 kg ⋅ m/s K = 12 mυ 2 = 12 (0.145 kg)(45 m/s) 2 = 150 J 2K = m

(b) (i) K = 12 mυ 2 so υ =

(ii) p = mυ so υ =

2(150 J) = 73 m/s 0.057 kg

p 6.5 kg ⋅ m/s = = 110 m/s m 0.057 kg

Reflect: K depends on υ 2 and p depends on υ so a smaller increase in v is required for the lighter ball to have the same kinetic energy.

8.6. Set Up: p = mυ and K = 12 mυ 2 Solve: (a) υ = (b) υ =

2

p p2 ⎛ p⎞ so K = 12 m ⎜ ⎟ = m 2m ⎝m⎠

2K 2K so p = m = m m

2mK

(c) p = 2mK = (2)(1.15 kg)(15.0 J) = 5.87 kg ⋅ m/s (d) K =

p 2 (0.220 kg ⋅ m/s) 2 = = 6.91 × 10−3 J 2m 2(3.50 kg)

8.7. Set Up: p = mυ. From Problem 8.6, p = 2mK . Solve: (a) p = mcυc . pd = mdυd . υd = υc and md = 3mc so pd = 3mcυc = 3 p (b) p =

2mc K c . pd =

2md K d . K d = K c and md = 3mc so pd = 2(3mc ) K c = 3 p

p 3 p υc p and υd = d = = . When the mass is increased the speed must be decreased to mc md 3mc 3 keep the same kinetic energy.

Reflect: In (b), υc =

8.8. Set Up: Consider the system consisting of the two skaters. m A = (625 N)/(9.80 m/s 2 ) = 63.8 kg. mB = (725 N)/(9.80 m/s 2 ) = 74.0 kg. Assume the heavier skater (B) travels to the right and take this to be the + x direction. Solve: (a) There is no horizontal external force so Pi, x = Pf, x . The skaters are initially at rest so Pi, x = 0. 0 = m A (υ A,f ) x + mB (υ B ,f ) x

(υ A,f ) x = −

mB (υ B,f ) x mA

=−

(74.0 kg)(1.50 m/s) = −1.74 m/s 63.8 kg


Momentum

8-3

The lighter skater travels to the left at 1.74 m/s. (b) Ki = 0. K f = 12 m Aυ A,f 2 + 12 mBυ B,f 2 = 12 (63.8 kg)(1.74 m/s)2 + 12 (74.0 kg)(1.50 m/s)2 = 180 J

The kinetic energy of the system was produced by the work the two skaters do on each other.

*8.9. Set Up: mE = 5.98 × 1024 kg. Consider the person and the earth to be an isolated system. Use coordinates where + y is upward, in the direction the person jumps. Solve: Pi, y = Pf, y . Pi, y = 0. The earth recoils in the − y direction with speed υ E , so 0 = mpersonυperson − mEυE .

⎛ ⎞ ⎛ mperson ⎞ 75 kg −23 ⎟⎟ (2.0 m/s) = 2.5 × 10 m/s ⎟υperson = ⎜⎜ 24 5 98 10 kg . × ⎝ mE ⎠ ⎝ ⎠

υE = ⎜

8.10. Set Up: m A = 0.150 kg. mB = 0.250 kg. Use coordinates where + x is in the direction of the initial motion of A. All motion is along the x-axis. Solve: (a) Pi, x = mA (υ A,i ) x + mB (υ B ,i ) x = (0.150 kg)(1.20 m/s) + 0 = 0.180 kg ⋅ m/s (b) There is no external force in the x direction so Pf, x = Pi, x = 0.180 kg ⋅ m/s (c) Pi, x = Pf, x gives 0.180 kg ⋅ m/s = ( mA + mB )(υf ) x .

(υf ) x =

0.180 kg ⋅ m/s = 0.450 m/s 0.150 kg + 0.250 kg

(d) Ki = 12 (0.150 kg)(1.20 m/s)2 = 0.108 J.

K f = 12 (0.150 kg + 0.250 kg)(0.450 m/s)2 = 0.0.0405 J K f < Ki ; kinetic energy is not conserved. Kinetic energy cannot be conserved in a collision where the objects stick

together.

8.11. Set Up: Use coordinates where + x is in the direction of the motion of the ball before it is caught. The figure below gives before and after sketches for the system of ball plus player. (vb,i ) x = 100 mph = 44.7 m/s

Solve: There is no external force in the x-direction, so Pi, x = Pf, x . This gives mbυ b,i = ( mb + mp )υf .

⎛ mb ⎞ ⎛ ⎞ 0.145 kg (44.7 m/s) = 9.9 cm/s ⎟υ = ⎜ mb + mp ⎟ b,i ⎝⎜ 0.145 kg + 65 kg ⎠⎟ ⎝ ⎠ Reflect: There is a large increase in the mass that is moving so to maintain constant total momentum there is a large decrease in speed.

υf = ⎜

8.12. Set Up: Let 1x be in the direction the ball is traveling initially. m A = 0.400 kg (ball). mB = 70.0 kg (you). Solve: (a) Pi, x = Pf, x gives (0.400 kg)(10.0 m/s) = (0.400 kg + 70.0 kg)υf and υf = 0.0568 m/s. (b) Pi, x = Pf, x gives (0.400 kg)(10.0 m/s) = (0.400 kg)(−8.00 m/s) + (70.0 kg)υ B ,f and υ B ,f = 0.103 m/s.

*8.13. Set Up: Let + x be to the right. Solve: (a) Pi, x = Pf, x says (0.250)υ A,i = (0.250 kg)(−0.120 m/s) + (0.350 kg)(0.650 m/s) and υ A,i = 0.790 m/s © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8-4

Chapter 8

(b) Ki = 12 (0.250 kg)(0.790 m/s)2 = 0.0780 J

K f = 12 (0.250 kg)(0.120 m/s)2 + 12 (0.350 kg)(0.650 m/s) 2 = 0.0757 J and ΔK = K f − Ki = −0.0023 J. Reflect: The total momentum of the system is conserved but the total kinetic energy decreases.

8.14. Set Up: Apply conservation of momentum and conservation of energy. Let + x be to the right. When the system is released, block A moves to the left and block B moves to the right. Solve: (a) Pi, x = Pf, x gives 0 = − (1.00 kg)υ A,f + (3.00 kg)(1.20 m/s) and υ A,f = 3.60 m/s. (b) Conservation of energy gives U el,i = K f = 12 (1.00 kg)(3.60 m/s) 2 + 12 (3.00 kg)(1.20 m/s)2 = 8.64 J.

*8.15. Set Up: Each horizontal component of momentum is conserved. K = 12 mv 2 . Let +x be the direction of Rebecca’s initial velocity and let the +y axis make an angle of 36.9D with respect to the direction of her final velocity. vD1x = vD1 y = 0. vR1x = 13.0 m/s; vR1 y = 0.

vR 2 x = (8.00 m/s)cos53.1D = 4.80 m/s; vR 2 y = (8.00 m/s)sin 53.1D =

6.40 m/s. Solve for vD2 x and vD2 y . Solve: (a) P1x = P2 x gives mR vR1x = mR vR 2 x + mDvD2 x .

vD2 x =

mR (vR1x − vR 2 x ) (45.0 kg)(13.0 m/s − 4.80 m/s) = = 5.68 m/s. 65.0 kg mD

P1 y = P2 y gives 0 = mR vR 2 y + mDvD2 y . vD2 y = 2

⎛ 45.0 kg ⎞ mR vR 2 y = 2⎜ ⎟ (6.40 m/s) = 24.43 m/s. mD ⎝ 65.0 kg ⎠

vD2 y 4.43 m/s G G G and The directions of v R1, v R 2 , and v D2 are sketched in the figure below. We have tan θ = = vD2 x 5.68 m/s

θ = 38.0°. vD = vD2 x 2 + vD2 y 2 = 7.20 m/s. (b) K1 = 12 mR vR12 = 12 (45.0 kg)(13.0 m/s)2 = 3.80 × 103 J.

K 2 = 12 mR vR 2 2 + 12 mDvD2 2 = 12 (45.0 kg)(8.00 m/s) 2 + 12 (65.0 kg)(7.20 m/s)2 = 3.12 × 103 J. ΔK = K 2 − K1 = −680 J. Reflect: Each component of momentum is separately conserved. The kinetic energy of the system decreases.

8.16. Set Up: Let + x be the direction you are moving. Before you catch it, the flour sack has no momentum along the x-axis. The total mass of you and your skateboard is 60 kg. You, the skateboard, and the flour sack are all moving with the same velocity, after the catch. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Momentum

8-5

Solve: (a) In the absence of a horizontal force, we know that Pi, x = Pf, x . Thus, we have

(60 kg)(4.5 m/s) = (62.5 kg)υf, x . Solving for the final velocity we obtain υf, x = 4.3 m/s. (b) To bring the flour sack up to your speed, you must exert a horizontal force on it. Consequently, it exerts an equal and opposite force on you, which slows you down. (c) Since you exert a vertical force on the flour sack, your horizontal speed does not change and remains at 4.3 m/s. Since the flour sack is only accelerated in the vertical direction, its horizontal velocity-component remains at 4.3 m/s as well. Reflect: Unless you or the flour sack are deflected by an outside force, you will need to be ready to catch the flour sack as it returns to your arms!

8.17. Set Up: Use coordinates where + x is in the direction of motion of the bullet. The figure below gives before and after sketches for the system of bullet and block.

Solve: There is no external force in the x direction, so Pi, x = Pf, x . This gives

mbulletυbullet,i = mbulletυbullet,f + mblockυblock,f . ⎛ 4.25 × 10−3 kg ⎞ ⎛ mbullet ⎞ ⎟⎟ (375 m/s − 122 m/s) = 0.960 m/s ⎟ (υbullet,i − υbullet,f ) = ⎜⎜ ⎝ mblock ⎠ ⎝ 1.12 kg ⎠ Reflect: The momentum lost by the bullet equals the momentum gained by the block.

υblock,f = ⎜

8.18. Set Up: Before and after diagrams are sketched in the figure below. m A = 0.600 kg, mB = 0.400 kg. υ A,i, x = 0, υ B ,i, x = 0.250 m/s. υ B ,f = 0.200 m/s, θ = 36.9°.

Solve: (a) Conservation of momentum gives Pi, x = Pf, x and Pi, y = Pf, y . Pi, x = Pf, x says mBυ B ,i, x = mBυ B ,f, x + m Aυ A,f, x . (0.400 kg)(0.250 m/s) = (0.400 kg)(0.200 m/s)cos36.9° + (0.600 kg)υ A,f, x and υ A,f, x = 0.0600 m/s

Pi, y = Pf, y says 0 = mBυ B ,f, y + m Aυ A,f, y 0 = (0.400 kg)(0.200 m/s)sin 36.9° + (0.600 kg)υ A,f, y and υ A,f, y = −0.0800 m/s

υ A,f = (υ A,f, x ) 2 + (υ A,f, y ) 2 = 0.100 m/s. tan φ =

0.0800 m/s and φ = 53.1° 0.0600 m/s

(b) Ki = 12 (0.400 kg)(0.250 m/s)2 = 0.0125 J

K f = 12 (0.400 kg)(0.200 m/s)2 + 12 (0.600 kg)(0.100 m/s)2 = 0.0110 J ΔK = K f − Ki = −0.0015 J © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8-6

Chapter 8

*8.19. Set Up: Use coordinates where + x is to the right and + y is upward. Solve: Collision: There is no external horizontal force during the collision so

Pi, x = Pf, x . This gives

(5.00 kg)(12.0 m/s) = (10.0 kg)υf and υf = 6.0 m/s. Motion after the collision: Only gravity does work and the initial kinetic energy of the combined chunks is converted entirely to gravitational potential energy when the chunk reaches its maximum height h above the valley floor. Conservation of energy gives 2 2 1 m υ 2 = m gh and h = υ = (6.0 m/s) = 1.8 m tot 2 tot 2 g 2(9.8 m/s 2 ) Reflect: After the collision the energy of the system is

1 m υ2 2 tot

= 12 (10.0 kg)(6.0 m/s) 2 = 180 J when it is all kinetic

energy and it is mtot gh = (10.0 kg)(9.8 m/s 2 )(1.8 m) = 180 J when it is all gravitational potential energy. Mechanical energy is conserved during the motion after the collision. But before the collision the total energy of the system is 1 (5.0 kg)(12.0 m/s) 2 = 360 J; 50% of the mechanical energy is dissipated during the inelastic collision of the two chunks. 2

8.20. Set Up: The collision occurs over a short time interval and the block moves very little during the collision, so the spring force during the collision can be neglected. Use coordinates where 1x is to the right. Solve: Collision: There is no external horizontal force during the collision and Pi, x = Pf, x . This gives

(3.00 kg)(8.00 m/s) = (15.0 kg)υblock,f − (3.00 kg)(2.00 m/s) and υblock,f = 2.00 m/s.

Motion after the collision: When the spring has been compressed the maximum amount, all the initial kinetic energy of the block has been converted into potential energy 12 kd 2 that is stored in the compressed spring. Conservation of energy gives

1 (15.0 2

kg)(2.00 m/s) 2 = 12 (500.0 kg) d 2 and d = 0.346 m.

*8.21. Set Up: Let x be the direction of motion. Let each boxcar have mass m. Solve: (a) Pi, x = Pf, x says (3m)(20.0 m/s) = (4m)υf, x and υf, x = 15.0 m/s. (b) Ki = 12 (3m)(20.0 m/s)2 = 600m J/kg; K f = 12 (4m)(15.0 m/s) 2 = 450m J/kg

ΔK −150m J/kg = = −0.250; 600m J/kg Ki 25% of the original kinetic energy is dissipated. Kinetic energy is converted to other forms by work done by the forces during the collision. ΔK = −150m J/kg and

8.22. Set Up: Let + x be to the right. Solve: (a) Pi, x = Pf, x gives (0.500 kg)(−3.00 m/s) + (0.250 kg)(1.20 m/s) = (0.750 kg)υf, x and υf, x = −1.60 m/s.

The final velocity is 1.60 m/s, to the left. (b) Ki = 12 (0.500 kg)(3.00 m/s)2 + 12 (0.250 kg)(1.20 m/s)2 = 2.43 J. K f = 12 (0.750 kg)(1.60 m/s)2 = 0.96 J.

ΔK = K f − Ki = −1.47 J. Work done by nonconservative forces during the collision converts mechanical energy to

other forms.

8.23. Set Up: There is no net external force on the system of astronaut plus canister, so the momentum of the system is conserved. Let object A be the astronaut and object B be the canister. Assume the astronaut is initially at rest. After the collision she must be moving in the same direction as the canister. Let +x be the direction in which the canister is traveling initially, so v A1x = 0, v A2 x = +2.40 m/s, vB1x = +3.50 m/s, and vB 2 x = +1.20 m/s. Solve for mB . Solve: (a) P1x = P2 x . m Av A1x + mB vB1x = m Av A2 x + mB vB 2 x . mB =

m A (v A2 x − v A1x ) (78.4 kg)(2.40 m/s − 0) = = 81.8 kg. 3.50 m/s − 1.20 m/s vB1x − vB 2 x


Momentum

8-7

(b) We can now find the initial and final kinetic energy of the system: we have 1 Ki = (81.8 kg)(3.50 m/s) 2 = 501 J and 2 1 1 K f = (81.8 kg)(1.20 m/s) 2 + (78.4 kg)(2.40 m/s) 2 = 285 J. Thus, we obtain 2 2 ΔK = K f − Ki = 285 J − 501 J = 2216 J. Reflect: She must exert a force on the canister in the −x direction to reduce its velocity component in the +x direction. By Newton’s third law, the canister exerts a force on her that is in the +x direction and she gains velocity in that direction.

8.24. Set Up: The forces the two players exert on each other during the collision are much larger than the horizontal forces exerted by the slippery ground and it is a good approximation to assume momentum conservation. Each component of momentum is separately conserved. Let +x be east and +y be north. After the collision the two G players have velocity v 2 . Let the linebacker be object A and the halfback be object B, so v A1x = 0, v A1 y = 8.8 m/s, vB1x = 7.2 m/s, and vB1 y = 0. Solve for v2 x and v2 y .

Solve: P1x = P2 x gives m Av A1x + mB vB1x = (m A + mB )v2 x . v2 x =

m Av A1x + mB vB1x (85 kg)(7.2 m/s) = = 3.14 m/s. m A + mB 110 kg + 85 kg

P1 y = P2 y gives m Av A1 y + mB vB1 y = ( m A + mB )v2 y . v2 y =

m Av A1 y + mB vB1 y m A + mB

=

(110 kg)(8.8 m/s) = 4.96 m/s. 110 kg + 85 kg

v = v2 x 2 + v2 y 2 = 5.9 m/s. tan θ =

v2 y v2 x

=

4.96 m/s and θ = 58D. 3.14 m/s

The players move with a speed of 5.9 m/s and in a direction 58D north of east. Reflect: Each component of momentum is separately conserved.

*8.25. Set Up: Apply conservation of momentum to the collision and the work-energy equation to the motion after the collision. Let + x be the direction the bullet was traveling before the collision. Solve: Motion after the collision: Wother + U i + Ki = K f + U f . There is no change in height so U i = U f . The block stops so K f = 0. n = mg , f k = μk mg and Wother = W f = 2μk mgs. Let V be the speed of the block immediately after the collision, so Ki =

1 mV 2 . 2

Therefore, −μk mgs + 12 mV 2 = 0 and

V = 2μ k gs = 2(0.20)(9.80 m/s 2 )(0.230 m) = 0.950 m/s. Collision: mbullet vbullet, i = mV , where m = 1.20 kg + 5.00 × 10−3 kg = 1.205 kg.

υ bullet,i =

(1.205 kg)(0.950 m/s)

= 229 m/s 5.00 × 10−3 kg Reflect: Our analysis neglects the friction force exerted by the surface on the block during the collision, since we assume momentum is conserved in the collision. The collision forces between the bullet and block are much larger than the friction force, so this approximation is justified.

8.26. Set Up: Each component of horizontal momentum is conserved. Let +x be east and +y be north. We know that vS1 y = vA1x = 0, vS 2 x = (6.00 m/s) cos 37.0° = 4.79 m/s, vS 2 y = (6.00 m/s) sin 37.0° = 3.61 m/s, v A2 x = (9.00 m/s) cos 23.0° = 8.28 m/s and v A2 y = − (9.00 m/s)sin 23.0D = −3.52 m/s.

Solve: (a) P1x = P2 x gives mSvS1x = mSvS2 x + mA vA2 x . vS1x =

mSvS2 x + mA vA2 x (80.0 kg)(4.79 m/s) + (50.0 kg)(8.28 m/s) = = 9.97 m/s. mS 80.0 kg


8-8

Chapter 8

Sam’s speed before the collision was 9.97 m/s. P1 y = P2 y gives mA vA1 y = mSvS2y + mA vA2 y . vA1 y =

mSvS2y + mA vA2 y mS

=

(80.0 kg)(3.61 m/s) + (50.0 kg)(−3.52 m/s) = 2.26 m/s. 50.0 kg

Abigail’s speed before the collision was 2.26 m/s. (b) ΔK = 12 (80.0 kg)(6.00 m/s)2 + 12 (50.0 kg)(9.00 m/s) 2 − 12 (80.0 kg)(9.97 m/s)2 − 12 (50.0 kg)(2.26 m/s) 2 . ΔK = 2639 J. Reflect: The total momentum is conserved because there is no net external horizontal force. The kinetic energy decreases because the forces between the objects do negative work during the collision.

8.27. Set Up: Use coordinates where +x is east and +y is south. Let the big fish be A and the small fish be B. The system of the two fish before and after the collision is sketched in the figure below.

Solve: There are no external forces on the fish so Pi, x = Pf, x and Pi, y = Pf, y . Pi, x = Pf, x gives (11.5 kg)(0.750 m/s) = (12.75 kg)υf, x so υf, x = 0.676 m/s

Pi, y = Pf, y gives (1.25 kg)(3.60 m/s) = (12.75 kg)υf, y so υf, y = 0.353 m/s

υ = υf, x 2 + υf, y 2 = 0.763 m/s. υf, x = υf cos φ so cos φ =

0.676 m/s and φ = 27.6°. The large fish has velocity 0.763 m/s

0.763 m/s in a direction 27.6° south of east. Reflect: Momentum is a vector and we must treat each component separately.

8.28. Set Up: The system before and after the collision is sketched in the figure below. Use the coordinates shown. The collision forces are large so gravity can be neglected during the collision.

Solve: There is no external force on the system so Pi, x = Pf, x and Pi, y = Pf, y . Pi, x = Pf, x gives (1.5 kg)(9.0 m/s) = (1.5 kg)υ raven,f cos φ and υ raven,f cos φ = 9.0 m/s. Pi, y = Pf, y gives (0.600 kg)(20.0 m/s) = (0.600 kg)(−5.0 m/s) + (1.5 kg)υ raven,f sin φ and υ raven,f sin φ = 10.0 m/s. Combining these two equations gives tan φ =

10.0 m/s and φ = 48°. 9.0 m/s


Momentum

8-9

*8.29. Set Up: Use coordinates where +x is east and +y is south. The system of two cars before and after the collision is sketched in the figure below. Neglect friction from the road during the collision. The enmeshed cars have mass 2000 kg + 1500 kg = 3500 kg.

Solve: There are no external horizontal forces during the collision, so Pi, x = Pf, x and Pi, y = Pf, y . (a) Pi, x = Pf, x gives (1500 kg)(15 m/s) = (3500 kg)υf sin 65° and υf = 7.1 m/s (b) Pi, y = Pf, y gives (2000 kg)υ A,i = (3500 kg)υf cos65°. And then with υf = 7.1 m/s, υ A,i = 5.2 m/s. Reflect: Momentum is a vector and we must treat each component separately.

8.30. Set Up: The before and after diagrams are given in the figure below and the choice of coordinates is indicated. Each puck has mass m.

Solve: (a) Pi, x = Pf, x gives mv A,i = mυ A,f cos30.0° + mυ B ,f cos 45.0°. 40.0 m/s = 0.866υ A,f + 0.707υ B ,f and 0.707υ B ,f = 40.0 m/s − 0.866υ A,f . Pi, y = Pf, y gives 0 = mυ A,f sin 30.0° − mυ B,f sin 45.0° and 0.500υ A,f = 0.707υ B,f . Combining these two equations gives 0.500υ A,f = 40.0 m/s − 0.866υ A,f and υ A,f = 29.3 m/s. Then

⎛ 0.500 ⎞

(29.3 m/s) = 20.7 m/s. υ B ,f = ⎜ ⎝ 0.707 ⎟⎠

(b) Ki = 12 mυ A,i 2 . K f = 12 mυ A,f 2 + 12 mυ B ,f 2 . 2 2 K f υ A,f + υ B,f (29.3 m/s) 2 + (20.7 m/s) 2 = = = 0.804. Ki υ A,i 2 (40.0 m/s) 2

ΔK K f − Ki K f = = − 1 = −0.196. Ki Ki Ki 19.6% of the original kinetic energy is dissipated during the collision.

*8.31. Set Up: For an elastic collision with B initially stationary, the final velocities are ⎛ m A − mB ⎞ ⎛ 2m A ⎞ ⎟υ and υ B = ⎜ ⎟υ . ⎝ m A + mB ⎠ ⎝ m A + mB ⎠

υA = ⎜


8-10

Chapter 8

Apply these equations with m A = 0.300 kg, mB = 0.150 kg, and υ = 0.80 m/s.

⎛ 0.300 kg − 0.150 kg ⎞ Solve: (a) υ A = ⎜ ⎟ (0.80 m/s) = 0.27 m/s, ⎝ 0.300 kg + 0.150 kg ⎠ ⎛

⎞ 2[0.300 kg] ⎟ (0.80 m/s) = 1.07 m/s. ⎝ 0.300 kg + 0.150 kg ⎠

υB = ⎜

The 0.300 kg glider moves to the right at 0.27 m/s and the 0.150 kg glider moves to the right at 1.07 m/s.

(b) K A,f = 12 m Aυ A2 = 12 (0.300 kg)(0.27 m/s)2 = 0.011 J;

K B ,f = 12 mBυ B 2 = 12 (0.150 kg)(1.07 m/s)2 = 0.086 J Reflect: The relative velocity after the collision has magnitude 1.07 m/s − 0.27 m/s = 0.80 m/s, the same as before the collision. The initial kinetic energy of the system is K A,i = 12 (0.300 kg)(0.80 m/s) 2 = 0.096 J. The final kinetic energy is K A,f + K B ,f = 0.097 J, the same as the initial kinetic energy, apart from a slight difference due to rounding.

8.32. Set Up: Take the original direction of the penny (object A) as the positive direction. Since this is a onedimensional elastic collision, with the nickel (object B) initially stationary, we can use equations 8.11 and 8.12. The ⎛ m − mB ⎞ ⎛ 2m A ⎞ final velocities are given by υ A = ⎜ A ⎟υ and υ B = ⎜ ⎟υ . + m m B⎠ ⎝ A ⎝ m A + mB ⎠ Apply these equations with m A = 2.50 × 10−3 kg, mB = 5.00 × 1023 kg, and υ = 2.20 m/s.

⎛ 2.50 × 10−3 kg − 5.00 × 10−3 kg ⎞ Solve: υ A = ⎜ (2.20 m/s) = −0.733 m/s ⎜ 2.50 × 10−3 kg + 5.00 × 10−3 kg ⎟⎟ ⎝ ⎠ ⎡ ⎤ 2(2.50 × 10−3 kg) υB = ⎢ ⎥ (2.20 m/s) = + 1.47 m/s −3 −3 ⎢⎣ 2.50 × 10 kg + 5.00 × 10 kg ⎥⎦ Reflect: Since the equations we used involved a ratio of masses, there was no need to convert the masses into kilograms.

*8.33. Set Up: For an elastic collision with B initially stationary, the final velocities are ⎛ mA − mB ⎞ ⎛ 2m A ⎞ ⎟υ and υ B = ⎜ ⎟υ . + m m B⎠ ⎝ A ⎝ mA + mB ⎠

υA = ⎜

Apply these equations with m A = 1.67 × 10227 kg, mB = 6.65 × 10−27 kg, and υ = 258 km/s.

⎛ 1.67 × 10227 kg − 6.65 × 10227 kg ⎞ Solve: (a) υ A = ⎜ (258 km/s) = 2154 km/s ⎜ 1.67 × 10227 kg + 6.65 × 10227 kg ⎟⎟ ⎝ ⎠ ⎛ ⎞ 2[1.67 × 10227 kg] υB = ⎜ (258 km/s) = 104 km/s ⎜ 1.67 × 10227 kg + 6.65 × 10227 kg ⎟⎟ ⎝ ⎠ The proton recoils to the left at 154 km/s and the alpha particle travels to the right at 104 km/s. (b) The proton has initial kinetic energy

K A, i = 12 m Aυ 2 = 12 (1.67 × 10227 kg)(2.58 × 105 m/s) 2 = 5.56 × 10217 J and final kinetic energy K A,f = 12 m Aυ A2 = 12 (1.67 × 10−27 kg)(1.54 × 105 m/s) 2 = 1.98 × 10−17 J. The kinetic energy lost is K A,i − K A,f = 3.58 × 10−17 J.

(c) The kinetic energy gained by the alpha particle is 1 (6.65 × 10227 kg)(1.04 × 105 m/s) 2 2

= 3.60 × 10217 J.


Momentum

8-11

The energy lost by the proton is gained by the alpha particle. The total kinetic energy of the system is constant and the collision is elastic.

8.34. Set Up: Let +x be to the right. All velocities are x components but we will not write the x subscript. Use conservation of momentum and the relative velocity addition equation υ B ,f − υ A,f = − (υ B ,i − υ A,i ). m A = 0.400 kg and mB = 0.500 kg.

Solve: From conservation of momentum, (0.400 kg)(2.00 m/s) + (0.500 kg)(−3.00 m/s) = (0.400 kg)υ A,f + (0.500 kg)υ B,f 0.800υ A,f + υ B ,f = −1.40 m/s From the relative velocity equation, υ B , f − υ A, f = − (−3.00 m/s − 2.00 m/s) = 5.00 m/s. Combining these two equations gives 0.800(υ B ,f − 5.00 m/s) + υ B,f = −1.40 m/s and υ B ,f = 1.44 m/s. Then υ A,f = −3.55 m/s. the maximum amount the two blocks aren’t moving relative to each 8.35. Set Up: When the spring is compressed G

other and have the same velocity V relative to the surface. Apply conservation of momentum to find V and conservation of energy to find the energy stored in the spring. Since the collision is elastic, equations 8.11 and 8.12 give the final velocity of each block after the collision. Let +x be the direction of the initial motion of A. Solve: (a) Momentum conservation gives (2.00 kg) (2.00 m/s) = (12.0 kg)V and V = 0.333 m/s. Both blocks are moving at 0.333 m/s, in the direction of the initial motion of block A. Conservation of energy says the initial kinetic energy of A equals the total kinetic energy at maximum compression plus the potential energy U b stored in the

bumpers:

1 (2.00 2

kg)(2.00 m/s) 2 = U b + 12 (12.0 kg)(0.333 m/s) 2 and U b = 3.33 J.

⎛ m − mB ⎞ ⎛ 2.00 kg − 10.0 kg ⎞ (b) v A, f = ⎜ A ⎟ v A, i = ⎜ ⎟ (2.00 m/s) = −1.33 m/s. Block A is moving in the −x direction at 12.0 kg ⎝ ⎠ ⎝ mA + mB ⎠ 1.33 m/s. ⎛ 2m A ⎞ 2(2.00 kg) vB , f = ⎜ (2.00 m/s) = +0.667 m/s. Block B is moving in the +x direction at 0.667 m/s. ⎟ v A, i = 12.0 kg ⎝ mA + mB ⎠ Reflect: When the spring is compressed the maximum amount the system must still be moving in order to conserve momentum.

8.36. Set Up: m A = mB . Let one object be traveling in the + x direction; then the other is initially traveling in the − x direction, so υ A,i = +V and υ B ,i = −V . The initial momentum of the system is zero. Solve: (a) Pi, x = Pf, x gives 0 = (m A + mB )υf and υf = 0. The combined object is at rest after the collision. (b) Pi, x = Pf, x gives 0 = m Aυ A,f + mBυ B ,f and υ A,f + υ B ,f = 0

The relative velocity equation gives υ B ,f − υ A,f = − (υ B ,i − υ A,i ) = +2V . Adding these two equations gives 2υ B , f = + 2V and υ B ,f = V . Then υ A,f = −V . Each object rebounds in the opposite direction with unchanged speed.

*8.37. Set Up: Assume the ball is initially moving to the right, and let this be the + x direction. The ball stops, so its final velocity is zero. Solve: (a) J x = mυf, x − mυi, x = 0 − (0.145 kg)(36.0 m/s) = −5.22 kg ⋅ m/s. The magnitude of the impulse applied to the ball is 5.22 kg ⋅ m/s. (b) J x = Fx Δt so Fx =

J x 25.22 kg ⋅ m/s = = 2261 N Δt 20.0 × 1023 s

Reflect: The signs of J x and Fx show that both these quantities are to the left.

8.38. Set Up: Let + x be to the right. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8-12

Chapter 8

Solve: (a) Fx Δt = Pf, x − Pi, x .

Pf, x = Pi, x + Fx Δt = (2.50 kg)(8.00 m/s) + (5.00 N)(5.00 s) = 45.0 kg ⋅ m/s. Pf, x = mυf, x so υf, x =

45.0 kg ⋅ m/s = 18.0 m/s. The block is moving to the right at 18.0 m/s. 2.50 kg

(b) Pf, x = Pi, x + Fx Δt = (2.50 kg)(8.00 m/s) + ( −7.00 N)(5.00 s) = −15.0 kg ⋅ m/s.

υf, x =

−15.0 kg ⋅ m/s = − 6.00 m/s. 2.50 kg

The block is moving to the left at 6.00 m/s.

*8.39. Set Up: (a) Take the + x direction to be along the final direction of motion of the ball. The initial speed of the ball is zero. (b) Take the + x direction to be in the direction the ball is traveling before it is hit by the opponent. Solve: (a) J x = mυf, x − mυi, x = (57 × 10−3 kg)(73.14 m/s − 0) = 4.2 kg ⋅ m/s Fx =

Jx 4.2 kg ⋅ m/s = = 140 N Δt 30.0 × 10−3 s

(b) J x = mυf, x − mυi, x = (57 × 10−3 kg)(−55 m/s − 73.14 m/s) = −7.3 kg ⋅ m/s Fx =

J x −7.3 kg ⋅ m/s = = − 240 N Δt 30.0 × 10−3 s

Reflect: The signs of J x and Fx show their direction. 140 N = 31 lb. This very attainable force has a large effect

on the light ball. 140 N is 250 times the weight of the ball.

8.40. Set Up: Take + y to be upward. Use the motion of the ball after it leaves the racket to find its speed just after it is hit. After it leaves the racket a y = − g . At the maximum height υ y = 0. Solve: υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

υ0 y = 22a y ( y − y0 ) = 22(29.80 m/s 2 )(5.50 m) = 10.4 m/s For the interaction with the racket υi, y = 0 and υf, y = 10.4 m/s.

J y = mυf, y − mυi, y = (57 × 10−3 kg)(10.4 m/s − 0) = 0.593 kg ⋅ m/s

8.41. Set Up: The impulse is the area under the Fx versus t curve between ti and tf . Let the + x direction be to the left, so Fx is positive. Solve: (a) J x = area = Fx Δt = (4.0 × 103 N)(5.0 × 10−3 s − 2.0 × 10−3 s) = 12.0 N ⋅ s

J x is positive so the impulse is to the left. (b) J x = mυf, x − mυi, x

J x 12.0 N ⋅ s = = 80 m/s. m 0.150 kg The final velocity of the ball is 80 m/s, to the left. J (ii) υ i, x = −30.0 m/s, so υf, x = υi, x + x = − 30.0 m/s + 80.0 m/s = 50.0 m/s m The final velocity of the ball is 50 m/s, to the left. (i) υ i, x = 0, so υf, x =

Reflect: In both (i) and (ii) of part (b) the change in velocity is the same, 80 m/s. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Momentum

8-13

8.42. Set Up: Choose the positive x-axis in the direction of motion. We know that the impulse applied to the wagon will be equal to the wagon’s change in momentum: J x = Δpx . The total mass of your sister and her wagon is 33.5 kg. Solve: (a) Δpx = mυf, x − mυi, x = (33.5 kg)(1.80 m/s) − 0 = 60.3 kg ⋅ m/s. Thus, we have J x = 60.3 kg ⋅ m/s = 60.3 N ⋅ s. (b) Since J x = Fxt we have Fx =

J x 60.3 N ⋅ s = = 25.7 N. t 2.35 s

Reflect: The force calculated in part (b) is the net force acting on the wagon—which is the force that you must apply, assuming that no other forces act on the wagon.

*8.43. Set Up: With A = 1.5 × 1024 m 2 the maximum force without breaking the bone is (1.5 × 1024 m 2 )(1 × 108 N/m 2 ) = 1.5 × 104 N. Set the magnitude of the average force Fav during the collision equal to this value. Use coordinates where + x is in his initial direction of motion. Fx is opposite to this direction, so Fx = −1.5 × 104 N. Solve: J x = Fx Δt = (−1.5 × 104 N)(10.0 × 10−3 s) = − 150.0 N ⋅ s

J x = mυf, x − mυi, x and υf, x = 0.

υi, x = −

Jx −150 N ⋅ s =− = 2.1 m/s m 70 kg

8.44. Set Up: Let 1x be to the right and + y be upward. Solve: (a) J x = Δpx = mυf, x − mυi, x = (0.145 kg)( −[65.0 m/s]cos30° − 50.0 m/s) = −15.4 kg ⋅ m/s J y = Δp y = mυf, y − mυi, y = (0.145 kg)([65.0 m/s]sin30° − 0) = 4.71 kg ⋅ m/s The horizontal component is 15.4 kg ⋅ m/s, to the left and the vertical component is 4.71 kg ⋅ m/s, upward.

J y 4.71 kg ⋅ m/s J x −15.4 kg ⋅ m/s = = 2690 N = = − 8800 N. F y = − 3 Δt 1.75 × 10−3 s Δt 1.75 × 10 s The horizontal component is 8800 N, to the left, and the vertical component is 2690 N, upward.

(b) F x =

*8.45. Set Up: From the appendix we have mE = 5.97 × 1024 kg, mM = 7.35 × 1022 kg, and the distance between the earth and the moon is 3.84 × 108 m. Represent the earth and the moon as point masses located at their respective centers. Use coordinates with the origin at the center of the earth and with the moon on the + x axis; thus,

xE = 0 and xM = 3.84 × 108 m. Solve: xcm =

mE xE + mM xM (5.97 × 1024 kg)(0) + (7.35 × 1022 kg)(3.84 × 108 m) = = 4.67 × 106 m. mE + mM (5.97 × 1024 kg + 7.35 × 1022 kg)

The center of mass is 4.67 × 106 m from the center of the earth along the line that connects the center of the earth to that of the moon. Reflect: Since the radius of the earth is 6.38 × 106 m, the center of mass of the earth-moon system is roughly 1700 km below the earth’s surface.

8.46. Set Up: Since the beam is uniform, all its mass can be considered to be at its center. The mass of the beam is mb = 9.00 kg. Let m A = 8.00 kg and mC = 7.50 kg. For each choice of origin let the + x axis lie along the beam. Solve: (a) xcm =

m A x A + mC xC + mb xb . x A = 0, xC = 3.00 m − 0.60 m = 2.40 m, and xb = 1.50 m. mA + mC + mb

(7.50 kg)(2.40 m) + (9.00 kg)(1.50 m) = 1.29 m. 8.00 kg + 7.50 kg + 9.00 kg The center of mass is 1.29 m from the left-hand end of the beam.

xcm =


8-14

Chapter 8

(b) Now x A = 3.00 m, xC = 0.600 m, and xb = 1.50 m.

(8.00 kg)(3.00 m) + (7.50 kg)(0.600 m) + (9.00 kg)(1.50 m) = 1.71 m 8.00 kg + 7.50 kg + 9.00 kg The center of mass is 1.71 m from the right-hand end of the beam and therefore 1.29 m from the left-hand end. (c) We get the same result for either choice of coordinates.

xcm =

8.47. Set Up: Assume that the center of mass of the star and of the planet are located at their center, so all their mass can be considered as concentrated there. Use coordinates with the origin at the center of the star and with the + x

axis toward the planet. Our sun has mass mS = 1.99 × 1030 kg, Jupiter has mass mJ = 1.90 × 1027 kg, and earth has mass mE = 5.97 × 1024 kg. The orbit radius of Jupiter is rJ = 7.78 × 1011 m and the earth has orbit radius

rE = 1.50 × 1011 m. Solve: (a) xcm =

mS xS + (5mJ ) xJ 0 + 5(1.90 × 1027 kg)(7.78 × 1011 m) = = 3.70 × 109 m mS + 5mJ 1.99 × 1030 kg + 5(1.90 × 1027 kg)

The center of mass is 3.70 × 109 m from the center of the star. (b) xcm =

mS xS + mE xE 0 + (5.97 × 1024 kg)(1.50 × 1011 m) = = 4.47 × 105 m mS + mE 1.99 × 1030 kg + 5.97 × 1024 kg

The center of mass is 4.47 × 105 m from the center of the star. (c) For a large planet the star moves in an orbit of greater radius.

8.48. Set Up: m A = 0.300 kg, mB = 0.400 kg, mC = 0.200 kg Solve:

xcm = xcm =

(0.300 kg)(0.200 m) + (0.400 kg)(0.100 m) + (0.200 kg)(−0.300 m) = 0.0444 m 0.300 kg + 0.400 kg + 0.200 kg

ycm = ycm =

m A x A + mB xB + mC xC m A + mB + mC

m A y A + mB yB + mC yC m A + mB + mC

(0.300 kg)(0.300 m) + (0.400 kg)(−0.400 m) + (0.200 kg)(0.600 m) = 0.0556 m 0.300 kg + 0.400 kg + 0.200 kg

*8.49. Set Up: Use Eq. 8.18 to find the x and y coordinates of the center of mass of the machine part for each configuration of the part. In calculating the center of mass of the machine part, each uniform bar can be represented by a point mass at its geometrical center. Use coordinates with the axis at the hinge and the +x and +y axes along the horizontal and vertical bars in the figure in the problem. Let ( xi , yi ) and ( xf , yf ) be the coordinates of the bar before and after the vertical bar is pivoted. Let object 1 be the horizontal bar, object 2 be the vertical bar, and 3 be the ball. m x + m2 x2 + m3 x3 (4.00 kg)(0.750 m) + 0 + 0 Solve: xi = 1 1 = = 0.333 m. 4.00 kg + 3.00 kg + 2.00 kg m1 + m2 + m3

yi =

m1 y1 + m2 y2 + m3 y3 0 + (3.00 kg)(0.900 m) + (2.00 kg)(1.80 m) = = 0.700 m. m1 + m2 + m3 9.00 kg

xf =

(4.00 kg)(0.750 m) + (3.00 kg)(−0.900 m) + (2.00 kg)(−1.80 m) = 20.366 m. 9.00 kg

yf = 0. xf − xi = −0.700 m and yf − yi = −0.700 m. The center of mass moves 0.700 m to the right and 0.700 m upward. Reflect: The vertical bar moves upward and to the right so it is sensible for the center of mass of the machine part to move in these directions.


Momentum

8-15

8.50. Set Up: The leg in each position is sketched in Figure (a) and (b) below. Use the coordinates shown. The mass of each part of the leg may be taken as concentrated at the center of that part.

Solve: (a) xcm = 0,

ycm =

(23.0 cm)(8.60 kg) + (69.0 cm)(5.25 kg) = 40.4 cm 8.60 kg + 5.25 kg

The center of mass of the leg is 40.4 cm below the hip. (b) xcm =

(23.0 cm)(8.60 kg) + (46.0 cm)(5.25 kg) = 31.7 cm 8.60 kg + 5.25 kg

0 + (23.0 cm)(5.25 kg) = 8.7 cm. 8.60 kg + 5.25 kg The center of mass is a vertical distance of 8.7 cm below the hip and a horizontal distance of 31.7 cm from the hip.

ycm =

*8.51. Set Up: m A = 1200 kg, mB = 1800 kg. M = m A + mB = 3000 kg. Let + x be to the right and let the origin be at the center of mass of the station wagon. m x + mB xB 0 + (1800 kg)(40.0 m) = = 24.0 m Solve: (a) xcm = A A mA + mB 1200 kg + 1800 kg The center of mass is between the two cars, 24.0 m to the right of the station wagon and 16.0 m behind the lead car. (b) Px = mAυ A,i + mBυ B,i = (1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s) = 5.04 × 104 kg ⋅ m/s (c) υcm, x =

m Aυ A, x + mBυ B , x m A + mB

=

(1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s) = 16.8 m/s 1200 kg + 1800 kg

(d) Px = M υcm, x = (3000 kg)(16.8 m/s) = 5.04 × 104 kg ⋅ m/s, the same as in part (b).

8.52. Set Up: Let + x be to the right, with the origin at the initial position of the left-hand end of the canoe. m A = 45.0 kg, mB = 60.0 kg. The center of mass of the canoe is at its center. Solve: There is no net horizontal external force so υcm is constant. Initially, υcm = 0, so the center of mass doesn’t

move. Initially, xcm,i =

After she walks, xcm,f =

m A x A, i + mB xB ,i m A + mB m A x A,f + mB xB ,f m A + mB

.

.


8-16

Chapter 8

xcm,i = xcm,f gives m A x A,i + mB xB,i = m A x A,f + mB xB ,f . She walks to a point 1.00 m from the right-hand end of the canoe, so she is 1.50 m to the right of the center of mass of the canoe and x A,f = xB ,f + 1.50 m. (45.0 kg)(1.00 m) + (60.0 kg)(2.50 m) = (45.0 kg)( xB,f + 1.50 m) + (60.0 kg) xB ,f

(105.0 kg) xB ,f = 127.5 kg ⋅ m and xB,f = 1.21 m. xB ,f − xB ,i = 1.21 m − 2.50 m = − 1.29 m. The canoe moves 1.29 m to the left.

*8.53. Set Up: Use Eq. 8.25. Δm −0.0500 kg = − (1600 m/s) = + 80.0 N Δt 1.00 s (b) The absence of atmosphere would not prevent the rocket from operating. The rocket could be steered by ejecting the gas in a direction with a component perpendicular to the rocket’s velocity and braked by ejecting it in a direction parallel (as opposed to antiparallel) to the rocket’s velocity.

Solve: (a) F = − υex

8.54. Set Up: Use Eq. 8.26. Solve for Δm, with Δt = 1.00 s. υex Δm

Solve: a = 2

m Δt

.

Δm = −

maΔt

υex

=−

(6000 kg)(25.0 m/s 2 )(1.00 s) = − 75.0 kg. 2000 m/s

A mass of 75.0 kg must be ejected.

8.55. Set Up: Use Eq. 8.26. Solve: a = −

υex Δm m Δt

Δm m =− . Solve for υex . Δt 160

.

−15.0 m/s 2 a = = 2.40 × 103 m/s = 2.40 km/s ⎛ Δm ⎞ ⎛ m ⎞ /m⎟ ⎜ − ⎜⎝ ⎟ /m Δt ⎠ ⎝ 160 ⎠ Reflect: The acceleration is proportional to the speed of the exhaust gas and to the rate at which mass is ejected.

υex = −

8.56. Set Up: υex = 1200 m/s and Δm /Δt = − 135 kg/s. The maximum acceleration of the rocket occurs when the mass m is the smallest, just before the fuel runs out. Δm = − (1200 m/s)(−135 kg/s) = + 1.62 × 105 N Solve: (a) The thrust is F = − υex Δt υ Δm . Initially m = 75,000 kg and (b) a = − ex m Δt

⎛ 1200 m/s ⎞ 2 ainitial = − ⎜ ⎟ ( −135 kg/s) = 2.16 m/s 75,000 kg ⎝ ⎠ The maximum acceleration occurs when m = 25,000 kg so ⎛ 1200 m/s ⎞ 2 amax = − ⎜ ⎟ ( −135 kg/s) = 6.48 m/s ⎝ 25,000 kg ⎠ (c) After the fuel runs out the thrust suddenly becomes zero and the acceleration suddenly becomes zero. (d) The acceleration becomes zero so the speed is constant, with the value it had just before the fuel ran out. Δm Δm . Assume that is constant over the 5.0 s interval and that m Δt Δt Δm doesn’t change much during that interval. The thrust is F = − vex . Take m to have the constant value Δt

*8.57. Set Up: Use equation 8.24: ma = −vex


Momentum

110 kg + 70 kg = 180 kg. Note that the rate of change in mass,

8-17

Δm , is negative since the mass of the MMU Δt

decreases as gas is ejected. Δm m ⎛ 180 kg ⎞ (0.029 m/s 2 ) = −0.0106 kg/s. In 5.0 s the mass that is ejected is (0.0106 kg/s) Solve: (a) =− a = −⎜ ⎝ 490 m/s ⎟⎠ Δt vex (5.0 s) = 0.053 kg.

Δm = − (490 m/s)(−0.0106 kg/s) = 5.19 N. Δt Reflect: The mass change in the 5.0 s is a very small fraction of the total mass m, so it is accurate to take m to be constant. (b) F = − vex

8.58. Set Up: Let + x be the direction the balls travel. Pi, x = 0. Solve: Fav, x Δt = Pf, x − Pi, x .

(15)(1.13 × 10−4 kg)(88.5 m/s) = 0.150 N. Δt 1.00 s This is the average force on the paintballs. The average recoil force is 0.150 N, in the opposite direction. Reflect: We can only calculate the average recoil force since we do not have detailed information about the how the paintballs are fired.

Fav, x =

Pf, x

=

*8.59. Set Up: Momentum is conserved in the explosion. At the highest point the velocity of the boulder is zero. Since one fragment moves horizontally the other fragment also moves horizontally. Use projectile motion to relate the initial horizontal velocity of each fragment to its horizontal displacement. Use coordinates where +x is north. Since both fragments start at the same height with zero vertical component of velocity, the time in the air, t, is the same for both. Call the fragments A and B, with A being the one that lands to the north. Therefore, we have mB = 3m A. Solve: Apply P1x = P2 x to the collision: 0 = mAv Ax + mB vBx . vBx = −

the motion after the collision: x − x0 = v0 xt. Since t is the same,

mA v Ax = − v Ax /3. Apply projectile motion to mB

( x − x0 ) A ( x − x0 ) B and = v Ax vBx

⎛v ⎞ ⎛ −v /3 ⎞ ( x − x0 ) B = ⎜ Bx ⎟ ( x − x0 ) A = ⎜ Ax ⎟ ( x − x0 ) A = −(274 m)/3 = −91.3 m. The other fragment lands 91.3 m directly v ⎝ Ax ⎠ ⎝ v Ax ⎠ south of the point of explosion. Reflect: The fragment that has three times the mass travels one-third as far. υ2 . r The free-body diagram for the ornament plus missile is given in the figure below. Take + y to be upward, since that is the direction of the acceleration. Take the + x direction to be the initial direction of motion of the missile.

8.60. Set Up: After the collision the ornament moves in an arc of a circle and has acceleration arad =


8-18

Chapter 8

Solve: Apply conservation of momentum to the collision. Pi, x = Pf, x gives (3.00 kg)(12.0 m/s) = (8.00 kg)V , and

V = 4.50 m/s is the speed of the ornament immediately after the collision. Then ∑ Fy = ma y gives T − mtot g = mtot

υ2 r

.

2 ⎛ ⎛ υ2 ⎞ 2 [4.50 m/s] ⎞ T = mtot ⎜ g + ⎟⎟ = (8.00 kg) ⎜⎜ 9.80 m/s + ⎟ = 186 N ⎜ r ⎠ 1.50 m ⎟⎠ ⎝ ⎝

8.61. Set Up: Let + x be the initial direction of motion of the bullet and let + y be the direction of the final motion of the bullet. m A = 2.50 × 10−3 kg (bullet); mB = 0.100 kg (stone). The before and after diagrams are sketched in the figure below.

Solve: (a) Pi, x = Pf, x says m Aυ A,i = mBυ B ,f, x . (2.50 × 10−3 kg)(500 m/s) = (0.100 kg)υ B ,f, x and υ B ,f, x = 12.5 m/s.

Pi, y = Pf, y says 0 = mAυ A,f, y + mBυ B ,f, y . 0 = (2.50 × 10−3 kg)(300 m/s) + (0.100 kg)υ B ,f, y and υ B ,f, y = −7.5 m/s. Then

υ B ,f = (υ B ,f, x ) 2 + (υ B ,f, y )2 = (12.5 m/s) 2 + (−7.5 m/s)2 = 14.6 m/s. 7.5 m/s . θ = 31.0°. 12.5 m/s The stone rebounds at 14.6 m/s in a direction of 31.0° from the initial direction of motion of the bullet. tan θ =

(b) Ki = 12 (2.50 × 10−3 kg)(500 m/s)2 = 312 J and

K f = 12 (2.50 × 10−3 kg)(300 m/s)2 + 12 (0.100 kg)(14.6 m/s)2 = 123 J. K f < Ki and the collision is inelastic. Reflect: Each component of momentum is separately conserved. Kinetic energy is a scalar, not a vector, and we never consider its “components.”

8.62. Set Up: Let + y be downward. Use the heights to find υi, y and υf, y , the velocity of the ball just before and just after it strikes the slab. Solve: (a) 12 mυ 2 = mgh so υ = 2 gh .

υi, y = + 2(9.80 m/s 2 )(2.00 m) = 6.26 m/s. υf, y = + 2(9.80 m/s 2 )(1.60 m) = −5.60 m/s. J y = Δp y = m(υf, y − υi, y ) = (40.0 × 10−3 kg)( −5.60 m/s − 6.26 m/s) = −0.474 kg ⋅ m/s The impulse is 0.474 kg ⋅ m/s, upward. (b) Fy =

Jy Δt

=

−0.474 kg ⋅ m/s 2.00 × 10−3 s

= − 237 N. The average force on the ball is 237 N, upward.


Momentum

8-19

*8.63. Set Up: Apply conservation of momentum to the collision between the two people. Apply conservation of energy to the motion of the stuntman before the collision and to the entwined people after the collision. For the motion of the stuntman, we have y1 − y2 = 5.0 m. Let vS be the magnitude of his horizontal velocity just before the collision. Let V be the speed of the entwined people just after the collision. Let d be the distance they slide along the floor. Solve: (a) Motion before the collision: K1 + U1 = K 2 + U 2 . K1 = 0 and

1 mv 2 S 2

= mg ( y1 − y2 ). vS = 2 g ( y1 − y2 ) =

2(9.80 m/s 2 )(5.0 m) = 9.90 m/s. Collision: mSvS = mtotV . V =

⎛ 80.0 kg ⎞ mS vS = ⎜ ⎟ (9.90 m/s) = 5.28 m/s. mtot ⎝ 150.0 kg ⎠

(b) Motion after the collision: K1 + U1 + Wother = K 2 + U 2 gives 2

1m V2 2 tot 2

− μ k mtot gd = 0.

V (5.28 m/s) = = 5.7 m. 2 μk g 2(0.250)(9.80 m/s 2 ) Reflect: Mechanical energy is dissipated in the inelastic collision, so the kinetic energy just after the collision is less than the initial potential energy of the stuntman. d=

8.64. Set Up: Use coordinates with + x in the direction the ball is traveling initially. mb = 57 × 1023 kg and mp = 61 kg. (υ b,i ) x = 72 m/s, (υ b,f ) x = −45 m/s, and (υ p,i ) x = 0. The player is treated as a particle, with all her mass concentrated at her center of mass. Solve: (a) Pi, x = Pf, x gives (57 × 10−3 kg)(72 m/s) = (57 × 10−3 kg)(−45 m/s) + (61 kg)(υp,f ) x and (υ p,f ) x = 0.11 m/s (b) Apply the impulse-momentum relation to the ball: J x = Fx Δt and J x = mb (υ p,f ) x − mb (υ p,i ) x

Fx =

mb [(υb,f ) x − (υb,i ) x ]

(57 × 10−3 kg)(−45 m/s − 72 m/s)

= − 220 N 30.0 × 10−3 s The racket exerts a force of 220 N on the ball. Apply the impulse-momentum relation to the player: mp [(υp,f ) x − (υp,i ) x ] (61 kg)(0.11 m/s − 0) Fx = = = 220 N Δt 30.0 × 10−3 s Δt

=

The ball exerts a force on 220 N on the racket. These two forces form an action-reaction pair and Newton’s third law requires that they have the same magnitude.

*8.65. Set Up: Apply conservation of energy to the motion before and after the collision and apply conservation of momentum to the collision. Let v be the speed of the mass released at the rim just before it strikes the second mass. Let each object have mass m. Solve: Conservation of energy says

1 mv 2 2

= mgR; v = 2 gR .

Set Up: This is speed v1 for the collision. Let v2 be the speed of the combined object just after the collision. Solve: Conservation of momentum applied to the collision gives mv1 = 2mv2 so v2 = v1/2 = gR/2 Set Up: Apply conservation of energy to the motion of the combined object after the collision. Let y3 be the final

height above the bottom of the bowl. Solve:

y3 =

1 (2m)v2 2 2 2

= (2m) gy3.

v2 1 ⎛ gR ⎞ = ⎜ ⎟ = R /4. 2g 2g ⎝ 2 ⎠

Reflect: Mechanical energy is lost in the collision, so the final gravitational potential energy is less than the initial gravitational potential energy. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8-20

Chapter 8

8.66. Set Up: Let the system be the two masses and the spring. The system is sketched in the figure below, in its initial and final situations. Use coordinates where + x is to the right. Call the masses A and B.

Solve: Pi, x = Pf, x so 0 = (1.50 kg)(−υ A ) + (1.50 kg)(υ B ) and υ A = υ B . Energy conservation says the potential energy

originally stored in the spring is all converted into kinetic energy of the blocks, so

1 kx 2 2 i

= 12 mυ A2 + 12 mυ B 2 . υ A = υ B so

k 175 N/m = (0.200 m) = 1.53 m/s 2m 2[1.50 kg]

υ A = xi

Reflect: If the objects have different masses they will end up with different speeds. The lighter one will have the greater speed, since they end up with equal magnitudes of momentum.

8.67. Set Up: Let + x be to the right. Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. The total mass is m = mbullet + mblock = 1.00 kg. The spring has force constant

F 0.750 N = = 300 N/m. x 0.250 × 10−2 m Let V be the velocity of the block just after impact. Solve: (a) Conservation of energy for the motion after the collision gives Ki = U el,f . k=

V=x

1 mV 2 2

=

1 kx 2 2

and

k 300 N/m = (0.150 m) = 2.60 m/s m 1.00 kg

(b) Conservation of momentum applied to the collision gives mbulletυi = mV .

υi =

(1.00 kg)(2.60 m/s) mV = = 325 m/s mbullet 8.00 × 1023 kg

8.68. Set Up: Let + x be to the right. Apply conservation of momentum to the collision and conservation of energy to the motion of the block after the collision. Let the bullet be A and the block be B. Let V be the velocity of the block just after the collision. Solve: Motion of block after the collision: Ki = U grav,f . 12 mBV 2 = mB gh.

V = 2 gh = 2(9.80 m/s 2 )(0.450 × 10−2 m) = 0.297 m/s Collision: υ B ,f = 0.297 m/s. Pi, x = Pf, x gives m Aυ A,i = m Aυ A,f + mBυ B ,f .

υ A,f =

m Aυ A,i − mBυ B ,f

(5.00 × 10−3 kg)(450 m/s) − (1.00 kg)(0.297 m/s)

= 391 m/s 5.00 × 10−3 kg Reflect: We assume the block moves very little during the time it takes the bullet to through it. mA

=

*8.69. Set Up: Let + x be to the right. wA = 800 N, wB = 600 N and wC = 1000 N. Solve: Pi, x = Pf, x gives 0 = m Aυ A,f, x + mBυ B ,f, x + mCυC ,f, x . υC ,f, x =

m Aυ A,f, x + mBυ B ,f, x mC

=

wAυ A,f, x + wBυ B ,f, x wC

.

(800 N)( −[5.00 m/s]cos30.0°) + (600 N)( +[7.00 m/s]cos36.9°) = − 0.105 m/s 1000 N The sleigh’s velocity is 0.105 m/s, to the left.

υC ,f, x =


Momentum

8-21

Reflect: The vertical component of the momentum of the system consisting of the two people and the sleigh is not conserved, because of the net force exerted on the sleigh by the ice while they jump.

8.70. Set Up: Use coordinates where + x is in the direction the squid moves. Take the system to be the squid plus the water it expels. After it expels the water the squid has mass msquid = 4.75 kg. Solve: (a) Pi, x = Pf, x gives 0 = msquidυsquid − mwaterυ water

⎛ msquid ⎞ ⎛ 4.75 kg ⎞ ⎟υsquid = ⎜ ⎟ (2.50 m/s) = 6.79 m/s ⎝ 1.75 kg ⎠ ⎝ mwater ⎠ (b) The initial kinetic energy of the system is zero. The final kinetic energy is 1 (1.75 kg)(6.79 m/s) 2 + 1 (4.75 kg)(2.50 m/s) 2 = 55.2 J. 2 2

υwater = ⎜

This kinetic energy is produced by action of the squid.

*8.71. Set Up: The center of mass of each piece of length L is at its center. Solve: (a) From symmetry, the center of mass is on the vertical axis, a distance ( L /2)cos(α /2) below the apex. (b) The center of mass is on the vertical axis of symmetry, a distance 2( L/ 2)/3 = L/3 above the center of the horizontal segment. (c) Using the wire frame as a coordinate system, the coordinates of the center of mass are equal and each is equal to ( L/2)/2 = L/4. The center of mass is along the bisector of the angle, a distance L/ 8 from the corner. (d) By symmetry, the center of mass is at the center of the equilateral triangle, a distance ( L /2) tan 60° = L / 12

above the center of the horizontal segment.

8.72. Set Up: mbody = mMars = 6.42 × 1023 kg. The current mass of the earth is 5.97 × 1024 kg. In this scenario, this mass came from an original earth of mass mE combining with mMars and ejecting mass mMoon = 7.35 × 1022 kg. The mass mE is therefore given by the equation

mE + 6.42 × 1023 kg = 5.97 × 1024 kg + 7.35 × 1022 kg and mE = 5.40 × 1024 kg. The present orbital speed of the earth is its orbit circumference divided by its orbital period: 2π (1.50 × 1011 m) = 2.986 × 104 m/s (365.3 days)(86400 s/day) Treat the Mars-sized body and the earth as an isolated system. Use coordinates where + x is in the direction the earth is moving initially. Solve: Pi, x = Pf, x gives

υearth =

(5.40 × 1024 kg)(2.986 × 104 m/s) − (6.42 × 1023 kg)(11 × 103 m/s) = (5.40 × 1024 kg + 6.42 × 1023 kg)yf and

υf = 2.552 × 104 m/s. The earth slowed down by 2.986 × 104 m/s − 2.552 × 104 m/s = 4.3 × 103 m/s, a decrease of 14%. (b) The thermal energy created equals the decrease in kinetic energy as a result of the collision. Ki = 12 (5.40 × 1024 kg)(2.986 × 104 m/s) 2 + 12 (6.42 × 1023 kg)(11 × 103 m/s)2 = 2.45 × 1033 J

K f = 12 (5.40 × 1024 kg + 6.42 × 1023 kg)(2.552 × 104 m/s) 2 = 1.97 × 1033 J Ki − K f = 4.8 × 1032 J; this is the amount of thermal energy released.

(c)

4.8 × 1032 J 4.184 × 109 J/ton of TNT

= 1.1 × 1023 tons, of TNT so 1.1 × 1017 megaton bombs.

8.73. Set Up: The horizontal component of the momentum of the sand plus railroad system is conserved. As the sand leaks out it retains its horizontal velocity of 15.0 m/s.


8-22

Chapter 8

Solve: The horizontal component of the momentum of the sand doesn’t change when it leaks out so the speed of the railroad car doesn’t change; it remains 15.0 m/s. This result is similar to that of Exercise 8.16c. In contrast, in Exercise 8.16a the flour sack is falling vertically and initially has no horizontal component of momentum. Its momentum changes when you catch it. Therefore, in order to conserve the total horizontal momentum you and your skateboard must slow down. Reflect: The horizontal momentum of the sand does change when it strikes the ground, due to the force that is external to the system of sand plus railroad car.

8.74. Set Up: Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. Let + x be to the right. Solve: Collision: Pi, x = Pf, x gives (4.20 × 1023 kg)υ = (4.20 × 1023 kg + 2.50 kg)V

Motion after the collision: The kinetic energy of the block immediately after the collision is converted entirely to gravitational potential energy at the maximum angle of swing. The figure below shows that the maximum height h the block swings up is given by h = (0.750 m)(1 − cos34.7°) = 0.133 m.

Conservation of energy gives

1 m V2 2 tot

= mtot gh and V =

2 gh = 1.61 m/s

Then the conservation of momentum equation gives ⎛ 4.20 × 1023 kg + 2.50 kg ⎞ υ =⎜ ⎟⎟ (1.61 m/s) = 960 m/s ⎜ 4.20 × 1023 kg ⎝ ⎠ Reflect: The original speed of the bullet is nearly three times the speed of sound in air. Kinetic energy is not conserved in the collision. Our analysis assumes that the block moves very little during the collision, while the bullet is coming to rest relative to the block. This is a good approximation since the velocity the block gets in the collision is much less than the initial velocity of the bullet.

*8.75. Set Up: Apply conservation of energy to the motion before and after the collision. Apply conservation of momentum to the collision. First consider the motion after the collision. The combined object has mass G G mtot = 25.0 kg. Apply ∑ F = ma to the object at the top of the circular loop, where the object has speed v3. The acceleration is arad = v3 2 /R, downward. Solve: T + mg = m

v3 2 . R

The minimum speed v3 for the object not to fall out of the circle is given by setting T = 0. This gives v3 = Rg , where R = 3.50 m. Set Up: Next, use conservation of energy with point 2 at the bottom of the loop and point 3 at the top of the loop. Take y = 0 at point 2. Only gravity does work, so K 2 + U 2 = K3 + U 3. Solve:

1 m v 2 2 tot 2

Use v3 =

= 12 mtot v3 2 + mtot g (2 R).

Rg and solve for v2: v2 = 5 gR = 13.1 m/s.

Set Up: Now apply conservation of momentum to the collision between the dart and the sphere. Let v1 be the speed

of the dart before the collision. Solve: (5.00 kg)v1 = (25.0 kg)(13.1 m/s).

v1 = 65.5 m/s. Reflect: The collision is inelastic and mechanical energy is removed from the system by the negative work done by the forces between the dart and the sphere. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Momentum

8-23

8.76. Set Up: Eqs. 8.11, 8.12, and 8.13 give the outcome of the elastic collision. The blue puck is object A and the red puck is object B. Let +x be the direction of the initial motion of A. v A,i = 0.200 m/s, v A,f = 0.050 m/s and

vB,i = 0. Solve: (a) Eq. 8.13 gives vB,f = v A,f + v A,i = 0.250 m/s. ⎛ v A,i ⎞ ⎛ ⎡ 0.200 m/s ⎤ ⎞ (b) Eq. 8.12 gives mB = m A ⎜ 2 − 1⎟ = (0.0400 kg) ⎜ 2 ⎢ ⎥ − 1⎟ = 0.024 kg. ⎜ vB ,f ⎟ ⎝ ⎣ 0.250 m/s ⎦ ⎠ ⎝ ⎠ Reflect: We can that our results give K1 = K 2 and P1x = P2 x , as required in an elastic collision.

⎛ m A − mB ⎞ ⎛ 2m A ⎞ ⎟υ and υ B = ⎜ ⎟υ . ⎝ m A + mB ⎠ ⎝ mA + mB ⎠

8.77. Set Up: For an elastic collision with object B initially at rest, υ A = ⎜ A is the alpha particle and B is the gold nucleus. υ = 1.25 × 107 m/s.

⎛ 6.65 × 10−27 kg − 3.27 × 10−25 kg ⎞ Solve: υ A = ⎜ (1.25 × 107 m/s) = − 1.20 × 107 m/s ⎜ 6.65 × 10−27 kg + 3.27 × 10−25 kg ⎟⎟ ⎝ ⎠ −27 ⎛ ⎞ 2(6.65 × 10 kg) υB = ⎜ (1.25 × 107 m/s) = 4.98 × 105 m/s − − 27 25 ⎜ 6.65 × 10 kg + 3.27 × 10 kg ⎟⎟ ⎝ ⎠ Reflect: In an elastic collision of a light object with a stationary heavy object, the light object bounces back with its speed only slightly reduced.

8.78. Set Up: Take + y to be upward. Let A be the lower piece and B be the upper piece, so m A = mB /4. (a) Apply conservation of momentum to the explosion. (b) After the explosion the upper piece is in free-fall with a y = − g . At its maximum height υ y = 0. Solve: (a) Explosion: Pi, y = Pf, y gives ( m A + mB )(5.0 m/s) = m A (23.0 m/s) + mBυ B

mB = 4mA so (m A + 4m A )(5.0 m/s) = m A (23.0 m/s) + 4m Aυ B and υ B = 7.0 m/s (b) Motion after explosion: υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives

y − y0 =

υ y 2 − υ0 y 2 2a y

=

0 − (7.0 m/s) 2 2(29.8 m/s 2 )

= 2.5 m

*8.79. Set Up: Apply conservation of momentum. Take + x to be to the right. For (a) the collision is elastic. Let A be the cart that is initially moving to the right. For (b) the two carts stick together and the combined object has final speed υ f . Solve: (a)

Pi, x = Pf, x

gives

M υ0 + M (−υ0 ) = M υ A + M υ B

and

υ A = − υB .

For

an

elastic

collision

υ B ,f − υ A,f = − (υ B ,i − υ A,i ). This gives υ B − υ A = −( −υ0 − υ0 ) = 2υ0 . υ A = −υ B so υ B = υ0 and υ A = −υ0 . A rebounds with speed υ 0 and B rebounds with speed υ0 . (b) Pi, x = Pf, x gives M υ0 + M (−υ0 ) = (2 M )υf and υf = 0. Both carts are rest after the collision. Reflect: Collision (b) is totally inelastic. All the initial kinetic energy is converted to other forms. Any collision in which the objects stick together must be inelastic.

8.80. Set Up: Call the fragments A and B, with m A = 2.0 kg and mB = 5.0 kg. After the explosion fragment A moves in the + x direction with speed υ A and fragment B moves in the 2x direction with speed υ B . Solve: Pi, x = Pf, x gives 0 = mAυ A + mB (−υ B ) and ⎛ mB ⎞ ⎛ 5.0 kg ⎞ ⎟υ B = ⎜ ⎟υ B = 2.5υ B . ⎝ 2.0 kg ⎠ ⎝ mA ⎠

υA = ⎜


8-24

Chapter 8 2 1 (2.0 kg)(2.5υ ) 2 K A 12 mAυ A 12.5 B 2 = = = = 2.5. 1 (5.0 kg)υ 2 K B 1 mBυ B 2 5.0 B 2 2

K A = 100 J so K B = 250 J Reflect: In an explosion the lighter fragment receives the most of the liberated energy.

*8.81. Set Up: mn = 1.675 × 10227 kg, mp = 1.673 × 10227 kg and me = 9.109 × 10231 kg. After the decay the electron moves in the + x direction with speed υe and the proton moves in the − x direction with speed υp . Apply conservation of momentum. Solve: (a) Pi, x = Pf, x gives 0 = meυe + mp (2υ p ).

υe mp = = 1837 υp me

1mυ 2 1 m (1837υ ) 2 ⎛m ⎞ Ke e e p 2 e = 2 = = ⎜ e ⎟ (1837)2 = 1837. 2 2 1 1 ⎜ mp ⎟ Kp mυ mυ ⎝ ⎠ 2 p p 2 p p (c) The final speed of the electron is much greater than that of the proton.

(b)

8.82. Set Up: Use constant acceleration equations for the free-fall to find the speed of the acorn just before it collides with the bird. The system of the bird and acorn is sketched in the figure below, before and after the collision. Use the coordinates shown. Assume that gravity can be neglected during the collision.

Solve: (a) For the fall of the acorn, υ0 y = 0, a y = 9.80 m/s 2 and y − y0 = 17.5 cm.

υ y 2 = υ0 y 2 + 2a y ( y − y0 ) and υ y = 2a y ( y − y0 ) = 18.5 m/s. υacorn,i = 18.5 m/s Pi, x = Pf, x gives (135 g)(75.0 cm/s) = (135 g + 15.0 g)υf, x and υf, x = 67.5 cm/s. Pi, y = Pf, y gives (15.0 g)(1850 cm/s) = (135 g + 15.0 g)υf, y and υf, y = 185 cm/s. (b) υf = υf, x 2 + υf, y 2 = 197 cm/s; tan φ =

υf, x υf, y

=

67.5 cm/s so φ = 20.0° 185 cm/s

Reflect: Momentum is a vector and there are separate equations for conservation of its x and y components.

8.83. Set Up: The friction force is μk mtot g . Use energy considerations to find the velocity of the combined object immediately after the collision. Apply conservation of momentum to the collision. Use coordinates where + x is west and + y is south. Solve: Motion after collision: The negative work done by friction takes away all the kinetic energy that the combined object has just after the collision. 6.43 m tan φ = 5.39 m and φ = 50.0°. The wreckage slides 8.39 m in a direction 50.0° south of west. 1m V2 2 tot

= μ k mtot gd and V = 2 μk gd = 2(0.75)(9.80 m/s 2 )(8.39 m) = 11.1 m/s Vx = V cos φ = 7.13 m/s; V y = V sin φ = 8.50 m/s

Collision: Pi, x = Pf, x gives (2200 kg)υSUV = (1500 kg + 2200 kg)Vx and υSUV = 12 m/s. Pi, y = Pf, y gives (1500 kg)υsedan = (1500 kg + 2200 kg)Vy and υsedan = 21 m/s. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Momentum

8-25

Solutions to age Problems 8.84. Set Up: Choose the x-axis in the direction of the squirt. The momentum of a single squirt can be calculated from px = mυ x . Solve: px = mυ x = (0.25 kg)(6.00 m/s) = 1.5 kg ⋅ m/s. Thus, the correct answer is B.

*8.85. Set Up: Choose the squid and the ejected seawater (squirt) as the “system.” Assume that the initial momentum of the system is zero. Solve: Assuming that the external forces are negligible during the squirt, the total momentum of the squid/squirt system must be conserved. The momentum of the squid must be equal and opposite to that of the squirt. Thus, the correct answer is E. 8.86. Set Up: Since we know the momentum of the squid after the squirt, we can find its velocity using px = mυ x . Solve: υ x =

p x −1.5 kg ⋅ m/s = = −0.60 m/s = −60 cm/s. Thus, the correct answer is D. m 2.5 kg

*8.87. Set Up: Assume three identical squirts, each contributing −1.5 kg ⋅ m/s to the squid’s total momentum. Solve: The total momentum of the squid after three squirts is −4.5 kg ⋅ m/s, thus the squid’s velocity is

υx =

p x −4.5 kg ⋅ m/s = = −1.8 m/s = −180 cm/s. Thus, the correct answer is C. m 2.5 kg


9

ROTATIONAL MOTION

Answers to Multiple-Choice Problems 1. B 2. D 3. D 4. C 15. A 16. A

5. A

6. D

7. B

8. A, C 9. C

10. C 11. C

12. A

13. B

14. A

Solutions to Problems 9.1. Set Up: θ = Solve: θ =

s gives θ in radians. 2π rad = 360°. r

0.750 m ⎛ 360° ⎞ = 17.2° = 0.300 rad; (0.300 rad) ⎜ ⎝ 2π rad ⎟⎠ 2.50 m

9.2. Set Up: π rad = 180°. s = rθ , with θ in radians. Solve: (a) θ = (b) r =

s

θ

=

s 1.50 m = = 0.600 rad = 34.4° r 2.50 m

14.0 cm = 6.27 cm (128°)(π rad/180°)

(c) s = rθ = (1.50 m)(0.700 rad) = 1.05 m

*9.3. Set Up: For one revolution, Δθ = 2π rad. Assume constant angular velocity, so Δθ ω= .

Δt The second hand makes 1 revolution in 1 minute = 60.0 s. The minute hand makes 1 revolution in 1 h = 3600 s, and the hour hand makes 1 revolution in 12 h = 43,200 s. Solve: (a) second hand ω =

2π rad = 0.105 rad/s; 60 s

2π rad = 1.75 × 10−3 rad/s; 3600 s 2π rad hour hand ω = = 1.45 × 10−4 rad/s 43,200 s (b) The period is the time for 1 revolution. Second hand, 1 min; minute hand, 1 h; hour hand, 12 h. Reflect: When the angular velocity is constant, ω = ω av . minute hand ω =


9-1

9-2

Chapter 9

9.4. Set Up: 33 13 rpm = 0.5556 rev/s is the angular speed. ⎛ 2π rad ⎞ Solve: (a) (0.5556 rev/s) ⎜ = 3.49 rad/s ⎝ 1 rev ⎟⎠

(b) For 1 rev, t =

θ 1 rev = = 1.80 s ω 0.5556 rev/s

*9.5. Set Up: 1 rev = 2π rad Solve: (a) The period is the time for 1 rev, so period = 2.25 s. 2π rad = 2.79 rad/s 2.25 s t Reflect: The angular speed and period don’t depend on the radius of the wheel.

(b) ω =

θ

=

9.6. Set Up: 1 rev = 2π rad. (a) The period is 24 h = 86,400 s. (b) The period is 1 yr = 3.156 × 107 s. (c) The period is 27.3 days = 2.359 × 106 s. Solve: ω =

θ

t 2π rad (a) ω = = 7.27 × 10−5 rad/s 86,400 s 2π rad (b) ω = = 1.99 × 10−7 rad/s 7 3.156 × 10 s (c) ω =

2π rad 2.359 × 106 s

= 2.66 × 1026 rad/s

*9.7. Set Up: According the appendix, the distance between the earth and the moon is 3.84 × 108 m. Since the radius of both the earth and the moon are small in comparison, we will assume that this is roughly the distance, r , that the laser beam travels. The relation between the angular velocity of the beam and its linear velocity on the lunar surface is υ = rω . Solve: (a) The speed of the laser beam across the moon’s surface is

υ = rω = (3.84 × 108 m)(1.50 × 10−3 rad/s) = 5.76 × 105 m/s. (b) Assume that the laser light spreads out from a point source on the earth to a 6.00 km diameter spot on the moon. The angle of divergence (in radians) is the angle that subtends a 6.00 km arc on the lunar surface. Thus,

θ=

s 6.00 × 103 m = 1.56 × 10−5 rad = r 3.84 × 108 m

Reflect: The actual distance that the beam travels depends on the exact location where the beam leaves the earth and where it hits the lunar surface. Since the radius of the earth is roughly 0.06 × 108 m and the radius of the moon is

0.02 × 108 m, the distance traveled by the beam is known to an accuracy of only two significant figures. The answers should be rounded accordingly.

9.8. Set Up: 1 rev = 2π rad. The earth makes one revolution in 24 h = 8.64 × 104 s. Solve: (a) The period must be 24 h. θ 2π rad (b) ω = = = 7.27 × 10−5 rad/s t 8.64 × 104 s

9.9. Set Up: 1 rpm = (2π /60) rad/s. Period T = Solve: (a) ω = (1900)(2π rad/60 s) = 199 rad/s

2π

ω

. θ − θ0 = ω t .


Rotational Motion (b) 35° = (35°)(π /180°) = 0.611 rad. t =

9-3

θ − θ 0 0.611 rad = = 3.1 × 10−3 s ω 199 rad/s

⎛ 60 s ⎞ ⎛ 1 rev ⎞ = 172 rpm (c) (18 rad/s) ⎜ ⎝ 1 min ⎟⎠ ⎜⎝ 2π rad ⎟⎠ 2π rad = 0.032 s. ω 199 rad/s Reflect: The period is inversely proportional to the angular velocity.

(d) For the propeller rotating at 1900 rpm, T =

2π rad

=

9.10. Set Up: Aligned means θ1 = θ 2 + n(2π rad) for the same t, when hand 1 has turned through some integer number of revolutions more than hand 2. Solve: (a) The first alignment is when θ1 = θ 2 + 2π rad. θ1 = ω 1t and θ 2 = ω 2t . This gives ω 1t = ω 2t + 2π rad

and t =

2π rad 2π rad = = 240 s ω 1 − ω 2 0.0425 rad/s − 0.0163 rad/s

(b) The second alignment is when θ1 = θ 2 + 2(2π rad) and t =

2(2π rad) = 480 s ω1 − ω 2

⎛ 2π rad ⎞ ⎛ 1 min ⎞ = 8.17 rad/s ⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠

*9.11. Set Up: ω 0 = 0. ω = (78.0 rpm) ⎜

ω − ω0

8.17 rad/s − 0 = 2.33 rad/s 2 t 3.50 s ⎛ 360° ⎞ (b) θ − θ 0 = ω 0t + 12 α t 2 = 12 (2.33 rad/s 2 )(3.50 s) 2 = (14.27 rad) ⎜ ⎟ = 818° ⎝ 2π rad ⎠ Solve: (a) ω = ω0 + α t gives α =

=

9.12. Set Up: ω = 0 (stops). ω 0 = 78.0 rpm = 8.17 rad/s. θ − θ 0 = 10.5 rev = 65.97 rad Solve: (a) ω 2 = ω 02 + 2α (θ − θ 0 ) gives

α= (b) ω = ω 0 + α t gives t =

ω 2 − ω 0 2 0 − (8.17 rad/s) 2 = = − 0.506 rad/s 2 2(θ − θ 0 ) 2(65.97 rad)

ω − ω0 0 − 8.17 rad/s = = 16.1 s α 20.506 rad/s 2

*9.13. Set Up: 570 rpm = 59.7 rad/s; 1600 rpm = 167.6 rad/s ω − ω0 167.6 rad/s − 59.7 rad/s = = 0.0135 rad/s 2 Solve: ω = ω 0 + α t so α = (133 min)(60 s/1 min)

t

9.14. Set Up: Let the direction of the rotation of the blade be positive. ω 0 = 0. ω − ω 0 140 rad/s − 0 Solve: ω = ω 0 + ω 0t gives α = = = 23.3 rad/s 2 .

6.00 s ⎛ ω0 + ω ⎞ ⎛ 0 + 140 rad/s ⎞ (θ − θ 0 ) = ⎜ t =⎜ ⎟⎠ (6.00 s) = 420 rad ⎝ 2 ⎟⎠ ⎝ 2

t

9.15. Set Up: Let the direction the wheel is rotating be the positive sense of rotation. Since the wheel starts from rest, ω 0 = 0. 1 rev = 2π rad Solve: (a) ω = ω 0 + ω 0t gives t =

ω − ω0 α

(b) ω 2 = ω 02 + 2α (θ − θ 0 ) gives θ − θ 0 =

=

8.00 rad/s − 0 0.640 rad/s 2

ω 2 − ω 02 2α

=

= 12.5 s

(8.00 rad/s) 2 − 0 2(0.640 rad/s 2 )

= 50.0 rad = 7.96 rev.


9-4

Chapter 9

Reflect: We could also do part (b) by

⎛ ω0 + ω ⎞ ⎛ 0 + 8.00 rad/s ⎞ t=⎜ ⎟⎠ (12.5 s) = 50.0 rad. ⎝ 2 ⎟⎠ ⎝ 2

θ − θ0 = ⎜

9.16. Set Up: Let the direction of rotation of the fan be positive. Solve: (a) ω 0 = 500.0 rev/min = 8.33 rev/s and ω = 200.0 rev/min = 3.33 rev/s. ω = ω 0 + ω 0t gives ω − ω0

3.33 rev/s − 8.33 rev/s = = − 1.25 rev/s 2 . t 4.00 s ⎛ω + ω⎞ ⎛ 8.33 rev/s + 3.33 rev/s ⎞ θ − θ0 = ⎜ 0 t=⎜ (4.00 s) = 23.3 rev ⎝ 2 ⎠⎟ ⎝ ⎠⎟ 2

α=

(b) ω 0 = 3.33 rev/s, ω = 0, α = − 1.25 rev/s 2 . ω = ω 0 + ω 0t gives

t=

ω − ω 0 0 − 3.33 rev/s = = 2.66 s α 21.25 rev/s 2

*9.17. Set Up: 500.0 rpm = 8.33 rev/s. Let the direction of rotation of the flywheel be positive. ⎛ω + ω⎞ Solve: (a) (θ − θ 0 ) = ⎜ 0 t gives ⎝ 2 ⎟⎠ 2(θ − θ 0 ) 2(200.0 rev) − ω0 = − 8.33 rev/s = 5.00 rev/s = 300 rpm t 30.0 s (b) Use information in part (a) to find α : ω = ω 0 + ω 0t gives

ω=

ω − ω0

5.00 rev/s − 8.33 rev/s = = − 0.111 rev/s 2 . t 30.0 s Then, with ω = 0 and ω 0 = 8.33 rev/s the equation ω = ω 0 + ω 0t gives

α=

ω − ω 0 0 − 8.33 rev/s = = 75.0 s. α −0.111 rev/s 2 ⎛ ω 0 + ω ⎞ ⎛ 8.33 rev/s + 0 ⎞ t=

θ − θ0 = ⎜

⎟t = ⎜ ⎟ (75.0 s) = 312 rev. 2 ⎠ ⎝ 2 ⎠ ⎝ Reflect: The angular acceleration is negative because the wheel is slowing down.

9.18. Set Up: Let the direction of rotation of the flywheel be positive. 2(θ − θ 0 ) 2(162 rad) ⎛ω + ω⎞ Solve: (a) θ − θ 0 = ⎜ 0 −ω = − 108 rad/s = − 27.0 rad/s. ⎟ t gives ω 0 = ⎝

2 ⎠ t 4.00 s At the start of the interval the flywheel was rotating at 27.0 rad/s, in the opposite direction. ω − ω0 108 rad/s + 27 rad/s = = 33.8 rad/s 2 . (b) ω = ω 0 + ω 0t gives α = 4.00 s t

9.19. Set Up: Let the direction of rotation of the flywheel be positive. Solve: θ − θ 0 = ω 0t + 12 αt 2 gives θ − θ0

60.0 rad 1 − 12 α t = − 2 (2.25 rad/s 2 )(4.00 s) = 10.5 rad/s. t 4.00 s Reflect: At the end of the 4.00 s interval, ω = ω 0 + α t = 19.5 rad/s.

ω0 =

⎛ ω 0 + ω ⎞ ⎛ 10.5 rad/s + 19.5 rad/s ⎞ t= ⎟⎠ (4.00 s) = 60.0 rad, ⎝ 2 ⎟⎠ ⎜⎝ 2

θ − θ0 = ⎜ which checks.

9.20. Set Up: Eq. 9.7 is ω = ω 0 + ω 0t. Eq. 9.11 is θ = θ0 + ω0t + 12 α t 2 . © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Rotational Motion

Solve: t =

ω − ω0 α

9-5

so 2

1 ⎛ ω − ω0 ⎞ 1 ⎛ ω − ω0 ⎞ + α = θ0 + (ω − ω 0 )(2ω 0 + ω − ω 0 ) ⎝ α ⎟⎠ 2 ⎜⎝ α ⎟⎠ 2α

θ = θ0 + ω0 ⎜ and

θ = θ0 +

1 1 (ω − ω 0 )(ω + ω 0 ) = θ 0 + (ω 2 − ω 0 2 ). 2α 2α

ω 2 − ω 02 = 2α (θ − θ 0 ).

*9.21. Set Up: First, convert all distance measurements into feet. Use υ = rω to find υ. r = 12.0 in = 1.00ft. From the appendix, we have 1 mi/h = 1.466 ft/s.

⎛ mi ⎞ ⎛ 1.466 ft/s ⎞ Solve: Convert the speed of the car: υ = ⎜ 63 ⎟ ⎜ ⎟ = 92 ft/s. Find the angular velocity of the tires: ⎝ h ⎠ ⎝ 1 mi/h ⎠

ω=

υ r

=

92 ft/s = 92 rad/s. 1.00 ft

Reflect: υ = rω requires that ω be in rad/s.

2π rad/s. 60 Solve: (a) 620 rpm = 64.9 rad/s. υ = (0.0750 m)(64.9 rad/s) = 4.87 m/s

9.22. Set Up: υ = rω , where ω must be in rad/s. 1 rpm =

(b) ω =

υ r

=

0.600 m/s = 15 rad/s = 143 rpm 0.0400 m

9.23. Set Up: υ = rω , with ω in rad/s. Solve: (a) arad = rω 2 = (0.500 m)(6.00 rad/s) 2 = 18.0 m/s 2 (b) υ = rω = (0.500 m)(6.00 rad/s) = 3.00 m/s. arad =

υ2 r

=

(3.00 m/s) 2 = 18.0 m/s 2 0.500 m

9.24. Set Up: arad = rω 2 , with ω in rad/s. 1 rpm = (2π /60) rad/s Solve: ω =

arad = r

(400,000)(9.80 m/s 2 ) = 1.25 × 104 rad/s = 1.20 × 105 rpm 0.0250 m

*9.25. Set Up: r1 = 4.0 in., r2 = 1.5 in. and ω 2 = 75 rpm. The 3 in. sprocket and 24 in. wheel are mounted on the same axle and turn at the same rate. Solve: All points on the chain have the same speed so υ 1 = υ 2; points on the rim of each sprocket move at the same tangential speed. υ = rω gives r1ω1 = r2ω 2 . ⎛ r2 ⎞ ⎛ 1.5 in. ⎞ ⎟ω 2 = ⎜ ⎟ (75 rpm) = 28 rpm r ⎝ 4.0 in. ⎠ ⎝ 1⎠ Reflect: The large sprocket turns at a slower rate than the small sprocket.

ω1 = ⎜

9.26. Set Up: arad = rω 2 . Since arad depends on the rotation rate, it changes as the flywheel accelerates. atan = rα and is constant. Since arad and atan are in perpendicular directions, the resultant acceleration is a = atan 2 + arad 2 .

Solve: (a) ω = 0 so arad = 0. atan = rα = (0.300 m)(0.600 rad/s 2 ) = 0.180 m/s 2 . a = 0.180 m/s 2 .


9-6

Chapter 9

(b) 60° = (π /3) rad. ω 2 = ω 0 2 + 2α (θ − θ 0 ). ω 0 = 0 so ω 2 = 2α (θ − θ 0 ) and arad = 2rα (θ − θ 0 ) = 2(0.300 m)(0.600 rad/s 2 )(π /3 rad) = 0.377 m/s 2 .

a = (0.180 m/s 2 ) 2 + (0.377 m/s 2 ) 2 = 0.418 m/s 2 (c) 120° = (2π /3) rad. arad = 2(0.300 m)(0.600 rad/s 2 )(2π /3) = 0.754 m/s 2

a = (0.180 m/s 2 ) 2 + (0.754 m/s 2 )2 = 0.775 m/s2

9.27. Set Up: We have v = rω and arad = rω 2 = v 2 /r. Note that 2π rad = 1 rev, so π rad/s = 30 rev/min. Solve: (a) ω r = (1250 rev/min) (b)

⎛

rad/s ⎜ 12.7 × 10 ( 30πrev/min ) ⎜⎝ 2

−3

m⎞ ⎟⎟ = 0.831 m/s. ⎠

v2 (0.831 m/s) 2 = = 109 m/s 2 . r (12.7 × 10−3 m)/2

Reflect: In v = rω , ω must be in rad/s.

9.28. Set Up: 360° = 2π rad. In one back and forth cycle the head turns through 140° = 2.443 rad. Solve: (a) In 1 min = 60 s, the head turns through 7600(2.443 rad) = 1.857 × 104 rad.

1.857 × 104 rad = 310 rad/s t 60 s (b) υ = rω = (0.55 cm)(310 rad/s) = 170 cm/s = 1.7 m/s

ω=

θ

=

(c) Estimate that the toothbrush turns back and forth through 45° five times per second.

⎛ 2π rad ⎞ 5(90°) ⎜ ⎝ 360° ⎟⎠ ω= = = 7.9 rad/s and υ = rω = (0.55 cm)(7.9 rad/s) = 4 cm/s. t 1s

θ

G

G

*9.29. Set Up: Use Eq. (9.15) to relate ω to arad and use ∑ F = ma to relate arad to Frad . Use Eq. (9.13) to relate ω to v, where v is the tangential speed. Solve: (a) arad = rω 2 and Frad = marad = mrω 2 Frad,2 Frad,1

(b) Using v = rω we have

2

⎛ ω ⎞ ⎛ 640 rev/min ⎞2 =⎜ 2 ⎟ =⎜ ⎟ = 2.29 ⎝ ω1 ⎠ ⎝ 423 rev/min ⎠

v2 ω 2 640 rev/min = = = 1.51 v1 ω1 423 rev/min

⎛ 1 min ⎞ ⎛ 2π rad ⎞ (c) Using v = rω and ω = (640 rev/min) ⎜ = 67.0 rad/s we have ⎝ 60 s ⎟⎠ ⎜⎝ 1 rev ⎟⎠ v = rω = (0.235 m)(67.0 rad/s) = 15.7 m/s.

arad 1060 m/s 2 = = 108; a = 108 g . g 9.80 m/s 2

arad = rω 2 = (0.235 m)(67.0 rad/s) 2 = 1060 m/s 2 so

Reflect: In parts (a) and (b), since a ratio is used the units cancel and there is no need to convert ω to rad/s. In part (c), v and arad are calculated from ω , and ω must be in rad/s.

9.30. Set Up: According to Eq. (9.16), the moment of inertia for the entire object is equal to that of the rod plus that of the end caps. The object is shown in the figure below. 1 ML2 + 2( m)( L/2) 2 = 1 M + 1 m L2 Solve: We have I = I rod + 2 I cap . Thus, I = 12 12 2

(

)


Rotational Motion

9-7

Reflect: Table 9.2 was used for I rod . For the end caps we used I = mr 2 , since they are treated as point particles.

*9.31. Set Up: I for the object is the sum of the values of I for each part. For the bar, for an axis perpendicular to the bar, use the appropriate expression from Table 9.2. For a point mass, we have I = mr 2 , where r is the distance of the mass from the axis. 2

Solve: (a) I = I bar + I balls =

1 ⎛ L⎞ M bar L2 + 2mballs ⎜ ⎟ . ⎝ 2⎠ 12

I=

1 (4.00 kg)(2.00 m)2 + 2(0.500 kg)(1.00 m) 2 = 2.33 kg ⋅ m 2 12

1 1 (b) I = mbar L2 + mball L2 = (4.00 kg)(2.00 m) 2 + (0.500 kg)(2.00 m) 2 = 7.33 kg ⋅ m 2 3 3 (c) We have I = 0 because all the masses are on the axis, where r = 0. Reflect: The moment of inertia for an object depends on both the location and orientation of the axis of rotation in relation to the object. The axis of rotation with the lowest moment inertia is always one of those that through the object’s center of mass.

9.32. Set Up: (a) I = 121 ML2 (b) I = 13 ML2 (c) I = 12 MR 2 1 Solve: (a) I = 12 (0.0420 kg)(1.50 m)2 = 7.88 × 10−3 kg ⋅ m 2

(b) I = 13 (0.0420 kg)(1.50 m)2 = 3.15 × 10−2 kg ⋅ m2 (c) I = 12 (0.0420 kg)(0.150 × 10−2 m) 2 = 4.72 × 10−8 kg ⋅ m 2

*9.33. Set Up: (a) Each mass is a distance 2(0.400 m) 2 0.400 m = 2 2

from the axis. (b) each mass is 0.200 m from the axis. (c) two masses are on the axis and two are

0.400 m from the axis. 2

2

⎛ 0.400 m ⎞ 2 Solve: (a) I = ∑ mr 2 = 4(0.200 kg) ⎜ ⎟ = 0.0640 kg ⋅ m ⎝ 2 ⎠ (b) I = 4(0.200 kg)(0.200 m) 2 = 0.0320 kg ⋅ m 2 2

⎛ 0.400 m ⎞ 2 (c) I = 2(0.200 kg) ⎜ ⎟ = 0.0320 kg ⋅ m ⎝ 2 ⎠ Reflect: The value of I depends on the location of the axis.

9.34. Set Up: For a point mass, I = mR 2 , where R is its distance from the axis. Use Table 9.2 for the objects that have a distribution of mass. 1 Solve: (a) I = I bar + 2mR 2 = 12 M bar L2 + 2mR 2 1 I = 12 (2.50 kg)(1.50 m) 2 + 2(1.25 kg)(0.75 m) 2 = 1.88 kg ⋅ m 2


9-8

Chapter 9

(b) The mass at the axis makes no contribution to I. I = I bar + mR 2 = 13 M bar L2 + mR 2 = 13 (2.50 kg)(1.50 m) 2 + (1.25 kg)(1.50 m) 2 = 4.69 kg ⋅ m 2 (c) I = I hoop + 6 I spokes = M hoop R 2 + 6

(

1M R2 3 spoke

) = (M

hoop

+ 2 M spoke ) R 2

I = (0.725 kg + 2[0.112 kg])(0.300 m)2 = 0.0854 kg ⋅ m 2

9.35. Set Up: K = 12 I ω 2 . Use Table 9.2 to calculate I. I = 25 MR 2 . For the moon, M = 7.35 × 1022 kg and R = 1.74 × 106 m. The moon moves through 1 rev = 2π rad in 27.3 d. 1 d = 8.64 × 104 s. Solve: (a) I = 52 (7.35 × 1022 kg)(1.74 × 106 m) 2 = 8.90 × 1034 kg ⋅ m 2 .

ω=

2π rad 4

(27.3 d)(8.64 × 10 s/d)

= 2.664 × 1026 rad/s.

K = 12 I ω 2 = 12 (8.90 × 1034 kg ⋅ m 2 )(2.664 × 10−6 rad/s)2 = 3.16 × 1023 J. 3.16 × 1023 J

= 126 years. Considering the expense involved in tapping the moon’s rotational energy, this does 5(5.0 × 1020 J) not seem like a worthwhile scheme for only 126 years worth of energy. Reflect: The moon has a very large amount of kinetic energy due to its motion. The earth has even more, but changing the rotation rate of the earth would change the length of a day. (b)

9.36. Set Up: The rim is a thin-walled hollow cylinder with I = mrim R 2 and each of the 8 spokes can be treated as a slender rod with the axis at one end, so for each spoke I = 13 mspoke R 2 . Solve: I = mrim R 2 + 8

(

1 m R2 3 spoke

) = [1.40 kg +

8 (0.280 3

kg)][0.300 m]2 = 0.193 kg ⋅ m 2

*9.37. Set Up: K = 12 I ω 2 . Use Table 9.2 to relate I to the mass M of the disk. 45.0 rpm = 4.71 rad/s. For a uniform solid disk we have I = 12 MR 2 . Solve: (a) I =

2K 2

ω

=

2(0.250 J) (4.71 rad/s) 2

(b) I = 12 MR 2 and M =

2I

2(0.0225 kg ⋅ m 2 )

= 0.500 kg. R (0.300 m) 2 Reflect: No matter what the shape is, the rotational kinetic energy is proportional to the mass of the object. 2

=

= 0.0225 kg ⋅ m 2 .

9.38. Set Up: For a solid disk and an axis through its center, I = 12 MR 2 . K = 12 I ω 2 , where ω must be in rad/s. 1 rpm =

2π rad/s 60

Solve: (a) I = 12 MR 2 = 12 (3.00 kg)(0.100 m) 2 = 0.0150 kg ⋅ m 2 . ω = 2200 rpm = 230 rad/s.

K = 12 I ω 2 = 12 (0.0150 kg ⋅ m 2 )(230 rad/s) 2 = 397 J (b) In free fall, gravitational potential energy mgy is converted to kinetic energy. K = mgy and

y=

K 397 J = = 13.5 m mg (3.00 kg)(9.80 m/s 2 )

*9.39. Set Up: K = 12 I ω 2 , with ω in rad/s. 1 rev/min = (2π /60) rad/s. ΔK = − 500 J Solve: ω i = 650 rev/min = 68.1 rad/s. ω f = 520 rev/min = 54.5 rad/s.

ΔK = K f − Ki = 12 I (ωf 2 − ωi 2 ) and I=

2(ΔK ) 2

ωf − ω i

2

=

2(−500 J) (54.5 rad/s)2 − (68.1 rad/s)2

= 0.600 kg ⋅ m 2


Rotational Motion

9-9

9.40. Set Up: K = 12 Iω 2 . Use Table 9.2 to calculate I. I = 121 ML2 . 1 rpm = 0.1047 rad/s 1 Solve: (a) I = 12 (117 kg)(2.08 m) 2 = 42.2 kg ⋅ m 2 .

⎛ 0.1047 rad/s ⎞ = 251 rad/s. ⎝ 1 rev/min ⎠⎟

ω = (2400 rev/min) ⎜

K = 12 I ω 2 = 12 (42.2 kg ⋅ m 2 )(251 rad/s) 2 = 1.33 × 106 J . 1 1 (b) K1 = 12 M1L12ω1 2 , K 2 = 12 M 2 L2 2ω 22 . L1 = L2 and K1 = K 2 , so M1ω 12 = M 2ω 2 2 .

ω 2 = ω1

M1 = (2400 rpm) M2

M1 = 2770 rpm 0.750M1

Reflect: The rotational kinetic energy is proportional to the square of the angular speed and directly proportional to the mass of the object.

9.41. Set Up: The volume of a hollow cylinder of inner radius R1, outer radius R2 , and length L is V = π L( R22 − R12 ). I = 12 M ( R12 + R22 ). Let the density be ρ ; M = ρV . Solve: (a) ω =

1 rev = 0.571 rev/s = 3.59 rad/s 1.75 s

K = 12 I ω 2 = L=

(

1 1 M [R 2 1 2 2

4K

ρπ (R 2 4 − R14 )ω 2

=

)

+ R22 ] ω 2 = 14 ρπ L( R22 − R12 )( R12 + R2 2 )ω 2 = 14 ρπ L( R2 4 − R14 )ω 2 4(2.5 × 106 J) (2.20 × 103 kg/m3 )π ([1.50 m]4 − [0.500 m ]4 )(3.59 rad/s) 2

= 22.5 m

(b) Length L is proportional to 1/ω 2 , so doubling ω reduces L to (22.5 m)/4 = 5.63 m. Reflect: The length could also be reduced by using material of larger density.

9.42. Set Up: Use coordinates where +y is upward. Take the origin at the final position of the stone, so for the stone yf = 0 and yi = 2.50 m. The cylinder has no change in gravitational potential energy. The cylinder has rotational kinetic energy and the stone has translational kinetic energy. Let m be the mass of the stone and let M be the mass of the cylinder. For the cylinder I = 12 MR 2 . The speed of the stone and the angular speed ω of the cylinder are related by υ = Rω . Solve: Conservation of energy says U i + Ki = U f + K f . Ki = 0 and U f = 0, so U i = K f . The conservation of

energy expression becomes mgyi = 12 mυ 2 + 12 I ω 2 . 1 Iω 2 2

M=

=

(

1 1 MR 2 2 2

2m(2 gyi − υ 2 )

υ

2

=

) (υ /R)

2

= 14 M υ 2 , so mgyi = 12 mυ 2 + 14 M υ 2 and

2(3.00 kg)(2[9.80 m/s 2 ][2.50 m] − [3.50 m/s]2 ) (3.50 m/s)2

= 18.0 kg

*9.43. Set Up: The speed υ of the weight is related to ω of the cylinder by υ = Rω , where R = 0.325 m. Use coordinates where +y is upward and yi = 0 for the weight. yf = − h, where h is the unknown distance the weight descends. Let m = 1.50 kg and M = 3.25 kg. For the cylinder I = 12 MR 2 . Solve: (a) Conservation of energy says Ki + U i = K f + U f . Ki = 0 and U i = 0. U f = mgyf = − mgh.

K f = 12 mυ 2 + 12 I ω 2 = 12 mυ 2 +

(

1 1 MR 2 2 2

)

2

⎛ υ⎞ ⎜⎝ ⎟⎠ = R

( 12 m + 14 M ) υ 2

( 12 m + 14 M ) υ 2 − mgh = 0

h=

( 12 m + 14 M ) υ2 = ⎡⎣ 12 (1.50 kg) + 14 (3.25 kg)⎤⎦ (2.50 m/s)2 = 0.664 m mg

(1.50 kg)(9.80 m/s 2 )


9-10 (b) ω =

Chapter 9

υ R

=

2.50 m/s = 7.69 rad/s 0.325 m

Reflect: The net work done by the rope that connects the cylinder and weight is zero. The speed v of the weight equals the tangential speed at the outer surface of the cylinder, and this gives υ = Rω .

9.44. Set Up: I for the compound disk is the sum of I of the solid disk and of the ring. For the solid disk we have I = 12 md rd 2 . For the ring we have I r = 12 mr (r1 2 + r2 2 ), where r1 = 50.0 cm, r2 = 70.0 cm. The mass of the disk and ring is their area times their area density. Solve: I = I d + I r .

Disk: md = (3.00 g/cm 2 )π rd 2 = 23.56 kg. I d =

1 md rd 2 = 2.945 kg ⋅ m 2 . 2

1 Ring: mr = (2.00 g/cm 2 )π (r2 2 − r1 2 ) = 15.08 kg. I r = mr (r1 2 + r2 2 ) = 5.580 kg ⋅ m 2 . 2

I = I d + I r = 8.52 kg ⋅ m 2 . Reflect: Even though mr < md , I r > I d since the mass of the ring is farther from the axis.

9.45. Set Up: υcm = Rω . ω = 0.50 rev/s = 3.14 rad/s. I = 12 MR 2 with R = 0.50 m. Solve: (a) K tot = K cm + K rot with K cm = 12 M υcm 2 and K rot = 12 I cmω 2 .

υcm = Rω = (0.50 m)(3.14 rad/s) = 1.57 m/s. K cm = 12 (75 kg)(1.57 m/s)2 = 92.4 J. K rot = 12 I cmω 2 = 14 MR 2ω 2 = 14 M υcm 2 = 46.2 J. K tot = 92.4 J + 46.2 J = 140 J (b)

K rot 46.2 J = = 33, K tot 140 J

9.46. Set Up: The wheel has I = MR 2 , with M = 2.25 kg and R = 0.425 m. Rolling without slipping means υcm = Rω for the wheel. Initially the wheel has υcm,i = 11.0 m/s. Use coordinates where +y is upward and y = 0 at the bottom of the hill, so yi = 75.0 m and yf = 0. Solve: (a) Conservation of energy gives Ki + U i = K f + U f . 2

⎛υ ⎞ K = 12 mυcm 2 + 12 I cmω 2 = 12 mυcm 2 + 12 (mR 2 ) ⎜ cm ⎟ = mυcm 2 . ⎝ R ⎠

Ki = mυcm,i 2 , K f = mυcm,f 2 . U i = mgyi , U f = mgyf = 0 so mgyi + mυcm,i 2 = mυcm,f 2 .

υcm,f = υcm,i 2 + gyi = (11.0 m/s) 2 + (9.80 m/s 2 )(75.0 m) = 29.3 m/s (b) K f = mυcm,f 2 = (2.25 kg)(29.3 m/s) 2 = 1.93 × 103 J

*9.47. Set Up: Since there is rolling without slipping, vcm = Rω . The kinetic energy is given by Eq. (9.19). We have ω = 3.00 rad/s and R = 0.600 m. For a hoop rotating about an axis at its center we have I = MR 2 . Solve: (a) vcm = Rω = (0.600 m)(3.00 rad/s) = 1.80 m/s. (b) K = 12 Mvcm 2 + 12 I ω 2 = 12 Mvcm 2 + 12 ( MR 2 )(vcm /R 2 ) = Mvcm 2 = (2.20 kg)(1.80 m/s) 2 = 7.13 J Reflect: For the special case of a hoop, the total kinetic energy is equally divided between the motion of the center of mass and the rotation about the axis through the center of mass. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Rotational Motion

9-11

9.48. Set Up: The solid sphere has I cm = 52 mR 2 ; the uniform spherical shell has I cm = 23 mR 2 . υcm = Rω . Use coordinates where +y is upward and y = 0 at the bottom of the hill. Then yi = h and yf = 0. Use the notation that the solid sphere has υcm = υsolid and the hollow shell has υcm = υshell at the bottom of the hill. Solve: (a) Conservation of energy gives Ki + U i = K f + U f . Ki = 0, U f = 0 so U i = K f .

mgh = 12 mυcm 2 + 12 I cmω 2 solid sphere: I cm = 25 mR 2 so

1 I ω 2 2 cm cm

7 mυsolid 2 and υsolid = Then mgh = 10

hollow sphere: I cm = 23 mR 2 so

10 7

=

6 5

) ⎛⎜⎝ υ R

solid ⎞

2

2 1 ⎟⎠ = 5 mυsolid

gh = 1.20 gh .

1 I ω 2 2 cm cm

Then mgh = 56 mυshell 2 and υshell =

(

1 2 mR 2 2 5

=

(

1 2 mR 2 2 3

) ⎛⎜⎝ υ R

shell ⎞

2

2 1 ⎟⎠ = 3 mυshell

gh = 1.10 gh .

(b) The solid sphere is moving faster when it reaches the bottom and arrives there before the hollow sphere.

*9.49. Set Up: The ball has moment of inertia I cm = 23 mR 2 . Rolling without slipping means υcm = Rω . Use coordinates where +y is upward and y = 0 at the bottom of the hill, so yi = 0 and yf = h = 5.00 m. Solve: (a) Conservation of energy gives Ki + U i = K f + U f . U i = 0, K f = 0 (stops). Therefore Ki = U f and 1 mυcm 2 2

+ 12 I cmω 2 = mgh. 1 I ω2 2 cm

(

=

1 2 mR 2 2 3

υcm =

6 gh = 5

and

ω=

) ⎛⎜⎝ υR ⎞⎟⎠ cm

2

= 13 mυcm 2 so

5 mυcm 2 6

= mgh.

6(9.80 m/s 2 )(5.00 m) = 7.67 m/s 5

υcm R

=

7.67 m/s = 67.9 rad/s. 0.113 m

(b) K rot = 12 I ω 2 = 13 mυcm 2 = 13 (0.426 kg)(7.67 m/s) 2 = 8.35 J Reflect: Its translational kinetic energy at the base of the hill is

1 mυcm 2 = 32 K rot 2 2

= 12.52 J. Its total kinetic energy is

20.9 J. This equals its final potential energy: mgh = (0.426 kg)(9.80 m/s )(5.00 m) = 20.9 J

9.50. Set Up: The marble is a solid sphere and has I cm = 52 mR 2 . Since the marble rolls without slipping, υcm = Rω . The block of ice has only translational kinetic energy. At the bottom of the hill, the marble has speed υcm = υ m and the block has speed υcm = υ b . Use coordinates where +y is upward and y = 0 at the bottom of the hill. Then yi = H and yf = 0 for each object. Solve: (a) Conservation of energy gives Ki + U i = K f + U f . Ki = 0, U f = 0 so U i = K f .

marble: mgH = 12 mυcm 2 + 12 I cmω 2 1 I ω 2 2 cm cm 7 mυ m 2 and υ m = so mgH = 10

10 7

=

(

1 2 mR 2 2 5

) ⎛⎜⎝ υR ⎞⎟⎠ m

2

= 15 mυ m 2

gh = 1.20 gH .

block of ice: mgH = 12 mυ b 2 and υ b = 2 gH = 1.41 gH . (b) The ice is moving faster at the bottom. (c) For each object, K f = U i = mgH . They have the same kinetic energy at the bottom. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9-12

Chapter 9

*9.51. Set Up: Apply Eq. (9.19). For an object that is rolling without slipping we have vcm = Rω . Solve: The fraction of the total kinetic energy that is rotational is (1/2) I cmω 2 1 1 = = (1/2) Mvcm 2 + (1/2) I cmω 2 1 + ( M /I cm )vcm 2 /ω 2 1 + ( MR 2 /I cm ) (a) I cm = (1/2) MR 2 , so the above ratio is 1/3. (b) I cm = (2/5) MR 2 so the above ratio is 2/7. (c) I cm = (2/3) MR 2 so the ratio is 2/5. (d) I cm = (5/8) MR 2 so the ratio is 5/13. Reflect: The moment of inertia of each object takes the form I = β MR 2 . The ratio of rotational kinetic energy to

total kinetic energy can be written as

1 β = . The ratio increases as β increases. 1 + 1/β 1 + β

9.52. Set Up: Only gravity does work, so Wother = 0 and conservation of energy gives Ki + U i = K f + U f . Let yf = 0, so U f = 0 and yi = 0.750 m. The hoop is released from rest so Ki = 0. K f = 12 M υcm 2 + 12 I cmω 2 .

υ cm = Rω . For a hoop with an axis at its center, I cm = MR 2 . Solve: Conservation of energy gives U i = K f . K f = 12 MR 2ω 2 + 12 ( MR 2 )ω 2 = MR 2ω 2 , so MR 2ω 2 = Mgyi .

ω=

gyi = R

(9.80 m/s 2 )(0.750 m) = 33.9 rad/s 0.0800 m

9.53. Set Up: Solve this problem using energy conservation: ΔK = −ΔU . The change in potential energy of the cart is ΔU = Mg Δy , where the total mass is M = 150.0 kg + 4(45.0 kg) + 750.0 kg = 1080.0 kg and the vertical displacement of the cart is given by Δy = − (16.0 m)sin 7.50D = −2.09 m. The kinetic energy of the cart consists of its translational kinetic energy and the rotational kinetic energy of its four identical wheels: K =

1 ⎛1 ⎞ M υ 2 + 4 ⎜ Iω 2 ⎟ . ⎝ ⎠ 2 2

1 The moment of inertia of each wheel is that of a solid cylinder: I = mr 2 . 2 Solve: ΔU = Mg Δy = (1080.0 kg)(9.80 m/s 2 )(−2.09 m) = −2.21 × 104 J. 1 υ Substitute I = mr 2 and ω = into the expression for K to obtain: 2 r

K=

2 ⎛1⎛1 1 ⎞⎛ υ ⎞ ⎞ M υ 2 + 4 ⎜ ⎜ mr 2 ⎟⎜ ⎟ ⎟ = ⎜ 2⎝ 2 2 ⎠⎝ r ⎠ ⎟⎠ ⎝

4 ⎛1 ⎞ 2 2 ⎜⎝ (1080.0 kg) + (45.0)⎟⎠ υ = (585.0 kg)υ . 2 4

Thus, using energy conservation we obtain (585.0 kg)υ 2 = 2.21 × 104 J. Finally, solve for υ to obtain

υ=

2.21 × 104 J = 6.15 m/s. 585.0 kg

Reflect: If we ignore the rotational energy of the wheels, we would get υ = 2 gh =

2(9.80 m/s 2 )(2.09 m) = 6.40 m/s

for the speed of the cart at the bottom of the slope. The actual speed of the cart is only slightly slower than this due to the relatively small moment of inertia of the wheels.


Rotational Motion

9-13

9.54. Set Up: Apply conservation of energy to the motion of the marble. We have K = 12 mv 2 + 12 Iω 2 , with I = 25 MR 2 . Also, we know vcm = Rω for no slipping. Let y = 0 at the bottom of the bowl. The marble at its initial

and final locations is sketched in the figure below. 1 1 Solve: (a) Motion from the release point to the bottom of the bowl: mgh = mv 2 + I ω 2 . 2 2 2

1 1⎛ 2 10 ⎞⎛ v⎞ mgh = mv 2 + ⎜ mR 2 ⎟ ⎜ ⎟ and v = gh . ⎝ ⎠ ⎝ ⎠ 2 2 5 7 R Motion along the smooth side: The rotational kinetic energy does not change, since there is no friction torque on the 10 gh 1 5 v2 marble, mv 2 + K rot = mgh′ + K rot . h′ = = 7 = h 2 2g 2g 7 (b) mgh = mgh′ so h′ = h. Reflect: (c) With friction on both halves, all the initial potential energy gets converted back to potential energy. Without friction on the right half some of the energy is still in rotational kinetic energy when the marble is at its maximum height.

*9.55. Set Up: The linear acceleration a of the elevator equals the tangential acceleration of a point on the rim of the shaft. a = 0.150 g = 1.47 m/s. For the shaft, R = 0.0800 m.

Solve: atan = Rα so a = Rα and α =

1.47 m/s 2 = 18.4 rad/s 2 0.0800 m

Reflect: In atan = Rα , α is in units of rad/s 2 .

9.56. Set Up: Apply conservation of energy to the motion of the wheel. The wheel at points 1 and 2 of its motion is shown in the figure below. Take the point y = 0 at the center of the wheel when it is at the bottom of the hill.

The wheel has both translational and rotational motion so its kinetic energy is K = 12 I cmω 2 + 12 Mvcm 2 . Solve: K1 + U1 + Wother = K 2 + U 2

Wother = Wfric = 23500 J (the friction work is negative) K1 = 12 I ω1 2 + 12 Mv1 2 ; v = Rω and I = 0.800MR 2 so K1 = 12 (0.800) MR 2ω 1 2 + 12 MR 2ω 1 2 = 0.900MR 2ω 1 2 K 2 = 0, U1 = 0, U 2 = Mgh

Thus 0.900MR 2ω 1 2+ Wfric = Mgh


9-14

Chapter 9

M = w/g = 392 N/(9.80 m/s 2 ) = 40.0 kg h= h=

0.900MR 2ω 1 2+ Wfric Mg (0.900)(40.0 kg)(0.600 m)2 (25.0 rad/s)2 − 3500 J

(40.0 kg)(9.80 m/s2 ) Reflect: Friction does negative work and reduces h.

= 11.7 m

*9.57. Set Up: The distance d the car travels equals the arc length s traveled by a point on the rim of the tire, so d = rθ . The odometer reading depends on the angle through which the wheels have turned. Solve: (a) d = 0.10 mi = 528 ft. r = 12 in. @ 1 ft. θ =

d 528 ft = = 528 rad = 84.0 rev. 1 ft r

(b) d = rθ . 5000 rev = 3.14 × 104 rad. d = (1 ft)(3.14 × 104 rad) = 3.14 × 104 ft = 5.95 mi (c) With proper 24 in. diameter tires the angular displacement for d = 500 mi = 2.64 × 106 ft is

θ=

d 2.64 × 106 ft = = 2.64 × 106 rad. 1 ft r

With 28 in. tires this θ corresponds to d = rθ =

( 1412 ft ) (2.64 × 106 rad) = 3.08 × 106 ft = 583 mi.

Reflect: In s = rθ the angle θ must be expressed in radians.

9.58. Set Up: The speed υ of the car equals the tangential speed υ tan of a point on the rim of the tire, so υ = rω . In this equation, ω must be expressed in rad/s. 55 mph = 80.7 ft/s. The speedometer reading depends on the angular speed with which the wheels are turning. υ 80.7 ft/s ⎛ 1 rev ⎞ ⎛ 60 s ⎞ Solve: (a) ω = = = (80.7 rad/s) ⎜ = 770 rpm ⎝ 2π rad ⎟⎠ ⎜⎝ 1 min ⎟⎠ 1 ft r (b) 500 rpm = 52.4 rad/s. υ = rω = (1 ft)(52.4 rad/s) = 52.4 ft/s = 36 mph (c) When the speedometer reads 50 mph = 73.3 ft/s the wheels have angular speed

ω=

υ r

=

73.3 ft/s = 88.0 rad/s. 10 ft 12

The speedometer reads the υ for this ω if standard tires were used. For ω = 88.0 rad/s and r = 1 ft, υ = rω = 88.0 ft/s = 60 mph.

*9.59. Set Up: The translational kinetic energy is K = 12 mv 2 and the kinetic energy of the rotating flywheel is K = 12 I ω 2 . Use the scale speed to calculate the actual speed v. From that calculate K for the car and then solve for

ω that gives this K for the flywheel. Solve: (a)

vtoy vscale

=

Ltoy Lreal

⎛ Ltoy ⎞ ⎛ 0.150 m ⎞ vtoy = vscale ⎜ ⎟ = (700 km/h) ⎜ ⎟ = 35.0 km/h ⎝ 3.0 m ⎠ ⎝ Lreal ⎠ vtoy = (35.0 km/h)(1000 m/1 km)(1 h/3600 s) = 9.72 m/s (b) K = 12 mv 2 = 12 (0.180 kg)(9.72 m/s) 2 = 8.50 J (c) K = 12 I ω 2 gives that ω =

2K 2(8.50 J) = = 652 rad/s I 4.00 × 10−5 kg ⋅ m 2

Reflect: K = 12 I ω 2 gives ω in rad/s. 652 rad/s = 6200 rev/min so the rotation rate of the flywheel is very large.


Rotational Motion

9-15

9.60. Set Up: K = 12 I ω 2 , with ω in rad/s. For a solid cylinder, I = 12 MR 2 . 1 rev/min = (2π /60) rad/s Solve: ω = 3000 rev/min = 314 rad/s. I = 12 (1000 kg)(0.900 m)2 = 405 kg ⋅ m 2

K = 12 (405 kg ⋅ m 2 )(314 rad/s) 2 = 2.00 × 107 J

9.61. Set Up: I = 25 mR 2 . The earth has m = 5.97 × 1024 kg and R = 6.38 × 106 m. Its rotational period is 24 h = 8.64 × 104 s. The earth has orbit radius r = 1.50 × 1011 m and orbit period 365.3 days = 3.156 × 107 s. Solve: (a) K = 12 mυ 2

υ=

2π r 3.156 × 107 s

=

2π (1.50 × 1011 m) 3.156 × 107 s

= 2.986 × 104 m/s

K = 12 (5.97 × 1024 kg)(2.986 × 104 m/s)2 = 2.66 × 1033 J (b) K rot = 12 I ω 2

I = 52 mR 2 = 52 (5.97 × 1024 kg)(6.38 × 106 m) 2 = 9.72 × 1037 kg ⋅ m 2

ω=

θ t

=

2π rad 8.64 × 104 s

= 7.27 × 10−5 rad/s

K rot = 12 (9.72 × 1037 kg ⋅ m 2 )(7.27 × 10−5 rad/s) 2 = 2.57 × 1029 J (c)

2.57 × 1029 J 2.66 × 1033 J + 2.57 × 1029 J

= 9.66 × 10−5 = 9.66 × 10−3,

9.62. Set Up: 200 rpm = 20.94 rad/s. 500 rpm = 52.36 rad/s. The outer radius is 6.05 × 10−2 m. The angular speed is greatest when r is smallest. 74 min = 4440 s. Solve: (a) υ tan = rω . ω = 20.94 rad/s when r = 6.05 × 1022 m so υ tan = 1.27 m/s. (b) Find r when ω = 52.36 rad/s: r =

ω − ω0

υ tan 1.27 m/s = = 0.0242 m. The diameter is 4.84 cm. 52.36 rad/s ω

20.94 rad/s − 52.36 rad/s = = − 7.1 × 10−3 rad/s 2 4440 s t α is negative since the rotation rate slows as the tracking spirals outward.

(c) ω = ω 0 + α t gives α =

*9.63. Set Up: All points on the belt (which is shown in the figure below) move with the same speed. Since the belt doesn’t slip, the speed of the belt is the same as the speed of a point on the rim of the shaft and on the rim of the wheel, and these speeds are related to the angular speed of each circular object by v = rω . Solve: (a) v1 = r1ω 1

ω1 = (60.0 rev/s)(2π rad/1 rev) = 377 rad/s v 1 = r 1ω 1 = (0.45 × 10−2 m)(377 rad/s) = 1.70 m/s (b) v 1 = v 2

r 1ω 1 = r 2ω 2

ω 2 = ( r1 /r 2)ω 1 = (0.45 cm/2.00 cm)(377 rad/s) = 84.8 rad/s Reflect: The wheel has a larger radius than the shaft so it turns slower to have the same tangential speed for points on the rim.


9-16

Chapter 9

9.64. Set Up: I cm = 23 mR 2 . When it rolls without slipping, υcm = Rω . When there is no friction the angular speed of rotation is constant. Take +y upward and let y = 0 in the valley. Solve: (a) Find the speed υcm in the level valley: Ki + U i = K f + U f . yi = H 0 , yf = 0. Ki = 0, U f = 0. Therefore,

U i = Kf . mgH 0 = 12 mυcm 2 + 12 I cmω 2 1 I ω2 2 cm

=

(

1 2 mR 2 2 3

)

2

⎛ υcm ⎞ 2 1 ⎜⎝ ⎟ = 3 mυcm , R ⎠

6 gH 0 . 5 Find the height H it goes up the other side. Its rotational kinetic energy stays constant as it rolls on the frictionless surface. 1 mυcm 2 + 12 I cmω 2 = 12 I cmω 2 + mgH 2

so mgH 0 = 65 mυcm 2 and υcm 2 =

υcm 2

= 53 H 0 2g (b) Some of the initial potential energy has been converted into rotational kinetic energy so there is less potential energy at the second height H than at the first height H 0 . H=

*9.65. Set Up: My total mass is m = 90 kg. I model my head as a uniform sphere of radius 8 cm. I model my trunk and legs as a uniform solid cylinder of radius 12 cm. I model my arms as slender rods of length 60 cm. ω = 72 rev/min = 7.5 rad/s. Solve: (a) I tot = 25 (0.070m)(0.080 m) 2 + 12 (0.80m)(0.12 m)2 + 2

( 13 ) (0.13m)(0.60 m)2 = 3.3 kg ⋅ m2

(b) K rot = 12 I ω 2 = 12 (3.3 kg ⋅ m 2 )(7.5 rad/s) 2 = 93 J Reflect: According to these estimates about 85% of the total I is due to the outstretched arms. If the initial translational kinetic energy 12 mυ 2 of the skater is converted to this rotational kinetic energy as he goes into a spin, his initial speed must

be 1.4 m/s.

9.66. Set Up: I cm = 52 mR 2 . When it rolls without slipping, υcm = Rω . When there is no friction (on the smooth ice), ω and the rotational kinetic energy are constant. Use coordinates where +y is upward and y = 0 at the bottom of the hill. Solve: Motion along the rough part: Let υ 1 be the translational speed at the bottom of the rough part. Ki + U i = K f + U f . Ki = 0. yi = 50.0 m; yf = 25.0 m mgyi = 12 mυ12 + 12 I cmω 2 + mgyf 1 I ω2 2 cm

=

(

1 2 mR 2 2 5

)

2

⎛ υ1 ⎞ 2 2 7 1 ⎜⎝ ⎟⎠ = 5 mυ1 , so g ( yi − yf ) = 10 υ1 and R

υ1 =

10 (9.80 7

m/s 2 )(25.0 m/s) = 18.7 m/s

Motion along the smooth part: υ i = υ1. yi = 25.0 m; yf = 0. Ki + U i = K f + U f gives 1 mυ12 2

+ 12 I cmω 2 + mgyi = 12 mυf 2 + 12 I cmω 2

υf = υ12 + 2 gyi = (18.7 m/s) 2 + 2(9.80 m/s2 )(25.0 m) = 29.0 m/s

9.67. Set Up: Treat the V like two thin 0.160 kg bars, each 25 cm long. For a slender bar with the axis at one end we have I = 13 mL2 .


Rotational Motion

9-17

⎛1 ⎞ ⎛ 1⎞ Solve: I = 2 ⎜ mL2 ⎟ = 2 ⎜ ⎟ (0.160 kg)(0.250 m)2 = 6.67 × 1023 kg ⋅ m 2 ⎝3 ⎠ ⎝ 3⎠ Reflect: The value of I is independent of the angle between the two sides of the V; the angle 70.0D didn't enter into the calculation.

9.68. Set Up: The solid sphere has I = 52 MR 2 and the hollow spherical shell has Ishell = 23 M shell R 2 . Solve: K = 12 I ω 2 so same K and ω means I = Ishell .

2 MR 2 5

= 23 M shell R 2 and M shell = 53 M

*9.69. Set Up: I cm = 52 mR 2 . If the stone rolls without slipping, υcm = Rω . If there is no friction the stone slides without rolling and has no rotational kinetic energy. Use coordinates where 1y is upward and y = 0 at the bottom of the hill. Solve: After the stone is launched into the air its translational kinetic energy is converted to potential energy. At its maximum height h, mgh = 12 mυcm 2 , where υcm is its translational speed at the bottom of the hill. (a) Apply conservation of energy to the motion down the hill: mgH = 12 mυcm 2 and υcm 2 = 2 gH . Then

mgh = 12 mυcm 2 = 12 m(2 gH ) and h = H . (b) Now the initial potential energy is converted to both translational and rotational kinetic energy as the stone rolls down the hill: mgH = 12 mυcm 2 + 12 I cmω 2 1 I ω2 2 cm

This gives mgh = 12 m

=

(

1 2 mR 2 2 5

)

2

⎛ υcm ⎞ 2 2 2 7 10 1 ⎜⎝ ⎟ = 5 mυcm so mgH = 10 mυcm and υcm = 7 gH R ⎠

( 107 gH ) and h = 75 H .

(c) In (a) all the initial gravitational potential energy is converted back to final gravitational potential energy; the kinetic energy at the final maximum height is zero. In (b) the stone gains rotational kinetic energy as it rolls down the hill and it still has this rotational kinetic energy at its maximum height; not all the initial potential energy is converted into the final potential energy. Reflect: Both answers do not depend on the mass or radius of the stone. But the answer to (b) depends on how the mass is distributed; the answer would be different for a hollow rolling sphere.

9.70. Set Up: Use conservation of energy, Ki + U i = K f + U f , to find the speed υf of the ball just as it leaves the top of the cliff. Let y = 0 at the bottom of the cliff, so yi = 0, U i = 0 and yf = 28.0 m. K = 12 I cmω 2 + 12 mυ 2 . I cm = 52 mR 2 . Rolling without slipping gives υ = Rω . Consider the projectile motion of the ball after it goes over the

cliff to find its horizontal range and final speed. Solve: (a) While rolling without slipping, K = 7 mυi 2 10

(

1 2 mR 2 2 5

)ω

2

7 + 12 mυ 2 = 10 mυ 2 . Conservation of energy gives

7 = mgyf + 10 mυf 2 . υf = υ i 2− 10 gyf = 15.3 m/s. 7

Projectile motion: Use the vertical motion to find the time in the air. Let +y be downward. υ 0 y = 0, a y = 9.80 m/s 2 and y − y0 = 28.0 m. y − y0 = υ 0 yt + 12 a yt 2 gives t=

2( y − y0 ) = 2.39 s. ay

During this time the ball travels horizontally x − x0 = υ0 xt = (15.3 m/s)(2.39 s) = 36.6 m.

υ y = υ0 y + a yt = 0 + (9.80 m/s 2 )(2.39 s) = 23.4 m/s. υ x = υ0 x = 15.3 m/s and υ = υ x 2 + υ y 2 = 28.0 m/s


9-18

Chapter 9

(b) The rotation rate of the ball is less at the top of the hill than it was initially. While it moves through the air the rotation rate stays the same, so when the ball returns to the ground, the amount of energy in rotational kinetic energy is less than it was initially, before it started to roll up the hill. The total energy is the same, so at the end more energy is in translational motion than initially and the final translational speed is greater.

9.71. Set Up: For a uniform bar pivoted about one end, I = 13 mL2 . υ = 5.0 km/h = 1.4 m/s. (a) 60° = (π /3) rad. The average angular speed of each arm and leg is

π /3 rad 1s

= 1.05 rad/s.

(b) I = 13 marm Larm 2 + 13 mleg Lleg 2 = 13 [(0.13)(75 kg)(0.70 m) 2 + (0.37)(75 kg)(0.90 m) 2 ]

I = 9.08 kg ⋅ m 2 K rot = 12 I ω 2 = 12 (9.08 kg ⋅ m2 )(1.05 rad/s)2 = 5.0 J (c) K tran = 12 mυ 2 = 12 (75 kg)(1.4 m/s) 2 = 73.5 J and K tot = K tran + K rot = 78.5 J (d)

K rot 5.0 J = = 6.4, K tran 78.5 J

Reflect: If you swing your arms more vigorously more of your energy input goes into the kinetic energy of walking and it is more effective exercise.

9.72. Set Up: Now υ = 12 km/h = 3.33 m/s. I tot = 9.08 kg ⋅ m 2 as in Problem 9.71. π /3 rad Solve: (a) ω av = = 2.1 rad/s 0.5 s

(b) (c) (d)

K tot = 12 I ω 2 = 12 (9.08 kg ⋅ m 2 )(2.1 rad/s) 2 = 20 J K tran = 12 mυ 2 = 12 (75 kg)(3.33 m/s)2 = 416 J; K =

436 J

K rot 20 J = = 4.6, K tot 436 J

*9.73. Set Up: Apply conservation of energy to the system consisting of blocks A and B and the pulley. The system at points 1 and 2 of its motion is sketched in the figure below.

Use the work-energy relation K1 + U1 + Wother = K 2 + U 2 . Use coordinates where +y is upward and where the origin is at the position of block B after it has descended. The tension in the rope does positive work on block A and negative work of the same magnitude on block B, so the net work done by the tension in the rope is zero. Both blocks have the same speed. Solve: Gravity does work on block B and kinetic friction does work on block A. Therefore Wother = W f = − μk mA gd . K1 = 0 (system is released from rest)


Rotational Motion

9-19

U1 = mB gyB1 = mB gd ; U 2 = mB gy B 2 = 0

K 2 = 12 m Av2 2 + 12 mB v2 2 + 12 I ω 22 . But v(blocks) = Rω (pulley), so ω 2 = v2 /R and K 2 = 12 (m A + mB )v2 2 + 12 I (v2 /R ) 2 = 12 ( m A + mB + I /R 2 )v2 2

Putting all this into the work-energy relation gives mB gd − μk m A gd = 12 (m A + mB + I /R 2 )v2 2 (m A + mB + I /R 2 )v2 2 = 2 gd (mB − μ k m A )

v2 =

2 gd (mB − μk m A ) m A + mB + I /R 2

Reflect: If mB >> m A and I /R 2 , then v2 = 2 gd ; block B falls freely. If I is very large, v2 is very small. Must

have mB > μk m A for motion, so the weight of B will be larger than the friction force on A. I /R 2 has units of mass and is in a sense the “effective mass” of the pulley.

Solutions to age Problems 9.74. Set Up: Let us consider counterclockwise rotations to be positive and clockwise rotations to be negative. Thus, ω 0 = −30 rad/s and ω = +30 rad/s. Since the angular acceleration is constant, we may use equation 9.8: ω +ω ω av = 0 . 2

Solve: ω av =

ω0 + ω 2

=

−30 rad/s + 30 rad/s = 0 rad/s. The correct answer is A. 2

*9.75. Set Up: Let us consider counterclockwise rotations to be positive and clockwise rotations to be negative. Thus, ω 0 = −30 rad/s and ω = +30 rad/s. Since the angular acceleration is constant, we may use equation 9.7: ω = ω 0 + α t. ω − ω 0 (+30 rad/s) − (−30 rad/s) = = 6 rad/s 2 . The correct answer is D. Solve: α = t

10 s

9.76. Set Up: Let us consider counterclockwise rotations to be positive and clockwise rotations to be negative. Thus, ω 0 = −30 rad/s and ω = +30 rad/s. Since the angular acceleration is constant, we may use equation 9.10: 1 Δθ = (ω 0 + ω )t. 2 1 1 Solve: Δθ = (ω 0 + ω )t = (−30 rad/s + 30 rad/s)(10 s) = 0 rad. The correct answer is A. 2 2

*9.77. Set Up: Set ω 0 = 230 rad/s. Since the angular acceleration is constant, we may use equation 9.11: 1 Δθ = ω 0t + α t 2 . We know that α = 6 rad/s 2 from Problem 9.75. 2 1 1 Solve: The displacement from 0 s to 20 s is: Δθ = ω 0t + α t 2 = (−30 rad/s)(20 s) + (6 rad/s 2 )(20 s) 2 = 600 rad. 2 2 Since the displacement from 0 s to 10 s is zero, this is also the displacement from 10 s to 20 s. Alternatively, we may 1 1 set ω 0 = 130 rad/s and t = 10 s to obtain Δθ = ω 0t + α t 2 = ( +30 rad/s)(10 s) + (6 rad/s 2 )(10 s) 2 = 600 rad. In 2 2 either case, the correct answer is D.


10

DYNAMICS OF ROTATIONAL MOTION

Answers to Multiple-Choice Problems 1. C 2. C 15. D

3. E

4. B

5. B

6. D

7. A

8. D

9. D

10. B

11. C

12. D

13. C

14. C

Solutions to Problems *10.1. Set Up: Let counterclockwise torques be positive. τ = Fl with l = rsinφ. Solve: (a) τ = + (10.0 N)(4.00 m)sin 90.0° = 40.0 N ⋅ m, counterclockwise. (b) τ = + (10.0 N)(4.00 m)sin 60.0° = 34.6 N ⋅ m, counterclockwise. (c) τ = + (10.0 N)(4.00 m)sin 30.0° = 20.0 N ⋅ m, counterclockwise. (d) τ = − (10.0 N)(2.00 m)sin 60.0° = − 17.3 N ⋅ m, clockwise. (e) τ = 0 since the force acts on the axis and l = 0. (f) τ = 0 since the line of action of the force es through the location of the axis and l = 0. Reflect: The torque of a force depends on the direction of the force and where it is applied to the object.

10.2. Set Up: Let counterclockwise torques be positive. τ = Fl with l = rsinφ. Solve: τ1 = − F1l1 = − (8.00 N)(5.00 m) = − 40.0 N ⋅ m. τ 2 = + F2l2 = (12.0 N)(2.00 m)sin 30.0° = + 12.0 N ⋅ m. ∑ τ = τ1 + τ 2 = − 28.0 N ⋅ m. The net torque is 28.0 N ⋅ m, clockwise. *10.3. Set Up: Use τ = Fl = rF sin φ to calculate the magnitude of each torque and use the right-hand rule to determine the direction of each torque. Add the torques to find the net torque. Let counterclockwise torques be positive. For the 11.9 N force ( F1) we have r = 0. For the 14.6 N force ( F2 ), r = 0.350 m and φ = 40.0D. For the 8.50 N force ( F3 ), r = 0.350 m and φ = 90.0D. Solve: τ1 = 0. τ 2 = − (14.6 N)(0.350 m)sin 40.0D = −3.285 N ⋅ m.

τ 3 = + (8.50 N)(0.350 m)sin 90.0D = +2.975 N ⋅ m. ∑ τ = −3.285 N ⋅ m + 2.975 N ⋅ m = −0.31 N ⋅ m. The net torque is 0.31 N ⋅ m and is clockwise. G G Reflect: If we treat the torques as vectors, τ 2 is into the page and τ 3 is out of the page.

10.4. Set Up: Let counterclockwise torques be positive. τ = Fl


10-1

10-2

Chapter 10

Solve: (a) τ A = + (50 N)(0.20 m)sin 60° = + 8.7 N ⋅ m, counterclockwise.

τ B = 0. τ C = − (50 N)(0.20 m) sin 30° = −5.0 N ⋅ m, clockwise. τ D = − (50 N)(0.20 m)sin 90° = − 10.0 N ⋅ m, clockwise. ∑ τ = τ A + τ B + τ C + τ D = −6.3 N ⋅ m, clockwise.

10.5. Set Up: Let counterclockwise torques be positive. τ = Fl Solve: τ1 = − F1l1 = − (18.0 N)(0.090 m) = − 1.62 N ⋅ m. τ 2 = + F2l2 = + (26.0 N)(0.090 m) = +2.34 N ⋅ m. τ 3 = + F3l3 = + (14.0 N)(0.127 m) = + 1.78 N ⋅ m. ∑ τ = τ1 + τ 2 + τ 3 = 2.50 N ⋅ m, counterclockwise. Reflect: It is important to take into the direction of each torque when computing the net torque.

10.6. Set Up: Let the direction the wheel rotates be positive. Solve: ∑ τ = (40.0 N)(0.250 m) = 10.0 N ⋅ m. α=

∑τ 10.0 N ⋅ m = = 2.00 rad/s 2 I 5.00 kg ⋅ m 2

10.7. Set Up: I = 121 mL2 . The applied force has moment arm l = 1.15 m. 1 Solve: I = 12 (24.0 kg)(2.50 m)2 = 12.5 kg ⋅ m 2 ; τ = Fl = (35.0 N)(1.15 m) = 40.2 N ⋅ m

τ = I α gives α=

τ I

=

40.2 N ⋅ m 12.5 kg ⋅ m 2

= 3.22 rad/s 2

10.8. Set Up: I = 12 mR 2 . R = 0.125 m. m = 0.750 kg. ω 0 = 220 rpm = 23.04 rev/s. Assume the wheel is initially turning clockwise and take this to be the positive sense of rotation. Solve: Use the information about the motion to find α : ω = 0 (stops)

ω = ω 0 + α t gives α=

ω − ω0 t

=

0 − 23.04 rad/s = − 0.512 rad/s 2 45.0 s

I = 12 (0.750 kg)(0.125 m) 2 = 5.86 × 10−3 kg ⋅ m 2

τ = I α = (5.86 × 10−3 kg ⋅ m 2 )(−0.512 rad/s) = − 3.00 × 10−3 N ⋅ m Friction exerts a torque of 3.00 × 10−3 N ⋅ m, in the direction opposite to the motion.

*10.9. Set Up: Let the direction the grindstone is rotating be positive. From Table 9.2, I = 12 mR 2 . Use the information about the motion to calculate α . In oτ = I α , α must be in rad/s 2 . Solve: ω 0 = 850 rev/min = 89.0 rad/s. ω = 0. t = 7.50 s. ω = ω 0 + α t gives

α=

ω − ω0 t

=

0 − 89.0 rad/s = − 11.9 rad/s 2 . 7.50 s

The net torque is due to the kinetic friction force f k between the ax and the grindstone. ∑ τ = I α gives − f k R =

fk =

(

1 mR 2 2

)α and

− mRα (50.0 kg)(0.260 m)(−11.9 rad/s 2 ) =− = 77.4 N. 2 2

f k 77.4 N = = 0.484 n 160 N Reflect: ∑ τ = I α relates the forces on the rotating object to the motion of the object. f k = μk n so μk =

10.10. Set Up: Use a constant angular acceleration equation to calculate α and then apply ∑ τ = Iα to the motion of the cylinder. We know that f k = μk n and I = 12 mR 2 = 12 (8.25 kg)(0.0750 m) 2 = 0.02320 kg ⋅ m 2 . Let the direction the

cylinder is rotating be positive. We also know that ω 0 = 220 rpm = 23.04 rad/s; ω = 0; θ − θ 0 = 5.25 rev = 33.0 rad. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Dynamics of Rotational Motion

10-3

Solve: ω 2 = ω 0 2 + 2α (θ − θ 0 ) gives α = −8.046 rad/s 2 . Στ = τ f = − f k R = − μk nR. Then ∑ τ = I α gives − μk nR = I α

and n = 2

Iα = 7.47 N. μk R

Reflect: The friction torque is directed opposite to the direction of rotation and therefore produces an angular acceleration that slows the rotation.

*10.11. Set Up: For the pulley I = 12 M ( R12 + R22 ), where R1 = 30.0 cm and R2 = 50.0 cm. The free-body diagrams

G G for the stone and for the pulley are given in the figure below. Apply oF = ma to the stone. The stone accelerates

downward, so use coordinates with + y downward. Apply ∑ τ = I α to the rotation of the pulley, with clockwise as the positive sense of rotation. n is the normal force applied to the pulley by the axle. a = R2α .

Solve: (a) ∑ Fy = ma y applied to the stone gives mg − T = ma. ∑ τ = I α applied to the pulley gives TR2 = I α . I = 12 M ( R12 + R2 2 ) and α = a /R2 , so

T = 12 M (1 + [ R1 /R2 ]2 )a Combining the two equations to eliminate T gives g 9.80 m/s 2 = a= 1 + 12 ( M /m)(1 + [ R1 /R2 ]2 ) 1 + 12 (10.0 kg/2.00 kg)(1 + [30 cm/50 cm]2 ) a = 2.23 m/s 2 (b) T = 12 (10.0 kg)(1 + [30 cm/50 cm]2 )(2.23 m/s 2 ) = 15.2 N (c) α = a /R2 = (2.23 m/s 2 )/(0.500 m) = 4.46 rad/s 2 Reflect: The tension in the wire is less than the weight of the stone (19.6 N). For the stone, the downward force is

greater than the upward force and the stone accelerates downward.

10.12. Set Up: Apply oFy = ma y to the suitcase. Let + y be downward. Apply ∑ τ = I α to the wheel. Let the counterclockwise sense of rotation be positive. The angular velocity ω and angular acceleration α of the wheel are related to the linear velocity v and linear acceleration a of the suitcase by υ = Rω and a = Rα . υ 3.50 m/s = 8.75 rad/s Solve: (a) ω = = R

0.400 m

(b) For the suitcase, y − y0 = 4.00 m, υ0 y = 0, υ y = 3.50 m/s. υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives ay =

υ y 2 − υ0 y 2 2( y − y0 )

α=

ay R

=

=

(3.50 m/s) 2 − 0 = + 1.53 m/s 2 . 2(4.00 m)

1.53 m/s 2 = 3.82 rad/s 2 . 0.400 m


10-4

Chapter 10

The free-body diagram for the suitcase is given in Figure (a) below.

∑ Fy = ma y gives mg − T = ma and T = m( g − a ) = (15.0 kg)(9.80 m/s 2 − 1.53 m/s 2 ) = 124 N. The free-body

diagram for the wheel is given in Figure (b) above. ∑ τ = I α gives TR = I α and I=

TR

α

=

(124 N)(0.400 m) 3.82 m/s 2

= 13.0 kg ⋅ m 2

*10.13. Set Up: For the pulley I = 12 MR 2 . The elevator has 22,500 N = 2300 kg. 9.80 m/s 2 G G The free-body diagrams for the elevator, the pulley, and the counterweight are given in the figure below. Apply oF = ma to the elevator and to the counterweight. For the elevator take +y upward and for the counterweight take +y downward, in each case in the direction of the acceleration of the object. Apply ∑ τ = I α to the pulley, with clockwise as the positive sense of rotation. n is the normal force applied to the pulley by the axle. The elevator and counterweight each have acceleration a. a = Rα . m1 =

Solve: (a) and (b) Calculate the acceleration of the elevator: y − y0 = υ0 yt + 12 a yt 2 gives

a=

2( y − y0 ) t

2

=

2(6.75 m) (3.00 s)2

= 1.50 m/s 2

∑ Fy = ma y for the elevator gives T1 − m1g = m1a and T1 = m1 (a + g ) = (2300 kg)(1.50 m/s 2 + 9.80 m/s2 ) = 2.60 × 104 N

∑ τ = I α for the pulley gives (T2 − T1 ) =

(

1 MR 2 2

)α. With α = a /R this becomes T

2

− T1 = 12 Ma.

T2 = T1 + 12 Ma = 2.60 × 104 N + 12 (875 kg)(1.50 m/s 2 ) = 2.67 × 104 N

∑ Fy = ma y for the counterweight gives m2 g − T2 = m2a and

m2 =

2.67 × 104 N T2 = = 3.22 × 103 kg g − a 9.80 m/s 2 − 1.50 m/s 2

and w = 3.16 × 104 N.



10-5

Reflect: The tension in the cable must be different on either side of the pulley in order to produce the net torque on the pulley required to give it an angular acceleration. The tension in the cable attached to the elevator is greater than the weight of the elevator and the elevator accelerates upward. The tension in the cable attached to the counterweight is less than the weight of the counterweight and the counterweight accelerates downward.

10.14. Set Up: The free-body diagram for the disk is shown in the figure below. R = 0.150 m and F = 100.0 N. For a uniform disk I cm = 12 MR 2 . Use the coordinates shown and take clockwise rotation to be positive.

Solve: (a) Apply ∑ Fx = max to the motion of the center of mass: F = Macm and

acm =

F 100.0 N = = 25.0 m/s 2 . M 4.00 kg

Apply ∑ τ = I α to the rotation about the center of mass: FR = I cmα FR =

(

1 MR 2 2

)α

and α =

2(100.0 N) = 333 rad/s 2 (4.00 kg)(0.150 m)

Note that α is not equal to acm /R but is twice that. F = 167 rad/s 2 . acm = 25.0 m/s 2 as before. In this case α = acm /R, but that relation is MR established by the force and torque equations and not by a restriction on the motion. (b) I cm = MR 2 and α =

10.15. Set Up: Use a constant acceleration equation to calculate α z and then apply ∑ τ = Iα . I = 23 MR 2 + 2mR 2 , where M = 8.40 kg, m = 2.00 kg, so I = 0.600 kg ⋅ m 2 .

ω 0 = 75.0 rpm = 7.854 rad/s; ω = 50.0 rpm = 5.236 rad/s; t = 30.0 s. Solve: ω = ω 0 + α t gives α = −0.08726 rad/s 2 . τ = I α = −0.0524 N ⋅ m Reflect: The torque is negative because its direction is opposite to the direction of rotation, which must be the case for the speed to decrease.

10.16. Set Up: Let + x be down the incline and let the shell be turning in the positive sense of rotation. Apply G G ∑ F = ma to the translational motion of the center of mass and ∑ τ = I α to the rotation about the center of mass. The free-body diagram for the shell is given in the figure below. From Table 9.2, I cm = 23 mR 2 .


10-6

Chapter 10

Solve: ∑ Fx = ma x gives mg sin β − f = macm . ∑ τ = I α gives fR =

(

2 mR 2 3

)α.

With α = acm /R this becomes

f = 23 macm . Combining the equations gives mg sin β − 23 macm = macm and acm =

3g sin β 3(9.80 m/s 2 )(sin 38.0°) = = 3.62 m/s 2 . 5 5

f = 23 macm = 23 (2.00 kg)(3.62 m/s 2 ) = 4.83 N. The friction is static since there is no slipping at the point of . (b) The acceleration is independent of m and doesn’t change. The friction force is proportional to m so will double; f = 9.66 N.

1 2

*10.17. Set Up: Follow the method of Example 10.5 except use I = MR 2 for the rolling cylinder rather than 2 MR 2 for the rolling sphere. Choose the positive x-axis pointing up the ramp, so the x-component of gravity and 5 the acceleration of the disk are both negative. Since the disk is slowing down, both τ and α are negative. If the disk does not slip, its rotational rate must be decreasing as its speed decreases. This implies that the force of friction at the point where the disk s the ramp must act up the ramp to oppose the rotation of the disk. Solve: Translation: ∑ Fx = − Mg sin θ + fs = Macm, x I=

Rotation: ⎛1 ⎞ ∑ τ = − fs R = I cmα = ⎜ MR 2 ⎟ α ⎝2 ⎠

Start with the rotation equations. Eliminate α by using the equation acm = Rα , which is valid since the disk is not slipping. ⎛ acm, x ⎞ 1 1 Thus, we obtain fs R = − MR 2 ⎜ ⎟ which simplifies to f s = − Macm, x . Next, substitute this into the translation 2 R 2 ⎝ ⎠ 1 2 equation to obtain − Mg sin θ − Macm, x = Macm, x . Finally, solve for acm, x to obtain: solve for acm, x = − g sin θ = 2 3 2 − (9.80 m/s 2 )sin10.0D = −1.135 m/s 2 . 3 To determine the time that it takes for the disk to stop: set υ0, x = +2.50 m/s, υ x = 0, and use υ x = υ0, x + axt. Thus, we have t =

υ x − υ0, x

=

0 − 2.50 m/s

= 2.20 s. −1.135 m/s 2 Reflect: It is surprising that—in order to create the necessary change in rotational speed—the frictional force acts up the slope regardless of whether the disk is rolling up or down the slope. Note that you can also solve this problem using energy conservation methods. ax



10-7

⎛ 2π rad ⎞ ⎛ 1 min ⎞ = 503 rad/s. 1 hp = 746 W. ⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠

10.18. Set Up: P = τω , where ω must be in rad/s. (4800 rev/min) ⎜ Solve: P = (4.30 N ⋅ m)(503 rad/s) = 2.16 × 103 W = 2.90 hp

*10.19. Set Up: (a) Let the direction of rotation of the merry-go-round be positive. Apply ∑ τ = I α to the merrygo-round to calculate α and then use a constant acceleration equation to calculate ω after 20.0 s. (b) W = τ Δθ . Calculate Δθ = θ − θ 0 from a constant acceleration equation. (c) Pav = Solve: (a) ∑ τ = I α gives

α=

W . Δt

∑ τ FR (25.0 N)(4.40 m) = = = 4.49 rad/s 2 . I I 24.5 kg ⋅ m 2

ω = ω 0 + α t = 0 + (4.49 rad/s 2 )(20.0 s) = 89.8 rad/s (b) θ − θ 0 = ω 0t + 12 α t 2 = 12 (4.49 rad/s 2 )(20.0 s)2 = 898 rad

W = τ Δθ = (25.0 N)(4.40 m)(898 rad) = 9.88 × 104 J W 9.88 × 104 J = = 4.94 × 103 W Δt 20.0 s Reflect: The power applied when ω = 89.8 rad/s, at the end of the 20.0 s time interval, is (c) Pav =

P = τω = (25.0 N)(4.40 m)(89.8 rad/s) = 9.88 × 103 W. At t = 0, ω = 0 and P = 0. ω and hence P increase linearly in time, so Pav =

Pf − Pi 2

=

9.88 × 103 W = 4.94 × 103 W. 2

This agrees with our answer in part (c).

10.20. Set Up: Let the direction of rotation of the flywheel be positive. W = τ Δθ , where Δθ = θ − θ 0 must be in radians. Solve: (a) ∑ τ = I α gives ∑ τ 2000.0 N ? m α= = = 3.45 rad/s 2 . 2 I

2

580 kg ? m

2

(b) θ − θ 0 = 4.00 rev = 25.1 rad. ω 0 = 0. ω = ω 0 + 2α (θ − θ 0 ) gives

ω = 2α (θ − θ 0 ) = 2(3.45 rad/s2 )(25.1 rad) = 13.2 rad/s (c) W = τ Δθ = (2000.0 N ⋅ m)(25.1 rad) = 5.02 × 104 J

*10.21. Set Up: P = τω , with ω in rad/s. ω = 4000.0 rev/min = 419 rad/s. The speed υ of the weight is υ = Rω . Since ω and υ are constant, the tension in the rope equals the weight of the object. 150 × 103 W = 358 N ⋅ m 419 rad/s ω τ 358 N ⋅ m (b) τ = wR so w = = = 1790 N R 0.200 m (c) υ = Rω = (0.200 m)(419 rad/s) = 83.8 m/s Solve: (a) τ =

P

=

Reflect: The rate at which the weight gains gravitational potential energy, wv, equals the power output of the motor, so P 150 × 103 W υ= = = 83.8 m/s, 1790 N w which agrees with the result we calculated for part (c). © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

10-8

Chapter 10

10.22. Set Up: For a solid uniform sphere rotating about a diameter, Table 9.2 gives I = 52 MR 2 . Solve: I = 25 (14.0 kg)(0.120 m) 2 = 0.0806 kg ⋅ m 2 .

L = I ω = (0.0806 kg ⋅ m 2 )(6.00 rad/s) = 0.484 kg ⋅ m 2 /s K = 12 I ω 2 = 12 Lω = 12 (0.484 kg ⋅ m 2 /s)(6.00 rad/s) = 1.45 J

10.23. Set Up: L = I ω . For a particle of mass m moving in a circular path at a distance r from the axis, I = mr 2 and v = rω. For a uniform sphere of mass M and radius R and an axis through its center, I = 52 MR 2 . The earth has mass mE = 5.97 × 1024 kg, radius RE = 6.38 × 106 m and orbit radius r = 1.50 × 1011 m. The earth completes one rotation on its axis in 24 h = 86,400 s and one orbit in 1 y = 3.156 × 107 s.

⎛ 2π rad ⎞ Solve: (a) L = I ω = mr 2ω = (5.97 × 1024 kg)(1.50 × 1011 m)2 ⎜ = 2.67 × 1040 kg ⋅ m 2 /s. ⎝ 3.156 × 107 s ⎠⎟ The radius of the earth is much less than its orbit radius, so it is very reasonable to model it as a particle for this calculation. ⎛ 2π rad ⎞ (b) L = I ω = 25 MR 2 ω = 52 (5.97 × 1024 kg)(6.38 × 106 m) 2 ⎜ = 7.07 × 1033 kg ⋅ m 2 /s ⎝ 86,400 s ⎟⎠

(

)

Reflect: The angular momentum associated with each of these motions is very large.

10.24. Set Up: l A = 1.50 m, lB = 0 and lC = 0.30 m Solve: (a) L = myl (i) LA = (0.300 kg)(3.50 m/s)(1.50 m) = 1.58 kg ⋅ m 2 /s, counterclockwise (ii) LB = 0

(iii) LC = (0.300 kg)(3.50 m/s)(3.0 m) = 0.32 kg ⋅ m 2 /s, clockwise (b) The answers are all the same as in (a).

*10.25. Set Up: l = 1.6 m Solve: (a) υ = 0 so L = 0 (b) Find υ : υ y 2 = υ0 y 2 + 2a y ( y − y0 ) gives υ = 2 g ( y − y0 ) = 2(9.80 m/s 2 )(2.5 m) = 7.0 m/s L = mυl = (4.0 kg)(7.0 m/s)(1.6 m) = 44.8 kg ⋅ m 2 /s, counterclockwise

Reflect: The angular momentum increases as the brick falls because of the external torque of the gravity force.

10.26. Set Up: L = I ω . ω = 0.50 rev/s = 3.14 rad/s. I = I disk + I woman . I disk = 12 mdisk R 2 and I woman = 12 mwoman R 2 . Solve: I = (55 kg + 50.0 kg)(4.0 m) 2 = 1680 kg ⋅ m 2 . L = (1680 kg ⋅ m 2 )(3.14 rad/s) = 5.28 × 103 kg ⋅ m 2 /s

*10.27. Set Up: The figure below shows the force on the drawbridge. The weight acts at the center of the drawbridge and this force has moment arm l. Take clockwise rotation to be positive.



10-9

Solve: (a) τ = wl = (15,000 N)(6.0 m)(cos60.0°) = 45,000 N ⋅ m ΔL = oτ = 45,000 kg ⋅ m 2 /s 2 Δt Reflect: The torque changes as the angle of the bridge above the horizontal changes.

(b)

10.28. Set Up: For a point mass, I = mR 2 . The total moment of inertia is I tot = I w + I d , where I w = 5.00 kg ⋅ m 2 is the moment of inertia of the woman and I d = 2mR 2 is the moment of inertia of the two dumbbells. Li = Lf , where L = I ω . K = 12 I ω 2 . Solve: (a) Li = Iiω i. I i = I tot,i = 5.00 kg ⋅ m 2 + 2(5.00 kg)(0.60 m) 2 = 8.60 kg ⋅ m 2 .

Li = (8.60 kg ⋅ m 2 )(3.00 rad/s) = 25.8 kg ⋅ m 2 /s. (b) I f = I tot,f = 5.00 kg ⋅ m 2 + 2(5.00 kg)(0.20 m) 2 = 5.40 kg ⋅ m 2 . Li = Lf gives I iω i = I f ω f and

⎛ 8.60 kg ⋅ m 2 ⎞ ⎛ Ii ⎞ (3.00 rad/s) = 4.78 rad/s. ⎟ ωi = ⎜⎜ 2⎟ ⎟ ⎝ If ⎠ ⎝ 5.40 kg ⋅ m ⎠

ωf = ⎜

(c) Ki = 12 I iω i 2 = 12 (8.60 kg ⋅ m 2 )(3.00 rad/s) 2 = 38.7 J

K f = 12 I f ω f 2 = 12 (5.40 kg ⋅ m 2 )(4.78 rad/s) 2 = 61.7 J

The kinetic energy increases because of the positive work done by the woman as she pulls the weights in.

*10.29. Set Up: For a thin-walled hollow cylinder I = mR 2 . For a slender rod rotating about an axis through its 1 ml 2 . center, I = 12 1 Solve: Li = Lf so Iiω i = I f ω f . I i = 0.40 kg ⋅ m 2 + 12 (8.0 kg)(1.8 m)2 = 2.56 kg ⋅ m 2 . I f = 0.40 kg ⋅ m 2 + (8.0 kg)

(0.25 m) 2 = 0.90 kg ⋅ m 2 . ⎛ Ii ⎝ If

ωf = ⎜

⎛ 2.56 kg ⋅ m 2 ⎞ ⎞ (0.40 rev/s) = 1.14 rev/s ⎟ ω i = ⎜⎜ 2⎟ ⎟ ⎠ ⎝ 0.90 kg ⋅ m ⎠

10.30. Set Up: For a particle, I = mR 2 and L = I ω = mR 2ω . K = 12 I ω 2 . Solve: (a) The tension in the cord is directed toward the axis of rotation and produces no torque. There is no net external torque and the angular momentum of the block is conserved. (b) Li = Lf gives mRi 2ω i = mRf 2ω f and 2

2 ⎛ Ri ⎞ ⎛ 0.300 m ⎞ ⎟ ωi = ⎜ ⎟ (1.75 rad/s) = 7.00 rad/s ⎝ 0.150 m ⎠ ⎝ Rf ⎠

ωf = ⎜

(c) Ki = 12 I iω i 2 = 12 mi Ri 2ω i 2 = 12 (0.0250 kg)(0.300 m) 2 (1.75 rad/s) 2 = 3.44 × 10−3 J

K f = 12 I f ω f 2 = 12 mf Rf 2ω f 2 = 12 (0.0250 kg)(0.150 m) 2 (7.00 rad/s)2 = 1.38 × 10−2 J ΔK = K f − Ki = 1.04 × 10−2 J

(d) The work done equals the increase in kinetic energy, 1.04 × 10−2 J.


10-10

Chapter 10

10.31. Set Up: The system before and after the collision is sketched in the figure below. The gate has I = 13 ML2 . Take counterclockwise torques to be positive.

Solve: (a) The gravity forces exert no torque at the moment of collision and angular momentum is conserved. L1 = L2 . mυ1l = − mυ2l + I gateω with l = L /2.

ω=

m(υ1 + υ2 )l 1 ML2 3

=

3m(υ1 + υ2 ) 3(1.1 kg)(5.0 m/s + 2.0 m/s) = = 1.71 rad/s 2ML 2(4.5 kg)(1.5 m)

(b) Linear momentum is not conserved; there is an external force exerted by the pivot. But the force on the pivot has zero torque. There is no external torque and angular momentum is conserved. Reflect: K1 = 12 (1.1 kg)(5.0 m/s)2 = 13.8 J.

K 2 = 12 (1.1 kg)(2.0 m/s) 2 + 12 ( 13 [4.5 kg][1.5 m/s]2 )(1.71 rad/s) 2 = 7.1 J This is an inelastic collision and K 2 < K1.

10.32. Set Up: Apply conservation of angular momentum to the diver. The number of revolutions she makes in a certain time is proportional to her angular velocity. The ratio of her untucked to tucked angular velocity is (3.6 kg ⋅ m 2 )/(18 kg ⋅ m 2 ). Solve: If she had tucked, she would have made (2 rev)(3.6 kg ⋅ m2 )/(18 kg ⋅ m 2 ) = 0.40 rev in the last 1.0 s, so she would have made (0.40 rev)(1.5/1.0) = 0.60 rev in the total 1.5 s. Reflect: Untucked she rotates slower and completes fewer revolutions.

*10.33. Set Up: Li = Lf . I child = mr 2 , where r is his distance from the axis. Ti = 6.00 s so 2π rad ωi= = 1.047 rad/s. 6.00 s

Solve: I i = 1200 kg ⋅ m 2 , since the child is at r = 0. I f = 1200 kg ⋅ m 2 + (40.0 kg)(2.00 m) 2 = 1360 kg ⋅ m 2

I iω i = I f ω f gives ⎛ 1200 kg ⋅ m 2 ⎞ ⎞ (1.047 rad/s) = 0.924 rad/s. ⎟ ω i = ⎜⎜ 2⎟ ⎟ ⎠ ⎝ 1360 kg ⋅ m ⎠ Reflect: When some of the mass moves farther from the axis, the moment of inertia increases and the angular velocity decreases. ⎛ Ii ⎝ If

ωf = ⎜

10.34. Set Up: Li = Lf . I person = mR 2 . I tt = 12 MR 2 . Solve: I i = I tt = 12 (120 kg)(2.00 m) 2 = 240 kg ⋅ m 2 .

I f = I tt + I person = 240 kg ⋅ m 2 + (70.0 kg)(2.00 m)2 = 520 kg ⋅ m 2 © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


10-11

I iω i = I f ω f gives ⎛ 240 kg ⋅ m 2 ⎞ ⎛ Ii ⎞ (3.00 rad/s) = 1.38 rad/s ⎟ ω i = ⎜⎜ 2⎟ ⎟ ⎝ If ⎠ ⎝ 520 kg ⋅ m ⎠

ωf = ⎜

10.35. Set Up: An object is in translational equilibrium if the net force is zero. An object is in rotational equilibrium about an axis if the net torque about that axis is zero. Solve: (a) Only rotational equilibrium. (b) Neither. (c) Only translational equilibrium. (d) Only translational equilibrium. (e) Both. 10.36. Set Up: An object is in rotational equilibrium about an axis if ∑ τ = 0. An object is in translational

G equilibrium if ∑ F = 0.

Solve: (a) ∑ τ = 0 if (50 N)(30 cm) = F (10 cm), so F = 150 N. Not in translational equilibrium. (b) ∑ τ = 0 if (50 N)(10 cm) = F (30 cm), so F = 16.7 N. Not in translational equilibrium. (c) ∑ τ = 0 if (50 N)(30 cm) + (40 N)(20 cm) = F (10 cm), so F = 230 N. Not in translational equilibrium. (d) ∑ τ = 0 if (50 N)(10 cm) = F (30 cm), so F = 16.7 N. Not in translational equilibrium.

(e) ∑ τ = 0 if (100 N)(15 cm) = (50 N)(30 cm) + F (10 cm), so F = 0. Not in translational equilibrium.

*10.37. Set Up: The force diagram for the leg is given in the figure below. The weight of each piece acts at the center of mass of that piece. The mass of the upper leg is mul = (0.215)(37 kg) = 7.955 kg. The mass of the lower leg is mll = (0.140)(37 kg) = 5.18 kg. Use the coordinates shown, with the origin at the hip and the x axis along the leg.

Solve: xcm =

xul mul + xll mll + xcast mcast mul + mll + mcast xcm =

(18.0 cm)(7.955 kg) + (69.0 cm)(5.18 kg) + (78.0 cm)(5.50 kg) = 49.9 cm 7.955 kg + 5.18 kg + 5.50 kg

10.38. Set Up: The free-body diagram for the board is given in the figure below. Since the board is uniform its center of gravity is 1.50 m from each end. Apply ∑ Fy = 0, with + y upward. Apply ∑ τ = 0 with the axis at the end where the first person applies a force and with counterclockwise torques positive.


10-12

Chapter 10

Solve: ∑ Fy = 0 gives F1 + F2 − w = 0 and F2 = w − F1 = 160 N − 60 N = 100 N. ∑ τ = 0 gives F2 x − w(1.50 m) = 0

and ⎛ w⎞ ⎛ 160 N ⎞ x = ⎜ ⎟ (1.50 m) = ⎜ ⎟ (1.50 m) = 2.40 m. F 100 N ⎠ ⎝ ⎝ 2⎠ The second person lifts with a force of 100 N at a point 2.40 m from the end where the first person lifts.

10.39. Set Up: The free-body diagram is given in the figure below. Ff is the force on each foot and Fh is the force on each hand. Use coordinates as shown. Take the pivot at his feet and let counterclockwise torques to be positive.

Solve: ∑ τ = 0 gives + (2 Fh )(1.70 m) − w(1.15 m) = 0

Fh = w

1.15 m = 0.338w = 272 N 2(1.70 m)

∑ Fy = 0 gives 2 Ff + 2 Fh − w = 0 and Ff = 12 w − Fh = 402 N − 272 N = 130 N Reflect: His center of mass is closer to his hands than to his feet, so his hands exert a greater force.

10.40. Set Up: Apply ∑ Fy = 0, with + y upward, and apply ∑ τ = 0, with the axis at the left-hand end and counterclockwise torques positive. Solve: (a) The free-body diagram for the board is given in the figure below.

(b) ∑ Fy = 0 gives F1 + F2 − wm = 0 and wm = 1000 N. (c) ∑ τ = 0 gives F2 (2.00 m) − wm x = 0.

⎛F ⎞ ⎛ 600.0 N ⎞ x = ⎜ 2 ⎟ (2.00 m) = ⎜ ⎟ (2.00 m) = 1.20 m ⎝ 1000 N ⎠ ⎝ wm ⎠ The center of gravity of the motor is 1.20 m from the end where the 400.0 N force is applied.

*10.41. Set Up: Apply the first and second conditions of equilibrium to the shelf. The free-body diagram for the shelf is given in the following figure. Take the axis at the left-hand end of the shelf and let counterclockwise torque be positive. The center of gravity of the uniform shelf is at its center.



10-13

Solve: (a) ∑ τ = 0 gives − wt (0.200 m) − ws (0.300 m) + T2 (0.400 m) = 0.

(25.0 N)(0.200 m) + (50.0 N)(0.300 m) = 50.0 N 0.400 m ∑ Fy = 0 gives T1 + T2 − wt − ws = 0 and T1 = 25.0 N. The tension in the left-hand wire is 25.0 N and the tension in

T2 =

the right-hand wire is 50.0 N. Reflect: We can that ∑ τ = 0 is zero for any axis, for example for an axis at the right-hand end of the shelf.

10.42. Set Up: The free-body diagram for the beam is given in the figure below. H v and H h are the vertical and horizontal components of the force exerted on the beam at the wall (by the hinge). Since the beam is uniform, its center of gravity is 2.00 m from each end. The angle θ has cos θ = 0.800 and sin θ = 0.600. The tension T has been replaced by its x and y components.

Solve: (a) H v , H h and Tx = T cos θ all produce zero torque. ∑ τ = 0 gives − w(2.00 m) − wload (4.00 m) + T sin θ (4.00 m) = 0 and

T=

(150 N)(2.00 m) + (300 N)(4.00 m) = 625 N. (4.00 m)(0.600)

(b) ∑ Fx = 0 gives H h = T cos θ = 0 and H h = (625 N)(0.800) = 500 N. ∑ Fy = 0 gives H v − w − wload + T sin θ = 0 and

H v = w + wload − T sin θ = 150 N + 300 N − (625 N)(0.600) = 75 N. Reflect: For an axis at the right-hand end of the beam, only w and H v produce torque. The torque due to w is

counterclockwise so the torque due to H v must be clockwise. To produce a counterclockwise torque, H v must be upward, in agreement with our result from ∑ Fy = 0.


10-14

Chapter 10

*10.43. Set Up: The free-body diagram for the boom is given in the figure below. Let the length of the boom be L.

Solve: (a) ∑ τ = 0 gives T ( L sin 60.0°) − wload ( L cos60.0°) − w(0.35L cos60.0°) = 0 and

T=

wload cos60.0° + w(0.35cos60.0°) (5000 N)cos 60.0° + (2600 N)(0.35cos60.0°) = = 3.41 × 103 N sin 60.0° sin 60.0°

(b) ∑ Fx = 0 gives Fh − T = 0 and Fh = 3410 N. ∑ Fy = 0 gives Fv − w − wload = 0 and Fv = 5000 N + 2600 N = 7600 N

10.44. Set Up: There is a normal force nw at the wall. At the floor there is a normal force n and a friction force f . Let the length of the ladder be L. The center of gravity of the ladder is a distance L /2 from each end. Let + y be upward and let + x be horizontal, toward the wall. Let the pivot be at the foot of the ladder and let counterclockwise torques be positive. Solve: (a) The free-body diagram is given in the figure below.

(b) ∑ Fy = 0 gives n − w = 0 and n = 250 N. ∑ τ = 0 gives

⎛L ⎞ nw L cos35° − w ⎜ sin 35°⎟ = 0 ⎝2 ⎠ nw = 12 w

sin 35° 1 = w tan 35° = 87.5 N cos35° 2

(c) ∑ Fx = 0 gives f − nw = 0 and f = nw = 87.5 N Reflect: The friction force required increases when the angle between the ladder and the wall increases, when the ladder becomes closer to horizontal. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


10-15

10.45. Set Up: There is a normal force nw at the wall. At the floor there is a normal force n and a friction force f . The center of gravity of the ladder is 3.5 m from each end. Let + y be upward and + x be horizontal, toward the wall. Let the pivot be at the foot of the ladder and let counterclockwise torques be positive. Solve: (a) The free-body-diagram is given in the figure below.

(b) ∑ τ = 0 gives nw (7.0 m)(sin 60.0°) − wp (2.75 m)(cos60.0°) − wL (3.5 m)(cos60.0°) = 0 nw =

(750 N)(1.375 m) + (450 N)(1.75 m) = 300 N 6.062 m

(c) ∑ Fx = 0 gives nw − f = 0 and f = 300 N

∑ Fy = 0 gives n − wp − wL = 0 and n = wp + wL = 750 N + 450 N = 1200 N

10.46. Set Up: The distance along the beam from the hinge to where the cable is attached is 3.0 m. The angle φ that the cable makes with the beam is given by sin φ =

4.0 m 5.0 m

so φ = 53.1°. The center of gravity of the beam is 4.5 m from the hinge. Use coordinates with + y upward and + x to the right. Take the pivot at the hinge and let counterclockwise torque be positive. Express the hinge force as components H v and H h . Assume H v is downward and that H h is to the right. If one of these components is actually in the opposite direction we will get a negative value for it. Set the tension in the cable equal to its maximum possible value, T = 1.00 kN. Solve: (a) The free-body diagram is given in the figure below. T has been resolved into its x and y components.


10-16

Chapter 10

(b) ∑ τ = 0 gives (T sin φ )(3.0 m) − w(4.5 m) = 0

w=

(T sin φ )(3.0 m) (1.00 kN)(sin 53.1°)(3.0 m) = = 530 N 4.5 m 4.5 m

(c) ∑ Fx = 0 gives H h − T cos φ = 0 and H h = (1.00 kN)(cos53.1°) = 600 N

∑ Fy = 0 gives T sin φ − H v − w = 0 and H v = (1.00 kN)(sin 53.1°) − 530 N = 270 N Reflect: T cos φ , H v and H h all have zero moment arms for a pivot at the hinge and therefore produce zero

torque. If we consider a pivot at the point where the cable is attached we can see that H v must be downward to produce a torque that opposes the torque due to w.

*10.47. Set Up: The center of gravity of the beam is 2.0 m from the hinge. Use coordinates with + y upward and + x to the right. Take the pivot at the hinge and let counterclockwise torque be positive. Solve: (a) The free-body diagram for the beam is shown in the figure below. The tension T in the wire has been replaced by its x and y components.

(b) ∑ τ = 0 gives (T cos60.0°)(4.0 m) − wb (2.0 m) − w(2.5 m) = 0

T=

(2500 N)(2.00 m) + (3500 N)(2.50 m) = 6880 N 2.00 m

(c) ∑ Fx = 0 gives H h − T sin 60.0° = 0 and H h = T sin 60.0° = 5960 N

∑ Fy = 0 gives H v + T cos60.0° − wb − w = 0 and H v = wb + w − T cos60.0° = 2560 N

10.48. Set Up: The free-body diagram for the leg is given in the figure below. Take the pivot at the hip t and let counterclockwise torque be positive. There are also forces on the leg exerted by the hip t but these forces produce no torque and aren’t shown.



10-17

Solve: (a) ∑ τ = 0 gives T (10 cm)(sin θ ) − w(44 cm)(cos θ ) = 0

T= and for θ = 60°,

4.4wcosθ 4.4w = sinθ tanθ

4.4(15 kg)(9.80 m/s 2 ) = 370 N tan 60° (b) For θ = °, T = 7400 N. The tension is much greater when he just starts to raise his leg off the ground. T=

(c) T → ∞ as θ → 0. The person could not raise his leg. If the leg is horizontal so θ is zero, the moment arm for T is zero and T produces no torque to rotate the leg against the torque due to its weight.

*10.49. Set Up: The system of the person and diving board is at rest so the two conditions of equilibrium apply. The freebody diagram for the diving board is given in the figure below. Take the origin of coordinates at the left-hand end of the G G board (point A). F1 is the force applied at the point and F2 is the force at the end that is held down.

Solve: (a) ∑ τ A = 0 gives + F1 (1.0 m) − (500 N)(3.00 m) − (280 N)(1.50 m) = 0

F1 =

(500 N)(3.00 m) + (280 N)(1.50 m) = 1920 N 1.00 m

(b) ∑ Fy = ma y

F1 − F2 − 280 N − 500 N = 0 F2 = F1 − 280 N − 500 N = 1920 N − 280 N − 500 N = 1140 N Reflect: We can check our answers by calculating the net torque about some point and checking that τ z = 0 for that point

also. Net torque about the right-hand end of the board: (1140 N)(3.00 m) + (280 N)(1.50 m) − (1920 N)(2.00 m) = 3420 N ⋅ m + 420 N ⋅ m − 3840 N ⋅ m = 0, which checks.

10.50. Set Up: Apply ∑ Fy = 0 and ∑ τ = 0 to the board. Let + y be upward. Let x be the distance of the center of gravity of the motor from the end of the board where the 400 N force is applied. Solve: (a) If the board is taken to be massless, the weight of the motor is the sum of the applied forces, 1000 N. The (2.00 m)(600 N) motor is a distance = 1.20 m from the end where the 400 N force is applied, and so is 0.800 m from (1000 N) the end where the 600 N force is applied. (b) The weight of the motor is 400 N + 600 N − 200 N = 800 N. Applying ∑ τ = 0 with the axis at the end of the board where the 400 N acts gives (600 N)(2.00 m) = (200 N)(1.00 m) + (800 N) x and x = 1.25 m. The center of

gravity of the motor is 0.75 m from the end of the board where the 600 N force is applied. Reflect: The motor is closest to the end of the board where the larger force is applied.


10-18

Chapter 10

*10.51. Set Up: Let the forearm be at an angle φ below the horizontal. Take the pivot at the elbow t and let counterclockwise torques be positive. Let + y be upward and let + x be to the right. Each forearm has mass G marm = 12 (0.0600)(72 kg) = 2.16 kg. The weight held in each hand is w = mg , with m = 7.50 kg. T is the force the G biceps muscle exerts on the forearm. E is the force exerted by the elbow and has components Ev and Eh . Solve: (a) The free-body diagram for the forearm is given in the figure below.

(b) ∑ τ = 0 gives T (5.5 cm)(cos θ ) − warm (16.0 cm)(cos θ ) − w(38.0 cm)(cos θ ) = 0

T=

16.0 warm + 38.0 w 16.0(2.16 kg)(9.80 m/s 2 ) + 38.0(7.50 kg)(9.80 m/s 2 ) = = 569 N 5.5 5.5

(c) ∑ Fx = 0 gives Eh = 0. ∑ Fy = 0 gives T − Ev − warm − w = 0, so Ev = T − warm − w = 569 N − (2.16 kg)(9.80 m/s 2 ) − (7.50 kg)(9.80 m/s2 ) = 474 N

Since we calculate Ev to be positive, we correctly assumed that it was downward when we drew the free-body diagram. (d) The factor cos θ divides out of the ∑ τ = 0 equation in part (b), so the force T stays the same as she raises her arm.

10.52. Set Up: Take the pivot at the shoulder t and let counterclockwise torques be positive. Let + y be upward and let + x be to the right. wu and wf are the weights of his upper arm and forearm, respectively. w = (2.50 kg) (9.80 m/s 2 ) = 24.5 N is the weight held in his hand. Solve: (a) The free-body diagram is given in the figure below.

(b) ∑ τ = 0 gives T sin φ (15.0 cm) − wu (13.0 cm) − wf (43.0 cm) − w(60.0 cm) = 0

T=

[(2.63 kg)(13.0 cm) + (2.44 kg)(43.0 cm) + (2.50 kg)(60.0 cm)][9.80 m/s 2 ] = 781 N (15.0 cm)sin14.0°



10-19

Reflect: The force T from the muscle has a small perpendicular component and a small level arm, so its magnitude is much larger than the total weight being ed.

10.53. Set Up: Apply the first and second conditions for equilibrium to the table. Label the legs as shown in Figure (a) below. Legs A and C are 3.6 m apart. Let the weight be placed closest to legs C and D. By symmetry, A = B and C = D. Redraw the table as viewed from the AC side. The free-body diagram in this view is given in Figure (b) below. Solve: ∑ τ (about right end) = 0 gives 2 A(3.6 m) = (90.0 N)(1.8 m) + (1500 N)(0.50 m) and A = 130 N = B. ∑ Fy = 0 gives A + B + C + D = 1590 N. Using A = B = 130 N and C = D gives C = D = 670 N. By Newton’s third law of motion, the forces A, B, C, and D on the table are the same magnitude as the forces the table exerts on the floor. Reflect: As expected, the legs closest to the 1500 N weight exert a greater force on the floor.

10.54. Set Up: The precession angular velocity is Ω = 0.0250 kg = 0.165 kg. Ω=

wr . The total mass of the gyroscope is mr + mf = 0.140 kg + Iω

2π rad 2π rad = = 2.856 rad/s. T 2.20 s

Solve: (a) Fp = wtot = (0.165 kg)(9.80 m/s 2 ) = 1.62 N

wr (0.165 kg)(9.80 m/s 2 )(0.0400 m) = = 189 rad/s = 1.80 × 103 rev/min I Ω (1.20 × 10−4 kg ⋅ m 2 )(2.856 rad/s) G G (c) If the figure in the problem is viewed from above, τ is in the direction of the precession and L is along the axis of the rotor, away from the pivot. (b) ω =

*10.55. Set Up: According to section 10.7 the direction of the angular momentum vector is given by the right-hand rule. For counterclockwise rotations in the plane of this page (positive rotations), the angular momentum vector is perpendicular to the page and points toward you. For clockwise rotations in the plane of this page (negative rotations), the rotation vector is perpendicular to the page and points away from you. Solve: (a) As you see it, the minute hand of the clock appears to move clockwise, so its angular momentum vector is perpendicular to the face of the clock and directed away from you (i.e., inward). (b) If you stand facing the right side of the backward moving car, the car will be moving to your left and its right front tire will appear to rotate counterclockwise. Thus, its angular momentum vector points toward you (i.e., perpendicular to the wheel and pointed outwards from the right side of the car). (c) Assuming that the ice skater is spinning clockwise (as seen from above) about an axis that is perpendicular to the ice, the angular momentum vector of the ice skater will point downward. (d) As viewed from above its north pole, the earth appears to rotate counterclockwise. Thus, its angular momentum vector points toward you (i.e., aligned with the earth’s axis and pointing upward from the north pole). Note: If you switched your view to above the south pole of the earth, the earth would appear to rotate clockwise. Thus, its angular momentum vector would point away from you—however, this still means that it points out of the earth’s north pole. Reflect: For simple problems it is sufficient to follow the convention that angular momentum is taken to be positive for counterclockwise rotations and negative for clockwise rotations; however, in more complex problems it is often better to treat angular momentum as a vector.


10-20

Chapter 10

10.56. Set Up: The cylinder and sphere are drawn in the figure below. The mass of each object can be considered as located at its center of mass, at its geometrical center. Use coordinates that have the origin at the center of the cylinder and the x axis horizontal.

Solve: (a) Before pregnancy, xcm,i = 0. The center of mass of the pregnant woman is at

xcm,f =

0 + (20 cm)(10 kg) = 2.9 cm. 60 kg + 10 kg

The center of mass moves a horizontal distance of 2.9 cm. (b) The person must shift her upper body backward to keep her center of mass from extending past her feet. (c) The unnatural posture and curved back strains the back muscles.

*10.57. Set Up: Apply ∑ Fy = 0 with + y upward and apply ∑ τ = 0 with the pivot at the point of suspension for each rod. Solve: (a) The free-body diagram for each rod is given in the figure below.

(b) ∑ τ = 0 for the lower rod: (6.0 N)(4.0 cm) = wA (8.0 cm) and wA = 3.0 N.

∑ Fy = 0 for the lower rod: S3 = 6.0 N + wA = 9.0 N ⎛ 5.0 ⎞ ∑ τ = 0 for the middle rod: wB (3.0 cm) = (5.0 cm) S3 and wB = ⎜ ⎟ (9.0 N) = 15.0 N. ⎝ 3.0 ⎠ ∑ Fy = 0 for the middle rod: S2 = 9.0 N + S3 = 24.0 N ⎛ 2.0 ⎞ ∑ τ = 0 for the upper rod: S2 (2.0 cm) = wC (6.0 cm) and wC = ⎜ (24.0 N) = 8.0 N. ⎝ 6.0 ⎟⎠ ∑ Fy = 0 for the upper rod: S1 = S2 + wC = 32.0 N



10-21

10.58. Set Up: The free-body diagram for the arm is given in the figure below. Take the pivot at the shoulder t and let counterclockwise torques be positive. Use coordinates as shown. Let F be the force exerted by the muscle. There are also the weight of the arm and forces at the shoulder t, but none of these forces produce any torque when the arm is in this position. The forces F and T have been replaced by their x and y components.

Solve: ∑ τ = 0 gives ( F sin12.0°)(15.0 cm) − (T cos35°)(64.0 cm) = 0

F=

(36.0 N)(cos35°)(64.0 cm) = 605 N (sin12.0°)(15.0 cm)

10.59. Set Up: Apply the first and second conditions of equilibrium to the pole. The free-body diagram for the pole is given in the figure below. Note that n and f are the vertical and horizontal components of the force the ground exerts on the pole.

Solve: (a) Assuming that the pole is not accelerating, we have ∑ Fx = max = 0, which implies f = 0. Thus, the

force exerted by the ground has no horizontal component. Using ∑ τ A = 0, we obtain +T (7.0 m)cos θ − mg (4.5 m)cosθ = 0, so T = mg (4.5 m/7.0 m) = (4.5/7.0)(5700 N) = 3700 N Using ∑ Fy = 0 we obtain n + T − mg = 0, so n = mg − T = 5700 N − 3700 N = 2000 N The force exerted by the ground is vertical (upward) and has magnitude 2000 N. (b) In the ∑ τ A = 0 equation the angle θ divided out. All forces on the pole are vertical and their moment arms are all proportional to cos θ .


10-22

Chapter 10

Reflect: We can that ∑ τ = 0 for an axis at the cg of the pole. We find that T > n since T acts at a point closer to the cg and therefore has a smaller moment arm for this axis than n does.

10.60. Set Up: The free-body diagram for the drawbridge is given in the figure below. H v and H h are the components of the force the hinge exerts on the bridge.

Solve: (a) ∑ τ = 0 with the axis at the hinge gives − w(7.0 m)(cos37°) + T (3.5 m)(sin 37°) = 0 and

T = 2w

cos37° (45,000 N) = = 1.19 × 105 N sin 37° tan 37°

(b) ∑ Fx = 0 gives H h = T = 1.19 × 105 N. ∑ Fy = 0 gives H v = w = 4.50 × 104 N. H = H h 2 + H v 2 = 1.27 × 105 N. tan θ =

Hv and θ = 20.7°. The hinge force has magnitude 1.27 × 105 N and is Hh

directed at 20.7° above the horizontal. (c) T → 0. For an axis at the hinge, ∑ τ = w(7.0 m)(cos37°) and

⎛ 45,000 N ⎞ I = 13 mL2 = 13 ⎜ (14.0 m) 2 = 3.0 × 105 kg ⋅ m 2 . ⎝ 9.80 m/s 2 ⎟⎠ ∑ τ = I α gives

α=

∑ τ (45,000 N)(7.0 m)cos37° = = 0.839 rad/s 2 I 3.0 × 105 kg ⋅ m 2

*10.61. Set Up: Apply ∑ τ = 0 to the slab. The free-body diagram is given in Figure (a) below. We have tan β = 3.75 m so β = 65.0D. Also, 20.0D + β + α = 90.0D so we have α = 5.0D. The distance from the axis to the center of the 1.75 m 2

block is

2

⎛ 3.75 m ⎞ ⎛ 1.75 m ⎞ ⎜⎝ ⎟⎠ + ⎜⎝ ⎟ = 2.07 m. 2 2 ⎠

Solve: (a) w(2.07 m)sin 5.0D − T (3.75 m)sin 52.0D = 0. T = 0.061w. Each worker must exert a force of 0.012w,

where w is the weight of the slab. (b) As θ increases, the moment arm for w decreases and the moment arm for T increases, so the worker needs to exert less force. (c) Note that T → 0 when w es through the point. This situation is sketched in Figure (b) below. (1.75 m)/2 tan θ = and θ = 25.0D. If θ exceeds this value the gravity torque causes the slab to tip over. (3.75 m)/2 Reflect: The moment arm for T is much greater than the moment arm for w, so the force the workers apply is much less than the weight of the slab.



10-23

10.62. Set Up: Apply the first and second conditions of equilibrium to the gate. The free-body diagram for the gate is given in the figure below.

G G Use coordinates with the origin at B. Let H A and H B be the forces exerted by the hinges at A and B. The problem states G G that the force H A has no horizontal component. Replace the tension T by its horizontal and vertical components. Solve: (a) ∑ τ B = 0 gives + (T sin 30.0°)(4.00 m) + (T cos30.0°)(2.00 m) − w(2.00 m) = 0, which reduces to T (2sin 30.0° + cos30.0°) = w. Solving for T we obtain

T=

w 500 N = = 268 N 2sin 30.0° + cos30.0° 2sin 30.0° + cos30.0°

(b) Using ∑ Fx = max gives H Bh − T cos30.0° = 0, so H Bh = T cos30.0° = (268 N)cos30.0° = 232 N. (c) Using ∑ Fy = ma y gives H Av + H Bv + T sin 30.0° − w = 0, so

H Av + H Bv = w − T sin 30.0° = 500 N − (268 N)sin 30.0° = 366 N. Reflect: T has a horizontal component to the left so H Bh must be to the right, as these are the only two horizontal

forces. Note that we cannot determine H Av and H Bv separately, we can only determine their sum.

*10.63. Set Up: A accelerates downward, B accelerates upward and the wheel turns clockwise. Apply ∑ Fy = ma y to blocks A and B. Let + y be downward for A and y be upward for B. Apply ∑ τ = I α to the wheel, with the clockwise sense of rotation positive. Each block has the same magnitude of acceleration, a, and a = Rα . Call the tension in the cord between C and A TA and the tension between C and B TB .


10-24

Chapter 10

Solve: For A, ∑ Fy = ma y gives m A g − TA = m Aa. For B, ∑ Fy = ma y gives TB − mB g = mB a. For the wheel, ⎛ I ⎞ TA R − TB R = I α = I (a /R) w and TA − TB = ⎜ 2 ⎟ a. Adding these three equations gives ⎝R ⎠

I ⎞ ⎛ (m A − mB ) g = ⎜ m A + mB + 2 ⎟ a. ⎝ R ⎠

⎛ ⎞ ⎛ ⎞ m A − mB 4.00 kg − 2.00 kg 2 2 a =⎜ g = ⎜⎜ 2⎟ ⎟ (9.80 m/s ) = 0.730 m/s ⎜ m + m + I /R 2 ⎟⎟ ⎝ 4.00 kg + 2.00 kg + (0.300 kg)/(0.120 m) ⎠ B ⎝ A ⎠

α=

a 0.730 m/s 2 = = 6.08 rad/s 2 R 0.120 m

TA = m A ( g − a) = (4.00 kg)(9.80 m/s 2 − 0.730 m/s 2 ) = 36.3 N TB = mB ( g + a ) = (2.00 kg)(9.80 m/s 2 + 0.730 m/s 2 ) = 21.1 N Reflect: The tensions must be different in order to produce a torque that accelerates the wheel when the blocks accelerate.

10.64. Set Up: Let the pivot be at point P and let counterclockwise torques be positive. Solve: (a) The free-body diagram is given in the figure below.

(b) ∑ τ = 0 gives w(11.0 cm)(sin40.0) − T (1.50 cm) = 0

T=

(4.50 kg)(9.80 m/s 2 )(11.0 cm)sin 40.0° = 208 N 1.50 cm

(a) below. Let the pivot be at the hip t and *10.65. Set Up: The free-body diagram for the leg is given in Figure G

take counterclockwise torques to be positive. There is a force F of unknown magnitude and direction that the hip t exerts on the leg, but this force produces zero torque.



10-25

Solve: (a) ∑ τ = 0 gives wleg (41.0 cm) − W (82.0 cm) = 0 and

W = 12 wleg = 12 (14.2 kg)(9.80 m/s 2 ) = 69.6 N (b) The free-body diagram for the pulley attached to the foot is sketched in Figure (b) above. ∑ Fx = 0 gives 2W cos θ − T = 0, so θ = 85.1°. (c) From (b), T = 2W cos θ and the maximum traction force T is T = 2W = 139 N, when θ = 0°.

10.66. Set Up: Her upper arm has weight wu = 22.8 N. Her forearm has weight wf = 21.1 N. The free-body

G diagram for the arm is given in the figure below. F is the force exerted on the arm by the shoulder t. Take the pivot at the shoulder t and let counterclockwise torques be positive.

wu (13.0 cm) + wf (26.0 cm) (22.8 N)(13.0 cm) + (21.1 N)(26.0 cm) = = 19.2 cm wu + wf 22.8 N + 21.1 N w (17.0 cm) (21.1 N)(17.0 cm) ycm = f = = 8.2 cm wu + wf 22.8 N + 21.1 N The center of mass is 19.2 cm to the right and 8.2 cm above the shoulder t. (b) ∑ τ = 0 gives W (26.0 cm) − wf (26.0 cm) − wu (13.0 cm) = 0

Solve: (a) xcm =

W = wf + 12 wu = 21.1 N + 11.4 N = 32.5 N G (c) All the other forces on the arm are vertical so F has no horizontal component and Fy = F . oFy = 0 gives F + W − wu − wf = 0 and F = wu + wf − W = 22.8 N + 21.1 N − 32.5 N = 11.4 N, upward.

10.67. Set Up: Her upper arm has weight wu = 22.8 N and her forearm has weight wf = 21.1 N. The free-body diagram for the arm is given in the figure below. Take the pivot at the shoulder t and let counterclockwise torques be positive. The forces F and W have been replaced by their x and y components.


10-26

Chapter 10

Solve: (a) xcm =

wu (13.0 cm) + wf (26.0 cm + [17.0 cm][cos55°]) wu + wf xcm =

(22.8 N)(13.0 cm) + (21.1 N)(35.8 cm) = 24.0 cm 22.8 N + 21.1 N

ycm =

w(17.0 cm)(sin 55°) (21.1 N)(13.9 cm) = = 6.7 cm wu + wf 22.8 N + 21.1 N

The center of mass is 24.0 cm to the right and 6.7 cm above the shoulder t. (b) ∑ τ = 0 gives (W sin 55°)(26.0 cm + 19.5 cm) − (W cos55°)(27.9 cm) 2 wf (26.0 cm + 9.75 cm) − wu (13.0 cm) = 0

W=

(22.8 N)(13.0 cm) + (21.1 N)(35.75 cm) = 49.4 N (45.5 cm)sin 55° − (27.9 cm)cos55°

(c) ∑ Fy = 0 gives Fv − wu − wf + W sin 55° = 0 Fv = wu + wf − W sin 55° = 22.8 N + 21.1 N − 40.5 N = 3.4 N

∑ Fx = 0 gives W cos55° − Fh = 0 and Fh = W cos55° = 28.3 N

10.68. Set Up: The free-body diagram for the foot is given in the figure below. Take the pivot at the ankle t and let counterclockwise torques be positive.

Solve: (a) If we consider the entire person, the upward force n exerted by the floor must equal the weight w of the person; n = 750 N. (b) ∑ τ = 0 gives n(16.0 m)(cos30°) − T (4.5 m) = 0 and T = 3.08n = 2310 N. (c) ∑ Fx = 0 gives T cos60° − Fh = 0 and Fh = T cos60° = 1160 N, to the left. oFy = 0 gives T sin 60° − Fv + n = 0 and Fv = T sin 60° + n = 2001 N + 750 N = 2750 N, downward.

F = Fh 2 + Fv 2 = 2980 N

*10.69. Set Up: Apply the conditions of equilibrium to the cylinder. The free-body diagram for the cylinder is given in the figure below. The center of gravity of the cylinder is at its geometrical center. The cylinder has radius R.

Solve: (a) T produces a clockwise torque about the center of gravity so there must be a friction force that produces a counterclockwise torque about this axis. (b) Applying ∑ τ = 0 to an axis at the center of gravity gives −TR + fR = 0 and T = f . The condition ∑ τ = 0 applied to an axis at the point of between the cylinder and the ramp gives −T (2R ) + MgR sin θ = 0. T = ( Mg /2)sin θ .



10-27

Reflect: We can show that ∑ Fx = 0 and ∑ Fy = 0, for the x- and y-axes that are parallel and perpendicular to the ramp (respectively), or for x- and y-axes that are horizontal and vertical (respectively).

10.70. Set Up: The weight of the beam acts 4.0 m from each end. Take the pivot at the hinge and let counterclockwise torques be positive. Represent the force exerted by the hinge by its horizontal and vertical components, H h and H v .

Solve: (a) The free-body diagram for the beam is given in Figure (a) below.

(b) The moment arm for T is sketched in Figure (b) above and is equal to (6.0 m)sin 40.0°. ∑ τ = 0 gives T (6.0 m)(sin 40.0°) − w(4.0 m)(cos30.0°) = 0 T=

(1500 kg)(9.80 m/s 2 )(4.0 m)(cos30.0°) = 1.32 × 104 N (6.0 m)(sin 40.0°)

(c) ∑ Fx = 0 gives H h − T cos10.0° = 0 and H h = T cos10.0° = 1.30 × 104 N ∑ Fy = 0 gives H v + T sin10.0° − w = 0 and H v = w − T sin10.0° = (1500 kg)(9.80 m/s 2 ) − 2.29 × 103 N = 1.24 × 104 N H = H h 2 + H v 2 = 1.80 × 104 N. This is the force the hinge exerts on the beam. By Newton’s third law the force the beam exerts on the wall has the same magnitude, so is 1.80 × 104 N.

*10.71. Set Up: Apply ∑ τ z = 0 to the wheel. Take torques about the upper corner of the curb.

G Solve: The force F acts at a perpendicular distance R − h and the weight acts at a perpendicular distance R 2 − ( R − h) 2 = 2 Rh − h2 . Setting the torques equal for the minimum necessary force, F = mg

2 Rh − h 2 . R−h


10-28

Chapter 10

G (b) The torque due to gravity is the same, but the force F acts at a perpendicular distance 2 R − h, so the minimum force is (mg ) 2 Rh − hv /(2 R − h).

G Reflect: (c) Less force is required when the force is applied at the top of the wheel, since in this case F has a larger moment arm.

10.72. Set Up: ∑ τ = I α . Let the direction the wheel is rotating be positive. Use the constant acceleration equations to relate α to the motion. 100 rev/min = 10.47 rad/s ω − ω 0 10.47 rad/s − 0 Solve: (a) ω = ω 0 + α t gives α = = = 5.23 rad/s 2 . t

I=

∑τ

=

2.00 s 5.00 N ⋅ m

= 0.956 kg ⋅ m 2

α 5.23 rad/s 2 (b) ω 0 = 10.47 rad/s, ω = 0, t = 125 s. ω = ω 0 + α t gives ω − ω 0 0 − 10.47 rad/s α= = = − 0.0838 rad/s 2 t

125 s

2

∑ τ = I α = (0.956 kg ⋅ m )(−0.0838 rad/s 2 ) = − 0.0801 N ⋅ m ⎛ω + ω⎞ ⎛ 10.47 rad/s + 0 ⎞ (c) θ = ⎜ 0 t =⎜ ⎟ ⎟⎠ (125 s) = 654 rad = 104 rev ⎝ 2 ⎠ ⎝ 2

10.73. Set Up: Apply conservation of angular momentum. For a uniform sphere and an axis through its center, I = 52 MR 2 .

Solve: The moment of inertia is proportional to the square of the radius, and so the angular velocity will be proportional to the inverse of the square of the radius, and the final angular velocity is 2

2

⎛R ⎞ ⎛ ⎞ ⎛ 7.0 × 105 km ⎞ 2π rad 3 ω2 = ω 1⎜ 1 ⎟ = ⎜ ⎟⎟ = 4.6 × 10 rad/s. ⎟ ⎜⎜ (30 d)(86,400 s/d) 16 km R ⎠⎝ ⎝ 2⎠ ⎝ ⎠

Reflect: K = 12 I ω 2 = 12 Lω . L is constant and ω increases by a large factor, so there is a large increase in the rotational kinetic energy of the star. This energy comes from potential energy associated with the gravity force within the star.

10.74. Set Up: The thermal energy developed equals the decrease in kinetic energy of the system. The angular momentum of the system is conserved. I B = 3I A . Solve: L i = L f gives I Aω 0 = ( I A + I B )ω f and ⎛

IA ⎞ IA ω0 = ω0 /4. ⎟ ω0 = 4I A ⎝ I A + IB ⎠

ωf = ⎜

Ki = 12 I Aω 0 2 . 2

⎛ω ⎞ K f = 12 ( I A + I B )ω f 2 = 12 (4 I A ) ⎜ 0 ⎟ = 14 Ki . ⎝ 4⎠

Ki − Kf = 34 Ki = 2400 J and Ki = 3200 J. Reflect: Angular momentum is conserved because there is no external torque due to the force that pushes the shafts together, but the friction forces in the clutch mechanism do negative work and remove kinetic energy.

*10.75. Set Up: For an axis about one edge, the door has ⎛ 750 N ⎞ I = 13 ml 2 = 13 ⎜ (1.25 m) 2 = 39.9 kg ⋅ m 2 . ⎝ 9.80 m/s 2 ⎟⎠ Exena applies a torque Fl = (220 N)(1.25 m) = 275 N ⋅ m.



10-29

Solve: ∑ τ = I α gives

α=

275 N ⋅ m 39.0 kg ⋅ m 2

= 6.89 rad/s 2 .

The door turns through π /2 rad and ω 0 = 0. θ − θ 0 = ω 0t + 12 α t 2 gives

t=

2(θ − θ 0 )

α

=

2(π /2) rad 6.89 rad/s 2

= 0.675 s

10.76. Set Up: Figure (a) below shows the distances and angles. θ + φ = 90°. θ = 56.3° and φ = 33.7°. The distances x1 and x2 are x1 = (90 cm)cos θ = 50.0 cm and x2 = (135 cm)cos φ = 112 cm. The free-body diagram for the person is given in Figure (b) below. wl = 277 N is the weight of his feet and legs and

wt = 473 N is the weight of his trunk. nf and f f are the total normal and friction forces exerted on his feet and nh and f h are those forces on his hands. The free-body diagram for his legs is given in Figure (c) below. F is the force exerted on his legs by his hip ts.

Solve: (a) Consider the force diagram of Figure (b) above. ∑ τ = 0 with the pivot at his feet and counterclockwise torques positive gives

nh (162 cm) − (277 N)(27.2 cm) − (473 N)(103.8 cm) = 0 nh = 350 N, so there is a normal force of 175 N at each hand. nf + nh − wl − wt = 0 so 750 N − 350 N = 400 N, so there is a normal force of 200 N at each foot.

nf = wl + wt − nh =

(b) Consider the force diagram of Figure (c) above. ∑ τ = 0 with the pivot at his hips and counterclockwise torques positive gives f f (74.9 cm) + wl (22.8 cm) − nf (50.0 cm) = 0 ff =

(400 N)(50.0 cm) − (277 N)(22.8 cm) = 83 N 74.9 cm


10-30

Chapter 10

There is a friction force of 92 N at each foot. oFx = 0 in Figure (b) above gives f h = f f , so there is a friction force of 92 N at each hand.

Reflect: In this position the normal forces at his feet and at his hands don’t differ very much.

10.77. Set Up: Apply ∑ τ = 0 to the beam. The free-body diagram for the beam is given in the figure below. Solve: ∑ τ = 0, axis at hinge, gives T (6.0 m)(sin 40°) − w(3.75 m)(cos30°) = 0 and T = 7600 N. Reflect: The tension in the cable is less than the weight of the beam. T sin 40° is the component of T that is perpendicular to the beam.

Solutions to age Problems 10.78. Set Up: The inner surface of the space station pushes on the astronaut’s feet and causes the inward radial acceleration, which is given by arad = rω 2 . We have that r = 240 m and ω = 0.20 rad/s.

Solve: arad = rω 2 = (240 m)(0.20 rad/s 2 ) 2 = 9.6 m/s 2 . Thus, the correct answer is C.

*10.79. Set Up: Let F be the thrust of a single rocket located at a distance r from the axis of the space station; its torque is given by τ = rF sin θ . In this case the rockets are located on the outside surface where r = 250 m and maximum torque occurs when θ = 90D (i.e., when the thrust is perpendicular to the radius r. If all the rockets are oriented to produce the same sense of rotation (i.e., clockwise or counterclockwise) their torques will add, so that the net torque will be 40 times the torque of a single rocket.

Solve: We have the net torque of all 40 rockets to be τ net = 40rF sin θ = 40(250 m)(1000 N) = 1.00 × 107 N ⋅ m. Thus, the correct answer is C.

τ 10.80. Set Up: The relation between torque and angular acceleration is τ = Iα so we have α = . Use the value for τ that was calculated in the previous problem.

Solve: α =

τ I

=

1.00 × 107 N ⋅ m 1012 kg ⋅ m 2

I

= 10−5 rad/s 2 . Thus, the correct answer is A.

*10.81. Set Up: We know that ω 0 = 0 and ω = 0.20 rad/s. From the previous problem we know that α = 10−5 rad/s 2 . Since α is constant, we can use ω = ω 0 + α t to find t. ω − ω 0 0.20 rad/s − 0 Solve: We have t = = = 2 × 104 s, which is about 5.6 hours. Thus, the correct answer is A. −5 2 α 10 rad/s


11

ELASTICITY AND PERIODIC MOTION

Answers to Multiple-Choice Problems 1. A 2. B 3. B 14. D 15. C

4. D

5. C

6. C

7. A, C, D

8. B

9. C

10. C

11. B

12. B

13. B

Solutions to Problems *11.1. Set Up: A = π r 2 , with r = 2.75 × 1024 m. The force FT applied to the end of the wire is mg = (25.0 kg) (9.80 m/s 2 ) = 245 N. stress = FT /A and strain = Δl /l0 . Y = stress/strain Solve: (a) stress = (b) strain = Δl /l0 =

245 N

π (2.75 × 10−4 m) 2

= 1.03 × 109 Pa

1.10 × 10−3 m = 1.47 × 10−3 0.750 m

stress 1.03 × 109 Pa = = 7.01 × 1011 Pa −3 strain 1.47 × 10 Reflect: Our result for Y is about the same size as the values in Table 11.1. In SI units the stress is very large. The strain is dimensionless and small, so Y is very large and has the same units (Pa) as stress. (c) Y =

11.2. Set Up: p = F/A and 1 cm 2 = 1 × 10−4 m 2 Solve: (a)

250 N 0.75 × 10−4 m 2

= 3.33 × 106 Pa.

(b) (3.33 × 106 Pa)(2)(200 × 10−4 m 2 ) = 133 kN. Reflect: The pressure in part (a) is over 30 times larger than normal atmospheric pressure.

11.3. Set Up: Apply Eq. (11.4). From Table 11.1, for steel, Y = 2.0 × 1011 Pa and for copper, Y = 1.1 × 1011 Pa. A = π (d 2 /4) = 1.77 × 10−4 m 2 . F⊥ = 4000 N for each rod. Solve: (a) The strain is

Δl (4000N) Δl F = = 1.1 × 10−4. Similarly, the strain = . For steel 11 l l0 YA (2.0 × 10 Pa)(1.77 × 10−4 m 2 ) 0

for copper is 2.1 × 10−4. (b) Steel: (1.1 × 10−4 )(0.750 m) = 8.3 × 10−5 m. Copper: (2.1 × 10−4 )(0.750 m) = 1.6 × 10−4 m. Reflect: Copper has a smaller Y and therefore a greater elongation.


11-1

11-2

Chapter 11

11.4. Set Up: For steel, Y = 2.0 × 1011 Pa. The force applied to the end of the lower wire is (10.0 kg)(9.80 m/s 2 ) = 98.0 N. The force applied to the end of the upper wire is (5.0 kg + 10.0 kg)(9.80 m/s 2 ) = 147.0 N. Tensile strain =

Δl F' = . l0 AY 98.0 N

Solve: (a) lower wire: tensile strain =

upper wire: tensile strain =

−7

(6.0 × 10 m 2 )(2.0 × 1011 Pa) 147.0 N

(6.0 × 10

−7

m 2 )(2.0 × 1011 Pa)

= 8.2 × 10−4.

= 1.2 × 10−3

(b) lower wire: Δl = l0 (strain) = (0.500 m)(8.2 × 10−4 ) = 4.1 × 10−4 m

upper wire: Δl = l0 (strain) = (0.500 m)(1.2 × 10−3 ) = 6.0 × 10−4 m

*11.5. Set Up: A = 50.0 cm 2 = 50.0 × 1024 m 2 . Y = Solve: relaxed: Y =

(0.200 m)(25.0 N) (50.0 × 10−4 m 2 )(3.0 × 10−2 m)

maximum tension: Y =

l0 F' A Δl

= 3.33 × 104 Pa

(0.200 m)(500 N) (50.0 × 10−4 m 2 )(3.0 × 10−2 m)

= 6.67 × 105 Pa

11.6. Set Up: A = π r 2 = π (3.5 × 10−3 m)2 = 3.85 × 10−5 m 2 . The force applied to the end of the rope is the weight of the climber: F' = (65.0 kg)(9.80 m/s 2 ) = 637 N. Solve: Y =

l0 F' (45.0 m)(637 N) = = 6.77 × 108 Pa A Δl (3.85 × 10−5 m 2 )(1.10 m)

*11.7. Set Up: For steel, Y = 2.0 × 1011 Pa. A = π d 2 /4. Solve: Y =

l F l0 F' (2.00 m)(400.0 N) gives A = 0 ' = = 1.6 × 10−6 m 2 Y Δl (2.0 × 1011 Pa)(0.25 × 10−2 m) A Δl d=

4A

π

=

4(1.6 × 10−6 m 2 )

π

= 1.43 mm

Reflect: The thinner the wire the more it stretches for the same applied force. The length of this wire changes by 0.12%.

11.8. Set Up: Y =

stress . A 5.0% elongation means Δl /l0 = 0.050. For a spring, FT = kx. strain

Solve: (a) stress = Y × strain = (1474 × 106 Pa)(0.050) = 7.4 × 107 Pa (b) stress = FT /A so FT = stress × A = (7.37 × 107 Pa)(78.1 × 10−6 m 2 ) = 5.76 × 103 N

The change in length is x = Δl = (0.050)(25 cm) = 1.25 cm. FT = kx gives k=

FT 5.76 × 103 N = = 4.6 × 105 N/m x 1.25 × 10−2 m

(c) F = 13mg = 13(75 kg)(9.80 m/s 2 ) = 9555 N and x =

*11.9. Set Up: Y =

9555 N FT = = 2.08 cm k 4.6 × 105 N/m

⎛ YA ⎞ FT /A YA so FT = ⎜ ⎟ Δl and k = . From Problem 11.8, k = 4.6 × 105 N/m for the natural Δl /l0 l0 ⎝ l0 ⎠

Achilles tendon. A = π r 2 © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Elasticity and Periodic Motion

Solve: (a) k =

11-3

YA (4.6 × 105 N/m)(0.25 m) kl so A = 0 = = 3.8 × 10−6 m 2 l0 Y 30 × 109 Pa

A = π r 2 so r =

A/π = 1.1 mm and the diameter is 2.2 mm.

(b) The natural tendon has r = (78.1 mm 2 )/π = 4.99 mm and diameter 10.0 mm. The artificial tendon’s diameter is

much smaller. Reflect: The artificial tendon has a larger Y and therefore a smaller diameter.

11.10. Set Up: stress = Solve: (a) 196 × 106 Pa = (b) 0.40 =

F' Δl , strain = . A l0 F'

π (25 × 10−6 m)2

and F' = 0.385 N.

l − l0 and l = 12 cm gives l0 = 8.6 cm. l0

11.11. Set Up: @

FT /A Δl /l0

Solve: (a) FT = 8mg = 5880 N.

Δl FT 5880 N = = = 0.24 = 24, l0 AY (10 × 10−4 m 2 )(24 × 106 Pa)

Δl 1 = (24,) = 12, l0 2 Reflect: Young’s modulus for cartilage is much smaller than typical values for metals and the fractional change in length is larger. (b) FT = 4mg so

V0Δp . Δp is positive when the pressure increases. B Solve: (a) The volume would increase slightly.

11.12. Set Up: Apply Eq. (11.7). ΔV = −

(b) The volume change would be twice as great. (c) The volume change is inversely proportional to the bulk modulus for a given pressure change, so the volume change of the lead ingot would be four times that of the gold. Reflect: For lead, B = 4.1 × 1010 Pa, so Δp /B is very small and the fractional change in volume is very small.

*11.13. Set Up: At the surface the pressure is 1.0 × 105 Pa, so Δp = 1.16 × 108 Pa. V0 = 1.00 m3 . Density = m /V . At the surface 1.00 m3 of water has mass 1.03 × 103 kg. Solve: (a) B = −

( Δp )V0 ( Δp )V0 (1.16 × 108 Pa)(1.00 m3 ) gives ΔV = − =− = − 0.0527 m3 ΔV B 2.2 × 109 Pa

(b) At this depth 1.03 × 103 kg of seawater has volume V0 + ΔV = 0.9473 m3 . The density is 1.03 × 103 kg 0.9473 m3

11.14. Set Up:

= 1.09 × 103 kg/m3 .

ΔV Δp = − . Δp = (1.0 × 104 Pa/m)d , where d is the depth below the surface. V0 B

Solve: (a) ΔV = −

Δp (1.0 × 104 Pa/m)(33 m) V0 = − (1.0 cm3 ) = − 1.5 × 10−4 cm3 B 2.2 × 109 Pa

One cubic centimeter of her blood decreases in volume by 1.5 × 10−4 cm3 .


11-4

(b)

Chapter 11

ΔV 1 = − gives Δp = B V0 2

( 12 ) = 1.1 × 109 Pa Δp

1.1 × 109 Pa

= 1.1 × 105 m = 110 km. The ocean is not this deep; the 1.0 × 104 Pa/m 1.0 × 104 Pa/m greatest depth in the ocean is an order of magnitude less than this, about 11 km. The depth would be d =

=

*11.15. Set Up: Use Eq. (11.10). Same material implies same S stress stress F|| /A so strain = and same forces implies same F|| . = S S strain For the smaller object, (strain)1 = F|| /A1S

Solve: S =

For the larger object, (strain)2 = F|| /A2 S (strain) 2 ⎛ F|| ⎞ ⎛ A1S ⎞ A1 =⎜ ⎟= ⎟⎜ (strain)1 ⎝ A2 S ⎠⎝ F|| ⎠ A2 (strain) 2 1 Larger solid has triple each edge length, so A2 = 9 A1, and = (strain)1 9 Reflect: The larger object has a smaller deformation.

11.16. Set Up:

ΔV Δp = − . 1 atm = 1.01 × 105 Pa. V0 B

Solve: (a) Δp = − B

ΔV = − (15 × 109 Pa)(−0.0010) = 1.5 × 107 Pa = 150 atm V0

(b) The depth for a pressure increase of 1.5 × 107 Pa is 1.5 km. Unprotected dives do not approach this depth so bone

compression is not a concern.

11.17. Set Up: For steel the shear modulus is S = 0.84 × 1011 Pa. S=

shear stress F|| /A = . shear strain shear strain

Solve: A is the area of each edge of the plate: A = (0.100 m)(0.0100 m) = 1.00 × 10−3 m 2 .

F|| = SA(shear strain) = (0.84 × 1011 Pa)(1.00 × 10−3 m 2 )(0.0400) = 3.4 × 106 N F|| . F|| is the component of the force tangent to the surface, so F|| = (1375 N)cos8.50° = Aφ 1360 N. φ must be in radians, φ = 1.24° = 0.0216 rad.

11.18. Set Up: S =

Solve: S =

1360 N (0.0925 m) 2 (0.0216 rad)

*11.19. Set Up: S =

= 7.36 × 106 Pa

F|| . F|| = F sin12°. φ is in radians. F = 8mg , with m = 10 kg. π rad = 180° Aφ

F|| 8mg sin12° = = 0.1494 rad = 8.6° AS (10 × 10−4 m 2 )(12 × 106 Pa) Reflect: The shear modulus of cartilage is much less than the values for metals given in Table 11.1.

Solve: φ =

11.20. Set Up: The tensile force F' equals the weight w of the object suspended from the wire. The maximum weight that can be ed is one that makes the stress equal to the breaking stress.



Solve: (a) (b) Y =

11-5

F⊥ = 3.60 × 108 Pa so w = (3.60 × 108 Pa)(0.040 × 10−4 m 2 ) = 1.4 × 103 N. A

l0 F⊥ l F (5.00 m)(1.44 × 103 N) = 9.0 × 10−3 m = 9.0 mm. so Δl = 0 ⊥ = A Δl YA (2.0 × 1011 Pa)(0.040 × 1024 m 2 )

(c) F⊥ /A = 11.0 × 108 Pa so w = (11.0 × 108 Pa)(0.040 × 10−4 m 2 ) = 4.4 × 103 N

F⊥ 1 = (2.40 × 108 Pa) = 0.80 × 108 Pa. The free-body diagram for the elevator is given in the figure A 3 below. F' is the tension in the cable.

11.21. Set Up:

Solve: F⊥ = A(0.80 × 108 Pa) = (3.00 × 10−4 m 2 )(0.80 × 108 Pa) = 2.40 × 104 N. ∑ Fy = ma y applied to the elevator

gives F⊥ − mg = ma and F' 2.40 × 104 N −g= − 9.80 m/s 2 = 10.2 m/s 2 m 1200 kg Reflect: The tension in the cable is about twice the weight of the elevator. a=

11.22. Set Up: The cross-sectional area of a hollow cylinder with inner radius R1 and outer radius R2 is A = π ( R2 2 − R12 ). The total weight Wtot his legs is his weight and the weight he lifts. Assume each leg s half of Wtot . Solve: 200 × 106 Pa =

1W 2 tot

A

=

1m g 2 tot 2

π ([0.018 m] − [0.0125 m]2 )

and mtot = 2.2 × 104 kg

He can lift a mass of 2.2 × 104 kg, which is a weight of 2.1 × 105 N.

*11.23. Set Up: The frequency f in Hz is the number of cycles per second. The angular frequency ω is ω = 2π f and has units of radians. The period T and frequency f are related by T = Solve: (a) T = (b) f =

1 . f

1 1 = = 2.15 × 10−3 s. ω = 2π f = 2π (466 Hz) = 2.93 × 103 rad/s f 466 Hz

1 1 = = 2.00 × 104 Hz. ω = 2π f = 1.26 × 105 rad/s T 50.0 × 10−6 s


11-6

Chapter 11

(c) f =

T=

ω 2.7 × 1015 Hz 4.7 × 1015 Hz so f ranges from = 4.3 × 1014 Hz to = 7.5 × 1014 Hz 2π 2π rad 2π rad

1 1 1 so T ranges from = 1.3 × 10−15 s to = 2.3 × 10−15 s f 7.5 × 1014 Hz 4.3 × 1014 Hz

1 1 = = 2.0 × 10−7 s and ω = 2π f = 2π (5.0 × 106 Hz) = 3.1 × 107 rad/s f 5.0 × 106 Hz Reflect: Visible light has much higher frequency than either sounds we can hear or ultrasound. Ultrasound is sound with frequencies higher than what the ear can hear. Large f corresponds to small T. (d) T =

11.24. Set Up: The period is the time for the hand to make one complete revolution. f = Solve: (a) T = 1.00 min = 60.0 s. f = 0.0167 Hz. ω = 2π (0.0167 Hz) = 0.105 rad/s

1 and ω = 2π f . T

(b) T = 1.00 hr = 3600 s. f = 2.78 × 1024 Hz. ω = 2π (2.78 × 10−4 Hz) = 1.75 × 10−3 rad/s

*11.25. Set Up: The amplitude is the maximum displacement from equilibrium. In one period the object goes from x = + A to x = − A and returns. Solve: (a) A = 0.120 m (b) 0.800 s = T /2 so the period is 1.60 s

(c) f =

1 = 0.625 Hz T

11.26. Set Up: The amplitude A is the maximum displacement from equilibrium. The period T is the time for 1 and ω = 2π f . T Solve: (a) A = 0.40 mm. (b) T = 2.0 ms. 1 cycle. f =

1 = 500 Hz. T (d) ω = 2π f = 3140 rad/s

(c) f =

11.27. Set Up: Note that 900 beats per minute is the frequency of the wing beat, which we will convert into hertz. We will use the relationship between frequency and period, and we will also use the relationship between frequency 1 and angular frequency: T = and ω = 2π f , respectively. f 1 = 0.00111 minutes = 0.0667 s. 900 beats/minute beats 1 minute (b) f = 900 ⋅ = 15.0 hertz. minute 60 seconds (c) ω = 2π f = 94.2 rad/s

Solve: (a) T =

Reflect: We could have calculated the answer to part (b) first, and then used T =

1 to find the answer to part (a). f

11.28. Set Up: Velocity and position are related by E = 12 kA2 = 12 mυx2 + 12 kx2. Acceleration and position are related by − kx = max . The maximum speed is at x = 0 and the maximum magnitude of acceleration is at x = ± A. Solve: (a) For x = 0,

2 1 mυ max 2

= 12 kA2 and υmax = A

k 450 N/m = (0.040 m) = 1.20 m/s 0.500 kg m



(b) υ x = ±

k m

A2 − x 2 = ±

450 N/m 0.500 kg

11-7

(0.040 m) 2 − (0.015 m) 2 = ± 1.11 m/s.

The speed is υ = 1.11 m/s. (c) For x = ± A, amax = (d) ax = −

⎛ 450 N/m ⎞ k 2 A=⎜ ⎟ (0.040 m) = 36 m/s m ⎝ 0.500 kg ⎠

kx (450 N/m)(−0.015 m) =− = + 13.5 m/s 2 0.500 kg m

(e) E = 12 kA2 = 12 (450 N/m)(0.040 m) 2 = 0.360 J

*11.29. Set Up: Velocity, position, and total energy are related by E = 12 kA2 = 12 mυ x 2 + 12 kx 2 . The maximum speed is at x = 0. Solve: (a) E = 12 mυ x 2 + 12 kx 2 = 12 (0.150 kg)(0.30 m/s)2 + 12 (300.0 N/m)(0.012 m) 2

E = 6.75 × 1023 J + 2.16 × 1022 J = 2.84 × 1022 J. (b) E = 12 kA2 so A = (c) For x = 0,

2 1 mυ max 2

2E = k

2(2.84 × 1022 J) = 0.0138 m. 300.0 N/m 2E = m

= E and υmax =

2(2.84 × 1022 J) = 0.615 m/s. 0.150 kg

Reflect: When x = 0.0120 m the system has both kinetic and elastic potential energy. When x = ± A the energy is all elastic potential energy and when x = 0 the energy is all kinetic energy.

11.30. Set Up: Energy conservation gives 12 mυ x 2 + 12 kx 2 = 12 kA2 . At t = 0, x = 0 and υ x = − 12.0 m/s. Solve: (a) x = 0 gives (b) ax = −

1 mυ 2 x 2

= 12 kA2 and A = − υ x

m 2.00 kg =− (−12.0 m/s) = 0.956 m k 315 N/m

⎛ 315 N/m ⎞ k k 2 x so amax = A = ⎜ ⎟ (0.956 m) = 151 m/s m m ⎝ 2.00 kg ⎠

(c) Fx = − kx so Fmax = kA = (315 N/m)(0.956 m) = 301 N. Or, we can use ∑ Fx = ma x to get Fmax = mamax = (2.00 kg)(151 m/s 2 ) = 302 N, the same result apart a small round-off effect.

11.31. Set Up: Velocity, position, and total energy are related by E = 12 kA2 = 12 mυ x 2 + 12 kx 2 . Acceleration and position are related by − kx = max . The maximum magnitude of acceleration is at x = ± A. Solve: (a) E = 12 mυ x 2 + 12 kx 2 = 12 (2.00 kg)(−4.00 m/s) 2 + 12 (315.0 N/m)(+0.200 m)2 .

E = 16.0 J + 6.3 J = 22.3 J. E = 12 kA2 and A= (b) amax =

2E = k

2(22.3 J) = 0.376 m. 315 N/m

⎛ 315 N/m ⎞ k 2 A=⎜ ⎟ (0.376 m) = 59.2 m/s m ⎝ 2.00 kg ⎠

(c) Fmax = mamax = (2.00 kg)(59.2 m/s 2 ) = 118 N. Or, Fx = − kx gives Fmax = kA = (315 N/m)(0.376 m) = 118 N,

which checks. Reflect: The maximum force and maximum acceleration occur when the displacement is maximum and the velocity is zero.


11-8

Chapter 11

11.32. Set Up: Conservation of energy says

1 mv 2 2

+ 12 kx 2 = 12 kA2 and Newton’s second law says − kx = max .

Let +x be to the right. Let the mass of the object be m. ⎛ 28.40 m/s 2 ⎞ ma = (14.0 s22 ) m. Solve: k = 2 x = 2m ⎜ ⎜ 0.600 m ⎟⎟ x ⎝ ⎠ ⎛ ⎞ m 2 A = x 2 + ( m/k )v 2 = (0.600 m) 2 + ⎜⎜ ⎟⎟ (2.20 m/s) = 0.840 m. The object will therefore travel 0.840 m − 22 [14 . 0 s ] m ⎝ ⎠ 0.600 m = 0.240 m to the right before stopping at its maximum amplitude. Reflect: The acceleration is not constant and we cannot use the constant acceleration kinematic equations.

*11.33. Set Up: K + U = E , with E = 12 kA2 and U = 12 kx 2 Solve: U = K says 2U = E . This gives 2

( 12 kx2 ) = 12 kA2 , so x = A/

2.

Reflect: When x = A/2 the kinetic energy is three times the elastic potential energy.

11.34. Set Up: The total energy is E = 12 kA2 . Solve: (a) A → 2 A says E increases by a factor of 22 = 4. The new energy will be 4 E0 . (b)

1 kA2 2

=3

(

1 kA02 2

) . The new amplitude A is A = A

0

3.

*11.35. Set Up: The period is the time for one cycle. A is the maximum value of x. Solve: (a) From the figure, T = 0.80 s.

1 = 1.25 Hz T (c) ω = 2π f = 7.85 rad/s (d) From the figure, A = 3.0 cm (b) f =

(e) T = 2π

2

2

m ⎛ 2π ⎞ ⎛ 2π ⎞ so k = m ⎜ ⎟ = (2.40 kg) ⎜ ⎟ = 148 N/m k ⎝ T ⎠ ⎝ 0.80 s ⎠

11.36. Set Up: When the fish hangs at rest the upward spring force Fx = kx equals the weight mg of the fish. Solve: (a) mg = kx so k = (b) T = 2π

mg (65.0 kg)(9.80 m/s 2 ) = = 5.31 × 103 N/m. x 0.120 m

m 65.0 kg = 2π = 0.695 s. Note that T depends only on m and k and is independent of the k 5.31 × 103 N/m

distance the fish is pulled down. 2

m k ⎛ 2π ⎞ , so we find k = m ⎜ A. ⎟ . Equation 11.18 gives υmax = k m ⎝ T ⎠ k From the analysis of Example 11.15 we know that amax = A. m

*11.37. Set Up: Equation 11.26 gives T = 2π

2

2

⎛ 2π ⎞ ⎛ 2π ⎞ Solve: (a) k = m ⎜ ⎟ = (0.355 kg) ⎜ ⎟ = 3.03 N/m. ⎝ T ⎠ ⎝ 2.15 s ⎠ (b) As x varies from − A to + A the glider moves a distance of 2 A = 1.80 m − 1.06 m = 0.74 m, thus A =

0.37 m. Finally, we calculate υmax =

0.74 m = 2

k 3.03 N/m A= (0.37 m) = 1.1 m/s m 0.355 kg



(c) amax =

11-9

k 3.03 N/m A= ⋅ (0.37 m) = 3.2 m/s 2 m 0.355 kg

Reflect: We measure the position of the mass in a harmonic oscillator from its equilibrium position. For an ideal spring, the equilibrium position is halfway between the two extreme positions of the vibrating mass.

11.38. Set Up: We have m = 0.20 kg, T = 0.45 s, and A = 0.15 m. Equation 11.18 states that υmax = equation 11.26 states that T = 2π

k A and m

m 2π : combining these two equations gives us υ max = A. k T

Solve: (a) We have a maximum velocity of υmax =

2π 2π A= ⋅ (0.15 m) = 2.09 m/s = 2.1 m/s. T 0.45 s

(b) Since the potential energy is zero and the velocity is maximum when the mass is at x = 0, we know that the total energy of the system is equal to the kinetic energy of the mass when υ = υ max . Thus, we have Etotal =

1 1 mυ max 2 = (0.20 kg)(2.09 m/s) 2 = 0.44 J. 2 2 (c) The general form for harmonic motion is x = Acos(ω t + φ0 ). Since we know that x = A when t = 0, we may conclude

that φ0 = 0. Also, we know that ω = 2π f =

2π 2π = = 14 rad/s. Thus, we may write x = (0.15 m)cos(14t ). T 0.45 s

Reflect: You could also find maximum velocity by using T = 2π

m k A. In that to find k, and then use υmax = k m

1 2 kA —since there is no kinetic energy when the mass has its 2 maximum displacement the total energy is simply the maximum potential energy stored in the spring.

case you could find the total energy from Etotal =

11.39. Set Up: T = 2π

m k

2

2

⎛ 2π ⎞ ⎛ 2π ⎞ 3 Solve: k = ⎜ ⎟ =⎜ ⎟ (0.600 kg) = 1.05 × 10 N/m ⎝ T ⎠ ⎝ 0.150 s ⎠

11.40. Set Up: T = 2π

m k 2

2

⎛ 2π ⎞ ⎛ 2π ⎞ Solve: (a) Consider the chair alone: k = ⎜ ⎟ m=⎜ ⎟ (35.4 kg) = 894 N/m ⎝ T ⎠ ⎝ 1.25 s ⎠ 2

2

⎛ T ⎞ ⎛ 2.23 s ⎞ (b) m = k ⎜ ⎟ = (894 N/m) ⎜ ⎟ = 113 kg. The mass of the astronaut is 113 kg − 35.4 kg = 77.6 kg. π 2 ⎝ ⎠ ⎝ 2π ⎠

*11.41. Set Up: The period is the time for 1 cycle; after time T the motion repeats. The graph shows υmax = 20.0 cm/s.

1 mυ 2 x 2

+ 12 kx 2 = 12 kA2

Solve: (a) T = 1.60 s 1 = 0.625 Hz T (c) ω = 2π f = 3.93 rad/s

(b) f =


11-10

Chapter 11

(d) υ x = υ max when x = 0 so

1 kA2 2

= 12 mυmax 2 . A = υmax

m 1 . f = k 2π

k so A = υ max /(2π f ). From the graph m

in the problem, υmax = 0.20 m/s. A = (0.20 m/s)/(2π )(0.625 Hz) = 0.0.051 m = 5.1 cm The mass is at x = ± A when υ x = 0, and this occurs at t = 0.4 s, 1.2 s, and 1.8 s. (e) − kx = max gives

kA = (2π f ) 2 A = (4π 2 )(0.625 Hz) 2 (0.051 m) = 0.79 m/s 2 = 79 cm/s 2 . m The acceleration is maximum when x = ± A and this occurs at the times given in (d). amax =

2

2

m ⎛ T ⎞ ⎛ 1.60 s ⎞ so m = k ⎜ ⎟ = (75 N/m) ⎜ ⎟ = 4.9 kg k ⎝ 2π ⎠ ⎝ 2π ⎠ Reflect: The speed is maximum at x = 0, when ax = 0. The magnitude of the acceleration is maximum at x = ± A, (f) T = 2π

where υ x = 0.

11.42. Set Up: f =

ω 2π

1 2π

=

1 k . T= . f m

1 1 = = 0.167 s f 6.00 Hz (b) ω = 2π f = 2π (6.00 Hz) = 37.7 rad/s Solve: (a) T =

(c) m =

k

ω

=

2

120 N/m (37.7 rad/s) 2

*11.43. Set Up: f = Solve: (a) fs =

1 2π

= 0.0844 kg

k m

1 2π

k 1 , fs + v = ms 2π fs + v ⎛ 1 =⎜ ⎜ fs ⎝ 2π

k ms + mv k ms + mv

⎞⎛ ms ⎞ π 2 ⎟⎜ ⎟= ⎟⎜ k ⎠⎟ ⎠⎝

ms 1 = ms + mv 1 + mv /ms

2

⎛ f ⎞ 1 . Solving for mv gives (b) ⎜ s + v ⎟ = + 1 f m v /ms ⎝ s ⎠ 2 ⎛ ⎞ 15 ⎛ ⎡ f ⎤2 ⎞ 216 ⎜ ⎡ 2.00 × 10 Hz ⎤ s ⎟ = 9.99 × 10215 g ⎜ ⎟ mv = ms ⎢ 1 − ⎥ ⎥ − 1 = (2.10 × 10 g) ⎜ ⎢ 14 ⎟ ⎜ ⎣ fs + v ⎦ ⎟ 2 87 10 Hz . × ⎢ ⎥ ⎦ ⎝ ⎠ ⎝⎣ ⎠

mV = 9.99 femtograms Reflect: When the mass increases, the frequency of oscillation increases.

11.44. Set Up: T = 2π Solve: T =

L g

85.0 s = 3.40 s so L = 25.0

*11.45. Set Up: T = 2π Solve: T = 2π

3.50 m 9.80 m/s 2

2

2

⎛ T ⎞ 2 ⎛ 3.40 s ⎞ g⎜ ⎟ = (9.80 m/s ) ⎜ ⎟ = 2.87 m ⎝ 2π ⎠ ⎝ 2π ⎠

L . The motion specified is one-half of a cycle so the time is t = T /2. g = 3.75 s, so t = 1.88 s



11-11

Reflect: The period of a simple pendulum does not depend on its amplitude, as long as the amplitude is small, and does not depend on the mass of the bob. The period depends only on the length of the pendulum and the value of g.

L m and Equation 11.32 gives T = 2π . Equate these two expressions g k

11.46. Set Up: Equation 11.26 gives T = 2π L and then solve for m to obtain m = k ⋅ . g Solve: We have m = k ⋅

L ⎛ 1.55 m ⎞ = (2.45 N/m) ⎜ ⎟ = 0.388 kg g ⎝ 9.80 m/s 2 ⎠

Reflect: Alternatively, we could find a numerical value for period of the pendulum from T = 2π

T = 2π

L , and then use g

m to find the required mass to add to the spring. k

11.47. Set Up: On earth, ge = 9.80 m/s 2 . Solve: T = 2π

L so T 2 g = 4π 2 L = constant. TM 2 g M = Te 2 g e and g ge 9.80 m/s 2 = (1.60 s) = 2.60 s. gM 3.71 m/s 2

TM = Te Reflect: Smaller g gives a larger period.

11.48. Set Up: The period is for the time for one cycle. The angular amplitude is the maximum value of θ . Solve: (a) From the graph, T = 1.60 s. f =

1 = 0.625 Hz. ω = 2π f = 3.93 rad/s. From the graph we determine T

that the amplitude is 6.0 degrees. 2

2

L ⎛ T ⎞ 2 ⎛ 1.60 s ⎞ so L = g ⎜ ⎟ = (9.80 m/s ) ⎜ ⎟ = 0.635 m g ⎝ 2π ⎠ ⎝ 2π ⎠ (c) No. The graph is unchanged if the mass of the bob is changed while the length of the pendulum and amplitude of swing are kept constant. The period is independent of the mass of the bob. (b) T = 2π

*11.49. Set Up: T = 2π

1 L and f = 2π g

g . L

Solve: (a) L → 2 L, so Tnew = 2T . (b) L =

g (2π f )2

. f → 3 f , so Lnew = L /9

(c) g → 10 g and T unchanged, so Lnew = 10 L (d) g → 10 g , so Tnew = T / 10 (e) The period does not depend on m, so Tnew = T .

11.50. Set Up: E = 12 kA2 Solve: (a) Ei =

1 2

kAi 2 = 12 (2.50 × 102 N/m)(0.060 m)2 = 0.45 J Ef = 12 kAf 2 = 12 (2.50 × 102 N/m)(0.035 m)2 = 0.15 J

The mechanical energy lost is Ei − Ef = 0.30 J.


11-12

Chapter 11

(b) The mechanical energy has been converted to other forms by air resistance and by dissipative forces within the spring.

*11.51. Set Up: As shown in the figure below, the height h above the lowest point of the swing is h = L − Lcosθ = L(1 − cos θ ).

Solve: (a) At the maximum angle of swing, K = 0 and E = mgh. Ei = mgL(1 − cosθi ) = (2.50 kg)(9.80 m/s 2 )(1.45 m)(1 − cos11°) = 0.653 J

Ef = mgL(1 − cosθf ) = (2.50 kg)(9.80 m/s 2 )(1.45 m)(1 − cos 4.5°) = 0.110 J The mechanical energy lost is Ei − Ef = 0.543 J. (b) The mechanical energy has been converted to other forms by air resistance and by dissipative forces within the rope. Reflect: After a while the rock will have come to rest and then all its initial mechanical energy has been “lost,” has been converted to other forms.

11.52. Set Up: E = 12 mυ x 2 + 12 kx 2 Solve: (a) When x is a maximum and the tangent to the curve is horizontal the speed of the mass is zero. This

occurs at t = 0, t = 1.0 s, t = 2.0 s, t = 3.0 s and t = 4.0 s. (b) At t = 0, υ x = 0 and x = 7.0 cm so E0 = 12 kx 2 = 12 (225 N/m)(0.070 m) 2 = 0.55 J (c) At t = 1.0 s, υ x = 0 and x = − 6.0 cm so E1 = 12 kx 2 = 12 (225 N/m)(− 0.060 m)2 = 0.405 J. At t = 4.0 s , υ x = 0

and x = 3.0 cm so E4 = 12 kx 2 = 12 (225 N/m)(0.030 m)2 = 0.101 J. The mechanical energy “lost” is E1 − E4 = 0.30 J. The mechanical energy lost was converted to other forms by nonconservative forces, such as friction, air resistance, and dissipative forces.

11.53. Set Up: E = K + U. K is maximum when U = 0 and at that point K max = E. U is a maximum when K = 0 and at that point U max = E. Problem 11.35 gives that k = 148 N/m and A = 3.0 cm. Solve: K max = U max = 12 kA2 = 12 (148 N/m)(0.030 m) 2 = 0.0666 J

K = K max when x = 0.

11.54. Set Up: Work in an inertial frame moving with the vehicle after the engines have shut off. The acceleration before engine shut-off determines the amount the spring is initially stretched. The initial speed of the ball relative to the vehicle is zero. Before the engine shut-off the ball has acceleration a = 5.00 m/s 2 . Solve: (a) Fx = − kx = ma x gives A =

ma (3.50 kg)(5.00 m/s 2 ) = = 0.0778 m. This is the amplitude of the subsequent k 225 N/m

motion. (b) f =

1 2π

k 1 = m 2π

225 N/m = 1.28 Hz 3.50 kg



1 kA2 2

(c) Energy conservation gives

11-13

k 225 N/m A= (0.0778 m) = 0.624 m/s. 3.50 kg m

= 12 mvmax 2 and vmax =

Reflect: During the simple harmonic motion of the ball its maximum acceleration, when x = ± A, continues to have

magnitude 5.00 m/s 2 .

*11.55. Set Up: T = 2π

m . M is the mass of the empty car and the mass of the loaded car is M + 250 kg. k M M + 250 kg . The period of the loaded car is TL = 2π . k k

Solve: The period of the empty car is TE = 2π

k=

(250 kg)(9.80 m/s 2 ) 4.00 × 10−2 m

= 6.125 × 104 N/m

2

2

⎛T ⎞ ⎛ 1.08 s ⎞ M = ⎜ L ⎟ k − 250 kg = ⎜ (6.125 × 104 N/m) − 250 kg = 1.56 × 103 kg. ⎝ 2π ⎠ ⎝ 2π ⎟⎠

TE = 2π

11.56. Set Up: T = 2π

1.56 × 103 kg 6.125 × 104 N/m

= 1.00 s

L . Relative to the accelerating rocket, the downward acceleration of an object that is g

dropped inside the rocket is a + 9.80 m/s 2 ; the rock is accelerated downward by gravity and the rocket accelerates upward toward the rock. Solve: Use the period T = 2.50 s when the rocket is at rest to calculate the length of the pendulum. 2

2

⎛ T ⎞ 2 ⎛ 2.50 s ⎞ L = g⎜ ⎟ = (9.80 m/s ) ⎜ ⎟ = 1.55 m ⎝ 2π ⎠ ⎝ 2π ⎠

When the rocket is accelerating, T = 2π 2

L . g+a

2

⎛ 2π ⎞ ⎛ 2π ⎞ 2 2 2 2 g + a = L⎜ ⎟ = (1.55 m) ⎜ ⎟ = 39.2 m/s and a = 39.2 m/s − 9.80 m/s = 29.4 m/s . T ⎝ ⎠ ⎝ 1.25 s ⎠ Reflect: If the rocket had a downward acceleration, as an elevator might, then the period decreases. If the rocket is in free-fall, with downward acceleration g, then the pendulum bob doesn’t swing at all and the period is infinite.

11.57. Set Up: E = 12 kA2 = K + U . U = 12 kx 2 Solve: x =

A so U = 2

(

1 1 kA2 4 2

)=

1 E. 4

K + U = E so K = 34 E .

11.58. Set Up: the motion from one position of υ = 0 to the next is half a period. The period depends only on g and g = G

m R2

, where m and R are the mass and radius of Newtonia. The circumference c is related to R by c = 2π R.

G = 6.673 × 10211 N ⋅ m 2 /kg 2 . Solve: T = 2(1.42 s) = 2.84 s. T = 2π

L gives g 2

2

⎛ 2π ⎞ ⎛ 2π ⎞ 2 g =⎜ ⎟ L=⎜ ⎟ (1.85 m) = 9.06 m/s . ⎝ T ⎠ ⎝ 2.84 s ⎠ R=

c 5.14 × 107 m = = 8.18 × 106 m. 2π 2π


11-14

Chapter 11

gR 2 (9.06 m/s 2 )(8.18 × 106 m) 2 = = 9.08 × 1024 kg G 6.673 × 10211 N ⋅ m 2 /kg 2 Reflect: The value of G on Newtonia is within 10% of the value on earth. The mass and radius are somewhat larger than those of earth. m=

*11.59. Set Up: The bounce frequency is given by Eq. (11.25) and the pendulum frequency by Eq. (11.31). Use the relation between these two frequencies that is specified in the problem to calculate the equilibrium length L of the spring, when the apple hangs at rest on the end of the spring. 1 k 1 g . For pendulum motion (small amplitude): f p = . The problem specifies Solve: For vertical SHM: f b = 2π m 2π L 1 g 1 1 k = . Solve for L: g /L = k /4m so L = 4 gm/k = 4w/k = 4(1.00 N)/1.50 N/m = 2.67 m. 2π L 2 2π m Reflect: This is the stretched length of the spring, its length when the apple is hanging from it. (Note: Small angle of swing means v is small as the apple es through the lowest point, so arad is small and the component of mg

that f p =

1 f 2 b

so

perpendicular to the spring is small. Thus the amount the spring is stretched changes very little as the apple swings back and forth.) Set Up: Use Newton’s second law to calculate the distance that the spring is stretched from its unstretched length when the apple hangs from it. The free-body diagram for the apple hanging at rest on the end of the spring is given in the figure below.

Solve: ∑ Fy = ma y

k ΔL − mg = 0 ΔL = mg /k = w/k = 1.00 N/1.50 N/m = 0.667 m Thus the unstretched length of the spring is 2.67 m − 0.67 m = 2.00 m. Reflect: The spring shortens to its unstretched length when the apple is removed.

11.60. Set Up: The maximum acceleration of the lower block can’t exceed the maximum acceleration that can be given to the other block by the friction force. Solve: For block m, the maximum friction force is fs = μs n = μs mg . ∑ Fx = ma x gives μs mg = ma and a = μs g . ⎛ k ⎞ Then treat both blocks together and consider their simple harmonic motion. amax = ⎜ ⎟ A. Set amax = a and ⎝M + m⎠ ⎛ k ⎞ μs g ( M + m) solve for A: μs g = ⎜ . ⎟ A and A = k ⎝M + m⎠ Reflect: If A is larger than this the spring gives the block with mass M a larger acceleration than friction can give the other block, and the first block accelerates out from underneath the other block.

*11.61. Set Up: Apply conservation of energy to the motion before and after the collision. Apply conservation of linear momentum to the collision. After the collision the system moves as a simple pendulum. If the maximum 1 g . In the motion before and after the collision there is energy angular displacement is small we have f = 2π L



11-15

conversion between gravitational potential energy mgh, where h is the height above the lowest point in the motion, and kinetic energy. Solve: Energy conservation during downward swing: m2 gh 0 = 12 m2v 2 and v = 2 gh 0 =

2(9.8 m/s 2 )(0.100 m) =

1.40 m/s. Momentum conservation during collision: m2v = ( m2 + m3 )V and V =

m2v (2.00 kg)(1.40 m/s) = = 0.560 m/s. m2 + m3 5.00 kg

1 (0.560 m/s) 2 = 0.0160 m = 1.60 cm. Energy conservation during upward swing: Mgh f = MV 2 and h f = V 2 /2 g = 2 2(9.80 m/s 2 )

The figure below shows how the maximum angular displacement is calculated from h f . We have cosθ = so θ = 14.5°. f =

1 2π

g 1 = l 2π

48.4 cm , 50.0 cm

9.80 m/s 2 = 0.705 Hz. 0.500 m

Reflect: Note that sin (14.5D ) = 0.250. Since the maximum angle, 14.5D = 0.253 rad, is small, the approximation

sin θ ≈ θ rad is valid. Thus, the condition for Eq. (11.32) to be valid is met.

11.62. Set Up: For steel, Y = 2.0 × 1011 Pa. ω = 2.00 rev/s = 12.6 rad/s. The elongation is a small fraction of the length, so use l0 = 0.50 m to calculate arad . Solve: Apply ∑ Fy = ma y to the mass when it is at the bottom of the circle. The acceleration is arad =

rω 2 = (0.50 m) (12.6 rad/s ) = 79.4 m/s 2 and is upward, so take + y upward. F⊥ is the tension in the wire. 2

F' − mg = marad . F⊥ = m( g + arad ) = (15.0 kg)(9.80 m/s 2 + 79.4 m/s 2 ) = 1.34 × 103 N. Y=

l F (0.50 m)(1.34 × 103 N) l0 F⊥ = 3.35 × 10−3 m = 3.35 mm. so Δl = 0 ⊥ = AY A Δl (0.010 × 10−4 m 2 )(2.0 × 1011 Pa)

Reflect: When the mass hangs at rest at the end of the wire, F⊥ = mg = 147 N and the wire is stretched 0.37 mm.

The radial acceleration is large and F⊥ is much greater than mg. ⎛ Δl ⎞ F⊥ = Y ⎜ ⎟. The height from which he jumps determines his speed at the ground. The acceleration A ⎝ l0 ⎠ as he stops depends on the force exerted on his legs by the ground. In considering his motion take + y downward.

11.63. Set Up:

⎛ Δl ⎞ Solve: (a) F⊥ = YA ⎜ ⎟ = (3.0 × 10−4 m 2 )(14 × 109 Pa)(0.010) = 4.2 × 104 N ⎝ l0 ⎠ (b) As he is stopped by the ground, the net force on him is Fnet = F⊥ − mg , where F⊥ is the force exerted on him by the ground. From part (a), F⊥ = 2(4.2 × 104 N) = 8.4 × 104 N and F = 8.4 × 104 N − (70 kg)(9.80 m/s 2 ) = 8.33 × 104 N. Fnet = ma gives a = 1.19 × 103 m/s 2 . a y = − 1.19 × 103 m/s 2 since the acceleration is upward. υ y = υ 0 y + a yt gives


11-16

Chapter 11

υ0 y = − a yt = ( −1.19 × 103 m/s 2 )(0.030 s) = 35.7 m/s. His speed at the ground therefore is υ = 35.7 m/s. This speed is related to his initial height h above the floor by h=

υ

2

2g

1 mυ 2 2

=

= mgh and

(35.7 m/s) 2 2(9.80 m/s 2 )

= 65 m.

11.64. Set Up: Y = F⊥l0 /A Δl (Eq. 11.4 holds since the problem states that the stress is proportional to the strain.) Thus Δl = F⊥l0 /AY . Use proportionality to see how changing the wire properties affects Δl . Solve: (a) Change l0 but F⊥ (same floodlamp), A (same diameter wire), and Y (same material) all stay the same.

Δl F⊥ Δl Δl = = constant, so 1 = 2 l0 AY l01 l02 Δl2 = Δl1(l02 /l01 ) = 2Δl1 = 2(0.18 mm) = 0.36 mm (b) A = π (d/2) 2 = 14 π d 2 , so Δl =

F⊥l0 1 π d 2Y 4 2

F⊥ , l0 , Y all stay the same, so Δl ( d ) = F⊥l0 /( 14 π Y ) = constant Δl1 (d1 2 ) = Δl2 (d 2 2 )

Δl2 = Δl1 (d1/d 2 ) 2 = (0.18 mm)(1/2) 2 = 0.045 mm (c) F⊥ , l0 , A all stay the same so ΔlY = F⊥l0 /A = constant

Δl1Y1 = Δl2Y2 Δl2 = Δl1 (Y1/Y2 ) = (0.18 mm)(20 × 1010 Pa/11 × 1010 Pa) = 0.33 mm Reflect: Greater l means greater Δl , greater diameter means less Δl , and smaller Y means greater Δl .

*11.65. Set Up: Y =

FT /A . The foot is in rotational equilibrium. Let counterclockwise torques be positive. Δl/l0

Solve: (a) The free-body diagram for the foot is given in the figure below, T is the tension in the tendon and A is the force exerted on the foot by the ankle. n = (75 kg) g , the weight of the person.

(b) Apply ∑τ = 0, with the pivot at the ankle: T (4.6 cm) − n(12.5 cm) = 0

⎛ 12.5 cm ⎞ 2 T =⎜ ⎟ (75 kg)(9.80 m/s ) = 2000 N, 2.72 times his weight. ⎝ 4.6 cm ⎠ (c) The foot pulls downward on the tendon with a force of 2000 N. 2000 N ⎛F ⎞ Δl = ⎜ T ⎟ l0 = (25 cm) = 4.4 mm 6 (1470 × 10 Pa)(78 × 1026 m 2 ) ⎝ YA ⎠



11-17

11.66. Set Up: Let + x be downward. The tension F⊥ in the wire equals the 980 N weight of the mass. Fx = − kx gives k = −

F⊥ −980 N =− = 1.63 × 105 N/m. x 6.0 × 10−3 m

Solve: (a) f = (b) Y =

1 2π

k 1 1.63 × 105 N/m = = 6.4 Hz 100 kg m 2π

l0 F⊥ (4.00 m)(980 N) = = 6.5 × 1010 Pa. A Δl (0.10 × 1024 m 2 )(6.0 × 1023 m)

*11.67. Set Up: The tension is the same at all points along the composite rod. Apply Eqs. (11.2) and (11.4) to relate the elongations, stresses, and strains for each rod in the compound. Solve: Each piece of the composite rod is subjected to a tensile force of 4.00 × 104 N. F l F l (a) Y = ⊥ 0 so Δl = ⊥ 0 A Δl YA F⊥l0,b F⊥l0,n Δlb = Δln gives that = (b for brass and n for nickel); l0,n = L Yb Ab Yn An But the F⊥ is the same for both, so

l0,n =

Yn An l0,b Yb Ab

⎛ 21 × 1010 Pa ⎞⎛ 1.00 cm 2 ⎞ L=⎜ (1.40 m) = 1.63 m ⎜ 9.0 × 1010 Pa ⎟⎜ ⎟⎜ 2.00 cm 2 ⎟⎟ ⎝ ⎠⎝ ⎠ (b) stress = F⊥ /A = T/A

brass: stress = T/A = (4.00 × 104 N)/(2.00 × 1024 m 2 ) = 2.00 × 108 Pa nickel: stress = T/A = (4.00 × 104 N)/(1.00 × 1024 m 2 ) = 4.00 × 108 Pa (c) Y = stress/strain and strain = stress/Y brass: strain = (2.00 × 108 Pa)/(9.0 × 1010 Pa) = 2.22 × 10−3 nickel: strain = (4.00 × 108 Pa)/(21 × 1010 Pa) = 1.90 × 10−3 Reflect: Larger Y means less Δl and smaller A means greater Δl , so the two effects largely cancel and the lengths don’t differ greatly. Equal Δl and nearly equal l means the strains are nearly the same. But equal tensions and A differing by a factor of 2 means the stresses differ by a factor of 2.

11.68. Set Up: Constant speed means a = 0 and the force he applies to her hair equals his weight. Y =

FT /A . Δl/l0

l0 = 20 m. A = π r 2 = π (1.0 cm) 2 = 3.14 cm 2 = 3.14 × 10−4 m 2 (60 kg)(9.80 m/s 2 ) ⎛F ⎞ ⎛ mg ⎞ (20 m) = 7.6 cm Solve: (a) Δl = ⎜ T ⎟ l0 = ⎜ ⎟ l0 = (3.14 × 1024 m 2 )(490 × 106 Pa) ⎝ AY ⎠ ⎝ AY ⎠ (b)

(196 × 106 Pa)(3.14 × 1024 m 2 ) FT mg = 6280 kg = = 196 × 106 Pa so m = A A 9.80 m/s 2

11.69. Set Up: As the oil ascends to the surface the surrounding pressure changes from 2.27 × 107 Pa to 1× 105 Pa; thus, Δp = 1 × 105 Pa − 2.27 × 107 Pa = −2.26 × 107 Pa. Equation 11.7 gives the relationship between pressure change and the volume change of the oil: B = −

Δp . ΔV /V0


11-18

Chapter 11

Solve: Solving Equation 11.7 for the change in volume we obtain ( −2.26 × 107 Pa)(35,600 barrels) (Δp )V0 = + 342 barrels. Thus, the volume of the oil at the surface is ΔV = − =− B 2.35 × 109 Pa

V = V0 + ΔV = 35,600 barrels + 342 barrels = 35,900 barrels. Reflect: The volume of the oil is greater where the pressure is lower. The very large decrease in pressure causes a noticeable (approximately 1%) increase in the volume of the oil as it ascends to the surface.


12

MECHANICAL WAVES AND SOUND

Answers to Multiple-Choice Problems 1. A 2. B 3. A 4. A 5. B 6. C 7. B, C 8. D 9. A 10. B 11. D 12. A 13. C 14. A 15. C

Solutions to Problems *12.1. Set Up: f1 = Solve: (a) υ =

v 326 m/s = = 494 Hz. 2 L 2(0.330 m)

6. 0 m υ 344 m/s = 1.7 cm. For f = 20 Hz, = 1.2 m/s. For f = 20,000 Hz, λ = = 5. 0 s f 20,000 Hz

λ=

υ f

=

344 m/s = 17 m. 20 Hz

The range of wavelengths is 1.7 cm to 17 m. (b) υ = c = 3.00 × 108 m/s. For λ = 700 nm,

f =

c

λ

=

3.00 × 108 m/s 700 × 10−9 m

= 4.3 × 1014 Hz.

For λ = 400 nm, f =

c

λ

=

3.00 × 108 m/s 400 × 10−9 m

= 7.5 × 1014 Hz.

The range of frequencies for visible light is 4.3 × 1014 Hz to 7.5 × 1014 Hz. υ 344 m/s (c) υ = 344 m/s. λ = = = 1.5 cm f 23 × 103 Hz (d) υ = 1480 m/s. λ =

υ

=

1480 m/s

= 6.4 cm 23 × 103 Hz Reflect: For a given υ , larger f corresponds to smaller λ . For the same f , λ increases when υ increases. f

12.2. Set Up: υ = c = 3.00 × 108 m/s. υ = f λ Solve: (a) λ = (b) f =

c

λ

=

c 3.00 × 108 m/s = = 2.87 m f 104.5 × 106 Hz

3.00 × 108 m/s 0.10 × 10−9 m

= 3.0 × 1018 Hz


12-1

12-2

(c) f =

Chapter 12

c

λ

=

3.00 × 108 m/s 1.1 × 10−3 m

= 2.7 × 1011 Hz

(d) λ =

c 3.00 × 108 m/s = = 30 nm f 1016 Hz

(e) λ =

c 3.00 × 108 m/s = = 21.0 cm f 1.43 × 109 Hz

(f) λ =

c 3.00 × 108 m/s = = 0.122 m f 2.45 × 109 Hz

12.3. Set Up: υ = f λ . f = 10.0 cycles/min = 0.167 cycles/s = 0.167 Hz. Solve: υ = f λ = (0.167 Hz)(13 km) = 2.2 km/s 12.4. Set Up: The time from the highest displacement to lowest displacement is T /2. The distance from highest displacement to lowest displacement is 2A. The distance between wave crests is λ . υ = f λ = λ /T . Solve: (a) T = 2(2.5 s) = 5.0 s. λ = 6.0 m. υ = (b) A = (0.62 m)/2 = 0.31 m

6.0 m = 1.2 m/s. 5. 0 s

(c) The amplitude becomes 0.15 m but the wavelength, period and wave speed are unchanged. F'

*12.5. Set Up: υ =

μ

. The mass per unit length of the wire is

μ= Solve: υ =

m 0.0600 kg = = 0.015 kg/m. 4.00 m L

1000 N = 258 m/s 0.0150 kg/m

12.6. Set Up: υ =

F'

μ

. μ = m/L. υ = f λ .

Solve: F' = μυ 2 = μ ( f λ ) 2 =

0.120 kg ([40.0 Hz][0.750 m]) 2 = 43.2 N 2.50 m

*12.7. Set Up: μ = 0.0550 kg/m. F' = mg , the weight of the hanging mass. The frequency of the waves is 120 Hz. υ = f λ . Solve: (a) υ = (b) λ =

υ f

=

F'

μ

=

(1.50 kg)(9.80 m/s 2 ) = 16.3 m/s 0.0550 kg/m

16.3 m/s = 0.136 m 120 Hz

(c) Doubling the mass increases both the wave speed and the wavelength by a factor of λ = 0.192 m.

2; υ = 23.1 m/s and

Reflect: The speed of the waves on the rope is much less than the speed of sound in air.

12.8. Set Up: The intensity is the rate of energy transfer per unit area and is proportional to the square of the amplitude of the wave. Solve: (a) Intensity is increased by a factor of 22 = 4. (b) The amplitude must be increased by a factor of

9 = 3.


Mechanical Waves and Sound

12.9. Set Up: υ =

FT

μ

Solve: FT = Mg . V =

=

12-3

FT L , where m is the mass of the wire. m

MgL = m

Mg

μ

(a) M → 2 M gives Vnew = 2V (b) μ is the same since an identical wire twice as long will have twice the mass. Vnew = V (c) m → 3m; Vnew = V/ 3 (d) L → 2 L while m stays the same; Vnew = 2V (e) L → 2 L while m stays the same and M → 10 M ; Vnew = 2(10)V = 2 5V Reflect: The speed increases when the tension in the wire increases and the speed decreases when the mass per unit length μ increases.

12.10. Set Up: The general form of y ( x, t ) for a wave traveling in the + x direction is x⎞ ⎛t y ( x, t ) = Asin2π ⎜ − ⎟ . T λ ⎝ ⎠

υ = f λ. Solve: Compare y ( x, t ) given in the problem to the general form. (a) A = 6.50 mm (b) λ = 0.280 m (c) T = 0.0360 s and f = 1/T = 27.8 Hz (d) υ = f λ = (27.8 Hz)(0.280 m) = 7.78 m/s (e) The wave is propagating in the + x direction.

*12.11. Set Up: υ = f λ . T = 1/f . Wave number k = 2π /λ . The general form for y ( x, t ) is x⎞ ⎛t y ( x, t ) = Asin2π ⎜ − ⎟ . ⎝T λ ⎠ υ 8.00 m/s 2π = 25.0 Hz. T = 1/f = 0.0400 s. k = Solve: (a) f = = = 19.6 rad/m. λ 0.320 m λ t x ⎛ ⎞ (b) y ( x, t ) = − (0.0700 m)sin 2π ⎜ − ⎟ ⎝ 0.0400 s 0.320 m ⎠

⎛ 0.150 s 0.360 m ⎞ (c) y (0.360 m, 0.150 s) = − (0.0700 m)sin 2π ⎜ − ⎝ 0.0400 s 0.320 m ⎟⎠ y = − (0.0700 m)sin (16.5 rad) = 0.0498 m

x⎞ ⎛t − ⎟. υ = λ f. ⎝T λ⎠

12.12. Set Up: For a wave traveling in the + x direction, y ( x, t ) = A sin 2π ⎜ Solve: By comparing to the general equation,

λ = 0.150 m.

2π 2π = 157 m −1, = 41.9 m −1 and A = 1.50 mm. T = 0.0400 s and T λ

1 = 25.0 Hz. A = 1.50 mm T (b) υ = f λ = (25.0 Hz)(0.150 m) = 3.75 m/s. The wave is traveling in the + x direction.

(a) λ = 0.150 m. f =

(c) y (0.135 m, 0.100 s) = (1.50 mm)sin ([157 s −1 ][0.100 s] − [41.9 m −1 ][0.135 m]) y (0.135 m, 0.100 s) = − 0.870 mm


12-4

Chapter 12

12.13. Set Up: For a wave traveling in the + x direction, x⎞ ⎛t y ( x, t ) = A sin 2π ⎜ − ⎟ . λ T ⎝ ⎠

υ = f λ and υ =

FT

μ

.

2π 2π = 415 s −1 and = 44.9 m −1. T = 0.0151 s and λ = 0.140 m. T λ λ 0.140 m υ = fλ = = = 9.27 m/s. T 0.0151 s F 4.00 N μ = T2 = = 4.65 × 10−2 kg/m = 0.465 g/cm υ (9.27 m/s) 2 Reflect: A 1-meter length of the string has a mass of 0.0465 kg and a weight of about 1.6 ounces. This is a reasonable value. Solve: By comparing to the general equation,

12.14. Set Up: The distance d the wave travels in time t is d = υt . The distance the wave travels is twice the depth of the ocean. Solve: d = (1531 m/s)(3.27 s) = 5006 m; the depth is 2500 m

*12.15. Set Up: The waves obey the principle of superposition: the resultant displacement of the string is the algebraic sum of the displacements due to each wave. Each wave travels 1 cm each second. Solve: The locations of each wave pulse and the resultant displacement of the string is shown at each time in Figures (a)–(d) below.



12-5

Reflect: When the pulses displace the string in the same direction they reinforce each other and give a larger pulse and when the pulses displace the string in opposite directions they tend to cancel and give a smaller pulse.

υ

12.16. Set Up: For a string fixed at each end, f1 =

2L

frequency f3 = 3 f1. The nth harmonic is f n = nf1. λ = Solve: (a) f1 = 2L =

. f 2 is the first overtone. f3 is the second overtone and has

υ f

.

48.0 m/s υ = 16.0 Hz. λ1 = = 2 L = 3.00 m. f1 2(1.50 m)

υ 2L ⎛υ ⎞ (b) f3 = 3 f1 = 3 ⎜ = = 1.00 m. ⎟ = 48.0 Hz. λ 3 = L 2 f3 3 ⎝ ⎠

υ L ⎛υ ⎞ (c) f 4 = 4 f1 = 4 ⎜ = = 0.750 m. ⎟ = 64.0 Hz. λ4 = f4 2 ⎝ 2L ⎠ F'

*12.17. Set Up: υ = Solve: (a) υ =

μ

F⊥ = m/L

where μ = m/L. f1 =

F⊥ L = m

2L

(800 N)(0.400 m) 3.00 × 10−3 kg

f1 = (b) n =

υ

υ 2L

=

. The nth harmonic has frequency f n = nf1.

= 327 m/s.

327 m/s = 409 Hz. 2(0.400 m)

10,000 Hz = 24.4. The 24th harmonic is the highest that could be heard. f1

12.18. Set Up: υ =

F⊥

μ

, where m/L. f1 =

υ 2L

.

Solve: (a) υ = 2 Lf1 = 2(0.800 m)(60.0 Hz) = 96.0 m/s (b) F⊥ = μυ 2 = mυ 2 /L = (0.0400 kg)(96.0 m/s)2 /(0.800 m) = 461 N © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12-6

Chapter 12

*12.19. Set Up: For the fundamental, f1 =

υ

, where x is the length of the portion of the string that is free to 2x vibrate. The wave speed υ depends on the tension and linear mass density of the string, so it is the same no matter where the finger is placed. (a) For x = L = 0.600 m, f1 = 440 Hz. υ = 2 xf1 = 2(0.600 m)(440 Hz) = 528 m/s. Then for f1 = 587 Hz,

x=

υ 2 f1

=

528 m/s = 0.450 m = 45.0 cm 2(587 Hz)

(b) For f1 = 392 Hz,

λ=

υ 2 f1

= 0.673 m.

The maximum length of the vibrating string, the distance between the bridge and the upper end of the fingerboard, is 0.600 m, so this is not possible. A G 4 note cannot be played without retuning. Reflect: The speed of the waves on the string is larger than the speed of sound in air.

12.20. Set Up: T = 1/f is the time for one cycle. For f = 440 Hz, T = 2.27 × 10−3 s and this is the distance along the time axis occupied by one cycle of this wave. For f = 880 Hz, T = 1.13 × 10−3 s. The net displacement is the algebraic sum of the displacements due to each wave. Solve: (a) and (b) The two waves and their sum are shown in the figure below.

12.21. Set Up: T = 1/f is the time for one cycle. For f = 370 Hz, T = 2.70 × 10−3 s and this is the distance along the time axis occupied by one cycle of this wave. For f = 740 Hz, T = 1.35 × 10−3 s and for f = 1110 Hz, T = 0.901 × 1023 s. The net displacement is the algebraic sum of the displacements due to each wave.



12-7

Solve: The three waves and their sum are shown in the figure below.

Reflect: The net sound wave is periodic, with period 2.70 ms, but it is not sinusoidal.

v . v = F /μ . A standing wave on a string with frequency f produces a 2L sound wave that also has frequency f. f1 = 245 Hz. L = 0.635 m.

12.22. Set Up: For the fundamental, f1 =

Solve: (a) v = 2 f1L = 2(245 Hz)(0.635 m) = 311 m/s. (b) The frequency of the fundamental mode is proportional to the speed and hence to the square root of the tension; (245 Hz) 1.01 = 246 Hz. (c) The frequency will be the same, 245 Hz. The wavelength will be λair = vair /f = (344 m/s)/(245 Ηz) = 1.40 m,

which is larger than the wavelength of standing wave on the string by a factor of the ratio of the speeds. Reflect: Increasing the tension increases the wave speed and this in turn increases the frequencies of the standing waves. The wavelength of each normal mode depends only on the length of the string and doesn't change when the tension changes.

*12.23. Set Up: An open end is a displacement antinode. A closed end is a displacement node. Adjacent nodes are a distance λ /2 apart. Adjacent antinodes are also a distance λ /2 apart, and the node to antinode distance is λ /4. Let x be the distance from the left-hand end of the pipe. Solve: (a) There is an antinode at x = 0. The first node is at x = λ /4 and each successive node is λ /2 farther to the

right. Each successive overtone adds one node. fundamental: λ1 = 2 L = 2.40 m. λ1/4 = 0.60 m. There is a node at x = 0.60 m. 1st overtone: λ2 = L = 2.40 m. λ2 /4 = 0.30 m. There are a nodes at x = 0.30 m, 0.90 m. 2nd overtone: λ3 = 2 L/3 = 0.80 m. λ3/4 = 0.20 m. There are a nodes at x = 0.20 m, 0.60 m, and 1.00 m. (b) There now is a node at x = 0. Successive nodes are λ /2 farther to the right. fundamental: λ1 = 4 L = 4.80 m. λ1/2 = 2.40 m. There only a node is at x = 0. 1st overtone: λ3 = 4 L/3 = 1.60 m. λ3/2 = 0.80 m. There are a nodes at x = 0 and x = 0.80 m.


12-8

Chapter 12

2nd overtone: λ5 = 4 L/5 = 0.96 m. λ5 /2 = 0.48 m. There are a nodes at x = 0, 0.48 m, and 0.96 m. Reflect: In each case the fundamental has one node, the 1st overtone has two nodes, and the 2nd overtone has three nodes.

12.24. Set Up: A pipe closed at one end is a stopped pipe. fundamental: λ1 = 2 L = 2.40 m. λ1/4 = 0.60 m. There is a node at x = 0.60 m. Solve: (a) f1 =

υ 2L

and f n = nf1. f1 =

(b) f1 =

υ 4L

344 m/s = 382 Hz. f 2 = 764 Hz, f3 = 1146 Hz, f 4 = 1528 Hz 2(0.450 m)

and f n = nf1, n = 1, 3, 5, . . . f1 =

(c) open pipe: n =

n = 103.

344 m/s = 191 Hz. f3 = 573 Hz, f5 = 955 Hz, f 7 = 1337 Hz 4(0.450 m)

f 20,000 Hz f 20,000 Hz = = 52. closed pipe: = = 104. But only odd n are present, so f1 f1 382 Hz 191 Hz

*12.25. Set Up: A pipe open at one end and closed at the other is a stopped pipe. υ 344 m/s Solve: (a) For an open pipe, f1 = (b) For a stopped pipe, f1 =

2L

=

2(4.88 m)

= 35.2 Hz.

υ 35.2 Hz = = 17.6 Hz. 4L 2

v v . For a stopped pipe, f1 = . v = f λ . v = 344 m/s. For a pipe, there 2L 4L must be a displacement node at a closed end and an antinode at the open end. v 344 m/s = = 0.290 m. Solve: (a) L = 2 f1 2(594 Hz) (b) There is a node at one end, an antinode at the other end and no other nodes or antinodes in between, so

12.26. Set Up: For an open pipe, f1 =

λ1 4

= L and λ1 = 4 L = 4(0.290 m) = 1.16 m.

(c) f1 =

v 1⎛ v ⎞ 1 = ⎜ ⎟ = (594 Hz) = 297 Hz. 4L 2 ⎝ 2L ⎠ 2

Reflect: We could also calculate f1 for the stopped pipe as f1 =

v

λ1

=

344 m/s = 297 Hz, which agrees with our result 1.16 m

in part (c).

12.27. Set Up: υ = 354 m/s. L = 8.0 cm. When the mouth is open it behaves as an open pipe and when it is closed it behaves as a closed pipe. 2L nυ Solve: (a) open λn = and f n = , with n = 1, 2, 3, . . . . 2L n λ1 = 16.0 cm, λ2 = 8.0 cm, λ3 = 5.3 cm, λ4 = 4.0 cm f1 =

υ 2L

=

354 m/s = 2210 Hz, f 2 = 2 f1 = 4420 Hz, f3 = 3 f1 = 6630 Hz, f 4 = 4 f1 = 8840 Hz 2(0.080 m)



4L nυ and f n = , with n = 1, 3, 5, . . . . 4L n λ1 = 32.0 cm, λ3 = 10.7 cm, λ5 = 6.4 cm, λ7 = 4.6 cm

(b) closed λn =

f1 =

12-9

υ 4L

=

354 m/s = 1110 Hz , f3 = 3 f1 = 3330 Hz, f5 = 5 f1 = 5550 Hz, f 7 = 7 f1 = 7770 Hz 4(0.080 m)

12.28. Set Up: For either type of pipe, stopped or open, the fundamental frequency is proportional to the wave speed v. According to Table 12.1, the speed of sound in air is 344 m/s and the speed of sound in helium is 999 m/s at 20D C. Solve: (a) For a given pipe, the fundamental frequency is proportional to the speed of sound. Thus,

f He υHe = , f air υair

⎛ 999 m/s ⎞ = 761 Hz. which gives us f He = (262 Hz) ⎜ ⎝ 344 m/s ⎟⎠ (b) No. In either case the frequency is proportional to the speed of sound in the gas. Reflect: The frequency is much higher for helium, since the rms speed is greater for helium.

*12.29. Set Up: For a stopped pipe, f1 = Solve: L =

υ

4 f1 typical adult.

=

υ 4L

. f1 = 220 Hz

344 m/s = 39.1 cm. This result is a reasonable value for the mouth to diaphragm distance for a 4(220 Hz)

Reflect: 1244 Hz is not an integer multiple of the fundamental frequency of 220 Hz, it is 5.65 times the fundamental. The production of sung notes is more complicated than harmonics of an air column of fixed length.

12.30. Set Up: There are displacement nodes at the closed end of an air column. Problem 12.28 shows that Eq. (12.7) applies to a pipe closed at both ends as well as one open at both ends. Solve: (a) There are displacement nodes at the floor and ceiling. υ 344 m/s (b) λ1 = 2 L = 4.90 m. f1 = = = 70.2 Hz. λ1 4.90 m (c) λ2 =

λ1 λ = 2.45 m. f 2 = 2 f1 = 140 Hz. λ3 = 1 = 1.63 m. f3 = 3 f1 = 211 Hz. 2 3

12.31. Set Up: For an open pipe the first overtone has frequency f 2 = 2 f1 = 2 overtone has frequency f3 = 3 f1 =

υ 2L

=

υ L

. For a stopped pipe the first

3υ . The mouthpiece acts as a closed end. 4L

3υ 3(344 m/s) = = 70 Hz 4L 4(3.7 m) (b) A pipe closed at both ends has the same overtone frequencies as a pipe open at both ends (Problem 12.28). υ 344 m/s f2 = = = 93 Hz L 3.7 m Reflect: The fundamental frequency for a stopped pipe is lower by a factor of two than the fundamental frequency for a pipe of the same length but closed at both ends. But for the stopped pipe the first overtone is the third harmonic and the first overtone is lower than that of the other pipe only by a factor of 4/3. Solve: (a) f3 =

12.32. Set Up: Use the standing wave pattern to relate the wavelength of the standing wave to the length of the air column and then use Eq. (12.1) to calculate f. There is a displacement antinode at the top (open) end of the air column and a node at the bottom (closed) end, as shown in the figure below.


12-10

Chapter 12

Solve: (a) From the figure below we see that λ /4 = L, so λ = 4 L = 4(0.140 m) = 0.560 m and

f =

v

λ

=

344 m/s = 614 Hz. 0.560 m

(b) Now the length L of the air column becomes

1 (0.140 2

m) = 0.070 m and λ = 4 L = 0.280 m.

v 344 m/s = = 1230 Hz λ 0.280 m Reflect: Smaller L means smaller λ which in turn corresponds to larger f. f =

*12.33. Set Up: The path difference for the two sources is d. For destructive interference, the path difference is a half-integer number of wavelengths. For constructive interference, the path difference is an integer number of wavelengths. λ = υ /f

υ

344 m/s = 0.474 m f 725 Hz (a) Will first produce destructive interference when d = λ /2 = 0.237 m. (b) Will next produce destructive interference when d = 3λ /2 = 0.711 m. (c) Will first produce constructive interference again when d = λ = 0.474 m. Solve: λ =

=

12.34. Set Up: f =

υ with υ = 344 m/s. There is destructive interference when the path difference (45.0 cm) is λ

a half-integer number of wavelengths. There is constructive interference when the path difference is an integer number of wavelengths. Solve: (a) For destructive interference, 45.0 cm = (n + 12 )λ , n = 0, 1, 2 for the lowest frequencies. f = (n + 12 )

υ 0.450 m

(

= n+

1 2

) (764 Hz)

and f = 382 Hz, 1146 Hz, and 1910 Hz. All three of these frequencies are audible. (b) For constructive interference 45.0 cm = nλ, n = 1, 2, 3 for the lowest frequencies. f =n

υ 0.450 m

= n(764 Hz)

and f = 764 Hz, 1528 Hz, and 2292 Hz. All three of these frequencies are audible.

*12.35. Set Up: One speaker is 4.50 m from the microphone and the other is 4.03 m from the microphone, so the path difference is 0.42 m. For constructive interference, the path difference is an integer number of wavelengths. For destructive interference, the path difference is a half-integer number of wavelengths. f = υ /λ

υ = 820 Hz; λ υ 2λ = 0.42 m gives λ = 0.21 m and f = = 1640 Hz; λ 3λ = 0.42 m gives λ = 0.14 m and f = υ = 2460 Hz, and so on.

Solve: (a) λ = 0.42 m gives f =

The frequencies for constructive interference are n(820 Hz), n = 1, 2, 3, . . . . © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Mechanical Waves and Sound (b) λ /2 = 0.42 m gives λ = 0.84 m and f =

12-11

υ = 410 Hz; λ

υ = 1230 Hz; λ υ 5λ /2 = 0.42 m gives λ = 0.168 m and f = = 2050 Hz, and so on. λ 3λ /2 = 0.42 m gives λ = 0.28 m and f =

The frequencies for destructive interference are (2n + 1)(410 Hz), n = 0, 1, 2, . . . . Reflect: The frequencies for constructive interference lie midway between the frequencies for destructive interference. ⎛ υ⎞ , n = 1, 3, 5, . . . . ⎝ 4 L ⎟⎠

12.36. Set Up: The auditory canal acts as a stopped pipe and has frequencies f n = n ⎜ Solve: (a) f1 = (b) λ1 =

υ f1

υ 4L

=

344 m/s = 3.6 kHz; f3 = 3 f1 = 11 kHz 4(0.024 m)

= 9.6 cm; λ3 =

υ f3

= 3.2 cm. λ1 = 4L and λ3 = 4 L/3

12.37. Set Up: υ = f λ υ 1531 m/s = 90 m Solve: (a) λ = =

f 17 Hz υ 1531 m/s (b) f = = = 102 kHz λ 0.015 m υ 344 m/s (c) λ = = = 1.4 cm f 25 × 103 Hz υ 344 m/s (d) For f = 78 kHz , λ = = = 4.4 mm. For f = 39 kHz, f 78 × 103 Hz υ 344 m/s = 8.8 mm. λ= = f 39 × 103 Hz The range of wavelengths is 4.4 mm to 8.8 mm. υ 1550 m/s (e) λ = 0.25 mm so f = = = 6.2 MHz λ 0.25 × 1023 m

12.38. Set Up: υ =

FT

μ

. f1 =

υ 2L

. λ=

υ

. The sound waves it produces have the same frequency as the standing

f

waves on the wire. The second overtone is the third harmonic, with f = 3 f1. Solve: (a) υ =

35.0 N (5.625 × 10−3 kg)/(0.750 m)

= 68.3 m/s.

f =

3υ 3(68.3 m/s) = = 137 Hz 2 L 2(0.750 m)

and

λ=

υ f

=

68.3 m/s = 0.50 m 137 Hz

=

344 m/s = 2.51 m 137 Hz

3λ ⎛ ⎞ = L⎟ ⎜⎝ Note: ⎠ 2 (b) f = 137 Hz, the same as for the wire.

λ=

υ f

λ is larger in air because υ is larger there. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12-12

Chapter 12

*12.39. Set Up: P = I (4π r 2 ) and

I1 r22 = I 2 r12

Solve: (a) P = I (4π r 2 ) = (6.50 W/m 2 )(4π )(2.50 m) 2 = 511 W 2

2 ⎛r ⎞ ⎛ 2.50 m ⎞ 2 (b) I 2 = I1 ⎜ 1 ⎟ = (6.50 W/m 2 ) ⎜ ⎟ = 0.829 W/m . r 7 00 m ⎝ ⎠ ⎝ 2⎠ (c) All the power radiated in one second is received by the walls of the room in one second. Each second the walls receive 511 J. ⎛ I ⎞ 12.40. Set Up: β = 10 log ⎜ ⎟ , with I 0 = 10212 W/m 2 . If x = log y then y = 10 x. ⎝ I0 ⎠ ⎛ I ⎞ I Solve: (a) β = 55.0 dB gives 5.50 = log ⎜ ⎟ and = 105.5 = 3.16 × 105. I I 0 ⎝ 0⎠

I = (3.16 × 105 )(1 × 10−12 W/m 2 ) = 3.16 × 10−7 W/m 2 ⎛ I ⎞ (b) β = 92.0 dB gives 9.20 = log ⎜ ⎟ and ⎝ I0 ⎠

I = (109.2 )(1 × 10−12 W/m 2 ) = 1.58 × 10−3 W/m 2 ⎛ I ⎞ (c) β = − 2.0 dB gives −0.20 = log ⎜ ⎟ and ⎝ I0 ⎠

I = (10−0.20 )(1 × 10−12 W/m 2 ) = 6.31 × 10213 W/m 2 .

β is negative when I < I 0 . ⎛ I ⎞ 212 W/m 2 . ⎟ , with I 0 = 10 I ⎝ 0⎠

*12.41. Set Up: β = 10log ⎜

⎛ 1.27 × 10−4 W/m 2 ⎞ = 81.0 dB (a) β = 10log ⎜ ⎜ 1 × 10−12 W/m 2 ⎟⎟ ⎝ ⎠ ⎛ 6.53 × 10−10 W/m 2 ⎞ = 28.1 dB (b) β = 10log ⎜ ⎜ 1 × 10−12 W/m 2 ⎟⎟ ⎝ ⎠ ⎛ 1.5 × 10−14 W/m 2 ⎞ = − 18.2 dB. β is negative when I < I 0 . (c) β = 10log ⎜ ⎜ 1 × 10−12 W/m 2 ⎟⎟ ⎝ ⎠

12.42. Set Up: From Example 12.9 Δβ = (10 dB)log( I 2 /I1 ) Set Δβ = 13.0 dB and solve for I 2 /I1. Solve: 13.0 dB = 10 dBlog( I 2 /I1 ) so 1.3 = log( I 2 /I1 ) and I 2 /I1 = 20.0. (b) Reflect: According to the equation in part (a) the difference in two sound intensity levels is determined by the ratio of the sound intensities. So you don’t need to know I1, just the ratio I 2 / I1.

⎛ I 2 ⎞ I 2 r12 = 2 ⎟. ⎝ I1 ⎠ I1 r2

12.43. Set Up: Example 12.10 shows that β 2 − β1 = (10 dB)log ⎜ ⎛I ⎞ I Solve: Δβ = + 40.0 dB. log ⎜ 2 ⎟ = 4.00 and 2 = 1.00 × 104 I1 I ⎝ 1⎠

r2 = r1

I1 1 = (15.0 m) = 15.0 cm I2 1.00 × 104



12-13

Reflect: A change of 102 in distance gives a change of 104 in intensity. Our analysis assumes that the sound spreads from the source uniformly in all directions.

12.44. Set Up: For a whisper, I = 1 × 10210 W/m 2 . The area of the tympanic membrane is A = π r 2 , with r = 4.2 × 10−3 m. Intensity is energy per unit time per unit area. (a) E = IAt = (1 × 10−10 W/m 2 )π (4.2 × 10−3 m) 2 (1 s) = 5.5 × 10−15 J (b) K = 12 mυ 2 so

υ=

2K = m

2(5.5 × 10−15 J) 2.0 × 10−6 kg

= 7.4 × 10−5 m/s = 0.074 mm/s

⎛ I ⎞ −12 2 ⎟ , with I 0 = 1 × 10 W/m . The I ⎝ 0⎠

*12.45. Set Up: Intensity is energy per unit time per unit area. β = 10log ⎜

area of the eardrum is A = π r 2 , with r = 4.2 × 10−3 m. Part (b) of Problem 12.42 gave y = 0.074 mm/s. ⎛ I ⎞ Solve: (a) β = 110 dB gives 11.0 = log ⎜ ⎟ and I = (1011 ) I 0 = 0.100 W/m 2 . ⎝ I0 ⎠ E = IAt = (0.100 W/m 2 )π (4.2 × 10−3 m)2 (1 s) = 5.5 μ J

(b) K = 12 mυ 2 so

2K = m

υ=

2(5.5 × 10−6 J) 2.0 × 10−6 kg

= 2.3 m/s.

This is about 31,000 times faster than the speed in Problem 12.46b. Reflect: Even though the sound wave intensity level is very high, the rate at which energy is delivered to the eardrum is very small, because the area of the eardrum is very small.

12.46. Set Up: Apply the relation β 2 − β1 = (10 dB)log( I 2 /I1 ) that is derived in Example 12.9. Solve: (a) Δβ = (10 dB)log

( 4II ) = 6.0 dB

(b) The total number of crying babies must be multiplied by four, for an increase of 12 kids. Reflect: For I 2 = α I1, where α is some factor, the increase in sound intensity level is Δβ = (10 dB)log α . For α = 4, Δβ = 6.0 dB.

*12.47. Set Up: Apply β = (10 dB)log( I /I 0 ), where I 0 = 10−12 W/m 2 . For part (b) use I = I 010( β / 10 dB) , which is given in Example 12.8. ⎛ 0.500 μ W/m 2 ⎞ Solve: (a) β = (10 dB) log ⎜ −12 ⎟ = 57 dB. ⎜ 10 W/m 2 ⎟⎠ ⎝ (b) I = I 010(β /10 dB) = (10−12 W/m 2 )1010.3 = 2.00 × 10−2 W/m 2 . Reflect: As expected, the sound intensity is larger for the jackhammer.

12.48. Set Up: The other player’s note can be either larger or smaller than 440 Hz. Solve: f beat = f1 − f 2 so the other player’s note could have frequency 440 Hz + 2.8 Hz = 442.8 Hz or 440 Hz − 2.8 Hz = 437.2 Hz. *12.49. Set Up: f beat = f1 − f 2 and f =

υ λ


12-14

Chapter 12

Solve: f1 =

υ 344 m/s υ 344 m/s = = 1014 Hz; f 2 = = = 1000 Hz λ1 0.3394 m λ2 0.3440 m

f beat = 1014 Hz − 1000 Hz = 14 Hz, so 14 beats each second.

12.50. Set Up: We have f beat = q f 1 − f 2 q and v = f λ , where v = 344 m/s. Let λ1 = 6.50 cm and λ2 = 6.52 cm. λ2 > λ1 so f1 > f 2 . ⎛ 1 1 ⎞ v(λ2 − λ1 ) (344 m/s)(0.02 × 10−2 m) = = 16 Hz. There are 16 beats per second. Solve: f1 − f 2 = v ⎜ − ⎟ = λ1λ2 (6.50 × 10−2 m)(6.52 × 10−2 m) ⎝ λ1 λ2 ⎠ Reflect: We could have calculated f1 and f 2 and subtracted, but doing it this way we would have to be careful to

retain enough figures in intermediate calculations to avoid round-off errors.

12.51. Set Up: f beat = f1 − f 2 . Tightening the string increases the wave speed for transverse waves on the string and this in turn increases the frequency. Solve: (a) If the beat frequency increases when she raises her frequency by tightening the string, it must be that her frequency is 433 Hz, 3 Hz above concert A. (b) She needs to lower her frequency by loosening her string.

12.52. Set Up: The positive direction is from listener to source. υ + υS 344 m/s − 25.0 m/s Solve: (a) υS = − 25.0 m/s. λ = = = 0.798 m. fS

400 Hz

344 m/s + 25.0 m/s = 0.922 m 400 Hz ⎛ υ ⎞ ⎛ ⎞ 344 m/s (c) υS = − 25.0 m/s. f L = ⎜ ⎟ fS = ⎜ ⎟ (400 Hz) = 431 Hz υ υ 344 m/s 25 0 m/s − . ⎝ ⎠ ⎝ + S⎠ (b) υS = + 25.0 m/s. λ =

⎛ ⎞ 344 m/s (d) υS = + 25.0 m/s. f L = ⎜ ⎟ (400 Hz) = 373 Hz 344 m/s 25 0 m/s + . ⎝ ⎠ Reflect: If the train were at rest, the wavelength of the sound from its whistle would be 344 m/s = 0.860 m. 400 Hz When the listener is in front of the source, the source is approaching. The frequency is increases and the wavelength is decreased. When the listener is behind the source, the source is moving away. The frequency is decreased and the wavelength is increased.

*12.53. Set Up: The positive direction is from listener to source. fS = 392 Hz. (a) υS = 0. υL = − 15.0 m/s. ⎛ υ + υL ⎞ ⎛ 344 m/s − 15.0 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (392 Hz) = 375 Hz υ υ 344 m/s + ⎝ ⎠ S⎠ ⎝ (b) υS = + 35.0 m/s. υL = + 15.0 m/s. ⎛ υ + υL ⎞ ⎛ 344 m/s + 15.0 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (392 Hz) = 371 Hz ⎝ 344 m/s + 35.0 m/s ⎠ ⎝ υ + υS ⎠ (c) f beat = f1 − f 2 = 4 Hz

12.54. Set Up: The positive direction is from listener to source. fS = 1200 Hz. f L = 1240 Hz. ⎛ υ + υL ⎞ ⎛ υ ⎞ Solve: f L = ⎜ ⎟ fS . υL = 0. υS = − 25.0 m/s. f L = ⎜ ⎟ fS gives ⎝ υ + υS ⎠ ⎝ υ + υS ⎠

υ=

υS f L

fS − f L

=

(−25 m/s)(1240 Hz) = 780 m/s. 1200 Hz − 1240 Hz



12-15

Reflect: f L > fS since the source is approaching the listener.

*12.55. Set Up: Choose the positive direction pointing from the motorcycle toward the car. The car is stationary, ⎛ v + vL ⎞ so vS = 0. Use f L = ⎜ ⎟ fS . ⎝ v + vS ⎠ ⎛ f ⎞ v + vL ⎛ 490 Hz ⎞ fS = (1 + vL /v) fS , which gives vL = v ⎜ L − 1⎟ = (344 m/s) ⎜ − 1⎟ = −19.8 m/s. You v + vS ⎝ 520 Hz ⎠ ⎝ fS ⎠ must be traveling at 19.8 m/s. Solve: We have f L =

Reflect: vL < 0 means that the listener is moving away from the source.

12.56. Set Up: Apply the Doppler effect formula, Eq. (12.19). Solve: (a) The positive direction is from the listener toward the source, as shown in the figure below.

⎛ v + vL ⎞ ⎛ 344 m/s + 18.0 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (262 Hz) = 302 Hz. ⎝ 344 m/s − 30.0 m/s ⎠ ⎝ v + vS ⎠ (b) The positive direction is from the listener toward the source, as shown in the figure below.

⎛ v + vL ⎞ ⎛ 344 m/s − 18.0 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (262 Hz) = 228 Hz. ⎝ 344 m/s + 30.0 m/s ⎠ ⎝ v + vS ⎠ Reflect: In part (a) the listener and source are moving together so f L > fS ; however, in part (b) the listener and

source are moving apart so f L < fS.

12.57. Set Up: ⎛ υ + υL ⎞ fL = ⎜ ⎟ fS . ⎝ υ + υS ⎠

To treat the reflection of the waves use Problem-Solving Strategy 3 in Section 12.12.


12-16

Chapter 12

Solve: (a) First consider the truck as the listener, as shown in Figure (a) below.

⎛ υ + υL ⎞ ⎛ 344 m/s − 20.0 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (2000 Hz) = 2064 Hz ⎝ 344 m/s − 30.0 m/s ⎠ ⎝ υ + υS ⎠

Now consider the truck as a source, with fS = 2064 Hz, and the fire engine driver as the listener (Figure (b) above). ⎛ υ + υL ⎞ ⎛ 344 m/s + 30.0 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (2064 Hz) = 2120 Hz υ υ + ⎝ 344 m/s + 20.0 m/s ⎠ S⎠ ⎝

The objects are getting closer together so the frequency is increased. (b) The driver detects a frequency of 2120 Hz and the waves returning from the truck move past him at 344 m/s + 30.0 m/s, so the wavelength he measures is

λ=

344 m/s + 30 m/s = 0.176 m. 2120 Hz

The wavelength of waves emitted by the fire engine when it is stationary is

λ=

344 m/s = 0.172 m. 2000 Hz

12.58. Set Up: ⎛ υ + υL ⎞ fL = ⎜ ⎟ fS . ⎝ υ + υS ⎠

To treat the reflection of the waves use Problem-Solving Strategy 3 in Section 12.12. Solve: (a) Since the frequency is increased the moving car must be approaching the police car. Let υc be the speed of the moving car. The speed υp of the police car is zero. First consider the moving car as the listener, as shown in Figure (a) below:



12-17

⎛ υ + υL ⎞ ⎛ υ + υc ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (1200 Hz) + υ υ ⎝ υ ⎠ S⎠ ⎝ Then consider the moving car as the source and the police car as the listener (Figure (b) above): ⎛ υ + υL ⎞ ⎛ υ ⎞ ⎛ υ + υc ⎞ fL = ⎜ ⎟ fS gives 1250 Hz = ⎜ ⎟⎜ ⎟ (1200 Hz). ⎝ υ + υS ⎠ ⎝ υ − υc ⎠ ⎝ υ ⎠

Solving for υc gives ⎛ 50 ⎞ ⎛ 50 ⎞ ⎟υ = ⎜ ⎟ (344 m/s) = 7.02 m/s ⎝ 2450 ⎠ ⎝ 2450 ⎠ (b) Repeat the calculation of part (a), but now υp = 20.0 m/s, toward the other car.

υc = ⎜

Waves received by the car (Figure (c) above): ⎛ υ + υc ⎞ ⎛ 344 m/s + 7 m/s ⎞ fL = ⎜ (1200 Hz) = 1300 Hz ⎟f = ⎜ υ − υp ⎟ S ⎝⎜ 344 m/s − 20 m/s ⎠⎟ ⎝ ⎠ Waves reflected by the car and received by the police car (Figure (d) above): ⎛ υ + υp ⎞ ⎛ 344 m/s + 20 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (1300 Hz) = 1404 Hz ⎝ 344 m/s − 7 m/s ⎠ ⎝ υ − υc ⎠ Reflect: The cars move toward each other with a greater relative speed in (b) and the increase in frequency is much larger there.

*12.59. Set Up: An observer on a stationary boat would measure the frequency of the waves as υ 16.5 m/s f = = = 0.4125 Hz —so for simplicity imagine that the waves are generated by a stationary source λ 40.0 m (υs = 0) with fs = f = 0.4125 Hz. Thus, we will apply Eq. (12.19) for the special case of a stationary source. As © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12-18

Chapter 12

usual, we choose the positive direction facing toward the source of the waves (i.e., westward), so that υL > 0 in part (a) and υL < 0 in part (b). Solve: (a)

⎛ 16.5 m/s + 5.00 m/s ⎞ ⎛ υ + υL ⎞ fL = ⎜ ⎟ (0.4125 Hz) = 0.5375 Hz. Thus, the time interval between ⎟ fS = ⎜ 16.5 m/s ⎝ υ ⎠ ⎝ ⎠

crests, as seen by an observer on the ship, is TL =

1 1 = = 1.86 s. f L 0.5375 Hz

(b) The analysis is the same as in part (a)—except that υL = 25.00 m/s :

⎛ 16.5 m/s + − 5.00 m/s ⎞ ⎛ υ + υL ⎞ fL = ⎜ f = ⎟ (0.4125 Hz) = 0.2875 Hz. Thus, in this case the time interval between ⎝ υ ⎟⎠ S ⎜⎝ 16.5 m/s ⎠ 1 1 crests, as seen by an observer on the ship, is TL = = = 3.48 s. f L 0.2875 Hz Reflect: In reality we do not know that the source of the waves is stationary. However, if the source of the waves were moving with a velocity υS it would need to generate a frequency fS so that a stationary boat would still observe ⎛ υ ⎞ the known frequency of 0.4125 Hz = ⎜ ⎟ fS. Thus, the frequency observed by a moving boat would be ⎝ υ + υS ⎠ ⎛ υ + υL ⎞ ⎛ υ + υL ⎞ ⎛ υ ⎞ ⎛ υ + υL ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ fS = ⎜ ⎟⎜ ⎟ (0.4125 Hz), which is the same as we obtained by assuming a ⎝ υ ⎠ ⎝ υ + υS ⎠ ⎝ υ ⎠ ⎝ υ + υS ⎠ stationary source.

12.60. Set Up: In this example we have a moving source and a stationary observer. Take the positive direction to ⎛ υ + υL ⎞ ⎛ υ ⎞ be from listener to source and apply f L = ⎜ ⎟ fS = ⎜ ⎟ fS , where fS = 260 Hz. Since the source is ⎝ υ + υS ⎠ ⎝ υ + υS ⎠ moving toward you, you will hear the moving car horn at a higher pitch than your horn, and the beat frequency will be given by f beat = f L − fS. Solve: We can determine f L from the beat frequency: f beat = 6.0 Hz = f L − fS = f L − 260 Hz; thus, f L = 266 Hz. Assuming that υ = 344 m/s, we obtain

⎛ 344 m/s ⎞ 266 Hz = ⎜ ⎟ (260 Hz). Solving for υS we obtain ⎝ 344 m/s + υS ⎠ ⎛ 260 Hz ⎞ υS = ⎜ − 1 (344 m/s) = −7.8 m/s. Thus, your friend is moving at 7.8 m/s toward you. ⎝ 266 Hz ⎟⎠ Reflect: What frequency will your friend hear? In this case, you have a stationary source (your horn) and a moving ⎛ 344 m/s + 7.8 m/s ⎞ observer (your friend). The positive direction points from listener to source. Thus, we have fL = ⎜ ⎟⎠ ⎝ 344 m/s (260 Hz) = 265.8 Hz ≈ 266 Hz. At low speeds, there is little difference in the Doppler shift of a moving source or

that of a moving observer.

*12.61. Set Up: fS = 1000 Hz. The positive direction is from the listener to the source. (a) υS = − (344 m/s)/2 = − 172 m/s, υL = 0.

⎛ υ + υL ⎞ ⎛ ⎞ 344 m/s fL = ⎜ ⎟ fS = ⎜ ⎟ (1000 Hz) = 2000 Hz ⎝ 344 m/s − 172 m/s ⎠ ⎝ υ + υS ⎠ (b) υS = 0, υL = + 172 m/s.

⎛ υ + υL ⎞ ⎛ 344 m/s + 172 m/s ⎞ fL = ⎜ ⎟ fS = ⎜ ⎟ (1000 Hz) = 1500 Hz υ υ + 344 m/s ⎝ ⎠ S⎠ ⎝ (c) The answer in (b) is much less than the answer in (a). It is the velocity of the source and listener relative to the air that determines the effect, not the relative velocity of the source and listener relative to each other. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


12.62. Set Up: Use the equation for the Doppler shift of light: f L =

12-19

c −υ fS . Require f R = 1.100 fS. Since we c +υ

know that f R > fS the star would be moving toward us and υ < 0, so υ = − υ . c = 3.00 × 108 m/s. Solve: f R =

c+ υ c+qυ q [(1.100) 2 − 1]c = (1.100) 2 . Solving for q υ q gives q υ q = = fS . f R = 1.100 fS gives c−qυ q c− υ 1 + (1.100) 2

0.0950c = 2.85 × 107 m/s. υ f − fS υ Δf Δf Reflect: = 9.5,. = R = 10.0,. and are approximately equal. c c fS fS fS

12.63. Set Up: Take the tension in the wire to be the weight of the suspended mass; neglect the small variation in tension along the wire that is due to the mass of the wire. The time is the distance (14.0 m) divided by the wave speed. Let m be the mass of the wire and let M be the mass of the suspended object. Solve: υ =

F⊥

μ

=

(7.50 kg)(9.80 m/s 2 )(14.0 m) = 35.9 m/s. 0.800 kg

MgL = m

t=

14.0 m = 0.390 s. 35.9 m/s

12.64. Set Up: The heart wall first acts as the listener and then as the source. The positive direction is from listener to source. The heart wall is moving toward the receiver so the Doppler effect increases the frequency and the final frequency received, f L2 , is greater than the source frequency, fS1. f L2 − fS1 = 85 Hz. Solve: Heart wall receives the sound: fS = fS1. f L = f L1. υS = 0. υL = − υwall .

Heart wall emits the sound: fS2

⎛ υ + υL ⎞ ⎛ υ − υ wall ⎞ fL = ⎜ ⎟ fS gives f L1 = ⎜ ⎟ fS1. υ υ υ + ⎝ ⎠ S⎠ ⎝ = f L1. υS = + υwall . υL = 0.

⎛ ⎞ ⎛ ⎞ ⎛ υ − υ wall ⎞ ⎛ υ − υ wall ⎞ υ υ f L2 = ⎜ ⎟ fS2 = ⎜ ⎟⎜ ⎟ fS1. ⎟ fS1 = ⎜ υ ⎠ ⎝ υ + υ wall ⎠ ⎝ υ + υ wall ⎠ ⎝ ⎝ υ + υ wall ⎠ ⎛ υ − υ wall ⎞ ⎛ 2υ wall ⎞ f L2 − fS1 = ⎜1 − ⎟ fS1 = ⎜ ⎟ fS1. ⎝ υ + υ wall ⎠ ⎝ υ + υ wall ⎠ ( f L2 − fS1 )υ υwall = . 2 fS1 − ( f L2 − fS1 )

fS1 W f L2 − fS1 and

υ wall =

( f L2 − fS1 )υ (85 Hz)(1500 m/s) = = 0.0319 m/s = 3.19 cm/s. 2 fS1 2(2.00 × 106 Hz)

Reflect: fS1 = 2.00 × 106 Hz and f L2 − fS1 = 85 Hz, so the approximation we made is very accurate. Within this

approximation, the frequency difference between the reflected and transmitted waves is directly proportional to the speed of the heart wall.

*12.65. Set Up: β = 10log

P I gives I if β is specified. Then I = relates I, r and P. I0 4π r 2

Solve: (a) β = 100 dB so log

I I = 10. = 1010 and I = (10−12 W/m 2 )(1010 ) = 1 × 10−2 W/m 2 . I0 I0 r=

P = 4π I

20.0 W 4π (1 × 10−2 W/m 2 )

= 12.6 m


12-20

Chapter 12

(b) β = 60 dB so log

I I 6 = 6. 10 and I = (10−12 W/m 2 )(106 ) = 1 × 10−6 W/m 2 I0 I0 r=

12.66. Set Up: f1 = Solve: (a) f1 =

1 2L

υ 2L

and υ =

20.0 W 4π (1 × 10−6 W/m 2 )

F⊥

μ

=

= 1.26 × 103 m = 1.26 km

F⊥ L . m

F⊥ L 1 F⊥ = . m 2 Lm F⊥ = (2 f1 ) 2 ( Lm) = (2[65.4 Hz]) 2 (0.600 m)(14.4 × 1023 kg) = 148 N

(b) For f1 = 73.4 Hz, 2

⎛ 73.4 Hz ⎞ F⊥ = ⎜ ⎟ (148 N) = 186 N ⎝ 65.4 Hz ⎠

The percentage increase in the tension is 186 N − 148 N = 0.257 = 26,. 148 N

*12.67. Set Up: The flute acts as a stopped pipe and its harmonic frequencies are given by Eq. (12.12). The resonant frequencies of the string are f n = nf1, n = 1, 2, 3,… The string resonates when the string frequency equals the flute frequency. For the string we have

f1s = 600.0 Hz. For the flute, the fundamental frequency is

v 344.0 m/s f1f = = = 800.0 Hz. Let nf label the harmonics of the flute and let ns label the harmonics of the 4 L 4(0.1075 m) string. Solve: For the flute and string to be in resonance, nf f1f = ns f1s , where f1s = 600.0 Hz is the fundamental frequency

for the string. ns = nf ( f1f /f1s ) = 43 nf . Thus, ns is an integer when nf = 3N , N = 1, 3, 5, … (the flute has only odd harmonics). nf = 3N gives ns = 4 N Flute harmonic 3 N resonates with string harmonic 4 N , N = 1, 3, 5, …. Reflect: We can check our results for some specific values of N. For N = 1, nf = 3 and f3f = 2400 Hz. For this N, ns = 4 and f 4s = 2400 Hz. For N = 3, nf = 9 and f9f = 7200 Hz, and ns = 12, f12s = 7200 Hz. Our general results do give equal frequencies for the two objects.

12.68. Set Up: The wall first acts as a listener and then as a source. The positive direction is from listener to source. The bat is moving toward the wall so the Doppler effect increases the frequency and the final frequency received, f L2 , is greater than the original source frequency, fS1. fS1 = 2000 Hz. f L2 − fS1 = 10.0 Hz. Solve: The wall receives the sound: fS = fS1. f L = f L1. υS = − υbat and υL = 0.

The wall receives the sound: fS2

⎛ υ + υL ⎞ ⎛ υ fL = ⎜ ⎟ fS gives f L1 = ⎜ ⎝ υ + υS ⎠ ⎝ υ − υbat = f L1. υS = 0 and υL = + υbat .

⎞ ⎟ fS1. ⎠

⎞ ⎛ υ + υbat ⎞ ⎛ υ + υbat ⎞ ⎛ υ + υbat ⎞ ⎛ υ f L2 = ⎜ ⎟ fS1 = ⎜ ⎟ fS1. ⎟ fS2 = ⎜ ⎟⎜ υ υ υ υ − ⎝ ⎠ ⎝ ⎠⎝ bat ⎠ ⎝ υ − υbat ⎠ ⎛ υ + υbat ⎞ ⎛ 2υbat ⎞ f L2 − fS1 = Δf = ⎜ − 1⎟ fS1 = ⎜ ⎟ fS1. υ υ − bat ⎝ ⎠ ⎝ υ − υbat ⎠ υ Δf (344 m/s)(10.0 Hz) υbat = = = 0.858 m/s. 2 fS1 + Δf 2(2000 Hz) + 10.0 Hz © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


12-21

*12.69. Set Up: According to convention the positive direction always points from the listener to the source. Thus, your velocity is positive when you walk toward a stationary source and negative when you walk away from a stationary source—and so we may write υL = ± qυLq to handle both cases. Since you are simultaneously moving toward one source and away from the other source, you will simultaneously hear each source Doppler shifted—one source will be shifted to a higher frequency, f1, and the other to a lower frequency, f2. These two frequencies will mix to produce beats with a beat frequency f beat = f1 − f2. ⎛ υ + qυ L q ⎞ Solve: (a) As you walk toward a speaker you have υL = 1qυLq so f1 = ⎜ ⎟ fS. Similarly, as you walk away ⎝ υ ⎠ ⎛ υ − qυ L q ⎞ from a speaker you have υL = −qυLq so f2 = ⎜ ⎟ fS . ⎝ υ ⎠ ⎛ υ + qυ L q ⎞ ⎛ υ − qυ L q ⎞ 2qυLq fS Also, we know f beat = f1 − f2 = ⎜ ⎟ fS − ⎜ ⎟ fS = υ υ υ ⎝ ⎠ ⎝ ⎠

This reduces to qυLq =

1 f beat ⋅υ = 2 fS

1 ⎛ 2.50 Hz ⎞ ⎜ ⎟ ⋅ (344 m/s) = 1.878 m/s = 1.88 m/s. 2 ⎝ 229 Hz ⎠

2(1.878 m/s) ⋅ (573 Hz) = 6.26 Hz. 344 m/s Reflect: Alternatively, consider the standing wave established between the two speakers with antinodes spaced by λ υ qυLqt = . As you walk for a time t and a distance qυ Lqt , you will cross antinodes. As a result, you will υ /2 fS 2 2 fS (b) According to the results of part (a) f beat =

number of antinodes crossed qυ L q 1 f beat . Thus, we have qυLq = = ⋅ υ, which is time 2 fS υ /2 fS the result we obtained in part (a) by using the Doppler effect.

perceive a beat frequency of f beat =

12.70. Set Up: The whale acts as the listener and then as the source. The direction from the listener to the source is positive. fS = 22.0 × 103 Hz. Solve: (a) λ =

υ f

=

1482 m/s 22.0 × 103 Hz

= 0.0674 m

(b) whale receives the sound: υS = 0, υL = + υ whale . f L1 = f L .

⎛ υ + υL ⎞ ⎛ υ + υ whale ⎞ f L1 = ⎜ ⎟ fS = ⎜ ⎟ fS υ υ υ + ⎝ ⎠ S⎠ ⎝

whale transmits the sound: υS = − υ whale , υL = 0. fS = fS2 = f L1. f L2 = f L . ⎛ ⎞ ⎛ υ + υ whale ⎞ ⎛ υ + υ whale ⎞ υ f L2 = ⎜ ⎟⎜ ⎟ fS . ⎟ fS = ⎜ υ ⎠ ⎝ υ − υ whale ⎠ ⎝ ⎝ υ − υ whale ⎠ ⎛ ⎞ ⎛ υ + υ whale ⎞ ⎛ 2υ whale ⎞ 2[ 4.95 m/s ] 3 Δf = f L2 − fS = ⎜ − 1 ⎟ fS = ⎜ ⎟ fS = ⎜ ⎟ (22.0 × 10 Hz) = 147 Hz ⎝ υ − υ whale ⎠ ⎝ υ − υ whale ⎠ ⎝ 1482 m/s − 4.95 m/s ⎠ Reflect: The speed of the whale is much less than the speed of the waves and the fractional shift in frequency is very small.

12.71. Set Up: In one octave the frequency doubles. 20 × 103 Hz = 1000 = 2 x , solve for x. x ln2 = ln 1000 and x = 10 20 Hz This person can hear about 10 octaves. Solve:


12-22

Chapter 12

⎛ υ + υL ⎞ ⎟ fS . Take the positive direction to be from the listener to the source. For reflection ⎝ υ + υS ⎠ off the bottom of the well the water surface first serves as a listener and then as a source. Solve: υ y 2 = υ0 y 2 + 2a y ( y − y0 ) so the speed of the siren just before it hits the water is

12.72. Set Up: f L = ⎜

2(9.80 m/s 2 )(125 m) = 49.5 m/s. (a) The situation is shown in Figure (a) below.

⎛ ⎞ ⎛ ⎞ υ 344 m/s fL = ⎜ ⎟ fS = ⎜ ⎟ (2500 Hz) = 2186 Hz ⎝ υ + 49.5 m/s ⎠ ⎝ 344 m/s + 49.5 m/s ⎠ The source is moving away and the frequency is lowered. υ 344 m/s λL = = = 0.157 m. f L 2186 Hz (b) The water serves as a listener (Figure (b) above). ⎛ ⎞ υ fL = ⎜ ⎟ fS = 2920 Hz 49 5 m/s υ − . ⎝ ⎠

The source and listener are approaching and the frequency is raised.

λL =

υ

= 0.118 m fL Both the person and the water are at rest so there is no Doppler effect when the water serves as a source and the person is the listener. The person detects sound with frequency 2920 Hz and wavelength 0.118 m. (c) f beat = f1 − f 2 = 2920 Hz − 2186 Hz = 734 Hz ⎛ v + vL ⎞ ⎟ fS . In the SHM the source moves toward and ⎝ v + vS ⎠ away from the listener, with maximum speed ω p Ap . The direction from listener to source is positive.

*12.73. Set Up: Apply the Doppler effect formula f L = ⎜

Solve: (a) The maximum velocity of the siren is ω P AP = 2π f P AP . You hear a sound with frequency

f L = fsiren v /

(v + vS ), where vS varies between +2π f P AP and −2π f P AP . f L − max = fsiren v /(v − 2π f P AP ) and f L − min = fsiren v / (v + 2π f P AP ). (b) The maximum (minimum) frequency is heard when the platform is ing through equilibrium and moving up (down). Reflect: When the platform is moving upward the frequency you hear is greater than fsiren and when it is moving

downward the frequency you hear is less than fsiren . When the platform is at its maximum displacement from equilibrium its speed is zero and the frequency you hear is fsiren .

12.74. Set Up: Compare the stick to a vibrating air column, for which we know the standing wave wavelengths. Solve: (a) The ends are free to move and are displacement antinodes.



12-23

(b) The stick has the same nodal structure as an open pipe. 2 L 6.00 m λn = = , n = 1, 2, 3, . . . . n n The wavelength of the first five harmonics are: λ1 = 6.00 m, λ 2 = 3.00 m, λ3 = 2.00 m, λ 4 = 1.50 m and λ5 = 1.20 m.

*12.75. Set Up: The greatest frequency shift from the Doppler effect occurs when one speaker is moving away and ⎛ υ + υL ⎞ one is moving toward the person. The speakers have speed υ0 = rω , where r = 0.75 m. f L = ⎜ ⎟ fS , with the ⎝ υ + υS ⎠ positive direction from the listener to the source. υ = 344 m/s

Solve: (a) f =

υ 344 m/s = = 1100 Hz. λ 0.313 m ⎛ 2π rad ⎞ ⎛ 1 min ⎞ = 7.85 rad/s ⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠

ω = (75 rpm) ⎜ and υ0 = (0.75 m)(7.85 rad/s) = 5.89 m Speaker A, moving toward the listener:

⎛ ⎞ υ f LA = ⎜ ⎟ (1100 Hz) = 1119 Hz. 5 89 m/s υ − . ⎝ ⎠ Speaker B, moving toward the listener: ⎛ ⎞ υ f LB = ⎜ ⎟ (1100 Hz) = 1081 Hz ⎝ υ + 5.89 m/s ⎠ f beat = f1 − f 2 = 1119 Hz − 1081 Hz = 38 Hz (b) A person can her individual beats only up to about 7 Hz and this beat frequency is much larger than that. Reflect: As the turntable rotates faster the beat frequency at this position of the speakers increases.

12.76. Set Up: υ L = 0. Let υcar be the speed of the car. Assume the car is traveling to the right. The direction from the listener to the source is positive. When the source is approaching, the frequency is increased and when the source is moving away, the frequency is decreased. Solve: (a) 21/12 = 1.059 ⎛ υ ⎞ (b) car approaching: The situation is sketched in Figure (a) below. υS = − υcar . f L1 = ⎜ ⎟ fS . ⎝ υ − υcar ⎠

⎛ υ ⎞ car moving away: The situation is sketched in Figure (b) above. υS = + υcar . f L2 = ⎜ ⎟ fS ⎝ υ + υcar ⎠ ⎛ υ ⎞ ⎛ υ ⎞ ⎛ υ ⎞ ⎛ υ + υcar ⎞ υ + υcar f L1 = 1.059 = ⎜ ⎟ fS / ⎜ ⎟ fS = ⎜ ⎟⎜ ⎟= f L2 ⎠ υ − υcar ⎝ υ − υcar ⎠ ⎝ υ + υcar ⎠ ⎝ υ − υcar ⎠ ⎝ υ © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12-24

Chapter 12

1.059υ − 1.059υcar = υ + υcar and

υcar =

0.059υ = 0.0287υ = (0.0287)(344 m/s) = 9.9 m/s 2.059 v , where we use v = 344 m/s. Let La = 1.14 m and 4L

12.77. Set Up: f beat = q f a − fbq. For a stopped pipe, f1 = Lb = 1.16 m. Lb > La so f1a > f1b .

v⎛ 1 1 ⎞ v( Lb − La ) (344 m/s)(2.00 × 10−2 m) Solve: f1a − f1b = ⎜ − ⎟= = = 1.3 Hz. There are 1.3 beats per second. 4 ⎝ La Lb ⎠ 4 La Lb 4(1.14 m)(1.16 m) Reflect: Increasing the length of the pipe increases the wavelength of the fundamental and decreases the frequency.

Solutions to age Problems 12.78. Set Up: For a pipe open at both ends we have f1 =

υ 2L

, where υ is the speed of sound. Note that the speed

of sound decreases by 2% so that υnew = 0.98υold . Solve: Assuming that L remains nearly fixed, we can compare the new fundamental frequency to the original f υ /2 L υnew = = 0.98. Thus the fundamental frequency decreases by 2%. The fundamental frequency: new = new υold /2 L υold f old correct answer is C. Note that, although most objects do expand when heated, we will show in Chapter 14 that it would take a large change in temperature to change the length of a typical string or pipe by an amount comparable to the given 2% change in the speed of sound.

*12.79. Set Up: For a string fixed at both ends we have f1 =

υ 2L

, where υ =

T

μ

is the speed of the wave on the string.

Solve: Assuming that L remains nearly fixed, both T and μ will remain nearly fixed. Thus, the fundamental

frequency does not change significantly. The correct answer is A. Note that, although most objects do expand when heated, we will show in Chapter 14 that it would take a large change in temperature to change the length of a typical string or pipe by an amount comparable to the given 2% change in the speed of sound.

12.80. Set Up: The beat frequency is the magnitude of the difference between the two frequencies: f beat = q f v − f f q. Solve: The fundamental frequency of the violin remains at f v = 262 Hz, but the fundamental frequency of the flute is

reduced to 0.98 f v . Thus, we have f beat = q f v − f f q = f v − 0.98 f v = 0.02 f v = 0.02(262 Hz) = 5 Hz. The correct answer is C.


13

FLUID MECHANICS

Answers to Multiple-Choice Problems 1. B

2. B

3. A, D

4. C

5. A

6. B

7. C

8. D

9. D

10. A

11. C

12. C

13. E

14. A

15. D

Solutions to Problems *13.1. Set Up: The density of gold is 19.3 × 103 kg/m3 . Solve: V = (5.0 × 10−3 m)(15.0 × 10−3 m)(30.0 × 10−3 m) = 2.25 × 10−6 m3 .

ρ=

0.0158 kg m = = 7.02 × 103 kg/m3 . V 2.25 × 10−6 m3

The metal is not pure gold.

13.2. Set Up: The density of platinum is 21.4 × 103 kg/m3 . Solve: ρ =

m m 40.0 kg so V = = = 1.87 × 10−3 m3 . V = L3 and L = 0.123 m = 12.3 cm. ρ 21.4 × 103 kg/m3 V

13.3. Set Up: According to Table 13.1 the density of air at 20° C is 1.20kg/m3. The volume of the room (which is a rectangular solid) is given by V = L × W × H . The mass of the air can be calculated from its density and volume, m = ρV, and the weight of the air can be calculated from its mass, W = mg = ρ gV . According to the Appendix, the conversion from newtons into pounds is 1 N = 0.2248 lb. Solve: V = (5.00 m)(4.50 m)(3.25 m) = 73.13 m3. Thus, the weight of the air that occupies the room is W = ρ gV = ⎛ 0.2248 lb ⎞ (1.20kg/m3 )(9.80 m/s 2 )(73.13 m3 ) = 860 N. Finally, we convert this into pounds: W = (860 N) ⎜ ⎟ = 193 lb. ⎝ 1N ⎠ Reflect: Since we are adapted to live submerged in air we tend not to think about its weight. Thus, the weight of the air in this room may be surprising—especially since the size of this room (16.4 × 14.8 × 10.6 feet ) is not extraordinarily large.

13.4. Set Up: ρ = m/V so m = ρV . w = mg . 1 gal = 3.788 L = 3.788 × 1023 m3 . ⎛ 3.788 × 10−3 m3 ⎞ −2 3 Solve: V = (11.5 gal) ⎜ ⎟⎟ = 4.36 × 10 m ⎜ 1 gal ⎝ ⎠ m = ρV = (737 kg/m3 )(4.36 × 10−2 m3 ) = 32.1 kg and w = mg = 315 N


13-1

13-2

Chapter 13

*13.5. Set Up: ρ = m/V . For a sphere, V = 43 π r 3. Solve: (a) The mass of the nugget is

$1.00 × 106 ⎛ 31.1035 g ⎞ 4 = (2344 ounces) ⎜ ⎟ = 7.29 × 10 g $426.6/ounce 1 ounce ⎝ ⎠

V=

m

ρ

=

1/3

⎛ 3V ⎞ r=⎜ ⎟ ⎝ 4π ⎠

7.29 × 104 g 19.3 g/cm3

= 3.78 × 103 cm3 1/3

⎡ 3(3.78 × 103 cm3 ) ⎤ =⎢ ⎥ 4π ⎣⎢ ⎦⎥

= 9.66 cm

and its diameter is 2r = 19.3 cm. (b) The platinum nugget would have mass m = ρV = (21.4 g/cm3 )(3.78 × 103 cm3 ) = 8.09 × 104 g = 2.60 × 103 troy ounces

and would be worth 2.29 million dollars. Reflect: The platinum nugget is worth more because a nugget of the same size has more mass, since platinum’s density is greater, and because it is worth more per ounce. The value if platinum is larger than the value if gold by (21.4/19.3)(879.00/426.60).

13.6. Set Up: The volume of a cube with side length L is L3 . The volume of a cylinder of radius r and length L is π r 2 L. Solve: (a) The volume of the metal left after the hole is drilled is the volume of the solid cube minus the volume of the cylindrical hole: V = L3 − π r 2 L = (5.0 cm)3 − π (1.0 cm)2 (5.0 cm) = 109 cm3 = 1.09 × 10−4 m3

The cube with the hole has mass

m=

7.50 N w = = 0.765 kg g 9.80 m/s 2

and density

ρ=

m 0.765 kg = = 7.02 × 103 kg/m3 V 1.09 × 1024 m3

(b) The solid cube has volume V = L3 = 125 cm3 = 1.25 × 10−4 m3 and mass m = ρV = (7.02 × 103 kg/m3 )(1.25 × 1024 m3 ) = 0.878 kg. The original weight of the cube was w = mg = 8.60 N.

*13.7. Set Up: ρ = m/V = m/L3 3

3

⎛L ⎞ ⎛ L ⎞ Solve: (a) m = ρ L3 so ρ1L13 = ρ2 L23 and ρ 2 = ρ1 ⎜ 1 ⎟ = ρ ⎜ 1 ⎟ = 8ρ ⎝ L2 ⎠ ⎝ L1/2 ⎠ 1/3

1/3

⎛ ρ1 ⎞ ⎛ ρ1 ⎞ (b) L2 = L1 ⎜ ⎟ = L ⎜ ⎟ = L/31/3 ⎜ ρ2 ⎟ ⎜ 3ρ1 ⎟ ⎝ ⎠ ⎝ ⎠ Reflect: If the volume is decreased while the mass is kept the same, then the density increases.

13.8. Set Up: w = mg and m = ρV. Find the volume V of the pipe. For a hollow cylinder with inner radius R1, outer radius R2 , and length L the volume is V = π ( R2 2 − R1 2 ) L. R1 = 1.25 × 10−2 m and R2 = 1.75 × 10−2 m. The density of copper is given in Table 13.1. Solve: V = π ([0.0175 m]2 − [0.0125 m]2 )(1.50 m) = 7.07 × 10−4 m3. m = ρV = (8.9 × 103 kg/m3 )(7.07 × 10−4 m3 ) = 6.29 kg. w = mg = 61.6 N. Reflect: The pipe weighs about 14 pounds. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Fluid Mechanics

13-3

13.9. Set Up: Apply ρ = m/V to relate the densities and volumes for the two spheres. For a sphere we have V = 43 πr3. For lead we have ρl = 11.3 × 103 kg/m3 and for aluminum, ρa = 2.7 × 103 kg/m3. 1/ 3

Solve: m = ρV = 43 π r 3 ρ. Since both spheres have the same mass we have ra3ρa = r13ρ1;

ra ⎛ ρ1 ⎞ =⎜ ⎟ r1 ⎜⎝ ρa ⎟⎠

=

1/ 3

⎛ 11.3 × 103 ⎞ = 1.6. ⎜⎜ 3 ⎟ ⎟ ⎝ 2.7 × 10 ⎠ Reflect: The aluminum sphere is larger, since its density is less.

13.10. Set Up: 1 mm Hg = 1 torr. 1 atm = 14.7 psi = 1.103 × 105 Pa. 1 Pa = 1 N/m 2 . ρ = 13.6 × 103 kg/m3 (Table 13.1). Solve: 1 mm Hg is the pressure p corresponding to h = 1 mm and ρ = ρ Hg in the equation p = ρ gh. So, 1 mm Hg corresponds to p = (13.6 × 103 kg/m3 )(9.80 m/s 2 )(1.00 × 10−3 m) = 133 Pa ⎞ ⎛ 133 Pa ⎞ ⎛ 1 atm (a) (120 mm Hg) ⎜ ⎟⎟ = 0.158 atm and 80 mm Hg = 0.105 atm, so the expression would be ⎟ ⎜⎜ 5 ⎝ 1 mm Hg ⎠ ⎝ 1.013 × 10 Pa ⎠ 0.158 . 0.105 120 (b) 1 mm Hg = 1 torr so the expression is the same, . 80

⎛ 133 Pa ⎞ 4 4 (c) (120 mm Hg) ⎜ ⎟ = 1.60 × 10 Pa and 80 mm Hg = 1.06 × 10 Pa, so the expression would be ⎝ 1 mm Hg ⎠ 1.60 × 104 1.06 × 104 (d) 1 Pa = 1 N/m 2 so the expression would be the same as in (c). (e) 120 mm Hg = 0.158 atm = 2.32 psi and 80 mm Hg = 0.105 atm = 1.54 psi, so the expression would be

2.32 . 1.54

*13.11. Set Up: 5L = 5 × 10−3 m3 . ρ = m /V. F = pA. 1 cm 2 = 10−4 m 2 . A sphere of radius r has volume V = 43 π r 3 . Specific gravity = 5.0 means ρ = 5.0 × 103 kg/m3 . Solve: (a) m = ρV = (1050 kg/m3 )(5 × 1023 m3 ) = 5.25 kg (b) F = pA = (13,000 Pa)(1.0 × 10−4 m 2 ) = 1.3 N (c) m = ρV = (5.0 × 103 kg/m3 )

( 43 )π (3.75 × 10−6 m)3 = 1.1 × 10−12 kg = 1.1 × 10−9 g

13.12. Set Up: The surface area of a hemisphere is A = 3π r 2 . On earth, pair = 1.01 × 105 Pa. F = pA. Solve: (a) F = (92)(1.01 × 105 Pa)(3π )(1.25 m) 2 = 1.4 × 108 N (b) On earth the pressure is less by a factor of 92, so F =

1.4 × 108 N = 1.5 × 106 N 92

13.13. Set Up: The density of seawater is 1.03 × 103 kg/m3 . Solve: (a) pgauge = p0 − patm = ρ gh = (1.03 × 103 kg/m3 )(9.80 m/s 2 )(250 m) = 2.52 × 106 Pa. (b) Fnet = pgauge A = (2.52 × 106 Pa)π (0.150 m)2 = 1.78 × 105 N Reflect: At the surface the gauge pressure is zero. At a depth of 250 m the force due to the water pressure is very large. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

13-4

Chapter 13

13.14. Set Up: 1 mm Hg = 133.3 Pa. F = pA. The surface area of a sphere of radius r is A = 4π r 2 . Solve:

⎛ 133.3 Pa ⎞ −2 2 ΔF = (Δ p ) A = (26 mm Hg) ⎜ ⎟ (4π )(1.25 × 10 m) = 6.8 N ⎝ 1 mm Hg ⎠

*13.15. Set Up: (a) Gauge pressure, Δp = p − patm , is related to depth according to Δ p = ρ gh. According to Table 13.1 the density of seawater is 1.03 × 103 kg/m3 , which is the same as that of freshwater to two significant figures. 1 atm = 1.01 × 105 Pa. (b) A uniform pressure p applied perpendicular to a surface of area A produces a force of pA on the surface. The lateral (side) area of a cylinder of radius r and length l is 2π rl , and the area of each end cap is π r 2 . Solve: (a) The gauge pressure at 400 meters is Δ p = ρ gh = (1.03 × 103 kg/m3 )(9.80 m/s 2 )(400 m) = 4 × 106 Pa, 1 atm ⎛ ⎞ which is equal to (4 × 106 Pa) ⎜ ⎟ = 40 atm. Similarly, at a depth of 3000 meters we have Δp = ⎝ 1.01 × 105 Pa ⎠

(1.03 × 103 kg/m3 )(9.80 m/s 2 )(3000 m) = 3 × 107 Pa = 300 atm. Thus, the pressure ranges from 40 atm to 300 atm. (b) We are asked to find the total inward pressure force acting on the surface of the whale at a depth of 3000 meters. Since the external pressure on all sides of the whale acts inward, we can simply multiply the external pressure times the total surface area of the whale. Modeling the whale as a cylinder, its total surface area is A = 2π r 2 + 2π rl = 2π (2 m) 2 + 2π (2 m)(16 m) = 226 m 2 . Thus, the total inward force is F = pA = (3 × 107 Pa)(226 m 2 ) = 7 × 109 N.

Reflect: According to Archimedes’s principle, the total upward buoyant force on the whale is equal to the weight of water that would occupy the whale’s volume. This is ρ gV = (1.03 × 103 kg/m3 )(9.80 m/s 2 )π (2 m)2 (16 m) = 2 × 106 N,

which is much smaller that the result in part (b) as we would expect.

13.16. Set Up: 1 atm = 1.013 × 105 Pa. The density of water is 1.00 × 103 kg/m3 . The gauge pressure must equal the pressure difference due to a column of water 1370 m − 730 m = 640 m tall. Solve: pgauge = p0 − patm = ρ gh = (1.00 × 103 m3 )(9.80 m/s 2 )(640 m) = 6.27 × 106 Pa = 61.9 atm

13.17. Set Up: 1 mm Hg = 133.3 Pa Solve: p = ρ gh = (1060 kg/m3 )(9.80 m/s 2 )(1.20 m) = 1.25 × 104 Pa = 93.5 mm Hg

13.18. Set Up: An inflation to 32.0 pounds means a gauge pressure of 32.0 lb/in.2 The area A with the pavement is related to the gauge pressure p − p0 in the tire and the force F⊥ the tire exerts on the pavement by F⊥ = ( p − p0 ) A. By Newton’s third law the magnitude of the force the tire exerts on the pavement equals the magnitude of the force the pavement exerts on the car, and this must equal the weight of the car. 14.7 lb/in.2 = 1.013 × 105 Pa = 1 atm. Assume p0 = 1 atm. Solve: (a) The gauge pressure is 32.0 lb/in.2 = 2.21 × 105 Pa = 2.18 atm. The absolute pressure is 46.7 lb / in.2 =

3.22 × 105 Pa = 3.18 atm. (b) No, the tire would touch the pavement at a single point and the area would be zero. F⊥ 9.56 × 103 N (c) F⊥ = mg = 9.56 × 103 N. A = = = 0.0433 m 2 = 433 cm 2 . p − p0 2.21 × 105 Pa Reflect: If the area is square, the length of each side for each tire is

433 cm 2 = 10.4 cm. This is a realistic 4

value, based on our observation of the tires of cars.


Fluid Mechanics

13-5

*13.19. Set Up: p = p0 + ρ gh. F = pA. For seawater, ρ = 1.03 × 103 kg/m3 Solve: The force F that must be applied is the difference between the upward force of the water and the downward forces of the air and the weight of the hatch. The difference between the pressure inside and out is the gauge pressure, so F = ( ρ gh) A − w = (1.03 × 103 kg/m3 )(9.80 m/s 2 )(30 m)(0.75 m 2 ) − 300 N = 2.27 × 105 N.

Reflect: The force due to the gauge pressure of the water is much larger than the weight of the hatch and would be impossible for the crew to apply it just by pushing.

13.20. Set Up: The added pressure is equal to ρ gh. Solve: (a) ρ gh = (1050 kg/m3 )(9.80 m/s 2 )(1.85 m) = 1.90 × 104 Pa (b) This additional pressure causes additional outward force on the walls of the blood vessels in your brain.

13.21. Set Up: 1 atm = 1.013 × 105 Pa. The pressure difference is Δp = ρ gh. Solve: (a) Δp = 0.125 atm = 1.27 × 104 Pa.

ρ=

Δp 1.27 × 104 Pa = = 864 kg/m3 gh (9.80 m/s 2 )(1.50 m)

(b) Δp = ρ g Mars h = (864 kg/m3 )(0.379)(9.80 m/s 2 )(1.50 m) = 4.81 × 103 Pa = 0.0475 atm Reflect: The pressure difference on Mars is less than it is on earth by a factor of 0.379. Our calculated density of oil is a reasonable value, somewhat less than the density of water.

13.22. Set Up: The density of seawater is 1.03 × 103 kg/m3 . The area of the eardrum is A = π r 2 , with r = 4.1 mm. The pressure increase with depth is Δp = ρ gh and F = pA. Solve: ΔF = (Δp ) A = ρ ghA h=

ΔF 1. 5 N = = 2. 8 m 3 3 ρ gA (1.03 × 10 kg/m )(9.80 m/s 2 )π (4.1 × 10−3 m)2

*13.23. Set Up: p0 = psurface + ρ gh where psurface is the pressure at the surface of a liquid and p0 is the pressure at a depth h below the surface. Solve: (a) For the oil layer, psurface = patm and p0 is the pressure at the oil-water interface. p0 − patm = pgauge = ρ gh = (600 kg/m3 )(9.80 m/s 2 )(0.120 m) = 706 Pa (b) For the water layer, psurface = 706 Pa + patm .

p0 − patm = pgauge = 706 Pa + ρ gh = 706 Pa + (1.00 × 103 kg/m3 )(9.80 m/s 2 )(0.250 m) = 3.16 × 103 Pa Reflect: The gauge pressure at the bottom of the barrel is due to the combined effects of the oil layer and water layer. The pressure at the bottom of the oil layer is the pressure at the top of the water layer.

13.24. Set Up: F = pA. 1 cm2 = 1 × 10−4 m2 . 1 torr = 1 mm Hg = 133.3 Pa. The pressure change with depth is Δp = ρ gh. The arm is above the heart, so the pressure in the arm is less than the pressure at the heart. Solve: (a) maximum F = (120 torr)(133.3 Pa/torr)(1.0 × 10−4 m 2 ) = 1.6 N

minimum F = (80 torr)(133.3 Pa/torr)(1.0 × 10−4 m 2 ) = 1.1 N (b) The pressure in the arm is less by an amount

ρ gh = (1050 kg/m3 )(9.80 m/s 2 )(0.300 m) = 3.09 × 103 Pa = 23 torr The blood pressure reading should be

120 − 23 97 = . 80 − 23 57


13-6

Chapter 13

13.25. Set Up: 1 torr = 1 mm Hg = 133.3 Pa. The pressure change with depth is Δp = ρ gh. The arm is above the heart, so the pressure in the arm is less than the pressure at the heart. On the moon, g = 1.67 m/s 2 . Solve: The pressure in the arm is less by an amount

ρ gh = (1050 kg/m3 )(1.67 m/s 2 )(0.25 m) = 438 Pa = 3.3 torr 120 − 3 117 = . 80 − 3 77 Reflect: The changes in blood pressure for points in the body above or below the heart is much less on the moon than on Earth, since g is much less on the moon.

The blood pressure reading should be

13.26. Set Up: 1 atm = 1.013 × 105 Pa. The force diagram for the piston is given in the figure below. p is the absolute pressure of the hydraulic fluid.

Solve:

pA − w − patm A = 0 and p − patm = pgauge =

w mg (1200 kg)(9.80 m/s 2 ) = = = 1.7 × 105 Pa = 1.7 atm A π r2 π (0.15 m) 2

*13.27. Set Up: Pascal’s law says the pressure is the same everywhere in the hydraulic fluid, so F1/A1 = F 2 /A 2 . F1 = 100 N and F = (3500 kg)(9.80 m/s 2 ) = 3.43 × 104 N, the weight of the car and platform that is being lifted. The

volume of fluid displaced at each piston when the pistons move distances d1 and d 2 is the same, so d1 A1 = d 2 A2 . Solve: (a) A = π r 2 so

F1

r12

=

F2 r2

. r2 = r1 2

F2 3.43 × 104 N = (0.125 m) = 2.32 m and the diameter is 4.64 m. 100 N F1

(b) d1 A1 = d 2 A2 . 2

d 2 = d1

A1 πr 2 ⎛ 0.125 m ⎞ = d1 1 2 = (50 cm) ⎜ ⎟ = 1.45 mm A2 π r2 ⎝ 2.32 m ⎠

Reflect: The work done by the man is F1 d1 = (100 N)(0.50 m) = 50 J. The work done on the car is F2 d 2 = (3.43 × 104 N)(1.43 × 10−3 m) = 50 J. Energy conservation requires that these two quantities be equal, and they are

equal.

13.28. Set Up: Apply p = p0 + ρ gh. For water we have ρ = 1.00 × 103 kg/m3. Solve:

p − pair = ρ gh = (1.00 × 103 kg/m3 )(9.80 m/s 2 )(6.1 m) = 6.0 × 104 Pa.

Reflect: The pressure difference increases linearly with depth.


Fluid Mechanics

13-7

*13.29. Set Up: The density of aluminum is 2.7 × 103 kg/m3 . The density of water is 1.00 × 103 kg/m3 . ρ = m /V. The buoyant force is FB = ρwater Vobj g . Solve: (a) T = mg = 89 N so m = 9.08 kg. V =

m

ρ

=

9.08 kg 2.7 × 103 kg/m3

= 3.36 × 10−3 m3 = 3.4 L.

(b) When the ingot is totally immersed in the water while suspended, T + FB − mg = 0.

FB = ρwater Vobj g = (1.00 × 103 kg/m3 )(3.36 × 1023 m3 )(9.80 m/s 2 ) = 32.9 N. T = mg − FB = 89 N − 32.9 N = 56 N. Reflect: The buoyant force is equal to the difference between the apparent weight when the object is submerged in the fluid and the actual gravity force on the object.

13.30. Set Up: The buoyant force exerted by the water is FB = ρw gVfish . When the fish is fully submerged the buoyant force on it must equal its weight. Solve: (a) The average density of the fish is very close to the density of water. (b) Before inflation, FB = w = (2.75 kg)(9.80 m/s 2 ) = 27.0 N. When the volume increases by a factor of 1.10, the buoyant force also increases by a factor of 1.10 and becomes (1.10)(27.0 N) = 29.7 N. (c) The water exerts an upward force 29.7 N and gravity exerts a downward force of 27.0 N so there is a net upward force of 2.7 N; the fish moves upward.

*13.31. Set Up: FB = ρfluid gVsub , where Vsub is the volume of the object that is below the fluid’s surface. Solve: (a) Floats, so the buoyant force equals the weight of the object: FB = mg . Using Archimedes’s principle gives

ρw gV = mg and V=

m

ρw

=

5750 kg 1.00 × 103 kg/m3

= 5.75 m3 .

(b) FB = mg and ρw gVsub = mg . Vsub = 0.80V = 4.60 m3 , so the mass of the floating object is

m = ρw Vsub = (1.00 × 103 kg/m3 )(4.60 m3 ) = 4600 kg.

He must throw out 5750 kg − 4600 kg = 1150 kg. Reflect: He must throw out 20% of the boat’s mass.

13.32. Set Up: FB = ρwater Vobj g . w = mg = 17.50 N and m = 1.79 kg. Solve: T + FB − mg = 0. FB = mg − T = 17.50 N − 11.20 N = 6.30 N. Vobj =

FB

ρwater g ρ=

=

6.30 N 3

(1.00 × 10 kg/m3 )(9.80 m/s 2 )

= 6.43 × 10−4 m3 .

m 1.79 kg = = 2.78 × 103 kg/m3 V 6.43 × 1024 m3

13.33. Set Up: FB = ρwater Vice g . mice = ρice V . ρice = 0.92 × 103 kg/m3 . Solve: The ice is floating, so FB = (mice + mwoman ) g . ρwaterVice = ρice Vice + mwoman . 45.0 kg = 0.562 m3 . 1000 kg/m3 − 920 kg/m3 Reflect: The buoyant force must both the slab of ice and the woman. If the slab of ice is 20 cm thick and square on its top and bottom surfaces, its top surface measures about 1.7 m on a side. The woman would also need to be concerned about the slab tipping to one side, because of unbalanced torques. Vice =

mwoman

ρwater − ρice

=


13-8

Chapter 13

4 3

13.34. Set Up: Treating Saturn as approximately spherical its volume is given by V = π r 3. Its average density can then be calculated from ρ = Solve: ρ =

M . According to the appendix r = 6.03 × 107 m and M = 5.68 × 1026 kg for Saturn. V

3M 3(5.68 × 1026 kg) M = = = 618 kg/m3. 3 V 4π r 4π (6.03 × 107 m)3

The average density of Saturn is about 62% that of water, which has a density of 1000 kg/m3. If Saturn were a solid body, it could float in a large enough body of water. Reflect:

*13.35. Set Up: FB = ρwater Vobj g . The net force on the sphere is zero. Solve: (a) FB = (1000 kg/m3 )(0.650 m3 )(9.80 m/s 2 ) = 6.37 × 103 N (b) FB = T + mg and

m=

FB − T 6.37 × 103 N − 900 N = = 558 kg. g 9.80 m/s 2

(c) Now FB = ρwater Vsub g , where Vsub is the volume of the sphere that is submerged. FB = mg . ρwaterVsub = mg and

Vsub =

m

ρwater

=

558 kg 1000 kg/m3

= 0.558 m3 .

Vsub 0.558 m3 = = 0.858 = 85.8, Vobj 0.650 m3

Reflect: When the sphere is totally submerged, the buoyant force on it is greater than its weight. When it is floating, it needs to be only partially submerged in order to produce a buoyant force equal to its weight.

13.36. Set Up: According to Archimedes’s principle, the buoyant force on the balloon is equal to the weight of the air displaced by the balloon: FB = mg = ρairVg . ⎛4 ⎞ Solve: We have FB = mg = ρairVg = (1.20 kg/m3 ) ⎜ π ⎟ (0.150 m)3 (9.80 m/s 2 ) = 0.1663 N. ⎝3 ⎠

⎛4 ⎞ (b) The mass of the helium in the balloon is ρHeV = (0.166 kg/m3 ) ⎜ π ⎟ (0.150 m)3 = 0.002347 kg = 2.34 grams. ⎝3 ⎠ The net force acting on the balloon is equal to the buoyant force minus the weight of the balloon and the helium that it contains, which is 0.1663 N − (0.00200 kg + 0.002347 kg)(9.80 m/s 2 ) = 0.124 N.

In this case the buoyant force on the balloon is significantly larger than the weight of the balloon. If 0.124 N released, the initial acceleration of the balloon would be a = = 28.5 m/s 2 . 0.00435

Reflect:

*13.37. Set Up: FB = ρfluid gVsub . For a floating object, FB = mg = ρobjVobj g . ρice = 0.92 × 103 kg/m3 . Solve: (a) FB = mg says ρsw gVsub = ρice gVice . Vsub = 0.90Vice , so 0.90 ρsw = ρice and

ρsw = (b) ρw gVsub = ρice gVice and

ρice

0.90

= 1.02 × 103 kg/m3 .

Vsub ρice 0.92 × 103 kg/m3 = = = 0.92 ρ w 1.00 × 103 kg/m3 Vice

92.0% is submerged and 8.0% is above the surface.


Fluid Mechanics

13.38. Set Up: 1 dyn/cm = 10−3 N/m. The excess pressure inside a liquid drop is Δp = 2(72.8 × 10−3 N/m)

Solve: (a) Δp = (b) Δp =

1.50 × 10−3 m

2(72.8 × 10−3 N/m) 0.0100 × 1023 m

13-9

2γ . R

= 97.1 Pa

= 1.46 × 104 Pa

*13.39. Set Up: 1 dyn/cm = 1023 N/m. The gauge pressure inside a bubble is 4γ pgauge = p − patm =

Solve:

pgauge =

4(25.0 × 10−3 N/m) 3.50 × 10−2 m

R

.

= 2.86 Pa.

13.40. Set Up: 1 dyn/cm = 1023 N/m. 1 atm = 1.013 × 105 Pa. The pressure difference for a liquid drop is 2γ Δp =

Solve:

R=

R

.

2γ 2(72.8 × 10−3 N/m) = = 7.19 × 10−5 m = 0.0719 mm. Δp (0.0200 atm)(1.013 × 105 Pa/atm)

13.41. Set Up: The gauge pressure p − patm in a drop is related to the surface tension γ of the liquid and the radius R of the drop by p − patm = 1/3

⎛ 3V ⎞ Solve: R = ⎜ ⎟ ⎝ 4π ⎠

4γ . The volume of a spherical drop is V = 43 π R3 . R 1/3

⎛ 4π ⎞ so p − patm = 4γ ⎜ ⎟ ⎝ 3V ⎠

. Same gauge pressure, so γ V −1/3 = constant and γ w Vw −1/3 = γ ct Vct −1/3 . 3

3

Vw ⎛ γ w ⎞ ⎛ 72.8 dynes/cm ⎞ =⎜ ⎟ =⎜ ⎟ = 20.0 Vct ⎝ γ ct ⎠ ⎝ 26.8 dynes/cm ⎠ Reflect: Water has a larger surface tension so the water drop is larger.

13.42. Set Up: The continuity equation is A1υ1 = A2υ2 . Solve: (18.5 m)(3.75 m)(2.50 cm/s) = (16.5 m) h(11.0 cm/s) so h = 0.96 m, the depth of the canal at the second point. *13.43. Set Up: υ1 A1 = υ2 A2 ⎛ 0.070 m 2 ⎞ ⎛A ⎞ = 2.33 m/s Solve: (a) υ2 = υ1 ⎜ 1 ⎟ = (3.50 m/s) ⎜ ⎜ 0.105 m 2 ⎟⎟ ⎝ A2 ⎠ ⎝ ⎠ ⎛ 0.070 m 2 ⎞ ⎛A ⎞ = 5.21 m/s (b) υ2 = υ1 ⎜ 1 ⎟ = (3.50 m/s) ⎜ ⎜ 0.047 m 2 ⎟⎟ ⎝ A2 ⎠ ⎝ ⎠

13.44. Set Up: The volume flow rate is υ1 A1 = υ2 A2 . Solve: (a) υ1 =

1.20 m3/s 1.20 m3/s = = 17.0 m/s A1 π (0.150 m)2

(b) υ1π r12 = υ2 π r2 2 .

r2 = r1

υ1 17.0 m/s = (0.150 m) = 0.317 m υ2 3.80 m/s


13-10

Chapter 13

*13.45. Set Up: Apply the equation of continuity, υ1 A1 = υ2 A2 , where A = π r 2 Solve: υ2 = υ1 ( A1/A2 ), where Α1 = π (0.80 cm) 2 , Α2 = 20π (0.10 cm) 2 . υ2 = (3.0 m/s)

π (0.80) 2 = 9.6 m/s. 20π (0.10) 2

Reflect: The total area of the showerhead openings is less than the cross-section area of the pipe, and the speed of the water in the showerhead opening is greater than its speed in the pipe.

13.46. Set Up: The equation of continuity states that A1υ1 = A2υ2 . The distance traveled by a projectile that is fired from a height h with an initial horizontal velocity υ is x = υt , where t =

2h . g

Solve: Since h is fixed, t does not change as we constrict the nozzle. Looking at the ratio of distances we obtain 2

2 ⎛ r2 ⎞ x1 υ1t A2 π r2 2 ⎛ 1.80 cm ⎞ (0 950 m) x = x = . = = = , which gives ⎜ ⎟ 1 2 r ⎜ ⎟ = 5.47 m. x2 υ2t A1 π r1 2 ⎝ 0.750 cm ⎠ ⎝ 1⎠ Reflect: A smaller constriction results in a higher exit velocity, which results in a greater range.

13.47. Set Up: Toricelli’s theorem says the speed of efflux is υ = 2 gh , where h is the distance of the small hole below the surface of the water in the tank. Solve: υ = 2(9.80 m/s 2 )(14.0 m) = 16.6 m/s

13.48. Set Up: Let point 1 be at the surface of the water in the tank. Let point 2 be in the emerging stream of water. Since the hole is small, υ1 ≈ 0. p1 − pair = 3.00 atm = 3.04 × 105 Pa. p2 = pair . y1 − y2 = 11.0 m. The density of seawater is ρ = 1.03 × 103 kg/m3 . Solve:

p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 . p1 − pair + ρ g ( y1 − y2 ) = 12 ρυ22 .

υ2 =

2( p1 − pair )

ρ

+ 2 g ( y1 − y2 ) =

2(3.04 × 105 Pa) 1.03 × 103 kg/m3

+ 2(9.80 m/s 2 )(11.0 m) = 28.4 m/s

*13.49. Set Up: Let point 1 be in the mains and point 2 be in the emerging stream at the end of the fire hose. υ1 ≈ 0. y1 − y2 = 0. p2 = pair . After the water emerges from the hose, with upward velocity, it moves in free fall. Solve:

p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 gives p1 − pair = 12 ρυ2 2 . The motion of a drop of water after it

leaves the hose gives

1 mυ22 2

= mgh.

υ2 2 = 2 gh = 2(9.80 m/s 2 )(15.0 m) = 294 m 2 /s 2 . p1 − pair = 12 (1000 kg/m3 )(294 m 2 /s 2 ) = 1.47 × 105 Pa Reflect: Alternatively, the problem can be solved entirely by using Bernoulli’s equation. In this case point 1 is in the mains and point 2 is at the maximum height of the stream. Thus, υ1 = υ2 = 0 and p1 + ρ gy1 = p2 + ρ gy2 . This

implies that p1 − pair = ρ g Δy, which gives the same answer as before. Note that this is the gauge pressure required to a column of water 15.0 m high.

13.50. Set Up: υ 1 A1 = υ 2 A2 relates υ2 and υ 1. r2 = 2r1. Solve:

p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 . We can subtract air pressure pair from both sides of the equation,

and the pressures become gauge pressures.

υ 2 = υ1

2

A1 πr 2 ⎛ 1⎞ = υ 1 1 2 = (3.00 m/s) ⎜ ⎟ = 0.750 m/s. y1 − y2 = + 11.0 m. ⎝ 2⎠ A2 πr2

p2 = p1 + ρ g ( y1 − y2 ) + 12 ρ (υ1 2 − υ 2 2 )


Fluid Mechanics

13-11

p2 = 4.00 × 104 Pa + (1000 kg/m3 )(9.80 m/s 2 )(11.0 m) + 12 (1000 kg/m3 )([3.00 m/s]2 − [0.750 m/s]2 ) p2 = 1.52 × 105 Pa

*13.51. Set Up: Let point 1 be at the top surface and point 2 be at the bottom surface. Neglect the ρ g ( y1 − y2 ) term in Bernoulli’s equation. In calculating the net force, ∑ Fy , take + y to be upward. Solve:

p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ 22 .

p2 − p1 = 12 ρ(υ1 2 − υ22 ) = 12 (1.20 kg/m3 )([70.0 m/s]2 − [60.0 m/s]2 ) = 780 Pa ∑ Fy = p2 A − p1 A − mg = (780 Pa)(16.2 m 2 ) − (1340 kg)(9.80 m/s 2 ) = − 500 N. The net force is 500 N, downward. Reflect: The pressure is lower where the fluid speed is higher.

13.52. Set Up: Let point 1 be where r1 = 4.00 cm and point 2 be where r2 = 2.00 cm. The volume flow rate υ A has the value 7200 cm3 /s at all points in the pipe. Apply the continuity equation (Eq. 13.10) to find the fluid speed at points 1 and 2 and then use Bernoulli’s equation for these two points to find p2 . Solve: υ 1 A1 = υ 1π r1 2 = 7200 cm3 /s, so υ1 = 1.43 m/s

We know that υ 2 A2 = υ 2π r2 2 = 7200 cm3 /s, so υ2 = 5.73 m/s We know that p1 + ρ gy1 + 12 ρυ 12 = p2 + ρ gy2 + 12 ρυ 22 , y1 = y2 and p1 = 2.40 × 105 Pa, so p2 = p1 + 12 ρ(υ 12 − υ 2 2 ) = 2.25 × 105 Pa. Reflect: Where the area decreases the speed increases and the pressure decreases.

*13.53. Set Up: y1 = y2 . υ 1 A1 = υ 2 A2 = 465 × 10−6 m3/s. Solve:

p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ 22 .

υ1 = Then υ 2 =

2

ρ

( p1 − p2 ) + υ1 2 =

465 × 10−6 m3 /s

π r1

2

2 1000 kg/m3

=

465 × 10−6 m3 /s

π (2.05 × 10−2 m) 2

= 0.352 m/s

(0.40 × 105 Pa) + (0.352 m/s) 2 = 8.95 m/s.

υ 1π r1 2 = υ 2 π r2 2 and r2 = r1

υ1 υ2

= (2.05 cm)

0.352 m/s = 4.07 mm. 8.95 m/s

13.54. Set Up: p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 . The two points are close together so we can neglect ρ g ( y1 − y2 ). ρ = 1.03 × 103 kg/m3 . A1υ 1 = A2υ 2 . Solve:

p1 − p2 + 12 ρυ1 2 = 12 ρυ 2 2

υ2 =

2( p1 − p2 )

ρ

+ υ1 2 =

2(1.20 × 104 Pa − 1.15 × 104 Pa) 3

1.05 × 10 kg/m

3

+ (0.300 m/s) 2

υ 2 = 1.0 m/s = 100 cm/s A2 υ1 30 cm/s = = = 0.30. A1 υ2 100 cm/s A2 = 0.30 A1, so 70% is blocked.


13-12

Chapter 13

13.55. Set Up: y1 = y2 . υ 1 A1 = υ 2 A2 . A2 = 2 A1. Solve: p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ 22 .

⎛ A1 ⎞ ⎛ A1 ⎞ ⎟ = (2.50 m/s) ⎜ ⎟ = 1.25 m/s. A ⎝ 2⎠ ⎝ 2 A1 ⎠

υ 2 = υ 1⎜

p2 = p1 + 12 ρ(υ1 2 − υ22 ) = 1.80 × 104 Pa + 12 (1000 kg/m3 )([2.50 m/s]2 − [1.25 m/s]2 ) = 2.03 × 104 Pa Reflect: The gauge pressure is higher at the second point because the water speed is less there.

13.56. Set Up: At the terminal speed, a = 0 and the forces balance. The viscous drag force is 6πη rυ t . The weight of the ball bearing is mg = 43 π r 3 ρ g . The density of steel is ρ = 7.8 × 103 kg/m3 . Solve: υ t =

2r 2 g ρ 2(1.00 × 10−3 m) 2 (9.80 m/s 2 )(7.8 × 103 kg/m3 ) = = 0.113 m/s. 9η 9(0.150 N ⋅ s/m 2 )

*13.57. Set Up: The viscous drag force is F = 6πη rυ t . The weight of the ball bearing is mg = 43 π r 3 ρ g . Solve: F = 14 mg gives 6πη rυ t = 13 π r 3 ρ g .

υt =

r 2 g ρ (3.00 × 1023 m) 2 (9.80 m/s 2 )(19.3 × 103 kg/m3 ) = = 0.0959 m/s = 9.59 cm/s. 18η 18(0.986 N ⋅ s/m 2 )

13.58. Set Up: At the terminal speed a = 0 and the forces balance. The viscous drag force is 6πη rυ t . The weight of the sphere is mg = 43 π r 3 ρ g . 1/3

⎛ 3m ⎞ Solve: The radius of the sphere is r = ⎜ ⎟ ⎝ 4πρ ⎠

η=

1/3

⎛ 3[0.20 × 10−3 kg] ⎞ =⎜ ⎜ 4π [8900 kg/m3 ] ⎟⎟ ⎝ ⎠

= 1.75 × 10−3 m. 6πη rυ t = mg and

mg (0.20 × 10−3 kg)(9.80 m/s 2 ) = = 0.990 N ⋅ s/m 2 . 6π rυ t 6π (1.75 × 10−3 m)(6.0 × 10−2 m/s)

*13.59. Set Up: The flow rate, ΔV /Δt , is related to the radius R or diameter D of the artery by Poiseuille’s law: ΔV π R 4 ⎛ p1 − p2 ⎞ π D 4 ⎛ p1 − p2 ⎞ = ⎜ ⎟= ⎜ ⎟. Δt 8η ⎝ L ⎠ 128η ⎝ L ⎠ Assume the pressure gradient ( p1 − p2 )/L in the artery remains the same. Solve: (ΔV /Δt )/D4 =

π ⎛ p1 − p2 ⎞ 4 4 ⎜ ⎟ = constant, so (ΔV /Δt )old /Dold = (ΔV /Δt ) new /Dnew . (ΔV/Δt ) new = 2( ΔV/Δt )old 128η ⎝ L ⎠

and Dold = D. This gives 1/4

Dnew

⎡ ( ΔV/Δt )new ⎤ = Dold ⎢ ⎥ ⎣ ( ΔV/Δt )old ⎦

= 21/ 4 D = 1.19 D.

Reflect: Since the flow rate is proportional to D 4 , a 19% increase in D doubles the flow rate.

13.60. Set Up: Apply Newton’s second law to the car. The buoyancy force is given by Archimedes’s principle. When the car starts to sink it is fully submerged and the buoyant force is equal to the weight of the car plus the water that is inside it.


Fluid Mechanics

13-13

(a) Solve: The free-body diagram for the floating car is given in the figure below. (Vsub is the volume that is

submerged.)

Using ∑ Fy = ma y we obtain B − mg = 0, so ρwVsub g − mg = 0. This gives Vsub = m /ρw = (900 kg)/(1000 kg/m3 ) = 0.900 m3. Thus, we have Vsub /Vobj = (0.900 m3 )/(3.0 m3 ) = 0.30 = 30,. Reflect: The average density of the car is (900 kg)/(3.0 m3 ) = 300 kg/m3 . ρcar /ρwater = 0.30; this equals Vsub /Vobj. (b) Solve: When the car is full submerged Vsub = V, the volume of the car and B = ρwaterVg = (1000 kg/m3 )(3.0 m3 ) (9.80 m/s 2 ) = 2.94 × 104 N. The weight of the car is mg = (900 kg)(9.80 m/s 2 ) = 8820 N. Thus the weight of the

water in the car when it sinks is the buoyant force minus the weight of the car itself: m water = (2.94 × 104 N − 8820 N)/ (9.80 m/s2 ) = 2.10 × 103 kg and Vwater = mwater /ρwater = (2.10 × 103 kg)/(1000 kg/m3 ) = 2.10 m3. The fraction this is of the total interior volume is (2.10 m3 )/(3.00 m3 ) = 0.70 = 70, Reflect: The average density of the car plus the water inside it is (900 kg + 2100 kg)/(3.0 m3 ) = 1000 kg/m3 , so

ρcar = ρwater when the car starts to sink.

13.61. Set Up: ρHg = 13.6 × 103 kg/m3 . Let x be the height of the mercury surface in the right arm above the level of the mercury-water interface in the left arm. Solve: (a) p − pair = ρwater gh = (1.00 × 103 kg/m3 )(9.80 m/s 2 )(0.150 m) = 1.47 × 103 Pa (b) The gauge pressure a distance x below the mercury surface in the right arm equals the gauge pressure at the mercury-water interface, since these two points are at the same height in the mercury. ρHg gx = 1.47 × 103 Pa and

x=

1.47 × 103 Pa (13.6 × 103 kg/m3 )(9.80 m/s 2 )

= 1.1 cm.

h + x = 15.0 cm, so h = 13.9 cm.

Reflect: The weight of the water pushes the mercury down in the left-hand arm. A 1.1 cm column of mercury produces the same pressure as a 15.0 cm column of water.

13.62. Set Up: Steel has density 7.8 × 103 kg/m3 . The maximum buoyant force is FB = ρwater Vbarge g . Solve: The volume of steel in the barge is (2 × 22 m × 12 m + 2 × 40 m × 12 m + 22 m × 40 m) × (0.040 m) = 94.7 m3 .

The mass of the barge is mbarge = ρ V = (7.8 × 103 kg/m3 )(94.7 m3 ) = 7.39 × 105 kg. FB = (1000 kg/m3 )(22 m × 40 m × 12 m)(9.80 m/s 2 ) = 1.035 × 108 N.

FB = (mbarge + mcoal ) g and mcoal =

FB − mbarge = 9.8 × 106 kg. g

The volume of this mass of coal is V=

m

ρ

=

9.8 × 106 kg 1500 kg/m3

= 6.5 × 103 m3 .

The volume of the barge is 22 m × 40 m × 12 m = 1.06 × 104 m3 ; this mass of coal easily fits into the barge. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

13-14

Chapter 13

*13.63. Set Up: The density of lead is 11.3 × 103 kg/m3 . The buoyant force when the wood sinks is FB = ρwater Vtot g , where Vtot is the volume of the wood plus the volume of the lead. ρ = m /V . Solve: Vwood = (0.600 m)(0.250 m)(0.080 m) = 0.0120 m3 . mwood = ρwood Vwood = (600 kg/m3 )(0.0120 m3 ) = 7.20 kg.

FB = ( mwood + mlead ) g . Using FB = ρwater Vtot g and Vtot = Vwood + Vlead gives mlead = ρ lead Vlead

ρwater (Vwood + Vlead ) g = (mwood + mlead ) g . then gives ρwaterVwood + ρwater Vlead = mwood + ρlead Vlead .

Vlead =

ρwaterVwood − mwood (1000 kg/m3 )(0.0120 m3 ) − 7.20 kg = = 4.66 × 10−4 m3 . ρlead − ρwater 11.3 × 103 kg/m3 − 1000 kg/m3

mlead = ρlead Vlead = 5.27 kg. Reflect: The volume of the lead is only 3.9% of the volume of the wood. If the contribution of the volume of the lead to FB is neglected, the calculation is simplified: ρwaterVwood g = (mwood + mlead ) g and mlead = 4.8 kg. The

result of this calculation is in error by about 9%.

13.64. Set Up: The buoyant force on the balloon must equal the total weight of the balloon fabric, the basket and its contents and the gas inside the balloon. We know that mgas = ρgasV and B = ρairVg. The total weight, exclusive of the gas inside the balloon, is 900 N + 1700 N + 3200 N = 5800 N (5800 N) Solve: 5800 N + ρgasVg = ρairVg and ρgas = 1.23 kg/m3 − = 0.96 kg/m3 . (9.80 m/s 2 )(2200 m3 ) Reflect: The volume of a given mass of gas increases when the gas is heated, and the density of the gas therefore decreases.

*13.65. Set Up: Seawater has density ρsw = 1.03 × 103 kg/m3 . FB = ρsw Vsub g . Vsub is the submerged volume: Vsub = Vlp + 0.80Vp , where Vlp = 0.0400 m3 is the volume of the life preserver and Vp is the volume of the person. Solve: FB = mtot g . mtot = mlp + mp = ρlp Vlp + mp .

Vp =

mp

ρp

=

75.0 kg 980 kg/m3

= 0.0765 m3 .

FB = mtot g gives ρsw (Vlp + 0.80Vp ) g = ( ρlp Vlp + mp ) g .

ρlp = ρlp =

ρsw (Vlp + 0.80Vp ) − mp Vlp

(1.03 × 103 kg/m3 )(0.0400 m3 + [0.80][0.0765 m3 ]) − 75.0 kg 0.0400 m3

= 731 kg/m3

13.66. Set Up: FB = ρ VA g . Apply Newton’s second law to the beaker, liquid, and block as a combined object and also to the block as a single object. Take + y upward. Solve: Forces on the combined object: FD + FE − ( wA + wB + wC ) = 0. wA = FD + FE − wB − wC . D and E read

mass rather than weight, so write the equation as m A = mD + mE − mB − mC . mD = FD /g is the reading in kg of scale D; a similar statement applies to mE . m A = 3.50 kg + 7.50 kg − 1.00 kg − 1.80 kg = 8.20 kg. Forces on A: FB + FD − wA = 0. ρ VA g + FD − m A g = 0. ρ VA + mD = m A .

ρ=

m A − mD 8.20 kg − 3.50 kg = = 1.24 × 103 kg/m3 VA 3.80 × 10−3 m3

(b) D reads the mass of A: 8.20 kg. E reads the total mass of B and C: 2.80 kg. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Fluid Mechanics

13-15

13.67. Set Up: The density of gold is ρg = 19.3 × 103 kg/m3 and the density of aluminum is ρal = 2.7 × 103 kg/m3 . FB = ρw Vtot g . Solve: Find the volume Vtot of the ingot: FB + T − w = 0. ρw Vtot g = W − T .

Vtot =

W−T 45.0 N − 39.0 N = = 6.12 × 10−4 m3 . ρw g (1000 kg/m3 )(9.80 m/s 2 )

The mass of the ingot is mtot = W /g = 4.59 kg. The total mass is the mass of gold, mg , plus the mass of aluminum, mal . The total volume is the volume of gold, Vg , plus the volume of aluminum, Val . mtot = mg + mal = ρg Vg + ρal Val = ρg Vg + ρal (Vtot − Vg ). Vg =

mtot − ρal Vtot 4.59 kg − (2700 kg/m3 )(6.12 × 10−4 m3 ) = = 1.77 × 10−4 m3 . ρg − ρal 19.3 × 103 kg/m3 − 2.7 × 103 kg/m3

wg = mg g = ρ g Vg g = (19.3 × 103 kg/m3 )(1.77 × 10−4 m3 )(9.80 m/s 2 ) = 33.5 N. Reflect: If the ingot had the same weight in air (45.0 N) and was pure gold, its volume would be 2.38 × 10−4 m3

and the balance would read 42.7 N when the ingot was submerged.

13.68. Set Up: The pressure at the surface of the earth is p = 1.013 × 105 Pa. The height of fluid in the barometer is related to air pressure p by p = ρ gh. Solve: (a) ρ =

p 1.013 × 105 Pa = = 4.73 × 103 kg/m3 gh (9.80 m/s 2 )(2.185 m)

(b) p = ρ gh = (4.73 × 103 kg/m3 )

( 14 ) (9.80 m/s2 )(0.725 m) = 8.40 × 103 Pa

*13.69. Set Up: Apply p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 , with point 1 at the surface of the acid in the tank and point 2 in the stream as it emerges from the hole. p1 = p2 = pair . Since the hole is small the level in the tank drops slowly and υ1 ≈ 0. After a drop of acid exits the hole the only force on it is gravity and it moves in projectile motion. For the projectile motion take + y downward, so ax = 0 and a y = + 9.80 m/s 2 . Solve: Bernoulli’s equation with p1 = p2 and υ1 = 0 gives

υ 2 = 2 g ( y1 − y2 ) = 2(9.80 m/s 2 )(0.75 m) = 3.83 m/s. projectile motion: Use the vertical motion to find the time in the air. υ0 y = 0, a y = + 9.80 m/s 2 , y − y0 = 1.4 m.

y − y0 = υ0 y t + 12 a y t 2 gives t=

2( y − y0 ) = ay

2(1.4 m) 9.80 m/s 2

= 0.535 s

The horizontal distance a drop travels in this time is x − x0 = υ0 x t + 12 ax t 2 = (3.83 m/s)(0.535 s) = 2.05 m. Reflect: If the depth of acid in the tank is increased, then the velocity of the stream as it emerges from the hole increases and the horizontal range of the stream increases.

13.70. Set Up: After leaving the tank, the water is in free-fall, with ax = 0 and a y = + g . From Section 13.7 we know that the speed of efflux is

2 gh .

Solve: (a) The time it takes any portion of the water to reach the ground is t =

2( H − h) , in which time the water g

travels a horizontal distance R = vt = 2 h( H − h). (b) Note that if h′ = H − h, h′ ( H − h′ ) = ( H − h)h, and so h′ = H − h gives the same range. A hole H − h below the water surface is a distance h above the bottom of the tank. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

13-16

Chapter 13

Reflect: For the special case of h = H /2, h = h′ and the two points coincide. For the upper hole the speed of efflux is less but the time in the air during the free-fall is greater.

*13.71. Set Up: For a spherical astronomical object, g = G

m R2

. p = pair + ρ gh.

Solve: (a) Find g on Europa:

g = (6.673 × 10−11 N ⋅ m 2 /kg 2 )

4.78 × 1022 kg (1.565 × 106 m) 2

= 1.30 m/s 2

The gauge pressure at a depth of 100 m would be p − pair = ρ gh = (1.00 × 103 kg/m3 )(1.30 m/s 2 )(100 m) = 1.30 × 105 Pa (b) Solve for h on earth that gives p − pair = 1.30 × 105 Pa:

p − pair 1.30 × 105 Pa = = 13.3 m ρg (1.00 × 103 kg/m3 )(9.80 m/s 2 ) Reflect: g on Europa is less than on earth so the pressure at a water depth of 100 m is much less on Europa than it would be on earth. h=

13.72. Set Up: The discharge rate is υ1 A1 = υ2 A2 . The density of mercury is 13.6 × 103 kg/m3 . Let point 1 be where A1 = 40.0 × 1024 m 2 and point 2 is where A2 = 10.0 × 1024 m 2 . y1 = y2 . Solve: (a) υ1 =

6.00 × 10−3 kg/m3 40.0 × 10−4 m 2

= 1.50 m/s. υ2 =

6.00 × 10−3 kg/m3 10.0 × 10−4 m 2

= 6.00 m/s

(b) p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 .

p1 − p2 = 12 ρ (υ2 2 − υ1 2 ) = 12 (1000 kg/m3 )([6.00 m/s]2 − [1.50 m/s]2 ) = 1.69 × 104 Pa (c) p1 − p2 = ρ gh and

h=

p1 − p2 1.69 × 104 Pa = = 0.127 m = 12.7 cm. ρg (13.6 × 103 kg/m3 )(9.80 m/s 2 )

13.73. Set Up: p − pair = ρ gh, A1υ1 = A2 υ2 and p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 . y1 = y2 Solve: (a) p1 = pair + ρ gh1 and p2 = pair + ρ gh2 , where h1 and h2 are the heights of the liquid in the two cylinders.

p1 − p2 = ρ g ( h1 − h2 ). p1 − p2 = Δp and h1 − h2 = h so Δp = ρ gh. (b) p1 − p2 = 12 g (υ2 2 − υ1 2 ).

⎡⎛ A ⎞ 2 ⎤ ⎛ A1 ⎞ 2 1 1 ⎟υ1 so Δp = 2 ρ ⎢⎜ ⎟ − 1⎥υ1 and υ1 = A A ⎢ ⎥ 2 2 ⎝ ⎠ ⎝ ⎠ ⎣ ⎦

υ2 = ⎜

2Δp

ρ ⎡⎣( A1 /A2 )2 − 1⎤⎦

.

Using Δp = ρ gh gives

υ1 =

2 gh ( A1 /A2 ) 2 − 1

.

⎛A ⎞ (c) Once υ1 is measured, υ2 = ⎜ 1 ⎟υ1. ⎝ A2 ⎠

13.74. Set Up: Let ρ1 and ρ2 be the density of the fluid at points 1 and 2. Solve: (a) The derivation of Eq. (13.10) leads to Δm1 = Δm2 . Δm1 = ρ1 A1υ1 Δt and Δm2 = ρ2 A2 υ 2 Δt so

ρ1 A1υ1 Δt = ρ2 A2 υ2 Δt . Δt divides out and ρ1 A1υ1 = ρ2 A2υ2 . (b) If ρ1 = ρ 2 then the density divides out and the usual continuity equation A1υ1 = A2 υ 2 is obtained.


Fluid Mechanics

13-17

Solutions to age Problems *13.75. Set Up: According to Bernoulli’s equation p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 . Solve: Since the veins are flexible and do not collapse, we may assume that the pressure in the venous system is higher than atmospheric pressure. According to Bernoulli’s equation, a large pressure differential would cause blood to exit a wound with a high velocity, which is contrary to experience. Thus, we may assume that the pressure inside an artery is only slightly greater than atmospheric pressure.

13.76. Set Up: According to Bernoulli’s equation p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 . According to the continuity equation A1υ1 = A2υ2 . Solve: According to the continuity equation, we expect υ to be constant if A does not change. In that case, Bernoulli’s equation gives Δp = − ρ g Δy. Thus, we expect a greater pressure at lower elevations. The correct answer is B.

*13.77. Set Up: According to Bernoulli’s equation p1 + ρ gy1 + 12 ρυ1 2 = p2 + ρ gy2 + 12 ρυ22 . According to the continuity equation A1υ1 = A2υ2 . Solve: According to the continuity equation, we expect υ to be constant if A does not change. In that case, Bernoulli’s equation gives Δp = − ρ g Δy. Assuming that Δp does not change as the acceleration of gravity increases from g1 to

g 2 , the maximum height that the heart can pump blood will decrease from Δy1 to Δy2 . Using ρ g earth Δyearth =

ρ g max Δyhead we find that g max = gearth

1.3 m = 2.6 g earth ≈ 3 g earth . Thus, the correct answer is B. 0.5 m

13.78. Set Up: According to the continuity equation A1υ1 = A2υ2 . Solve: The continuity equation gives υ 2 = υ1

3 cm 2 A1 = (30 cm/s) = 0.15 cm/s. The correct answer is D. A2 600 cm 2


14

TEMPERATURE AND HEAT

Answers to Multiple-Choice Problems 1. C 14. C

2. B, C 15. A

3. D

4. B

5. C

6. B

7. C

8. C

9. A

10. B

11. A

12. D

13. B

Solutions to Problems *14.1. Set Up and Solve: (a) TF = 95 (TC + 32°) = 95 (40.2°) + 32° = 104.4°F. Yes, you should be concerned. (b) TF = 95 (TC + 32°) = 95 (12°C) + 32° = 54°F. Yes, bring a jacket. TC = TK − 273.15 = 350 − 273.15 = 77°C.

T = 95 (77°) + 32° = 171°F. This would be unpleasantly warm.

14.2. Set Up: To convert a temperature between °C and K use TC = TK − 273.15. To convert from °F to °C, subtract 32° and multiply by 5/9. To convert from °C to °F, multiply by 9/5 and add 32°. To convert a temperature difference use that Celsius and Kelvin degrees are the same size and that 9 F° = 5C°. Solve : (a) TC = TK − 273.15 = 310 − 273.15 = 36.9°C; TF = 95 TC + 32° = 95 (36.9°) + 32° = 98.4°F (b) TK = TC + 273.15 = 40 + 273.15 = 313 K; TF = 95 TC + 32° = 95 (40°) + 32° = 104°F (c) 7 C° = 7 K; 7 C° = (7 C°)(9 F°/5 C°) = 13 F° (d) 4.0°C: TF = 95 TC + 32° = 95 (4.0°) + 32° = 39.2°F; TK = TC + 273.15 = 4.0 + 273.15 = 277 K

−160°C: TF = 95 TC + 32° = 95 ( −160°) + 32° = − 256°F; TK = TC + 273.15 = − 160 + 273.15 = 113 K (e) TC = 95 (TF − 32°) = 95 (105° − 32°) = 41°C; TK = TC + 273.15 = 41 + 273.15 = 314 K

14.3. Set Up: ΔT is the same on the Celsius and Kelvin scales. 180 F° = 100 C°, so 1 C° =

9 5

F°.

⎛ 1 C° ⎞ Solve: (a) ΔT = 49.0 F°. ΔT = (49.0 F°) ⎜ 9 ⎟ = 27.2 C° = 27.2 K. ⎜ F° ⎟ ⎝5 ⎠ ⎛ 1 C° ⎞ (b) ΔT = − 100 F°. ΔT = (−100.0 F°) ⎜ 9 ⎟ = − 55.6 C° = − 55.6 K ⎜ F° ⎟ ⎝5 ⎠

14.4. Set Up: To convert a temperature from °C to °F, multiply by 9/5 and add 32°. To convert from Kelvin to Celsius TC = TK − 273.15. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

14-1

14-2

Chapter 14

Solve: 5000°C: TF = 95 TC + 32° = 95 (5000°) + 32° = 9032°F

15 × 106 K: TC = TK − 273.15 = 1.5 × 107°C; TF = 95 TC + 32° = 95 (1.5 × 107°) + 32° = 2.7 × 107°F

*14.5. Set Up and Solve: (a) TF = 95 TC + 32. Set TF = TC = T . T = 95 T + 32.

4 T 5

= − 32 and T = − 40°.

−40°C = − 40°F.

(b) TK = TC + 273.15. It is not possible to have TK = TC . Reflect: Any Celsius temperature is less than the corresponding Kelvin temperature.

14.6. Set Up: Convert TK to TC and then convert TC to TF . TK = TC + 273.15 and TF = 95 TC + 32D. Solve: (a) TC = 400 − 273.15 = 127°C, TF = (9/5)(126.85) + 32 = 260°F (b) TC = 95 − 273.15 = 2178°C, TF = (9/5)( −178.15) + 32 = 2289°F (c) TC = 1.55 × 107 − 273.15 = 1.55 × 107°C, TF = (9/5)(1.55 × 107 ) + 32 = 2.79 × 107°F Reflect: All temperatures on the Kelvin scale are positive. TC is negative if the temperature is below the freezing

point of water.

*14.7. Set Up: (a) T0 = − 8.0°C and T = 40.0°C. L0 = 984 ft and we are asked to solve for ΔL. From Table 14.1, the coefficient of linear expansion is 1.2 × 10−5 (C°) −1. (b) Use 9 F° = 5 C°. Solve: (a) Δ L = α L0ΔT = [1.2 × 10−5 (C°)−1 ][984 ft][40.0°C − ( −8.0°C)] = 0.57 ft = 6.8 in. (b) α = [1.2 × 10−5 (C°)−1 ][5 C°/9 F°] = 6.7 × 10−6 (F°)−1 Reflect: In (a) note that the fractional change in length, ΔL/L0 , is very small. In (b), α is smaller in (F°) −1 than in

(C°) −1. A Fahrenheit degree is smaller than a Celsius degree and a temperature difference is a larger number when expressed in F° than in C°. So, to give the same α ΔT value, α is smaller in (F°) −1.

14.8. Set Up: The coefficient of linear expansion for steel is α = 1.2 × 10−5 (C°) −1. Solve: L = L(1 + α ΔT ) = (80.00 m)(1 + [1.2 × 10−5 (C°) −1][−12.0°C − 35.0°C]) = 79.95 m.

14.9. Set Up: Let L0 = 40.125 cm and T0 = 20.0°C. Solve: L = L0 (1 + α ΔT ). α=

L − L0 40.148 cm − 40.125 cm = = 2.3 × 1025 (C°)21 L0ΔT (40.125 cm)(25.0 C°)

14.10. Set Up: For steel, α = 1.2 × 1025 (C°)21 and Y = 2.0 × 1011 Pa. Solve: (a) ΔL = L0α ΔT = (12.0 m)(1.2 × 1025 (C°)21 )(35.0 C°) = 5.0 mm (b) stress = − Yα ΔT = − (2.0 × 1011 Pa)(1.2 × 10−5 (C°)−1 )(35.0 C°) = − 8.4 × 107 Pa. The minus sign means the stress

is compressive.

*14.11. Set Up: For ethanol, β = 75 × 1025 (C°)21. Solve: ΔV = V0 β ΔT = (1700 L)(75 × 1025 (C°)21)(−9.0 C°) = − 11.5 L. The air space will have volume 11 L.

14.12. Set Up: For copper, β = 5.1 × 10−5 (C°)−1. ΔV /V0 = 0.150 × 1022. Solve: ΔV = V0 β ΔT so ΔT =

ΔV/V0

β

=

0.150 × 10−2 5.1 × 10−5 (C°)21

= 29.4 C°. Tf = Ti + ΔT = 49.4°C.


Temperature and Heat

14-3

*14.13. Set Up: ΔV = βV0ΔT . Use the diameter at −15°C to calculate the value of V0 at that temperature. For a hemisphere of radius R, the volume is V = 23 π R3 . Table 17.2 gives β = 7.2 × 10−5 (°C)−1 for aluminum. Solve: V0 = 23 π R3 = 23 π (27.5 m)3 = 4.356 × 104 m3 .

ΔV = (7.2 × 10−5 (°C) −1)(4.356 × 104 m3 )(35°C − [−15°C]) = 160 m3 Reflect: We could also calculate R = R0 (1 + αΔT ) and calculate the new V from R. The increase in volume is V − V0 , but we would have to be careful to avoid round-off errors when two large volumes of nearly the same size are subtracted.

14.14. Set Up: Find the change ΔL in the diameter of the lid. The diameter of the lid expands according to Eq. (14.4). Assume iron has the same α as steel, so α = 1.2 × 10−5 (°C) −1. Solve: ΔL = α L0ΔT = (1.2 × 10−5 (C°) −1)(725 mm)(30.0 C°) = 0.26 mm Reflect: In Eq. (14.4), ΔL has the same units as L.

*14.15. Set Up: For mercury, β Hg = 18 × 10−5 (C°)−1. When heated, both the volume of the flask and the volume of the mercury increase. 8.95 cm3 of mercury overflows, so ΔVHg − ΔVglass = 8.95 cm3 . Solve: ΔVHg = V0 β Hg ΔT = (1000.00 cm3 )(18 × 10−5 (C°)−1 )(55.0 C°) = 9.9 cm3 .

ΔVglass = ΔVHg − 8.95 cm3 = 0.95 cm3 . ΔVglass

0.95 cm3

= 1.7 × 10−5 (C°) −1. (1000.00 cm3 )(55.0 C°) Reflect: The coefficient of volume expansion for the mercury is larger than for glass. When they are heated, both the volume of the mercury and the inside volume of the flask increase. But the increase for the mercury is greater and it no longer all fits inside the flask.

β glass =

V0 ΔT

=

14.16. Set Up: Apply L = L0 (1 + αΔT ) to the diameter d of the rivet. For aluminum, α = 2.4 × 10−5 (°C)−1. Let d 0 be the diameter at –78.0°C and d be the diameter at 23.0°C. Solve: d = d 0 + Δd = d 0 (1 + αΔT ) = (0.4500 cm)(1 + (2.4 × 1025 (C°)−1 )(23.0°C − [−78.0°C])). d = 0.4511 cm = 4.511 mm. Reflect: We could have let d 0 be the diameter at 23.0°C and d be the diameter at −78.0°C. Then ΔT = −78.0D C −23.0°C.

14.17. Set Up: L0 = 20.0 cm at 0.00°C. For aluminum, α al = 2.4 × 10−5 (C°) −1. For brass, α br = 2.0 × 1025 (C°)21. The distance between the marks at 100.0°C is Lal − Lbr . Solve: Lal = L0 (1 + α al ΔT ). Lbr = L0 (1 + α br ΔT ).

Lal − Lbr = L0 (α al − α br ) ΔT = (20.0 cm)(0.4 × 10−5 (C°) −1)(100.0 C°) = 8.0 × 1023 cm. Reflect: This section of both rulers increases in length, but the aluminum one increases more since its coefficient of linear expansion is greater.

14.18. Set Up: For water, c = 4.19 × 103 J/kg ⋅ K. For mercury, c = 0.14 × 103 J/kg ⋅ K. Solve: (a) Q = mc ΔT = (0.250 kg)(4.19 × 103 J/kg ⋅ K)(10.0 C°) = 1.05 × 104 J (b) ΔT =

1.05 × 104 J Q = = 300 C°. Tf = Ti + 300 C° = 320°C. mc (0.250 kg)(0.14 × 103 J/kg ⋅ K)


14-4

Chapter 14

*14.19. Set Up: For aluminum, ca = 0.91 × 103 J/kg ⋅ K. For iron, ci = 0.47 × 103 J/kg ⋅ K. Solve: Q = ma ca ΔT + mici ΔT

Q = [(1.60 kg)(0.91 × 103 J/kg ⋅ K) + (0.300 kg)(0.47 × 103 J/kg ⋅ K)][190 C°] = 3.03 × 105 J

14.20. Set Up: 1 W = 1 J/s. For water, c = 4.19 × 103 J/kg ⋅ K. Solve: (a) Q = mcΔT = (0.320 kg)(4.19 × 103 J/kg ⋅ K)(60.0 C°) = 8.04 × 104 J (b) t =

8.04 × 104 J = 402 s = 6.7 min 200.0 J/s

*14.21. Set Up: 0.50 L of air has mass 0.65 × 10−3 kg. Solve: (a) Q = mcΔT = (0.65 × 10−3 kg)(1020 J/kg ⋅ K)(57 C°) = 38 J (b) Q = (38 J/breath)(20 breaths/min)(60 min/h) = 4.6 × 104 J/h. Reflect: Air has a small density, so one liter of air has little mass. But the specific heat capacity of air is rather large, larger than for most metals and about one-fourth the value for water.

14.22. Set Up: K = 12 mυ 2 . For aluminum, c = 0.91 × 103 J/kg ⋅ K. Solve: The kinetic energy of the hammer before it strikes the nail is K = 12 mυ 2 = 12 (1.80 kg)(7.80 m/s)2 = 54.8 J.

Each strike of the hammer transfers 0.60(54.8 J) = 32.9 J, and with 10 strikes Q = 329 J. Q = mcΔT and ΔT =

Q 329 J = = 45.2 C° mc (8.00 × 10−3 kg)(0.91 × 103 J/kg ⋅ K)

14.23. Set Up: Apply Q = mcΔT . m = w/g . The temperature change is ΔT = 18.0 K. Q gQ (9.80 m/s 2 )(1.25 × 104 J) = = = 240 J/kg ⋅ K. (28.4 N)(18.0 K) mΔT wΔT Reflect: The value for c is similar to that for silver in Table 14.3, so it is a reasonable result. Solve: c =

14.24. Set Up: The work done by the brakes equals the initial kinetic energy of the train. Use the volume of the air to calculate its mass. Use Q = mcΔT applied to the air to calculate ΔT for the air. K = 12 mv 2 . m = ρV . Solve: The initial kinetic energy of the train is K = 12 (25,000 kg)(15.5 m/s) 2 = 3.00 × 106 J. Therefore, Q for the air

is 3.00 × 106 J. m = ρV = (1.20 kg/m3 )(65.0 m)(20.0 m)(12.0 m) = 1.87 × 104 kg. Q = mcΔT gives Q 3.00 × 106 J = = 0.157°C. mc (1.87 × 104 kg)(1020 J/kg ⋅ K) Reflect: The mass of air in the station is comparable to the mass of the train and the temperature rise is small. ΔT =

*14.25. Set Up: Q = mcΔT . The mass of n moles is m = nM . For iron, M = 55.845 × 10−3 kg/mol and c = 470 J/kg ⋅ K. Solve: (a) The mass of 3.00 mol is m = nM = (3.00 mol)(55.845 × 10−3 kg/mol) = 0.1675 kg. ΔT = Q/mc = (8950 J) /[(0.1675 kg)(470 J/kg ⋅ K)] = 114 K = 114 C°. (b) For m = 3.00 kg, ΔT = Q /mc = 6.35 C°. Reflect: (c) The result of part (a) is much larger; 3.00 kg is more material than 3.00 mol.

14.26. Set Up: The initial mechanical energy of the bird is U grav = mbird gh. For water, c = 4.19 × 103 J/kg ⋅ C°. Solve: U grav = Q so mbird gh = mwater c ΔT .

ΔT =

mbird gh (1.25 kg)(9.80 m/s 2 )(35.0 m) = = 2.05 × 10−3 C° mwater c (50.0 kg)(4.19 × 103 J/kg ⋅ C°)



14-5

*14.27. Set Up: Set the loss of kinetic energy of the bullet equal to the heat energy Q transferred to the water. From Table 14.3, the specific heat capacity of water is 4.19 × 103 J/kg ⋅ C°. Solve: The kinetic energy lost by the bullet is Ki − K f = 12 m(υi 2 − υf 2 ) = 12 (15.0 × 10−3 kg)([865 m/s]2 − [534 m/s]2 ) = 3.47 × 103 J, so for the water Q = 3.47 × 103 J. Q = mc ΔT gives Q 3.47 × 103 J = = 0.0613 C° mc (13.5 kg)(4.19 × 103 J/kg ⋅ C°) Reflect: The heat energy required to change the temperature of ordinary size objects is very large compared to the typical kinetic energies of moving objects. ΔT =

14.28. Set Up: 1 W = 1 J/s Solve: Find Q to raise the body temperature from 37°C to 44°C.

Q = mc ΔT = (70 kg)(3480 J/kg ⋅ K)(7 C°) = 1.7 × 106 J. t=

1.7 × 106 J = 1400 s = 23 min. 1200 J/s

14.29. Set Up: 1 W = 1 J/s Solve: (a) Q = (65.0 W)(120 s) = 7.8 × 103 J. Q = mc ΔT and

c=

Q 7.8 × 103 J = = 2.51 × 103 J/kg ⋅ K. m ΔT (0.780 kg)(3.99 C°)

(b) If some of the input energy doesn’t go into the liquid the Q for the calculation of c should be smaller and the correct c would be smaller than the value we calculated in (a), The result in (a) is an overestimate. Reflect: The specific heat capacity of the liquid is about 60% of the value for water.

14.30. Set Up: For water, c = 4.19 × 103 J/kg ⋅ K. The amount of mechanical energy that is converted to heat is mg Δy, where m is the mass of water that es over the waterfall and Δy = 100 m. Solve: Q = mg Δy and mc ΔT = mg Δy. ΔT =

g Δy (9.80 m/s 2 )(100 m) = = 0.234 C°. c 4.19 × 103 J/kg ⋅ K

*14.31. Set Up: From Table 14.4, for lead Lf = 24.5 × 103 J/kg and Lv = 871 × 103 J/kg and for water Lf = 334 × 103 J/kg. The temperatures given in the problem are the transition temperatures for each respective phase transition, so Q = mL. Solve: (a) Q = mLf = (0.150 kg)(24.5 × 103 J/kg) = 3.68 × 103 J (b) Q = mLv = (0.150 kg)(871 × 103 J/kg) = 1.31 × 105 J (c) Q = 3.68 × 103 J + 1.31 × 105 J = 1.35 × 105 J. Q = mLf so

m=

Q 1.35 × 105 J = = 0.404 kg Lf 334 × 103 J

14.32. Set Up: As the iron is cooled, the water is first warmed to100D C and then vaporized. The relevant equations are Q = mcΔT and Q = Lv m. The specific heat of iron is ciron = 0.47 × 103 J/(kg ⋅ K), the specific heat of water is cwater = 4.19 × 103 J/ (kg ⋅ K), and the heat of vaporization of water is Lv = 2256 × 103 J/kg.


14-6

Chapter 14

Solve: The iron cools: Qiron = mici ΔTi . The water warms and vaporizes: Qwater = cw mw ΔTw + mw Lv w =

mw (cw ΔTw + Lvw ). Assume that all of the heat lost by the iron is gained by the water so that Qwater = −Qiron . Equating the respective expressions for each Q and solving for mw we obtain mw =

− mici ΔTi −(1.20 kg)(0.47 × 103 J/kg ⋅ K)(120.0D C − 650.0D C) = = 0.114 kg cw ΔTw + Lv w (4.19 × 103 J/kg ⋅ K)(100.0D C − 15.0D C) + 2256 × 103 J/kg

Reflect: Note that only a relatively small amount of water is required to cause a very large temperature change in the iron. This is due to the high heat of vaporization and specific heat of water, and the relatively low specific heat capacity of iron.

14.33. Set Up: The heat that comes out of the person goes into the ice-water bath and causes some of the ice to melt. Normal body temperature is 98.6° F = 37.0°C, so for the person ΔT = − 5 C°. The ice-water bath stays at 0°C. A mass m of ice melts and Qice = mLf . From Table 14.4, for water Lf = 334 × 103 J/kg. Solve: Qperson = mc ΔT = (70.0 kg)(3480 J/kg ⋅ C°)(−5.0 C°) = − 1.22 × 106 J

Therefore, the amount of heat that goes into the ice is 1.22 × 106 J. mice Lf = 1.22 × 106 J and mice =

1.22 × 106 J 334 × 103 J/kg

= 3.7 kg.

Reflect: If less ice than this is used, all the ice melts and the temperature of the water in the bath rises above 0°C.

14.34. Set Up: For ice, c = 2.01 × 103 J/kg ⋅ K. For water, Lf = 3.34 × 105 J/kg. The system stays at 0.00°C until all the ice has melted. Solve: (a) Q to raise the temperature of ice to 0.00°C: Q = mc ΔT = (0.550 kg)(2.01 × 103 J/kg ⋅ K)(15.0 C°) = 1.66 × 104 J.

t=

1.66 × 104 J = 20.8 min. 800.0 J/min

(b) To melt all the ice requires Q = mLf = (0.550 kg)(3.34 × 105 J/kg) = 1.84 × 105 J.

1.84 × 105 J = 230 min. 800.0 J/min The total time after the start of the heating is 251 min. (c) A graph of T versus t is sketched in the figure below. t=

*14.35. Set Up: The asteroid’s kinetic energy is K = 12 mv 2 . To boil the water, its temperature must be raised to 100.0D C and the heat needed for the phase change must be added to the water. For water, c = 4190 J/kg ⋅ K and Lv = 2256 × 103 J/kg. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


14-7

Solve: K = 12 (2.60 × 1015 kg)(32.0 × 103 m/s) 2 = 1.33 × 1024 J. Q = mcΔT + mLv .

Q 1.33 × 1022 J = = 5.05 × 1015 kg. cΔT + Lv (4190 J/kg ⋅ K)(90.0 K) + 2256 × 103 J/kg Reflect: The mass of water boiled is 2.5 times the mass of water in Lake Superior. m=

14.36. Set Up: The density of water is 1000 kg/m3 . Solve: (a) The heat that goes into mass m of water to evaporate it is Q = + mLv . The heat flow for the man is

Q = mman c ΔT , where ΔT = − 1.00 C°. ∑ Q = 0 so mLv + mman c ΔT and m=−

mman c ΔT (70.0 kg)(3480 J/kg ⋅ K)(−1.00 C°) =− = 0.101 kg = 101 g. Lv 2.42 × 106 J/kg

0.101 kg = 1.01 × 10−4 m3 = 101 cm3 . 1000 kg/m3 This is about 28% of the volume of a soft-drink can.

(b) V =

m

ρ

=

14.37. Set Up: For water, c = 4.19 × 103 J/kg ⋅ K and Lf = 3.34 × 105 J/kg. Solve: Q = mc ΔT − mLf = (0.350 kg)([4.19 × 103 J/kg ⋅ K][ −18.0 C°] − 3.34 × 105 J/kg) = −1.43 × 105 J. The minus

sign says 1.43 × 105 J must be removed from the water. ⎛ 1 cal ⎞ 4 (1.43 × 105 J) ⎜ ⎟ = 3.42 × 10 cal = 34.2 kcal. ⎝ 4.186 J ⎠ Reflect: Q < 0 when heat comes out of an object the equation Q = mc ΔT puts in the correct sign automatically,

from the sign of ΔT = Tf − Ti . But in Q = ± mL we must select the correct sign.

14.38. Set Up: For ice, c = 2.01 × 103 J/kg ⋅ K. For water, c = 4.19 × 103 J/kg ⋅ K, Lf = 3.34 × 105 J/kg and Lv = 2.256 × 106 J/kg. Solve: Warm the ice to 0.0°C, melt the ice, heat the water from 0.0°C to 100.0°C, and boil the water. Q = m[cice ΔTice + Lf + mcw ΔTw + Lv ]. Q = (12.0 × 10−3 kg)([2.01 × 103 J/kg ⋅ K][10.0 C°] + 3.34 × 105 J/kg

+ [4.19 × 103 J/kg ⋅ K][100 C°] + 2.256 × 106 J/kg) = 3.63 × 104 J = 8.67 kcal.

*14.39. Set Up: For water, Lv = 2.256 × 106 J/kg and c = 4.19 × 103 J/kg ⋅ K. Solve: (a) Q = + m(− Lv + c ΔT )

Q = + (25.0 × 10−3 kg)(−2.256 × 106 J/kg + [4.19 × 103 J/kg ⋅ K][−66.0 C°]) = − 6.33 × 104 J (b) Q = mc ΔT = (25.0 × 10−3 kg)(4.19 × 103 J/kg ⋅ K)( −66.0 C°) = − 6.91 × 103 J. (c) The total heat released by the water that starts as steam is nearly a factor of ten larger than the heat released by water that starts at 100°C. Steam burns are much more severe than hot-water burns. Reflect: For a given amount of material, the heat for a phase change is typically much more than the heat for a temperature change.

14.40. Set Up: Energy is (power)(time); calculate the heat energy Q produced in one hour. The mass m of water that vaporizes is related to Q by Q = mLv . 1.0 kg of water has a volume of 1.0 L. Q 1.44 × 106 J = = 0.60 kg each hour. Lv 2.42 × 106 J/kg (b) (0.60 kg/h)(1.0 L/kg) = 0.60 L/h. The number of bottles of water is Solve: (a) Q = (0.80)(500 W)(3600 s) = 1.44 × 106 J. m =

0.60 L/h = 0.80 bottles/h. 0.750 L/bottle © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

14-8

Chapter 14

*14.41. Set Up: From Problem 14.40, Q = 1.44 × 106 J and m = 70 kg. Convert the temperature change in C° to F° using that 9 F° = 5 C°. Solve: (a) Q = mc ΔT so ΔT =

Q 1.44 × 106 J = = 5.9 C° mc (70 kg)(3500 J/kg ⋅ C°)

⎛ 9 F° ⎞ (b) ΔT = (5.9°C) ⎜ ⎟ = 10.6°F. T = 98.6°F + 10.6 F° = 109°F. A fever this high can be lethal (heat stroke). ⎝ 5 C° ⎠

14.42. Set Up: Calculate Q for each mass of water and set their algebraic sum equal to zero. Let the water you add have mass m. The 750 g of cold water has a temperature change of +65 C° and a heat flow Qc . The mass m of water has a temperature change of −25 C° and a heat flow Qh . Solve: Qc = mc ΔT = (0.750 kg)(4190 J/kg ⋅ C°)(65 C°) = 2.043 × 105 J

Qh = mc ΔT = m(4190 J/kg ⋅ C°)( −25 C°) = − (1.048 × 105 J/kg)m Qh + Qc = 0 so 2.043 × 105 J + (−1.048 × 105 J/kg)m = 0 and m = 1.95 kg = 1950 g

*14.43. Set Up: For water, c = 4.19 × 103 J/kg ⋅ K. Solve: (a) Qwater = mc ΔT = (1.00 kg)(4.19 × 103 J/kg ⋅ K)(2.0 C°) = + 8.38 × 103 J.

Qmetal = mc ΔT = (0.500 kg)c (−78 C°). ∑ Q = 0 says (0.500 kg)c (−78 C°) + 8.38 × 103 J = 0. 8.38 × 103 J = 215 J/kg ⋅ K. (0.500 kg)(78 C°) (b) Water has a much larger value of c so stores more heat for the same ΔT . (c) If some of the heat went into the Styrofoam™ then more would have to come out of the metal and c would be found to be greater. The value calculated in (a) would be too small. Reflect: The amount of heat that comes out of the metal when it cools equals the amount of heat that goes into the water to produce its temperature increase.

c=

14.44. Set Up: For copper, cc = 390 J/kg ⋅ K. For iron, ci = 470 J/kg ⋅ K. For water, cw = 4.19 × 103 J/kg ⋅ K. Solve: For the copper pot, Qc = mccc ΔTc = (0.500 kg)(390 J/kg ⋅ K)(T − 20.0°C) = (195 J/K)T − 3900 J.

For the block of iron, Qi = mi ci ΔTi = (0.250 kg)(470 J/kg ⋅ K)(T − 85.0°C) = (117.5 J/K)T − 9988 J. For the water, Qw = mw cw ΔTw = (0.170 kg)(4190 J/kg ⋅ K)(T − 20.0°C) = (712.3 J/K)T − 1.425 × 104 J

∑ Q = 0 gives (195 J/K)T − 3900 J + (117.5 J/K)T − 9988 J + (712.3 J/K)T − 1.425 × 104 J.

T=

2.814 × 104 J = 27.5°C. 1025 J/K

14.45. Set Up: The total mass of the coins is mc = 0.600 kg. For water, cw = 4.19 × 103 J/kg ⋅ K. Solve: For the coins, Qc = mccc ΔTc = (0.600 kg)(390 J/kg ⋅ K)(T − 100.0°C) = (234 J/K)T − 2.34 × 104 J.

For the water,

Qw = mw cw ΔTw = (0.240 kg)(4190 J/kg ⋅ K)(T − 20.0°C) = (1006 J/K)T − 2.01 × 104 J. ∑ Q = 0 gives (234 J/K)T − 2.34 × 104 J + (1006 J/K)T − 2.01 × 104 J = 0.

T=

4.35 × 104 J = 35.1°C. 1240 J/K



14-9

Reflect: We assumed that the water remained liquid. This is a correct assumption since the final temperature we calculated is in the range 0°C < T < 100°C.

14.46. Set Up: For water, cw = 4.19 × 103 J/kg ⋅ K. For copper, cc = 390 J/kg ⋅ K. Solve: For the water, Qw = mw cw ΔTw = (0.200 kg)(4.19 × 103 J/kg ⋅ K)(7.1 C°) = 5.95 × 103 J.

For the copper can, Qc = mccc ΔTc = (0.150 kg)(390 J/kg ⋅ K)(7.1 C°) = 415 J. For the sample, Qs = mscs ΔTs = (0.085 kg)cs ( −73.9 C°).

∑ Q = 0 gives (0.085 kg)( −73.9 C°)cs + 415 J + 5.95 × 103 J = 0. cs = 1.01 × 103 J/kg ⋅ K.

*14.47. Set Up: Since some ice remains, the ice and water from the melted ice remain at 0°C. For silver, cs = 230 J/kg ⋅ K. For water, Lf = 3.34 × 105 J/kg. Solve: For the silver, Qs = mscs ΔTs = (4.00 kg)(230 J/kg ⋅ K)( −750 C°) = − 6.90 × 105 J.

For the ice, Qi = + mLf , where m is the mass that melts.

∑ Q = 0 gives m(3.34 × 105 J/kg) − 6.90 × 105 J = 0 and m = 2.07 kg. Reflect: The heat that comes out of the ingot when it cools goes into the ice to produce a phase change.

14.48. Set Up: For ice, ci = 2.01 × 103 J/kg ⋅ K. For water, cw = 4.19 × 103 J/kg ⋅ K and Lf = 3.34 × 105 J/kg. Solve: For the water,

Qw = mw cw ΔTw = (0.250 kg)(4.19 × 103 J/kg ⋅ K)(−45.0 C°) = − 4.71 × 104 J. For the ice, Qi = mi (ci ΔTi + Lf + cw ΔTw ) Qi = ([2.01 × 103 J/kg ⋅ K][ 20.0 C°] + 3.34 × 105 J/kg + [4.19 × 103 J/kg ⋅ K ][30.0 C°]) = mi (5.00 × 105 J/kg)

∑ Q = 0 gives mi (5.00 × 105 J/kg) − 4.71 × 104 J = 0 and mi = 0.0942 kg = 94.2 g.

*14.49. Set Up: The final amount of ice is less than the initial mass of water, so water remains and the final temperature is 0°C. The ice added warms to 0°C and heat comes out of water to convert it to ice. Conservation of energy says Qi + Qw = 0, where Qi and Qw are the heat flows for the ice that is added and for the water that freezes. Let mi be the mass of ice that is added and mw is the mass of water that freezes. The mass of ice increases by 0.328 kg, so mi + mw = 0.328 kg. For water, Lf = 334 × 103 J/kg and for ice ci = 2100 J/kg ⋅ K. Heat comes out of the water when it freezes, so Qw = 2mLf Solve: Qi + Qw = 0 gives mici (15.0D C) + (− mw Lf ) = 0, mw = 0.328 kg − mi , so mici (15.0° C) + (−0.328kg + mi ) Lf = 0.

mi =

(0.328 kg) Lf

ci (15.0D C) + Lf

=

(0.328 kg)(334 × 103 J/kg) (2100 J/kg ⋅ K)(15.0 K) + 334 × 103 J/kg

= 0.300 kg.

0.300 kg of ice was added. Reflect: The mass of water that froze when the ice at −15.0D C was added was 0.778 kg − 0.450 kg − 0.300 kg = 0.028 kg.

14.50. Set Up: 1 day = 86,400 s. Solve: Q =

kA(TH − TC )t (0.075 W/m ⋅ K)(100 × 10−4 m 2 )(80 C°)(86,400 s) = = 1.73 × 105 J L 3.00 × 1022 m


14-10

Chapter 14

*14.51. Set Up: Apply Eq. (14.12) and solve for A. The area of each circular end of a cylinder is related to the diameter D by A = π R 2 = π ( D /2) 2 . For steel we have k = 50.2 W/m ⋅ K. The boiling water has T = 100°C, so ΔT = 300 K. Solve:

ΔT Q ⎛ 300 K ⎞ . This gives A = 4.98 × 10−3 m 2 , and D = 4 A/π = and 150 J/s = (50.2 W/m ⋅ K) A ⎜ = kA ⎝ 0.500 m ⎟⎠ t L

4(4.98 × 10−3 m 2 )/π = 8.0 × 10−2 m = 8.0 cm. Reflect: H increases when A increases.

14.52. Set Up: Apply H = kA

TH − TC and solve for k. L

HL (75 W)(0.75 × 10−3 m) = = 4.0 × 10−3 W/m ⋅ C°. A(TH − TC ) (2.0 m 2 )(37°C − 30.0°C) Reflect: This is a small value; skin is a poor conductor of heat. But the thickness of the skin is small, so the rate of heat conduction through the skin is not small. Solve: k =

14.53. Set Up: For water, Lv = 2.256 × 106 J/kg. For steel, k = 50.2 W/m ⋅ K. Solve: Q = mLv = (0.390 kg)(2.256 × 106 J/kg) = 8.80 × 105 J.

Q 8.80 × 105 J = = 4.89 × 103 W. t 180 s Q kAΔT (Q /t ) L (4.89 × 103 W)(8.50 × 10−3 m) = = = 5.52 C°. TH = TC + ΔT = 105.5°C. . ΔT t L kA (50.2 W/m ⋅ K)(0.150 m 2 )

14.54. Set Up: The heat current Q /t is the same through the wood as through the Styrofoam™. Solve : (a)

Q kA(TH − TC ) k A(T − [ −10.0°C]) ks A(19.0°C − T ) ⎛Q⎞ ⎛Q⎞ = = . and ⎜ ⎟ = ⎜ ⎟ gives w t L t t Lw Ls ⎝ ⎠ w ⎝ ⎠s

(0.080 W/m ⋅ K)(T + 10.0°C) (0.010 W/m ⋅ K)(19.0°C − T ) = . 0.030 m 0.022 m 26.7°C − 8.65°C 2.67(T + 10.0°C) = 0.455(19.0°C − T ) and T = = − 5.8°C. 23.125 k (TH − TC ) (0.080 W/m ⋅ K)(−5.8°C + 10.0°C) ⎛Q⎞ (b) ⎜ ⎟ = = = 11 W/m 2 0.030 m L ⎝ tA ⎠ w

⎛ Q ⎞ (0.010 W/m ⋅ K)(19.0°C − [ −5.8°C]) = 11 W/m 2 , which checks. Or, ⎜ ⎟ = . 0 022 m tA ⎝ ⎠s Reflect: k is much smaller for the Styrofoam™ so the temperature gradient across it is much larger than across the wood.

*14.55. Set Up: Assume the temperatures of the surfaces of the window are the outside and inside temperatures. Use the concept of thermal resistance. For part (b) use the fact that when insulating materials are in layers, the R values are additive. From Table 14.5 we obtain k = 0.8 W/m ⋅ K for glass. R = L /k . 5.20 × 10−3 m = 6.50 × 10−3 m 2 ⋅ K/W. 0.8 W/m ⋅ K A(TH − TC ) (1.40 m)(2.50 m)(39.5 K) = H= = 2.1 × 104 W 2 −3 R 6.50 × 10 m ⋅ K/W

Solve: (a) For the glass, Rglass =

0.750 × 10−3 m = 0.015 m 2 ⋅ K/W. The total R is R = Rglass + Rpaper = 0.0215 m 2 ⋅ K/W. 0.05 W/m ⋅ K A(TH − TC ) (1.40 m)(2.50 m)(39.5 K) = = 6.4 × 103 W. H= R 0.0215 m 2 ⋅ K/W Reflect: The layer of paper decreases the rate of heat loss by a factor of about 3. (b) For the paper, Rpaper =



14-11

14.56. Set Up: For water, Lf = 3.34 × 105 J/kg. Solve: The heat conducted by the rod in 10.0 min is Q = mLf = (8.50 × 10−3 kg)(3.34 × 105 J/kg) = 2.84 × 103 J. Q 2.84 × 103 J = = 4.73 W. 600 s t Q kA(TH − TC ) = . t L (Q/t ) L (4.73 W)(0.600 m) k= = = 227 W/m ⋅ K. A(TH − TC ) (1.25 × 10−4 m 2 )(100 C°)

Reflect: The heat conducted by the rod is the heat that enters the ice and produces the phase change.

*14.57. Set Up: Let the temperature of the fat-air boundary be T. A section of the two layers is sketched in the figure below. A Kelvin degree is the same size as a Celsius degree, so W/m ⋅ K and W/m ⋅ C° are equivalent units. At steady state the heat current through each layer is the same, equal to 50 W. The area of each layer is A = 4π r 2 , with r = 0.75 m.

Solve: (a) Apply H = kA

TH − TC to the fat layer and solve for TC = T . For the fat layer TH = 31°C. L

T = TH −

(b) Apply H = kA

HL (50 W)(4.0 × 10−2 m) = 31°C − = 31°C − 1.4°C = 29.6°C kA (0.20 W)(4π )(0.75 m)2

TH − TC to the air layer and solve for L = Lair . For the air layer TH = T = 29.6°C and L

TC = 2.7°C. L=

kA(TH − TC ) (0.024 W)(4π )(0.75 m) 2 (29.6°C − 2.7°C) = = 9.1 cm 50 W H

Reflect: The thermal conductivity of air is much lass than the thermal conductivity of fat, so the temperature gradient for the air must be much larger to achieve the same heat current. So, most of the temperature difference is across the air layer.

14.58. Set Up: We have H = eσ AT 4 . The emissivity of the stove is e = 0.85. The surface temperature of the stove is 205° C = 478 K. Solve: The stove has 6 rectangular sides which gives a total surface area of A = 2(lw + lh + wh) = 2[(0.75 m)(1.2 m) + (0.75 m)(0.40 m) + (1.2 m)(0.40 m)] = 3.36 m 2 . Thermal radiation from the stove is emitted at a rate of H = eσΑT 4 = (0.85)(5.67 × 10−8 W/m 2 ⋅ K 4 )(3.36 m 2 )(478 K) 4 = 8.5 × 103 W = 8.5 kW.

Reflect: The thermal emission has a strong temperature dependence. For example, a 5% increase in the kelvin temperature would cause a 20% increase in the thermal emission ( 1.054 = 1.2 ).


14-12

Chapter 14

14.59. Set Up: (a) M = 75 and h = 1.83. (b) and (c) e = 1.00. In H = Aeσ T 4 , T must be in kelvins. T = 30°C = 303 K. Ts = 20°C = 293 K.

Solve: (a) A = (0.202)(75)0.425 (1.83)0.725 = 1.96 m 2 (b) H = Aeσ T 4 = (1.96 m 2 )(1.00)(5.67 × 10−8 W/m 2 ⋅ K 4 )(303) 4 = 937 W (c) H net = Aeσ (T 4 − Ts 4 ) H net = (1.96 m 2 )(1.00)(5.67 × 10−8 W/m 2 ⋅ K 4 )([303 K]4 − [293 K]4 ) = 118 W

(d) The surface area of a sphere is 4π r 2 ; use this for your head. Treat each arm, each leg, and your trunk as cylinders; the

surface area of a cylinder of radius r and length l is 2π r 2 + 2π rl . Should you include the π r 2 area of each end of the cylinders? Reflect: Since the radiation heat current is proportional to T 4 , the net rate of heat loss increases rapidly when the air temperature decreases.

14.60. Set Up: The surface area of a sphere of radius r is A = 4π r 2 . Apply H = Aeσ T 4 and solve for A. Solve: H = Aeσ T 4 gives A= A = 4π r 2 gives r =

H eσ T 4

=

3.92 × 1026 W (1.00)(5.67 × 1028 W/m 2 ⋅ K 4 )(5800 K)4

= 6.11 × 1018 m.

A/4π = 6.97 × 108 m. The diameter is d = 2r = 1.39 × 109 m. The radius we calculated is in

excellent agreement with the value given in Appendix F.

*14.61. Set Up: In the radiation equation the temperatures must be in kelvins; T = 30°C = 303 K and Ts = 18°C = 291 K. Call the basal metabolic rate BMR. Solve: (a) H net = Aeσ (T 4 − Ts 4 ) H net = (2.0 m 2 )(1.00)(5.67 × 1028 W/m 2 ⋅ K 4 )([303 K]4 − [291 K]4 ) = 140 W (b) (0.80)BMR = 140 W, so BMR = 180 W

14.62. Set Up: The surface area of a sphere is 4π r 2 . Solve: H = Aeσ (T 4 − Ts 4 ) = 4π (0.0150 m) 2 (0.35)(5.67 × 10−8 W/m 2 ⋅ K 4 )([3000 K]4 − [290 K]4 ) = 4.54 × 104 W

14.63. Set Up and Solve: H = Aeσ T 4 . H

150 W

= 2.10 cm 2 . (0.35)(5.67 × 10 W/m 2 ⋅ K 4 )(2450 K) 4 eσ T Reflect: Even though the filament is small, its coil shape gives it a relatively large surface area. A=

4

=

28

14.64. Set Up: Assume that the walls of the coffee pot are of negligible thickness so that the surface temperature of the pot will be nearly the same as the water it contains. We can find the surface area of the pot from its known volume of 0.75 = 7.5 × 10−4 m3. We know that H net = eσ A(T 4 − Ts 4 ) where T = 95D = 368 K and Ts = 20.0D = 293 K. Finally, 4 we know that A = 4π R 2 and V = π R3. 3 Solve: First find the radius of the pot from its volume: 1/ 3

⎛ 3V ⎞ R=⎜ ⎟ ⎝ 4π ⎠

1/ 3

⎛ 3(7.5 × 10−4 m3 ) ⎞ =⎜ ⎟⎟ ⎜ 4π ⎝ ⎠

= 0.05636 m.

Next find the surface area of the pot: A = 4π R 2 = 4π (0.05636 m)2 = 0.0399 m 2 .



14-13

Finally, we determine the rate of thermal emission: H net = eσ A(T 4 − Ts 4 ) = (0.60)(5.67 × 1028 W/m 2 ⋅ K 4 )(0.0399 m 2 )[(368 K) 4 − (293 K) 4 ] = 15 W Reflect: According to Eq. (14.12), there must be a temperature difference between the outside and inside surface of the pot for heat conduction to occur. If the thickness of the pot is small, this temperature difference is also small and we can assume that the temperature of the two surfaces is the same. *14.65. Set Up: The heat energy generated by friction work equals the loss of kinetic energy. The heat Qice that

goes into the ice is related to the mass mmelts of ice that melts by Qice = mmelts Lf . For ice, Lf = 334 × 103 J/kg. The maximum amount of ice melts when the block loses all its kinetic energy and comes to rest. Solve: (a) For υi = 15.0 m/s and υf = 10.0 m/s, Q = Ki − K f = 12 m(υ i 2 − υ f 2 ) = 12 (8.5 kg)([15.0 m/s]2 − [10.0 m/s]2 ) = 531 J. Qice = 12 Q = 265 J. Then Qice = mLf gives m=

Qice 265 J = = 7.93 × 1024 kg = 0.793 g Lf 334 × 103 J/kg

Qice = 1.43 g Lf Reflect: Since the ice is at 0°C, any heat that goes into it causes a phase transition. 15.0 m/s is 34 mph. When brought to rest from this large initial speed only a very small fraction of the ice melts. Also note the irreversible nature of this; the water won’t refreeze and release energy to get the block moving again. ΔV/ΔT . Note that ΔV /ΔT is the 14.66. Set Up: Volume expansion: We know that ΔV = βV0 ΔT , so we have β = V0

(b) For υf = 0, Q = Ki = 956 J. Qice = 478 J and m =

slope of the graph of V versus T, which is the graph given in Figure 14.9 in the textbook. Slope of graph Solve: β = . Construct the tangent to the graph at 2°C and 8°C and measure the slope of this line. V0 At 2°C: Slope ≈ −

0.00010 cm3 0.00010 cm3 /3 C° ≈ −3 × 10−5 (C°) −1. Note and V ≈ 1.00 cm3. So we have β ≈ 2 3 C° 1.00 cm3

that the slope is negative, as the water contracts as it is heated. At 8°C: slope ≈

0.00024 cm3 and V ≈ 1.00 cm3. So 4 C°

0.00024 cm3 /4 C°

≈ 6 × 10−5 (C°) −1. The water now expands when heated. 1.00 cm3 Reflect: β > 0 when the material expands when it is heated and β < 0 when the material contracts when it is

we have β ≈

heated. The minimum volume is at about 4D C and β has opposite signs above and below this temperature.

*14.67. Set Up: Write the volume of the oceans as V = Ad , where d is the average depth and A is the average surface area of the oceans. Assume that only d changes when V changes, so ΔV = AΔd . Solve: ΔV = V0 β ΔT , ΔV = AΔd and V0 = Ad 0 , with d 0 = 4000 m. This gives Δd = d 0β ΔT = (4000 m)(0.207 × 10−3 (C°) −1)(3.5 C°) = 2.9 m.

Reflect: This rise in sea level would have serious effects on coastal areas. 14.68. Set Up: Use Eq. (14.4) to find the change in diameter of the sphere and the change in length of the cable. Set the sum of these two increases in length equal to 2.00 mm. α brass = 2.0 × 10−5 K −1 and α steel = 1.2 × 10−5 K −1. Solve: ΔL = (α brass L0,brass + α steel L0,steel )ΔT . ΔT =

2.00 × 10−3 m (2.0 × 10−5 K −1)(0.350 m) + (1.2 × 10−5 K −1 )(10.5 m)

= 15.0 C°.

T2 = T1 + ΔT = 35.0° C.

Reflect: The change in diameter of the brass sphere is 0.10 mm. This is small, but should not be neglected. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

14-14

Chapter 14

*14.69. Set Up: 1.00 L of water has a mass of 1.00 kg, so

9.46 L/min = (9.46 L/min)(1.00 kg/L)(1 min/60 s) = 0.158 kg/s. For water, c = 4190 J/kg ⋅ C°. Solve: Q = mc ΔT so H = (Q/t ) = ( m/t )c ΔT = (0.158 kg/s)(4190 J/kg ⋅ C°)(49°C − 10°C) = 2.6 × 104 W = 26 kW

Reflect: The power requirement is large, the equivalent of 260 light bulbs that are 100-W, but this large power is needed only for short periods of time.

14.70. Set Up: 1 food { calorie = 1 × 103 cal = 4.186 × 103 J. For water, 1 L has a mass of 1 kg. Solve: (a) The energy released when 1 lb of body fat is metabolized is (3500)(4.186 × 103 J) = 1.47 × 107 J. P = t=

Q so t

Q 1.47 × 107 J = = 1.13 × 104 s = 3.2 h. P 1290 J/s

Only a very fit person is capable of this. (b) The energy released is 1.47 × 107 J, so the heat produced is Q = (0.80)(1.47 × 107 J) = 1.18 × 107 J

(c) Q = mLv so m=

Q 1.18 × 107 J = = 4.88 kg, Lv 2.42 × 106 J/kg

so 4.88 L of water.

14.71. Set Up: Find the heat Q to raise the body temperature 1.0 C° and find the time it takes to produce this much heat energy at a rate of (290 W/m 2 )(1.8 m 2 ) = 522 J/s. Solve: (a) Q = mc ΔT = (68 kg)(3500 J/kg)(1.0 C°) = 2.38 × 105 J P=

Q 2.38 × 105 J Q so t = = = 456 s. t 522 J/s P

14.72. (a) Set Up: The diameter of the ring undergoes linear expansion (increases with T) just like a solid steel disk of the same diameter as the hole in the ring. Heat the ring to make its diameter equal to 2.5020 in. ΔL 0.0020 in. Solve: ΔL = α L0 ΔT so ΔT = = = 66.7 C° L0α (2.5000 in.)(1.2 × 10−5 (C°) −1 ) T = T0 + ΔT = 20.0°C + 66.7 C° = 87°C

(b) Set Up: Apply the linear expansion equation to the diameter of the brass shaft and to the diameter of the hole in the steel ring. Solve: L = L0 (1 + αΔT )

Want Ls (steel) = Lb (brass) for the same ΔT for both materials: L0s (1 + α s ΔT ) = L0b (1 + α b ΔT ) so L0s + L0sα s ΔT = L0b + L0bα b ΔT 2.5020 in. − 2.5000 in. L0b − L0s = L0sα s − L0bα b (2.5000 in.)(1.2 × 10−5 (C°) −1) − (2.5050 in.)(2.0 × 10−5 (C°)−1 ) 0.0020 ΔT = C° = −100 C° 3.00 × 10−5 − 5.00 × 10−5 T = T0 + ΔT = 20.0°C − 100 C° = −80°C ΔT =

Reflect: Both diameters decrease when the temperature is lowered but the diameter of the brass shaft decreases more since α b > α s ; ΔLb − ΔLs = 0.0020 in.



14-15

*14.73. Set Up: Let the cross sectional area of the cup be A. For the cup, V0,cup = A(10.0 cm). For the oil, V0,oil = A(9.9 cm). Oil starts to overflow when ΔVoil = ΔVcup + (0.100 cm) A.

Solve: ΔVcup = V0,cupβglass ΔT . ΔVoil = V0,oilβoil ΔT .

(9.9 cm) Aβoil ΔT = (10.0 cm) Aβglass ΔT + (0.100 cm) A. ΔT = ΔT =

0.100 cm (9.9 cm) β oil − (10.0 cm) β glass 0.100 cm

−4

−1

(9.9 cm)(6.8 × 10 (C°) ) − (10.0 cm)(2.7 × 10−5 (C°)−1 )

= 15.5 C°.

Tf = Ti + ΔT = 37.5 °C.

Reflect: The olive oil expands more than the capacity of the cup does because it has a much larger coefficient of volume expansion.

14.74. Set Up: For copper, cc = 390 J/kg ⋅ K. For lead, cl = 130 J/kg ⋅ K. For water, cw = 4.19 × 103 J/kg ⋅ K and Lf = 3.34 × 105 J/kg. The initial temperature of the ice and water mixture is 0.0°C. Assume all the ice melts. We will know that assumption is incorrect if the final temperature we calculate is less than 0.0° C. Solve : For the copper can, Qc = mccc ΔTc = (0.100 kg)(390 J/kg ⋅ K)(T − 0.0°C) = (39.0 J/K)T . For the water, Qw = mw cw ΔTw = (0.160 kg)(4.19 × 103 J/kg ⋅ K)(T − 0.0°C) = (670.4 J/K)T . For the ice, Qi = mi Lf + mi cw ΔTw 5

Qi = (0.018 kg)(3.34 × 10 J/kg ⋅ K) + (0.018 kg)(4.19 × 103 J/kg ⋅ K)(T − 0.0°C) = 6012 J + (75.4 J/K)T For the lead, Ql = mlcl ΔTl = (0.750 kg)(130 J/kg ⋅ K)(T − 255°C) = (97.5 J/K)T − 2.486 × 104 J ∑ Q = 0 gives (39.0 J/K)T + (670.4 J/K)T + 6012 J + (75.4 J/K) + (97.5 J/K)T − 2.486 × 104 J = 0. T=

1.885 × 104 J = 21.4°C. T > 0.0°C, which confirms that all the ice melts. 882.3 J

14.75. Set Up: The magnitude of the gravitational potential energy change, mgh, equals the heat Q that goes into the ice. Solve: (0.0100)mLf = mgh. h=

(0.010) Lf 3.34 × 103 J/kg = = 341 m g 9.80 m/s 2

14.76. Set Up: The mass if air in the room is m = ρV = (1.20 kg/m3 )(3200 m3 ) = 3840 kg. 1 W = 1 J/s.

Solve: (a) Q = (3000 s)(90 students)(100 J/s ⋅ student) = 2.70 × 107 J. (b) Q = mc ΔT . ΔT =

Q 2.70 × 107 J = = 6.89 C° mc (3840 kg)(1020 J/kg ⋅ K)

⎛ 280 W ⎞ (c) ΔT = (6.89 C°) ⎜ ⎟ = 19.3 C°. ⎝ 100 W ⎠

*14.77. Set Up: Use Q = McΔT to find Q for a temperature rise from 34.0D C to 40.0D C. Set this equal to Q = mLv and solve for m, where m is the mass of water the camel would have to drink. c = 3480 J/kg ⋅ K and Lv = 2.42 × 106 J/kg. For water, 1.00 kg has a volume 1.00 L. M = 400 kg is the mass of the camel. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

14-16

Chapter 14

Solve: The mass of water that the camel saves is McΔT (400 kg)(3480 J/kg ⋅ K)(6.0 K) m= = = 3.45 kg which is a volume of 3.45 L. Lv (2.42 × 106 J/kg) Reflect: This is nearly a gallon of water, so it is an appreciable savings.

14.78. Set Up: Heat Ql comes out of the lead when it solidifies and the solid lead cools to Tf . If mass ms of steam is produced, the final temperature is Tf = 100D C and the heat that goes into the water is Qw = mw cw (25.0D C) + ms Lv,w , where mw = 0.5000 kg. Conservation of energy says Ql + Qw = 0. Solve for ms . The mass that remains is 1.250 kg + 0.5000 kg − ms . For lead, Lf ,l = 24.5 × 103 J/kg, cl = 130 J/kg ⋅ K and the normal melting point of lead is 327.3D C. For water, cw = 4190 J/kg ⋅ K and Lv,w = 2256 × 103 J/kg.

Solve: Ql + Qw = 0. − ml Lf,l + mlcl (−227.3D C) + mw cw (25.0D C) + ms Lv,w = 0. ms = ms = ms =

ml Lf,l + mlcl (+227.3D C) − mw cw (25.0D C) Lv,w

.

+ (1.250 kg)(24.5 × 103 J/kg) + (1.250 kg)(130 J/kg ⋅ K)(227.3 K) − (0.5000 kg)(4190 J/kg ⋅ K)(25.0 K) 2256 × 103 J/kg

1.519 × 104 J 2256 × 103 J/kg

= 0.0067 kg. The mass of water and lead that remains is 1.743 kg.

Reflect: The magnitude of heat that comes out of the lead when it goes from liquid at 327.3D C to solid at 100.0D C is 6.76 × 104 J. The heat that goes into the water to warm it to 100D C is 5.24 × 104 J. The additional heat that goes into the water, 6.76 × 104 J − 5.24 × 104 J = 1.52 × 104 J converts 0.0067 kg of water at 100D C to steam.

14.79. Set Up: Neglect the effect of the increase in thickness of the ice on the rate of heat conduction, so H remains constant. For water, Lf = 3.34 × 105 J/kg ⋅ K. Consider an area A of ice. Ice has density ρ = 920 kg/m3 . The thermal conductivity of ice is k = 1.6 W/m ⋅ K.

Solve: The mass of the 2.00 mm thickness of ice is m = ρV = (920 kg/m3 )(2.0 × 10−3 m) A = (1.84 kg/m 2 ) A. Q = mLf = (1.84 kg/m 2 ) A(3.34 × 105 J/kg) = (6.146 × 105 J/m 2 ) A. The amount of heat conducted through the ice in time t is kA(TH − TC )t (1.6 W/m ⋅ K) A(10.0 C°)t Q = Ht = = = (64 J/m 2 ⋅ s) At . L 0.250 m Therefore, (6.146 × 105 J/m 2 ) A = (64 J/m 2 ⋅ s) At and t = 9.60 × 103 s = 160 min = 2.7 h.

14.80. Set Up: The emissivity of a human body is taken to be 1.0. In the equation for the radiation heat current the temperatures must be in kelvins. Solve: (a) Pjog = (0.80)(1300 W) = 1.04 × 103 J/s (b) H net = Aeσ (T 4 − Ts 4 ) H net = (1.85 m 2 )(1.00)(5.67 × 10−8 W/m 2 ⋅ K 4 )([306 K]4 − [313 K]4 ) = − 87.1 W The person gains 87.1 J of heat each second by radiation. (c) The total excess heat per second is 1040 J/s + 87 J/s = 1130 J/s

(d) In 1 min = 60 s the runner must dispose of (60 s)(1130 J/s) = 6.78 × 104 J. If this much heat goes to evaporate water, the mass m of water that evaporates in one minute is given by Q = mLv , so m=

Q 6.78 × 104 J = = 0.028 kg = 28 g. Lv 2.42 × 106 J/kg



14-17

(e) In a half-hour, or 30 minutes, the runner loses (30 min)(0.028 kg/min) = 0.84 kg. The runner must drink 0.84 L, 0.84 L = 1.1 bottles. 0.750 L/bottle Reflect: The person gains heat by radiation since the air temperature is greater than his skin temperature.

which is

*14.81. Set Up: Problem 14.83 calculates that the net rate of heat input to the person is 1130 W. 9 F° = 5 C°. Solve: (a) Q = Pt = (1130 W)(1800 s) = 2.03 × 106 J Q = mc ΔT so Q 2.03 × 106 J = = 8.6 C° mc (68 kg)(3480 J/kg ⋅ C°) (b) ΔT = (8.6 C°)(9 F°/5 C°) = 15.5 F°. T = 98.6°F + 15.5 F° = 114°F ΔT =

This body temperature is lethal.

14.82. Set Up: Assume all the ice melts. If we calculate Tf < 0, we will know this assumption is incorrect. For aluminum, ca = 910 J/kg ⋅ K. For water, Lf = 3.35 × 105 J/kg and cw = 4.19 × 103 J/kg ⋅ K. For ice, ci = 2.01 × 103 J/kg ⋅ K. The density of water is 1.00 × 103 kg/m3 , so 1.00 L of water has mass 1.00 kg.

Solve: For the soft drink, Qw = mw cw ΔTw = (2.00 kg)(4.19 × 103 J/kg ⋅ K)(T − 20.0°C) = (8380 J/K)T − 1.676 × 105 J. For the mug, Qa = ma ca ΔTa = (0.257 kg)(910 J/kg ⋅ K)(T − 20.0°C) = (234 J/K)T − 4.68 × 103 J. For the ice, Qi = mici ΔTi + mi Lf + micw ΔTw . Qi = (0.120 kg)([2010 J/kg ⋅ K ][15.0 C°] + 3.34 × 105 J/kg + [4.19 × 103 J/kg ⋅ K ][T − 0 C°]). Qi = 4.37 × 104 J + (503 J/K)T .

∑ Q = 0 gives (8380 J/K)T − 1.676 × 105 J + (234 J/K)T − 4.68 × 103 J + 4.37 × 104 J + (503 J/K)T = 0. 1.286 × 105 J = 14.1°C. 9117 J/K Reflect: T > 0°C, so our assumption that all the ice melts is correct. Note that the ice and the water from the melted T=

ice have different specific heat capacities.

*14.83. Set Up: Apply H = kA

ΔT and solve for k. H equals the power input required to maintain a constant L

interior temperature L (3.9 × 10−2 m) = (180 W) = 5.0 × 10−2 W / m ⋅ K. Solve: k = H AΔT (2.18 m 2 )(65.0 K)

Reflect: Our result is consistent with the values for insulating solids in Table 14.5.

14.84. Set Up: Heat Qi goes into the ice when it warms to 0D C, melts, and the resulting water warms to the final temperature Tf . Heat Qow comes out of the ocean water when it cools to Tf . Conservation of energy gives Qi + Qow = 0. For ice, ci = 2100 J/kg ⋅ K. For water, Lf = 334 × 103 J/kg and cw = 4190 J/kg ⋅ K. Let m be the total mass of the water on the earth’s surface. So mi = 0.0175m and mow = 0.975m.

Solve: Qi + Qow = 0 gives mici (30D C) + mi Lf + mi cwTf + mow cw (Tf − 5.00D C) = 0. Tf =

− mi ci (30D C) − mi Lf + mow cw (5.00D C) . (mi + mow )cw


14-18

Chapter 14

Tf =

−(0.0175m)(2100 J/kg ⋅ K)(30 K) − (0.0175m)(334 × 103 J/kg) + (0.975m)(4190 J/kg ⋅ K)(5.00 K) (0.0175m + 0.975m)(4190 J/kg ⋅ K) 1.348 × 104 J/kg

= 3.24°C. The temperature decrease is 1.76D C. 4.159 × 103 J/kg ⋅ K Reflect: The mass of ice in the ice caps is much less than the mass of the water in the oceans, but much more heat is required to change the phase of 1 kg of ice than to change the temperature of 1 kg of water 1D C, so the lowering of the temperature of the oceans would be appreciable. Tf =

14.85. Set Up: In H = Aeσ T 4 , T must be in kelvins. 0°C = 273.15 K. 3.5°C = 276.65 K. Solve: H1 = Aeσ T14 , H 2 = Aeσ T2 4 . The percent increase is ⎡⎛ T ⎞ 4 ⎤ H 2 − H1 T 4 −T4 × 100, = 2 4 1 × 100, = ⎢⎜ 2 ⎟ − 1⎥ × 100, = 5.2, H1 ⎢⎝ T1 ⎠ ⎥ T1 ⎣ ⎦

14.86. Set Up: The mass of the air is m = ρV . Solve: (a) Q = mc ΔT = ρVc ΔT = (1.29 kg/m3 )(0.0500 m3 )(1020 J/kg ⋅ K)(1.8 C°) = 118 J (b) m(24 J/kg)(0.10) = 118 J and m=

118 J = 49 g (0.10)(24 J/g)

*14.87. Set Up: Use Eq. (14.14) to find the net heat current into the can due to radiation. Use Q = Ht to find the heat that goes into the liquid helium, set this equal to mL and solve for the mass m of helium that changes phase. Solve: Calculate the net rate of radiation of heat from the can. H net = Aeσ (T 4 − Ts4 ). The surface area of the cylindrical can is A = 2π rh + 2π r 2 (see the figure below).

A = 2π r ( h + r ) = 2π (0.045 m)(0.250 m + 0.045 m) = 0.08341 m 2 .

H net = (0.08341 m 2 )(0.200)(5.67 × 10−8 W/m 2 ⋅ K 4 )((4.22 K)4 − (77.3 K) 4 ) H net = 20.0338 W (the minus sign says that the net heat current is into the can). The heat that is put into the can by radiation in one hour is Q = − ( H net )t = (0.0338 W)(3600 s) = 121.7 J. This heat boils a mass m of helium according to the equation Q = mLf , so m=

Q 121.7 J = = 5.82 × 10−3 kg = 5.82 g. Lf 2.09 × 104 J/kg

Reflect: In the expression for the net heat current into the can the temperature of the surroundings is raised to the fourth power. The rate at which the helium boils away increases by about a factor of (293/77) 4 = 210 if the walls surrounding the can are at room temperature rather than at the temperature of the liquid nitrogen.


15

THERMAL PROPERTIES OF MATTER

Answers to Multiple-Choice Problems 1. C 2. B, C 3. D 14. B, C 15. A, B, D

4. A, C

5. C

6. A

7. B

8. A

9. D

10. B

11. B

12. C

13. B, C, D

Solutions to Problems *15.1. Set Up: pV = nRT . ⎛ 0.110 m3 ⎞ ⎛V ⎞ Solve: nRT is constant so p1V1 = p2V2 . p2 = p1 ⎜ 1 ⎟ = (3.40 atm) ⎜ = 0.959 atm. ⎜ 0.390 m3 ⎟⎟ ⎝ V2 ⎠ ⎝ ⎠

15.2. Set Up: pV = nRT . T1 = 41.0°C = 314 K. R = 0.08206 L ⋅ atm/mol ⋅ K. Solve: n, R constant so

pV pV pV = nR is constant. 1 1 = 2 2 . T T1 T2

⎛ p ⎞⎛ V ⎞ T2 = T1 ⎜ 2 ⎟⎜ 2 ⎟ = (314 K)(2)(2) = 1.256 × 103 K = 983°C. ⎝ p1 ⎠⎝ V1 ⎠ pV (1.30 atm)(2.60 L) (b) n = = = 0.131 mol. RT (0.08206 L ⋅ atm/mol ⋅ K)(314 K) mtot = nM = (0.131 mol)(4.00 g/mol) = 0.524 g.

*15.3. Set Up: pV = nRT . T1 = 20.0°C = 293 K. Solve: (a) n, R, and V are constant.

p nR p p = = constant. 1 = 2 . T V T1 T2

⎛p ⎞ ⎛ 1.00 atm ⎞ T2 = T1 ⎜ 2 ⎟ = (293 K) ⎜ ⎟ = 97.7 K = − 175°C. ⎝ 3.00 atm ⎠ ⎝ p1 ⎠ (b) p2 = 1.00 atm, V2 = 3.00 L. p3 = 3.00 atm. n, R, and T are constant so pV = nRT = constant. p2V2 = p3V3 . ⎛p ⎞ ⎛ 1.00 atm ⎞ V3 = V2 ⎜ 2 ⎟ = (3.00 L) ⎜ ⎟ = 1.00 L. p ⎝ 3.00 atm ⎠ ⎝ 3⎠ Reflect: The final volume is one-third the initial volume. The initial and final pressures are the same, but the final temperature is one-third the initial temperature.

15.4. Set Up: 20.0 L = 20.0 × 10−3 m3 . M = 4.00 × 10−3 kg/mol. 1 atm = 1.013 × 105 Pa. 18.0°C = 291 K. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15-1

15-2

Chapter 15

Solve: (a) n = (b) p =

mtot 0.225 kg = = 56.3 mol. M 4.00 × 10−3 kg/mol

nRT (56.3 mol)(8.315 J/mol ⋅ K)(291 K) = = 6.81 × 106 Pa = 67.2 atm V 20.0 × 10−3 m3

15.5. Set Up: M = 32.0 × 10−3 kg/mol. T = 22.0°C = 295 K. p = 1.00 atm = 1.013 × 105 Pa. Solve: (a) n =

pV (1.013 × 105 Pa)(7.00 m)(8.00 m)(2.50 m) = = 5.78 × 103 mol RT (8.315 J/mol ⋅ K)(295 K)

(b) mtot = nM = (5.78 × 103 mol)(32.0 × 10−3 kg/mol) = 185 kg. Reflect: In the ideal gas law it is mandatory that the temperature be in kelvins. The weight of this mass of gas is about 410 lb.

15.6. Set Up: F = pA and pV = nRT . For a cube, V /A = L. Solve: (a) The force of any side of the cube is F = pA = ( nRT /V ) A = (nRT )/L, since the ratio of area to volume is

A/V = 1/L. For T = 20.0D C = 293.15 K, nRT (3 mol)(8.3145 J/mol ⋅ K)(293.15 K) = = 3.66 × 104 N. F= L 0.200 m (b) For T = 100.00°C = 373.15 K, nRT (3 mol)(8.3145 J/mol ⋅ K)(373.15 K) F= = = 4.65 × 104 N. L 0.200 m Reflect: When the temperature increases while the volume is kept constant, the pressure increases and therefore the force increases. The force increases by the factor T2 /T1.

*15.7. Set Up: pV = nRT . T1 = 300 K, T2 = 430 K. Solve: (a) n, R are constant so

pV pV pV = nR = constant. 1 1 = 2 2 . T T1 T2

⎛ 0.750 m3 ⎞ ⎛ 430 K ⎞ ⎛ V ⎞⎛ T ⎞ = 3.36 × 105 Pa. p2 = p1 ⎜ 1 ⎟⎜ 2 ⎟ = (1.50 × 105 Pa) ⎜ ⎜ 0.480 m3 ⎟⎟ ⎜⎝ 300 K ⎟⎠ V T ⎝ 2 ⎠⎝ 1 ⎠ ⎝ ⎠

15.8. Set Up: M CO2 = (12 + 2[16]) g/mol = 44 g/mol. M N 2 = 28 g/mol. ρ =

pM . R = 8.315 J/mol ⋅ K. T must RT

be in kelvins. Express M in kg/mol and p in Pa. 1 atm = 1.013 × 105 Pa Solve: (a) Mars: ρ =

Venus: ρ =

(650 Pa)(44 × 10−3 kg/mol) = 0.0136 kg/m3 (8.315 J/mol ⋅ K)(253 K) (92 atm)(1.013 × 105 Pa/atm)(44 × 10−3 kg/mol) = 67.6 kg/m3 (8.315 J/mol ⋅ K)(730 K)

Titan: T = −178 + 273 = 95 K

ρ=

(1.5 atm)(1.013 × 105 Pa/atm)(28 × 10−3 kg/mol) = 5.39 kg/m (8.315 J/mol ⋅ K)(95 K)

(b) Example 15.4 calculates the density of air at 20°C and p = 1 atm to be 1.20 kg/m3 . The density of the

atmosphere of Mars is much less, the density for Venus is much greater, and the density for Titan is somewhat greater.


Thermal Properties of Matter

15-3

15.9. Set Up: We are asked to compare two states. Use the ideal-gas law to obtain V1 in of V2 and the ratio of the temperatures in the two states. pV = nRT and n, R, p are constant so V /T = nR /p = constant and V1 /T1 = V2 /T2 Solve: T1 = (19 + 273) K = 292 K (T must be in kelvins)

V2 = V1(T2 /T1) = (0.600 L)(77.3 K/292 K) = 0.159 L Reflect: p is constant so the ideal-gas equation says that a decrease in T means a decrease in V.

15.10. Set Up: pV = nRT . Solve: n, R, T are constant, so pV = nRT = constant. p1V1 = p2V2 .

⎛V ⎞ ⎛ 6.00 L ⎞ p2 = p1 ⎜ 1 ⎟ = (1.00 atm) ⎜ ⎟ = 1.05 atm. ⎝ 5.70 L ⎠ ⎝ V2 ⎠

*15.11. Set Up: pV = nRT . T1 = 300 K. p1 = 1.01 × 105 Pa. p2 = 1.01 × 105 Pa + 2.72 × 106 Pa = 2.82 × 106 Pa. Solve: n, R are constant so

pV pV pV = nR = constant. 1 1 = 2 2 and T T1 T2

⎛ 2.82 × 106 Pa ⎞⎛ 46.2 cm3 ⎞ ⎛ p ⎞⎛ V ⎞ T2 = T1 ⎜ 2 ⎟⎜ 2 ⎟ = (300 K) ⎜ = 775.5 K = 502°C. ⎟⎜ 5 3 ⎟ ⎜ ⎟⎜ ⎟ ⎝ p1 ⎠⎝ V1 ⎠ ⎝ 1.01 × 10 Pa ⎠⎝ 499 cm ⎠ Reflect: Even though the pressures enter in a ratio, we must use absolute pressures. The ratio of the absolute pressures is different from the ratio of gauge pressures.

15.12. Set Up: pV = nRT . T1 = 277 K. T2 = 296 K. Assume the number of moles of gas in the bubble remains constant. Solve: (a) n, R are constant so

pV pV pV = nR = constant. 1 1 = 2 2 and T T1 T2

V2 ⎛ p1 ⎞⎛ T2 ⎞ ⎛ 3.50 atm ⎞⎛ 296 K ⎞ = ⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ = 3.74. V1 ⎝ p2 ⎠⎝ T1 ⎠ ⎝ 1.00 atm ⎠⎝ 277 K ⎠ (b) This increase in volume of air in the lungs would be dangerous.

15.13. Set Up: Use Eq. (15.4) and solve for p. ρ = pM /RT and p = RT ρ /M T = (−56.5 + 273.15) K = 216.6 K. For air M = 28.8 × 10−3 kg/mol (Example 15.3) Solve: p =

(8.3145 J/mol ⋅ K)(216.6 K)(0.364 kg/m3 ) −3

= 2.28 × 104 Pa

28.8 × 10 kg/mol Reflect: The pressure is about one-fifth the pressure at sea level.

15.14. Set Up: STP is T = 273 K and p = 1 atm. Set up a ratio using pV = nRT with nR constant. TV = 1003 + 273 = 1276 K. Solve: pV = nRT gives

p V p V pV = nR = constant, so E E = V V . T TE TV ⎛ p ⎞⎛ T ⎞ ⎛ 1 atm ⎞⎛ 1276 K ⎞ VV = VE ⎜ E ⎟⎜ V ⎟ = V ⎜ ⎟⎜ ⎟ = 0.0508V p T ⎝ 92 atm ⎠⎝ 273 K ⎠ ⎝ V ⎠⎝ E ⎠

*15.15. Set Up: The volume per mole for water at its critical point is (V /mole)c = 56.0 × 10−6 m3 /mole. 1 mole of water has mass 18.0 × 10−3 kg.


15-4

Chapter 15

Solve: V =

m

ρ

=

18.0 × 10−3 kg 998 kg/m3

= 1.80 × 10−5 m3 = 18.0 cm3 . This is about one-third of the volume of 1 mole at the

critical point.

15.16. Set Up: For water the triple-point pressure is 610 Pa and the critical-point pressure is 2.212 × 107 Pa. Solve: (a) To observe a solid to liquid (melting) phase transition the pressure must be greater than the triple-point pressure, so p1 = 610 Pa. For p < p1 the solid to vapor (sublimation) phase transition is observed. (b) No liquid to vapor (boiling) phase transition is observed if the pressure is greater than the critical-point pressure. p2 = 2.212 × 107 Pa. For p1 < p < p2 the sequence of phase transitions are solid to liquid and then liquid to vapor.

*15.17. Set Up: Figure 15.7 in the textbook shows that there is no liquid phase below the triple point pressure. Table 15.1 gives the triple point pressure to be 610 Pa for water and 5.17 × 105 Pa for CO2. Solve: The atmospheric pressure is below the triple point pressure of water, and there can be no liquid water on Mars. The same holds true for CO2. Reflect: On earth patm = 1 × 105 Pa, so on the surface of the earth there can be liquid water but not liquid CO2.

15.18. Set Up: N A = 6.022 × 1023 molecules/mol. M S = 32.064 × 10−3 kg/mol. M NH3 = (14.007 + 3[1.008]) × 10−3 kg/mol = 17.031 × 10−3 kg/mol Solve: M = N A m so m =

M NA

sulfur: m = ammonia: m =

32.064 × 10−3 kg/mol 6.022 × 1023 molecules/mol 17.031 × 10−3 kg/mol 23

6.022 × 10 molecules/mol

= 5.32 × 10−26 kg/molecule = 2.83 × 10−26 kg/molecule

*15.19. Set Up: The density of water is 1.00 × 103 kg/m3 . 1.00 L = 1.00 × 10−3 m3 . N A = 6.022 × 1023 molecules/mol. For water, M = 18 × 10−3 kg/mol. Solve: m = ρV = (1.00 × 103 kg/m3 )(1.00 × 10−3 m3 ) = 1.00 kg.

n=

m 1.00 kg = = 55.6 mol. M 18 × 10−3 kg/mol

N = nN A = (55.6 mol)(6.022 × 1023 molecules/mol) = 3.35 × 1025 molecules. Reflect: Note that we converted M to kg/mol.

15.20. Set Up: pV = NkT , with k = 1.381 × 10−23 J/molecule ⋅ K. TC = 95 (TF − 32°) = 95 (72° − 32°) = 22.2°C. T = 295 K. 1.0 cm3 = 1.0 × 10−6 m3 . Solve: N =

pV (1.013 × 105 Pa)(1.0 × 10−6 m3 ) = = 2.49 × 1019 molecules kT (1.381 × 10−23 J/molecule ⋅ K)(295 K)

N m RT = tot RT . We known that VA = VB and that TA > TB . NA M Solve: (a) p = nRT /V ; we don’t know n for each box, so either pressure could be higher.

*15.21. Set Up: pV = nRT =

⎛ N ⎞ pVN A (b) pV = ⎜ , where N A is Avogadro’s number. We don’t know how the pressures compare, ⎟ RT so N = RT N ⎝ A⎠ so either N could be larger. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


15-5

(c) pV = (mtot /M ) RT . We don’t know the mass of the gas in each box, so they could contain the same gas or

different gases. (d) 12 m(υ 2 )av = 32 kT . TA > TB and the average kinetic energy per molecule depends only on T, so the statement must be true. (e) υ rms = 3kT /m . We don’t know anything about the masses of the atoms of the gas in each box, so either set of

molecules could have a larger υ rms . Reflect: Only statement (d) must be true. We need more information in order to determine whether the other statements are true or false.

15.22. Set Up: We know that υrms =

3kT . The mass of a deuteron is m = mp + mn = 1.673 ×10−27 kg + 1.675 × 10−27 kg = m

3.35 × 10−27 kg. c = 3.00 × 108 m/s. k = 1.381 × 10−23 J/molecule ⋅ K.

3(1.381 × 10−23 J/molecule ⋅ K)(300 × 106 K)

Solve: (a) υrms =

υ rms c

3.35 × 10−27 kg

= 1.93 × 106 m/s.

= 6.43 × 10−3.

⎛ ⎞ 3.35 × 10−27 kg ⎛m⎞ 7 2 10 (b) T = ⎜ ⎟ (υrms ) 2 = ⎜ ⎟⎟ (3.0 × 10 m/s) = 7.3 × 10 K. 23 − ⎜ 3(1 381 10 J/molecule K) . × ⋅ ⎝ 3k ⎠ ⎝ ⎠ Reflect: Even at very high temperatures and for this light nucleus, vrms is a small fraction of the speed of light.

3kT . m

15.23. Set Up: M = 32.0 × 10−3 kg/mol. υrms = 3(8.315 J/mol ⋅ K)(300 K)

Solve: (a) υ rms = (b) K av = 32 kT =

32.0 × 10−3 kg

= 484 m/s

( 32 ) (1.381 × 10−23 J/molecule ⋅ K)(300 K) = 6.21 × 10−21 J.

Reflect: K av depends only on T and is independent of the mass of the molecules.

15.24. Set Up: To calculate υrms , take the square root of the average of υ 2 . Solve: (a) (υ 2 )av =

(10.0 m/s) 2 + (8.00 m/s) 2 + (7.00 m/s) 2 + (2.00 m/s) 2 = 54.25 m 2 /s 2 4

υ rms = (υ 2 )av = 7.37 m/s (b) (υ )av =

10.0 m/s + 8.00 m/s + 7.00 m/s + 2.00 m/s = 6.75 m/s 4

*15.25. Set Up: At STP, T = 273 K, p = 1.01 × 105 Pa. N = 6 × 109 molecules. Solve: pV = nRT .

V=

NkT (6 × 109 molecules)(1.381 × 10−23 J/molecule ⋅ K)(273 K) = = 2.24 × 10−16 m3 . p 1.01 × 105 Pa

L3 = V so L = V 1/3 = 6.1 × 10−6 m.

15.26. Set Up: For H 2 , M H 2 = 2.016 g/mol. M N 2 = 28.014 g/mol. υrms = Solve: υ rms, H 2 = υ rms, N 2 gives

3RTH 2

M H2

=

⎛ MN 2 TN 2 = TH 2 ⎜ ⎜ ⎝ M H2

3RTN 2

M N2

3RT . TH 2 = 293 K. M

.

⎞ ⎛ 28.014 g/mol ⎞ ⎟ = (293 K) ⎜ ⎟ = 4071 K = 3800°C. ⎟ ⎝ 2.016 g/mol ⎠ ⎠


15-6

Chapter 15

*15.27. Set Up: υrms = 3RT /M . T = 20°C = 293 K. M H2 = 2.02 × 10−3 kg/mol. M O2 = 32.0 × 10−3 kg/mol. M N 2 = 28.0 × 10−3 kg/mol Solve: (a) υ rms =

3(8.315 J/mol ⋅ K)(293 K) 2.02 × 10−3 kg/mol

= 1.90 × 103 m/s = 1.90 km/s

(b) υ rms < 11 km/s so no, the average H 2 molecule is not moving fast enough to escape. (c) O 2 : υ rms, O2 = υ rms, H 2

M H2 M O2

= (1.90 km/s)

N 2 : υ rms, N 2 = υrms, H 2

M H2 M N2

2.02 × 10−3 kg/mol 32.0 × 10−3 kg/mol = (1.90 km/s)

= 0.477 km/s

2.02 × 10−3 kg/mol 28.0 × 10−3 kg/mol

= 0.510 km/s

υrms for H 2 is about four times larger. (d) Figure 15.12 in Chapter 15 illustrates that some molecules move faster than υ rms , so some have a speed greater

than the escape speed and can escape. Since the rms speed for H 2 is greater than that of O 2 and N 2 , a higher percent of H 2 molecules are moving fast enough to escape. Reflect: At a given temperature all three species of molecules have the same average kinetic energy. To achieve this, the lighter H 2 molecules have a larger υ rms .

15.28. Set Up: K av = 32 kT . υrms =

3RT . M Ne = 20.180 g/mol, M Kr = 83.80 g/mol and M Rn = 222 g/mol. M

Solve: (a) K av = 32 kT depends only on the temperature so it is the same for each species of atom in the mixture. (b)

υrms, Ne M Kr 83.80 g/mol = = = 2.04. M Ne 20.18 g/mol υrms, Kr υ rms, Ne M Rn 222 g/mol = = = 3.32. M Ne 20.18 g/mol υ rms, Rn

υrms, Kr M Rn 222 g/mol = = = 1.63. M Kr 83.80 g/mol υrms, Rn

15.29. Set Up: Use pV = nRT to solve for V. Use R = 0.08206 L ⋅ atm/mol ⋅ K. T = 273.15 K. Solve: (a) V =

nRT (1.00 mol)(0.08206 L ⋅ atm/mol ⋅ K)(273.15 K) = = 22.4 L p 1.00 atm

⎛ p ⎞ ⎛ 1.00 atm ⎞ (b) pV = nRT = constant, so p1V1 = p2V2 . V2 = ⎜ 1 ⎟V1 = ⎜ ⎟ (22.4 L) = 0.243 L. p ⎝ 92 atm ⎠ ⎝ 2⎠ Reflect: For constant T, the volume of 1.00 mol is inversely proportional to the pressure.

15.30. Set Up: pV = NkT , with k = 1.381 × 10−23 J/molecule ⋅ K. p = 13 atm = 3.38 × 104 Pa. V = 0.50 L = 0.50 × 10−3 m3. N is the total number of molecules, of which 21% are O 2 and 78% are N 2 . T = −15°F = −26°C = 247 K. Solve: (a) In each breath,

N air =

(3.38 × 104 Pa)(0.50 × 10−3 m3 ) pV = = 4.95 × 1021 molecules kT (1.381 × 10−23 J/molecule ⋅ K)(247 K)

N = (0.21)(4.95 × 1021 molecules) = 1.0 × 1021 molecules



15-7

⎛ 1 atm ⎞ (b) At sea level p = 1 atm, so N O2 is larger by ⎜ = 3 and N O2 = 3.0 × 1021 molecules. The number at the ⎝ 1/3 atm ⎟⎠

top of Mt. Everest is 33% of the number at sea level.

*15.31. Set Up: 20.9, − 16.3, = 4.6, of the volume of air breathed in is O2 that is absorbed by the body. T = 20°C = 293 K. p = 1 atm = 1.013 × 105 Pa. 1.0 L = 1.0 × 10−3 m3 . Solve: (a) In one minute the person must absorb 14.5 L/hr = 0.242 L/min of O 2 . 60 min/hr Since the volume of O2 absorbed is 4.6% of the air breathed in, the person must have breathed in

0.242 L/min = 5.26 L/m of air. 0.046 The number of breaths per minute is 5.26 L/min = 10.5 breaths/min. 0.50 L/breath (b) With each breath the person inhales (0.209)(0.50 L) = 0.1045 L of O2 . pV = NkT gives the number of O2 molecules in this volume to be (1.013 × 105 Pa)(0.1045 × 10−3 m3 ) pV = = 2.62 × 1021 molecules N= kT (1.381 × 10−23 J/molecule ⋅ K)(293 K) NkT says that the volume of O2 is directly proportional to the number of O2 molecules. About 10 p breaths per minute is what we observe for a healthy, resting person; this result makes sense.

Reflect: V =

15.32. Set Up: For a monatomic ideal gas, CV = 32 R and for a diatomic ideal gas, CV = 52 R. For a constant volume process, Q = nCV ΔT . Solve: (a) Q = (2.50 mol)

( 32 ) (8.315 J/mol ⋅ K)(30.0 K) = 935 J.

(5/2) (935 J) = 1560 J. (3/2) (c) The pV diagram is given in the figure below. When the temperature increases at constant volume, the pressure increases. (b) Q =

*15.33. Set Up: The rigid container means the process occurs at constant volume. For a monatomic ideal gas CV = 32 R and for a diatomic ideal gas CV = 52 R. Solve: (a) Q = nCV ΔT so ΔT =

Q 1850 J = = 39.6 C°. nCV (2.25 mol) 5 (8.315 J/mol ⋅ K) 2

()

Tf = Ti + ΔT = 49.6°C © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15-8

Chapter 15

(b) Q = nCV ΔT = (2.25 mol)

( 32 ) (8.315 J/mol ⋅ K)(39.6 C°) = 1110 J

(c) V is constant. Then pV = nRT says p increases when T increases. The process for either gas is sketched in the

figure below.

Reflect: More heat is required for the diatomic gas because not all of the energy added by Q goes into the translational kinetic energy that determines the temperature. Some of the energy that flows into the gas goes into the rotational kinetic energy of the diatomic molecule. The answer in (b) is smaller than 1850 J by the ratio 3R 3 2 = . 5R 5 2

15.34. Set Up: M N 2 = 2(14.007 g/mol) = 28.014 × 10−3 kg/mol. C = Mc. For water, cw = 4190 J/kg ⋅ K. For N 2 , CV = 20.76 J/mol ⋅ K. Solve: cN 2 =

c 20.76 J/mol ⋅ K C = = 741 kg/mol. w = 5.65; cw is over five times larger. cN 2 M 28.014 × 10−3 kg/mol

15.35. Set Up: The heat Q added is related to the temperature increase ΔT by Q = nCV ΔT . For H 2 , CV ,H2 = 20.42 J/mol ⋅ K and for Ne (a monatomic gas), CV ,Ne = 12.47 J/mol ⋅ K. Q = constant, so CV ,H2 ΔTH2 = CV ,NeΔTNe . n ⎛ CV ,H2 ⎞ ⎛ 20.42 J/mol ⋅ K ⎞ (2.50D C) = 4.09D C. ΔTNe = ⎜ ΔT = ⎜ CV ,Ne ⎟⎟ H2 ⎜⎝ 12.47 J/mol ⋅ K ⎟⎠ ⎝ ⎠ Reflect: The same amount of heat causes a smaller temperature increase for H 2 since some of the energy input goes Solve: CV ΔT =

into the internal degrees of freedom.

15.36. Set Up: For a diatomic ideal gas, CV = 52 R. The number of moles is related to the total mass m of the gas by m = nM . M N 2 = 28.0 g/mol and M O2 = 32.0 g/mol. Solve: (a) n =

m 75.0 g = = 2.68 mol. M 28.0 g/mol Q = nCV ΔT = (2.68 mol)

( 52 ) (8.315 J/mol ⋅ K)(49.5°C − 12.1°C) = 2080 J

(b) The number of moles is changed by the factor

M N2 M O2

so the Q needed is changed by this same factor. So,

⎛ 28.0 g/mol ⎞ Q=⎜ ⎟ (2080 J) = 1820 J. ⎝ 32.0 g/mol ⎠

75.0 g of O2 is fewer moles than 75.0 g of N 2 , so Q is less.



15-9

*15.37. Set Up: The volume is constant. Solve: (a) The pV diagram is given in the figure below.

(b) Since ΔV = 0, W = 0. Reflect: For any constant volume process the work done is zero.

15.38. Set Up: For a constant pressure process, W = p ΔV . Solve: (a) The pV diagram is given in the figure below.

(b) W = (1.50 × 105 Pa)(0.0600 m3 − 0.0900 m3 ) = − 4500 J. W < 0 when the volume decreases.

15.39. Set Up: For a constant pressure process, W = p ΔV . If the process is also for an ideal gas, pV = nRT and p ΔV = nR ΔT . Solve: (a) The pV diagram is sketched in the figure below. When the gas is heated at constant pressure, the volume increases.

(b) W = nR ΔT = (2.00 mol)(8.315 J/mol ⋅ K)(80 K) = 1330 J. Reflect: Since the volume increases, the work done by the gas is positive.

15.40. Set Up: Apply pV = nRT to calculate T. For this constant pressure process, W = pΔV . Q = nC p ΔT . Use ΔU = Q − W to relate Q, W and ΔU . 2.50 atm = 2.53 × 105 Pa. For a monatomic ideal gas, CV = 12.47 J/mol ⋅ K and

C p = 20.78 J/mol ⋅ K.


15-10

Chapter 15

Solve: (a) T1 =

T2 =

pV1 (2.53 × 105 Pa)(3.20 × 10−2 m 2 ) = = 325 K. (3.00 mol)(8.314 J/mol ⋅ K) nR pV2 (2.53 × 105 Pa)(4.50 × 10−2 m 2 ) = = 456 K. nR (3.00 mol)(8.314 J/mol ⋅ K)

(b) W = pΔV = (2.53 × 105 Pa)(4.50 × 10−2 m3 − 3.20 × 10−2 m3 ) = 3.29 × 103 J (c) Q = nC p ΔT = (3.00 mol)(20.78 J/mol ⋅ K)(456 K − 325 K) = 8.17 × 103 J (d) ΔU = Q − W = 4.88 × 103 J Reflect: We could also calculate ΔU as ΔU = nCV ΔT = (3.00 mol)(12.47 J/mol ⋅ K)(456 K − 325 K) = 4.90 × 103 J,

which agrees with the value we calculated in part (d).

*15.41. Set Up: For a constant volume process, W = 0. For a constant pressure process, W = p ΔV . 1 atm = 1.013 × 105 Pa and 1 L = 10−3 m3 . The work done by the system in a process is positive when the volume increases and negative when the volume decreases. Solve: (a) Find W for each process in the cycle. 1 → 2 : W = p ΔV = (2.5 atm)(1.013 × 105 Pa/atm)(8 L − 2 L)(10−3 m3 /1 L) = 1.5 × 103 J 2 → 3 : W = 0 since ΔV = 0 3 → 4 : W = p ΔV = (0.5 atm)(1.013 × 105 Pa/atm)(2 L − 8 L)(10−3 m3 /1 L) = − 3.0 × 102 J 4 → 1: W = 0 since ΔV = 0

Wcycle = 1.5 × 103 J + 0 + (−3.0 × 102 J) + 0 = 1.2 × 103 J The area enclosed by the cycle is (2.0 atm)(6.0 L) = (12.0 L ⋅ atm)(10−3 m3 /L)(1.013 × 105 Pa/atm) = 1.2 × 103 J. The total work done does equal the area enclosed by the cycle. (b) For 1 → 4 → 3 → 2 → 1, Wcycle = − 1.5 × 103 J + 3.0 × 102 J = − 1.2 × 103 J. The negative work done for 2 → 1

is greater in magnitude than the positive work done in 4 → 3 and the total work for the cycle has the opposite sign from what it had in (a). (c) In the pV-diagram the cycle 3 → 4 → 2 → 3 is a triangle with area

The work done is

1 2

1 (2.0 2

atm)(6.0 L) = 6.0 L ⋅ atm = 6.0 × 102 J.

that done in the cycle of part (a). The positive work done in process 4 → 2 is larger in magnitude than

the negative work done in process 3 → 4 so the net work done in the cycle is positive. Reflect: For the cycles in parts (a) and (b), the initial and final states are the same but the work done is different. This illustrates the general result that the work done in a process depends not only on the initial and final states but also on the path.

15.42. Set Up: (a) The work done in a cycle is the area enclosed by the cycle. 1 mm of H 2 = 133.3 Pa. pgauge = p − pair . In calculating the enclosed area only changes in pressure enter and you can use gauge pressure. 1 L = 10−3 m3 . (b) Since pV = nRT and T is constant, the maximum number of moles of air in the lungs is when pV

is a maximum. In the ideal gas law the absolute pressure p = pgauge + pair must be used. pair = 760 mm of Hg. 1 mm of Hg = 1 torr. Solve: (a) By counting squares and noting that the area of 1 square is (1 mm of Hg)(0.1 L), we estimate that the area enclosed by the cycle is about 7.5 (mm of Hg) ⋅ L = 1.00 N ⋅ m. The net work done is positive



15-11

(b) The maximum pV is when p = 11 torr + 760 torr = 771 torr = 1.028 × 105 Pa and V = 1.4 L = 1.4 × 10−3 m3 . The

maximum pV is ( pV )max = 144 N ⋅ m. pV = nRT so nmax =

( pV ) max 144 N ⋅ m = = 0.059 mol. (8.315 J/mol ⋅ K)(293 K) RT

15.43. Set Up: ΔU = Q − W . Q > 0 when heat enters the system. W > 0 when the system expands and does work; W < 0 when work is done on the system. Solve: Q = + 254 J. W = − 73 J. ΔU = Q − W = 254 J − ( −73 J) = 327 J.

15.44. Set Up: ΔU = Q − W . Q = + 1.15 × 105 J, since heat enters the gas. For a constant pressure process, W = p ΔV . Solve: (a) W = p ΔV = (1.80 × 105 Pa)(0.320 m3 − 0.110 m3 ) = 3.78 × 104 J. (b) ΔU = Q − W = 1.15 × 105 J − 3.78 × 104 J = 7.72 × 104 J.

*15.45. Set Up: ΔU = Q − W . Q > 0 if heat flows into the gas. For a constant pressure process, W = p ΔV . Solve: (a) W = p ΔV = (2.30 × 105 Pa)(1.20 m3 − 1.70 m3 ) = − 1.15 × 105 J (b) Q = ΔU + W = − 1.40 × 105 J + (−1.15 × 105 J) = − 2.55 × 105 J. Q < 0 so this amount of heat flows out of the gas. Reflect: ΔV < 0 and W < 0. The work done on the gas adds energy to it, but more energy flows out as heat than is

added by the work and the internal energy decreases.

15.46. Set Up: The type of process is not specified. We can use ΔU = Q − W because this applies to all processes. Calculate ΔU and then from it calculate ΔT . Q is positive since heat goes into the gas; Q = 11200 J; W is positive since the gas expands; W = +2100 J. Solve: ΔU = 1200 J − 2100 J = 2900 J We can also use ΔU = n

( 32 R) ΔT

since this is true for any process for an ideal gas.

ΔT =

2 ΔU 2(−900 J) = = 214.4D C 3nR 3(5.00 mol)(8.3145 J/mol ⋅ K)

T2 = T1 + ΔT = 127°C − 14.4D C = 113°C Reflect: More energy leaves the gas in the expansion work than enters as heat. The internal energy therefore decreases, and for an ideal gas this means the temperature decreases. We didn’t have to convert ΔT to kelvins since ΔT is the same on the Kelvin and Celsius scales.

*15.47. Set Up: ΔU = Q − W . ΔU is path independent. Q > 0 when the system absorbs heat. Solve: (a) Use path acb to find ΔU = U b − U a . U b − U a = Q − W = 90.0 J − 60.0 J = 30.0 J. For path adb, ΔU = 30.0 J and Q = ΔU + W = 30.0 J + 15.0 J = 45.0 J. (b) ΔU = U a − U b = − 30.0 J. For process b → a, ΔV < 0 and W = − 35.0 J. Q = ΔU + W = −30.0 J + (−35.0 J) = −65.0 J. The system liberated 65.0 J of heat. (c) For process a → d : ΔU = U d − U a = 8.0 J. Wd → b = 0 so Wa → d = Wadb = 15.0 J.

Q = ΔU + W = 8.0 J + 15.0 J = 23.0 J. For process d → b : ΔU = U b − U d = 30.0 J − 8.0 J = 22.0 J and W = 0. Q = ΔU + W = 22.0 J. Reflect: Both Q and W depend on the path.

15.48. Set Up: Use information about the pressure and volume in the ideal gas law to determine the sign of ΔT , and from that we can determine the sign of Q.


15-12

Chapter 15

Solve: For constant p, Q = nC p ΔT

Since the gas is ideal, pV = nRT and for constant p, p ΔV = nR ΔT . ⎛ p ΔV Q = nC p ⎜ ⎝ nR

⎞ ⎛ ⎞ ⎟ p ΔV ⎟=⎜ ⎠ ⎝ R ⎠

Since the gas expands, ΔV > 0 and therefore Q > 0. Q > 0 means heat goes into gas. Reflect: Heat flows into the gas, W is positive and the internal energy increases. It must be that Q > W .

*15.49. Set Up: For an ideal gas, ΔU = nCV ΔT so for an isothermal process ΔU = 0. Use W = nRT ln (V2 /V1 ) to calculate W for the isothermal process and then apply ΔU = Q − W .

Solve : (a) W = nRT ln (V2 /V1 ) = (1.75 mol)(8.315 J/mol ⋅ K)(273 K)ln(1.35 L/4.20 L) = − 4510 J (b) ΔU = Q − W and ΔU = 0 says Q = W = − 4510 J. 4510 J of heat comes out of the gas. Reflect: When the gas is compressed, heat must be removed to keep the temperature constant.

15.50. Set Up: Since the gas is in a perfectly insulated container, Q = 0. Apply ΔU = Q − W . W is the work done by the gas, so W = − 457 J. Solve: ΔU = Q − W = 0 − (−457 J) = + 457 J. The internal energy increases.

15.51. Set Up: For a constant volume process, Q = nCV ΔT . For a constant pressure process, Q = nC p ΔT . For any process of an ideal gas, ΔU = nCV ΔT . From Table 15.3, for N 2 , CV = 20.76 J/mol ⋅ K and C p = 29.07 J/mol ⋅ K. Heat is added, so Q is positive and Q = 11557 J.

Solve: (a) ΔT = (b) ΔT =

Q 1557 J = = 125.0 K nCV (3.00 mol)(20.76 J/mol ⋅ K)

Q 1557 J = = 117.9 K nC p (3.00 mol)(29.07 J/mol ⋅ K)

(c) ΔU = nCV ΔT for either process, so ΔU is larger when ΔT is larger. The final internal energy is larger for the constant volume process in (a). Reflect: For constant volume W = 0 and all the energy added as heat stays in the gas as internal energy. For the constant pressure process the gas expands and W > 0. Part of the energy added as heat leaves the gas as expansion work done by the gas.

15.52. Set Up: The pV diagram shows that V is constant during the process. Solve: (a) W = 0 since ΔV = 0. ⎛V ⎞ (b) pV = nRT and T = ⎜ ⎟ p. When p increases while V and n are constant, then T must increase. In fact, if ⎝ nR ⎠

p triples the absolute temperature T triples. (c) ΔU = Q − W . Q = + 534 J and W = 0 gives ΔU = + 534 J. The internal energy increased by 534 J.

(d) Tb = 3Ta , as derived in part (b).

*15.53. Set Up: The pV diagram shows that in the process the volume decreases while the pressure is constant. 1 L = 10−3 m3 and 1 atm = 1.013 × 105 Pa

Solve: (a) pV = nRT . n, R, and p are constant so

V nR V V = = constant. a = b . T p Ta Tb

⎛T ⎞ ⎛ T /4 ⎞ Vb = Va ⎜ b ⎟ = (0.500 L) ⎜ a ⎟ = 0.125 L T ⎝ a⎠ ⎝ Ta ⎠



15-13

(b) For a constant pressure process, W = p ΔV = (1.50 atm)(0.125 L − 0.500 L) and ⎛ 10−3 m3 ⎞⎛ 1.013 × 105 Pa ⎞ W = (−0.5625 L ⋅ atm) ⎜ ⎟⎟ = − 57.0 J ⎜ 1 L ⎟⎜ ⎟⎜ 1 atm ⎝ ⎠⎝ ⎠ W is negative since the volume decreases. Since W is negative, work is done on the gas. (c) For an ideal gas, U = nCT so U decreases when T decreases. The internal energy of the gas decreases because

the temperature decreases. (d) For a constant pressure process, Q = nC p ΔT . T decreases so ΔT is negative and Q is therefore negative. Negative Q means heat leaves the gas. Reflect: W = nR ΔT and Q = nC p ΔT . C p > R, so more energy leaves as heat than is added by work done on the gas, and the internal energy of the gas decreases.

15.54. Set Up: Use pV = nRT to calculate T at each point. The work done in a process is the area under the curve in the pV diagram. ΔU = Q − W for all processes.

Solve: (a) pV = nRT so T =

pV nR

point a: Ta =

(2.0 × 105 Pa)(0.010 m3 ) = 535 K (0.450 mol)(8.315 J/mol ⋅ K)

point b: Tb =

(5.0 × 105 Pa)(0.070 m3 ) = 9350 K (0.450 mol)(8.315 J/mol ⋅ K)

(8.0 × 105 Pa)(0.070 m3 ) = 15,000 K (0.450 mol)(8.315 J/mol ⋅ K) (b) The work done by the gas is positive since the volume increases. The magnitude of the work is the area under the curve: W = 12 (2.0 × 105 Pa + 5.0 × 105 Pa)(0.070 m3 − 0.010 m3 ) = 2.1 × 104 J point c: Tc =

(c) ΔU = Q − W so Q = ΔU + W = 15,000 J + 2.1 × 104 J = 3.6 × 104 J Q is positive so heat energy goes into the gas.

*15.55. Set Up: Use pV = nRT to calculate Tc /Ta . Calculate ΔU and W and use ΔU = Q − W to obtain Q. For path ac, the work done is the area under the line representing the process in the pV-diagram. T pV (1.0 × 105 J)(0.060 m3 ) Solve: (a) c = c c = = 1.00. Tc = Ta . Ta paVa (3.0 × 105 J)(0.020 m3 )

(b) Since Tc = Ta , ΔU = 0 for process abc. For ab, ΔV = 0 and Wab = 0. For bc, p is constant and Wbc = pΔV = (1.0 × 105 Pa)(0.040 m3 ) = 4.0 × 103 J. Therefore, Wabc = 14.0 × 103 J. Since ΔU = 0,

Q = W = 14.0 × 103 J.

4.0 × 103 J of heat flows into the gas during process abc.

(c) W = 12 (3.0 × 105 Pa + 1.0 × 105 Pa)(0.040 m3 ) = +8.0 × 103 J. Qac = Wac = 18.0 × 103 J. Reflect: The work done is path dependent and is greater for process ac than for process abc, even though the initial and final states are the same.

15.56. Set Up: At each point pV = nRT , with T = 85 + 273 = 358 K. For an isothermal process of an ideal gas, W = nRT ln (V2 /V1). ΔU = nCV ΔT for any ideal gas process.

Solve: (a) At point b, p = 0.200 atm = 2.026 × 104 Pa and V = 0.100 m3 n=

pV (2.026 × 104 Pa)(0.100 m3 ) = = 0.681 moles RT (8.315 J/mol ⋅ K)(358 K)

(b) n, R, and T are constant so paVa = pbVb . ⎛p ⎞ ⎛ 0.200 atm ⎞ 3 Va = Vb ⎜ b ⎟ = (0.100 m3 ) ⎜ ⎟ = 0.0333 m p 0 600 atm . ⎝ ⎠ a ⎝ ⎠ © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15-14

Chapter 15

⎛ 0.100 m3 ⎞ = 2230 J (c) W = nRT ln (Vb /Va ) = (0.681 mol)(8.315 J/mol ⋅ K)(358 K)ln ⎜ ⎜ 0.0333 m3 ⎟⎟ ⎝ ⎠ W is positive and corresponds to work done by the gas. W is positive when the volume increases. (d) ΔU = nCV ΔT so for an isothermal process (ΔT = 0), ΔU = 0.

15.57. Set Up: For CO 2 , CV = 28.46 J/mol ⋅ K. For a constant pressure process of an ideal gas, W = p ΔV and pV = nRT combine to give W = nR ΔT . ΔU = nCV ΔT and ΔU = Q − W .

Solve: (a) The pV diagram is given in the figure below. When the temperature increases at constant pressure, the volume increases.

(b) W = p ΔV = nR ΔT = (0.250 mol)(8.315 J/mol ⋅ K)(100.0 K) = 208 J. (c) The work is done by the gas on the piston. (d) ΔU = nCV ΔT = (0.250 mol)(28.46 J/mol ⋅ K)(100.0 K) = 712 J. (e) Q = ΔU + W = 712 J + 208 J = 920 J. (f) W = nR ΔT so the work would be the same. If the pressure is halved, the volume change is doubled for the same ΔT , so the work done remains the same.

15.58. Set Up: 5.0 L = 5.0 × 1023 m3 . p = 1 atm = 1.013 × 105 Pa. For a constant pressure process, Q = nC p ΔT . For a constant volume process Q = nCV ΔT . Assume the air can be treated as an ideal gas. For N 2 , Table 15.4 gives C p = 29.07 J/mol ⋅ K and CV = 20.76 J/mol ⋅ K.

Solve: (a) T = 37°C = 310 K. pV = nRT so n=

pV (1.013 × 105 Pa)(5.0 × 1023 m3 ) = = 0.20 moles RT (8.315 J/mol ⋅ K)(310 K)

(b) The heating is done at constant pressure, so Q = nC p ΔT . − 10° F = 95 (−10° − 32°) = − 23.3°C, so ΔT = 60.3 K. Q = (0.20 mol)(29.07 J/mol ⋅ K)(60.3 K) = 350 J

(c) Now the heating is done at constant volume, so Q = nCV ΔT = (0.20 mol)(20.76 J/mol ⋅ K)(60.3 K) = 250 J

*15.59. Set Up: O2 is diatomic and if treated as ideal it has CV = 52 R and C p = 72 R. pV = nRT . 1 atm = 1.013 × 105 Pa. Q = nCV ΔT for ΔV = 0 and Q = nC p ΔT for Δp = 0.

Solve: (a) point a: T =

pV (0.60 atm)(1.013 × 105 Pa/atm)(0.10 m3 ) = = 665 K RT (1.10 mol)(8.315 J/mol ⋅ K)

pV = 4(665 K) = 2660 K RT pV 2660 K point c: T = = = 887 K; RT 3 point b: T =



point d : T =

(b) and (c) (i) ab Q = nC p ΔT = n

15-15

pV 887 K = = 222 K RT 4

( 72 R ) ΔT = (1.10 mol) ( 72 ) (8.315 J/mol ⋅ K)(2660 K − 665 K) = 6.39 × 104 J

Q > 0 so heat enters the gas.

(ii) bc Q = nCV ΔT = n

( 52 R ) ΔT = (1.10 mol) ( 52 ) (8.315 J/mol ⋅ K)(887 K − 2660 K) = − 4.05 × 104 J

Q < 0 so heat leaves the gas. (iii) cd Q = nC p ΔT = n

( 72 R ) ΔT = (1.10 mol) ( 72 ) (8.315 J/mol ⋅ K)(222 K − 887 K) = − 2.13 × 104 J

Q < 0 so heat leaves the gas.

(iv) da Q = nCV ΔT = n

( 52 R ) ΔT = (1.10 mol) ( 25 ) (8.315 J/mol ⋅ K)(665 K − 222 K) = 1.01 × 104 J

Q > 0 so heat enters the gas.

Reflect: The net heat flow for the complete cycle is Qab + Qbc + Qcd + Qda = 1.22 × 104 J. The work done in the cycle is positive and equal to the area enclosed by the cycle, so W = (0.40 atm)(1.013 × 105 Pa/atm)(0.30 m3 ) = 1.22 × 104 J. For a cycle, where the system returns to the initial state, ΔU = 0 so Q = W .

15.60. Set Up: For an adiabatic process, p1V1γ = p2V2γ and V1T1γ − 1 = V2T2γ − 1. γ = C p /CV . V2 = 1.50V1. Solve: (a) γ = 53 . γ

5/3 ⎛V ⎞ ⎛ 1 ⎞ p2 = p1 ⎜ 1 ⎟ = (4.00 atm) ⎜ ⎟ = 2.04 atm. ⎝ 1.50 ⎠ ⎝ V2 ⎠

γ −1

⎛V ⎞ T2 = T1 ⎜ 2 ⎟ ⎝ V1 ⎠ ⎛ 1 ⎞ (b) γ = 75 . p2 = (4.00 atm) ⎜ ⎟ ⎝ 1.50 ⎠

⎛ 1 ⎞ = (350 K) ⎜ ⎟ ⎝ 1.50 ⎠

2/3

= 267 K.

7/5

= 2.27 atm. ⎛ 1 ⎞ T2 = (350 K) ⎜ ⎟ ⎝ 1.50 ⎠

2/5

= 298 K.

*15.61. Set Up: ΔU = Q − W . Apply Q = nC p ΔT to calculate C p . Apply ΔU = nCV ΔT to calculate CV . γ = C p /CV . ΔT = 15.0D C = 15.0 K. Since heat is added, Q = 1970 J. Solve: (a) ΔU = Q − W = 1970 J − 223 J = 747 J (b) C p =

γ=

CV

=

ΔU Q 970 J 747 J = = 28.5 J/mol ⋅ K. = = 37.0 J/mol ⋅ K. CV = nΔT (1.75 mol)(15.0 K) nΔT (1.75 mol)(15.0 K) 37.0 J/mol ⋅ K = 1.30 28.5 J/mol ⋅ K

Reflect: The value of γ we calculated is similar to the values given in Tables 15.4 for polyatomic gases.

15.62. Set Up: ΔU = Q − W . For an ideal gas, ΔU = CV ΔT , and at constant pressure, W = p ΔV = nR ΔT . CV = 32 R for a monatomic gas. Solve: ΔU = n

( 32 R) ΔT = 32 p ΔV = 32 W . Then Q = ΔU + W = 52 W , so W /Q = 25 .


15-16

Chapter 15

Reflect: For diatomic or polyatomic gases, CV is a different multiple of R and the fraction of Q that is used for expansion work is different.

15.63. Set Up: Assume the expansion is adiabatic. The equation T1V1γ −1 = T2V2γ −1 relates V and T. Assume the air behaves as an ideal gas, so ΔU = nCV ΔT . Use pV = nRT to calculate n. For air, CV = 29.76 J/mol ⋅ K and γ = 1.40. V2 = 0.800V1. T1 = 293.15 K. p1 = 2.026 × 105 Pa . For a sphere, V = 43 π r 3 . γ −1

⎛V ⎞ ⎛ V1 ⎞ Solve: (a) T2 = T1 ⎜ 1 ⎟ = (293.15 K) ⎜ ⎟ ⎝ V2 ⎠ ⎝ 0.800V1 ⎠ 4π (b) V1 = 43 π r 3 = (0.1195 m)3 = 7.15 × 10−3 m3 . 3 n=

0.40

= 320.5 K = 47.4° C.

p1V1 (2.026 × 105 Pa)(7.15 × 10−3 m3 ) = = 0.594 mol. (8.314 J/mol ⋅ K)(293.15 K) RT1

ΔU = nCV ΔT = (0.594 mol)(20.76 J/mol ⋅ K)(321 K − 293 K) = 345 J.

Reflect: We could also use ΔU = W =

1 ( p1V1 − p2V2 ) to calculate ΔU , if we first found p2 from pV = nRT . γ −1

15.64. Set Up: pV = nRT relates p, V, and T. For an adiabatic process Q = 0. Then ΔU = Q − W says ΔU = − W . W = + 85.0 J; it is positive since the volume increases. Solve: (a) Q = 0. (b) ΔU = − W = − 85.0 J.

(c) T =

pV (1.013 × 105 Pa)(6.00 × 10−3 m3 ) = = 1.12 × 103 K (0.0650 mol)(8.315 J/mol ⋅ K) nR

*15.65. Set Up: p = 9.00 × 10−14 atm = 9.12 × 10−9 Pa. V = 1.00 cm3 = 1.00 × 10−6 m3 . Solve: pV = NkT .

N=

pV (9.12 × 10−9 Pa)(1.00 × 10−6 m3 ) = = 2.20 × 106 molecules. kT (1.38 × 10−23 J/molecule ⋅ K)(300 K)

15.66. Set Up: N A = 6.022 × 1023 molecules. M = 18.0 × 10−3 kg/mol. Solve: n =

m 65 kg = = 3.61 × 103 moles. N = nN A = 2.17 × 1027. N is the number of water molecules. M 18.0 × 1023 mol

Each molecule has three atoms, so the number of atoms is 3 N = 6.5 × 1027 atoms.

*15.67. Set Up: The pressure difference between two points in a fluid is Δp = ρ gh, where h is the difference in height of two points. Solve: (a) Δp = ρ gh = (1.2 kg/m3 )(9.80 m/s 2 )(1000 m) = 1.18 × 104 Pa

(b) At the bottom of the mountain, p = 1.013 × 105 Pa. At the top, p = 8.95 × 104 Pa. pV = nRT = constant so

pbVb = ptVt and ⎛ 1.013 × 105 Pa ⎞ ⎛p ⎞ Vt = Vb ⎜ b ⎟ = (0.50 L) ⎜ ⎟⎟ = 0.566 L. 4 ⎜ ⎝ pt ⎠ ⎝ 8.95 × 10 Pa ⎠ Reflect: The pressure variation with altitude is affected by changes in air density and temperature and we have neglected those effects. The pressure decreases with altitude and the volume increases. You may have noticed this effect: bags of potato chips “puff-up” when taken to the top of a mountain.

15.68. Set Up: V = 3.00 × 1023 m3 . T = 295 K. p = 2.03 × 10−8 Pa. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


15-17

Solve: (a) pV = nRT and n=

pV (2.03 × 10−8 Pa)(3.00 × 10−3 m3 ) = = 2.48 × 10−14 mol. RT (8.315 J/mol ⋅ K)(295 K)

m = nM = (2.48 × 10−14 mol)(28.0 × 10−3 kg/mol) = 6.95 × 10−16 kg.

(b) ρ =

m 6.95 × 10−16 kg = = 2.32 × 10−13 kg/m3 . V 3.00 × 10−3 m3

15.69. Set Up: In pV = nRT we must use the absolute pressure. T1 = 278 K. p1 = 2.72 atm. T2 = 318 K. Solve: n, R constant, so

pV pV pV = nR = constant. 1 1 = 2 2 and T1 T2 T

⎛ 0.0150 m3 ⎞ ⎛ 318 K ⎞ ⎛ V ⎞⎛ T ⎞ p2 = p1 ⎜ 1 ⎟⎜ 2 ⎟ = ( 2.72 atm ) ⎜ ⎟ = 2.94 atm. 3⎟ ⎜ ⎟⎜ ⎝ V2 ⎠⎝ T1 ⎠ ⎝ 0.0159 m ⎠ ⎝ 278 K ⎠ The final gauge pressure is 2.94 atm − 1.02 atm = 1.92 atm. Reflect: Since a ratio is used, pressure can be expressed in atm. But absolute pressures must be used. The ratio of gauge pressures is not equal to the ratio of absolute pressures.

15.70. Set Up: We are asked to compare two states. Use the ideal-gas law to obtain m2 in of m1 and the ratio of pressures in the two states. In the derivation of Equation 15.4 it was shown that pV = nRT can be written pV = (m/M ) RT —apply this equation to the initial state to calculate m1. T, V, M, R are all constant, so p /m = RT /MV = constant. So we have p1 /m1 = p2 /m2 , where m is the mass of the gas in the tank.

Solve: p1 = 1.30 × 106 Pa + 1.01 × 105 Pa = 1.40 × 106 Pa p2 = 2.50 × 105 Pa + 1.01 × 105 Pa = 3.51 × 105 Pa m1 = p1VM /RT ; V = hA = hπ r 2 = (1.00 m)π (0.060 m) 2 = 0.01131 m3 m1 =

(1.40 × 106 Pa)(0.01131 m3 )(44.1 × 1023 kg/mol) = 0.2845 kg (8.3145 J/mol ⋅ K)((22.0 + 273.15)K)

⎛ 3.51 × 105 Pa ⎞ ⎛p ⎞ Then m2 = m1 ⎜ 2 ⎟ = (0.2845 kg) ⎜ = 0.0713 kg. ⎜ 1.40 × 106 Pa ⎟⎟ ⎝ p1 ⎠ ⎝ ⎠ m2 is the mass that remains in the tank. The mass that has been used is m1 − m2 = 0.2845 kg − 0.0713 kg = 0.213 kg. Reflect: Note that we have to use absolute pressures. The absolute pressure decreases by a factor of four and the mass of gas in the tank decreases by a factor of four.

15.71. Set Up: υrms = 3kT/m where k = 1.381 × 10−23 J/molecule ⋅ K. m is the mass of a single atom. Solve: (a) υrms =

3(1.381 × 10−23 J/molecule ⋅ K)(5800 K) 1.67 × 10−27 kg

= 1.2 × 104 m/s

(b) υrms m = 3kT = constant, so υrms,1 m1 = υ rms,2 m2 and m2 = m1 (υrms,1 /υrms,2 )2 . If υrms,2 = 12 υrms,1, then m2 = 4m1 = 6.68 × 10−27 kg.

15.72. Set Up: pV = NkT gives

N p = . 1 atm = 1.013 × 105 Pa. TK = TC + 273.15. k = 1.381 × 10−23 J/molecule ⋅ K. V kT

Solve: (a) TC = TK − 273.15 = 94 K − 273.15 = −179D C (b)

N p (1.5 atm)(1.013 × 105 Pa/atm) = = = 1.2 × 1026 molecules/m3 V kT (1.381 × 10223 J/molecule ⋅ K)(94 K)


15-18

Chapter 15

(c) For the earth, p = 1.0 atm = 1.013 × 105 Pa and T = 22D C = 295 K. N (1.0 atm)(1.013 × 105 Pa/atm) = = 2.5 × 1025 molecules/m3. V (1.381 × 10−23 J/molecule ⋅ K)(295 K) The atmosphere of Titan is about five times denser than earth’s atmosphere.

Reflect: Though it is smaller than earth and has weaker gravity at its surface, Titan can maintain a dense atmosphere because of the very low temperature of that atmosphere.

*15.73. Set Up: For helium, CV = 12.47 J/mol ⋅ K and C p = 20.78 J/mol ⋅ K. Solve: (a) ΔT = 0. ΔU = 0 and Q = W = 300 J. (b) Q = 0. ΔU = − W = − 300 J. (c) Δp = 0. W = p ΔV = nR ΔT . Q = nC p ΔT =

⎛ 20.78 J/mol ⋅ K ⎞ W =⎜ ⎟ (300 J) = 750 J. R ⎝ 8.315 J/mol ⋅ K ⎠

⎛C ⎞ ΔU = Q − W = 750 J − 300 J = 450 J. This is also ⎜ V ⎟ W . ⎝ R⎠

Reflect: In any process of an ideal gas, ΔU = nCV ΔT . ΔU is different for each of these processes because they have different values of ΔT .

15.74. Set Up: pV = nRT . For the isobaric process, W = p ΔV = nR ΔT . For the isothermal process, ⎛V ⎞ W = nRT ln ⎜ f ⎟ . ⎝ Vi ⎠ Solve: (a) The pV diagram for these processes is sketched in the figure below.

(b) Find T2 . For process 1 → 2, n, R, and p are constant so

T p T T = = constant. 1 = 2 and V nR V1 V2

⎛V ⎞ T2 = T1 ⎜ 2 ⎟ = (355 K)(2) = 710 K. ⎝ V1 ⎠ (c) The maximum pressure is for state 3. For process 2 → 3, n, R, and T are constant. p2V2 = p3V3 and ⎛V ⎞ p3 = p2 ⎜ 2 ⎟ = (2.40 × 105 Pa)(2) = 4.80 × 105 Pa. ⎝ V3 ⎠ (d) process 1 → 2: W = p ΔV = nR ΔT = (0.250 mol)(8.315 J/mol ⋅ K)(710 K − 355 K) = 738 K. ⎛V ⎞ ⎛1⎞ process 2 → 3: W = nRT ln ⎜ 3 ⎟ = (0.250 mol)(8.315 J/mol ⋅ K)(710 K)ln ⎜ ⎟ = − 1023 J. ⎝2⎠ ⎝ V2 ⎠ process 3 → 1: ΔV = 0 and W = 0. The total work done is 738 J + (−1023 J) = − 285 J. This is the work done by the gas. The work done on the gas is 285 J.



15-19

*15.75. Set Up: The mass of one molecule is the molar mass, M, divided by the number of molecules in a mole, N A . The average translational kinetic energy of a single molecule is 12 m(υ 2 )av = 32 kT . Use pV = NkT to calculate N, the number of molecules. k = 1.381 × 10−23 J/molecule ⋅ K. M = 28.0 × 10−3 kg/mol. T = 295.15 K. The volume

of the balloon is V = 43 π (0.250 m)3 = 0.0654 m3. p = 1.25 atm = 1.27 × 105 Pa.

Solve: (a) m = (b)

1 m(υ 2 )av 2

(c) N =

M 28.0 × 10−3 kg/mol = = 4.65 × 10−26 kg N A 6.022 × 1023 molecules/mol

= 32 kT = 32 (1.381 × 10−23 J/molecule ⋅ K)(295.15 K) = 6.11 × 10−21 J

pV (1.27 × 105 Pa)(0.0654 m3 ) = = 2.04 × 1024 molecules kT (1.381 × 10−23 J/molecule ⋅ K)(295.15 K)

(d) The total average translational kinetic energy is N

(

1 m(υ 2 )av 2

) = (2.04 × 10

24

molecules) (6.11 × 10221 J/molecule) = 1.25 × 104 J.

Reflect: The number of moles is n = 3 (3.39 2

N 2.04 × 1024 molecules = = 3.39 mol. We have K tr = 32 nRT = N A 6.022 × 1023 molecules/mol

mol)(8.314 J/mol ⋅ K)(295.15 K) = 1.25 × 104 J, which agrees with our results in part (d).

15.76. Set Up: For helium, CV = 12.47 J/mol ⋅ K and C p = 20.78 J/mol ⋅ K. ΔU = Q − W . W is the area under the p versus V curve. ΔU = nCV ΔT for any process of an ideal gas.

Solve: (a) W = 12 (1.0 × 105 Pa + 3.5 × 105 Pa)(0.0060 m3 − 0.0020 m3 )

+ 12 (1.0 × 105 Pa + 3.5 × 105 Pa)(0.0100 m3 − 0.0060 m3 ) = 1800 J. Find ΔT = Tc − Ta . p is constant so ΔT = Then ΔU = nCV ΔT =

p ΔV (1.0 × 105 Pa)(0.0100 m3 − 0.0020 m3 ) = = 289 K. 1 mol (8.315 J/mol ⋅ K) nR

(3

)

( 13 mol) (12.47 J/mol ⋅ K)(289 K) = 1.20 × 103 J

Q = ΔU + W = 1.20 × 103 J + 1800 J = 3.00 × 103 J. Q > 0 so this heat is transferred into the gas.

(b) This process is isobaric, so Q = nC p ΔT =

( 13 mol) (20.78 J/mol ⋅ K)(289 K) = 2.00 × 103 J.

Q > 0 so this heat is

transferred into the gas. (c) Q is larger in part (a). ΔU is the same because the initial and final states are the same, but in (a) more work is done.

15.77. Set Up: pV = nRT . V = Ax, where x is the length of the air column and A is the cross-sectional area of the cylinder. The final absolute pressure is pf = 1.01 × 105 Pa + 2.76 × 105 Pa = 3.77 × 105 Pa. Solve: nRT is constant, so piVi = pf Vf . pi Axi = pf Axf and

⎛ 1.01 × 105 Pa ⎞ ⎛ p ⎞ xf = xi ⎜ i ⎟ = (36.0 cm) ⎜ ⎟⎟ = 9.64 cm. 5 ⎜ ⎝ pf ⎠ ⎝ 3.77 × 10 Pa ⎠

15.78. Set Up: The pressure at depth h in a fluid is p = p0 + ρ gh, where p0 is the pressure at the surface. p0 = pair = 1.013 × 105 Pa. The density of water is ρ = 1000 kg/m3 .

Solve: p1 = p0 + ρ gh = 1.013 × 105 Pa + (1000 kg/m3 )(9.80 m/s 2 )(25 m) = 3.463 × 105 Pa p2 = pair = 1.013 × 105 Pa. V1 = 1.0 mm3 . n, R, and T are constant so pV = nRT = constant. p1V1 = p2V2 and


15-20

Chapter 15

⎛ 3.463 × 105 Pa ⎞ ⎛ p ⎞ = 3.4 mm3 V2 = V1 ⎜ 1 ⎟ = (1.0 mm3 ) ⎜ ⎜ 1.013 × 105 Pa ⎟⎟ ⎝ p2 ⎠ ⎝ ⎠ This is a large change and would have serious effects.

*15.79. Set Up: ab is at constant volume, ca is at constant pressure and bc is at constant temperature. For ΔT = 0, ΔU = 0 and Q = W = nRT ln (Vc /Vb ). For ideal H 2 (diatomic), CV = 52 R and C p = 72 R. ΔU = nCV ΔT for any process of an ideal gas. Solve: (a) Tb = Tc . For states b and c, pV = nRT = constant so pbVb = pcVc and

⎛p ⎞ ⎛ 2.0 atm ⎞ Vc = Vb ⎜ b ⎟ = (0.20 L) ⎜ ⎟ = 0.80 L p ⎝ 0.50 atm ⎠ ⎝ c⎠ (b) Ta =

paVa (0.50 atm)(1.013 × 105 Pa/atm)(0.20 × 10−3 m3 ) = = 305 K nR (0.0040 mol)(8.315 J/mol ⋅ K)

Va = Vb so for states a and b,

T V T T = = constant so a = b . p nR pa pb

⎛p ⎞ ⎛ 2.0 atm ⎞ Tb = Tc = Ta ⎜ b ⎟ = ( 305 K ) ⎜ ⎟ = 1220 K; Tc = 1220 K ⎝ 0.50 atm ⎠ ⎝ pa ⎠

(c) ab: Q = nCV ΔT = n

( 52 R) ΔT = (0.0040 mol) ( 52 ) (8.315 J/mol ⋅ K)(1220 K − 305 K) = + 76 J

Q is positive and heat goes into the gas. ca: Q = nC p ΔT = n 72 R ΔT = (0.0040 mol)

( )

( 72 ) (8.315 J/mol ⋅ K)(305 K − 1220 K) = − 107 J

Q is negative and heat comes out of the gas. bc: Q = W = nRT ln (Vc /Vb ) = (0.0040 mol)(8.315 J/mol ⋅ K)(1220 K)ln(0.80 L / 0.20 L) = 56 J Q is positive and heat goes into the gas. (d) ab: ΔU = nCV ΔT = n 52 R ΔT = (0.0040 mol)

( )

( 25 ) (8.315 J/mol ⋅ K)(1220 K − 305 K) = + 76 J

The internal energy increased. bc: ΔT = 0 so ΔU = 0. The internal energy does not change. ca: ΔU = nCV ΔT = n

( 52 R) ΔT = (0.0040 mol) ( 52 ) (8.315 J/mol ⋅ K)(305 K − 1220 K) = − 76 J

The internal energy decreased. Reflect: The net internal energy change for the complete cycle a → b → c → a is ΔU tot = + 76 J + 0 + (−76 J) = 0. For any complete cycle the final state is the same as the initial state and the net internal energy change is zero. For the cycle the net heat flow is Qtot = + 76 J + (−107 J) + 56 J = + 25 J. ΔU tot = 0 so Qtot = Wtot . The net work done in the cycle is positive and this agrees with our result that the net heat flow is positive.

15.80. Set Up: ab is at constant pressure, bc is at constant volume, and ca is isothermal. He is a monatomic gas so CV = 32 R and C p = 52 R. For any process of an ideal gas, ΔU = nCV ΔT . For an isothermal process of an ideal gas,

ΔU = 0 so Q = W = nRT ln (V2 /V1 ). Solve: (a) Apply pV = nRT to states a and c. Ta = Tc so nRT is constant and paVa = pcVc .

⎛ 0.040 m3 ⎞ ⎛V ⎞ = 8.0 × 105 Pa pa = pc ⎜ c ⎟ = (2.0 × 105 Pa) ⎜ ⎜ 0.010 m3 ⎟⎟ V ⎝ a⎠ ⎝ ⎠ (b) Ta =

paVa (8.0 × 105 Pa)(0.010 m3 ) = = 296 K; nR (3.25 mol)(8.315 J/mol ⋅ K) Tb =

pbVb (8.0 × 105 Pa)(0.040 m3 ) = = 1184 K; nR (3.25 mol)(8.315 J/mol ⋅ K)



Tc =

(c) ab: Q = nC p ΔT = (3.25 mol) bc: Q = nCV ΔT = (3.25 mol)

15-21

pcVc (2.0 × 105 Pa)(0.040 m3 ) = = 296 K = Ta nR (3.25 mol)(8.315 J/mol ⋅ K)

( 52 ) (8.315 J/mol ⋅ K)(1184 K − 296 K) = 6.00 × 104 J; heat enters the gas.

( 32 ) (8.315 J/mol ⋅ K)(296 K − 1184 K) = − 3.60 × 104 J; heat leaves the gas.

⎛ 0.010 m3 ⎞ ⎛V ⎞ = − 1.11 × 104 J; heat leaves the gas ca: Q = nRT ln ⎜ a ⎟ = (3.25 mol)(8.315 J/mol ⋅ K)(296 K)ln ⎜ ⎜ 0.040 m3 ⎟⎟ V ⎝ c⎠ ⎝ ⎠ (d) ab: ΔU = nCV ΔT = (3.25 mol) bc: ΔU = nCV ΔT = (3.25 mol)

( 32 ) (8.315 J/mol ⋅ K)(1184 K − 296 K) = 3.60 × 104 J; the internal energy increased.

( 32 ) (8.315 J/mol ⋅ K)(296 K − 1184 K) = − 3.60 × 104 J; the internal energy decreased.

ca: ΔT = 0 so ΔU = 0

*15.81. Set Up: In part (a), apply pV = nRT to the ethane in the flask. The volume is constant once the stopcock is in place. In part (b) apply pV =

mtot RT to the ethane at its final temperature and pressure. 1.50 L = 1.50 × 10−3 m3. M

M = 30.1 × 10−3 kg/mol. Neglect the thermal expansion of the flask.

Solve: (a) p2 = p1 (T2 / T1 ) = (1.013 × 105 Pa)(300 K/ 380 K) = 8.00 × 104 Pa.

⎛ (8.00 × 104 Pa)(1.50 × 10−3 m3 ) ⎞ ⎛pV⎞ (30.1 × 10−3 kg/mol) = 1.45 g. (b) mtot = ⎜ 2 ⎟ M = ⎜ ⎜ (8.3145 J/mol ⋅ K)(300 K) ⎟⎟ ⎝ RT2 ⎠ ⎝ ⎠ Reflect: We could also calculate mtot with

p = 1.013 × 105 Pa and T = 380 K, and we would obtain the

same result. Originally, before the system was warmed, the mass of ethane in the flask was ⎛ 1.013 × 105 Pa ⎞ = 1.84 g. m = (1.45 g) ⎜ ⎜ 8.00 × 104 Pa ⎟⎟ ⎝ ⎠

15.82. Set Up: Since no heat is put into the air during the compression, that process is adiabatic. For air, γ = 1.40. Treat the air as an ideal gas. For any process of an ideal gas, ΔU = nCV ΔT . From Table 15.4, CV = 20.76 J/mol ⋅ K for N 2 and this is the value we will use for air. 10° F is the same temperature as Solve: (a) For an adiabatic process T1V1γ V2 = V10.4 (T1/T2 ). V2 = V1 (T1/T2 )1/0.40 =

−1

= T2V2γ

−1

5 (10° − 9

32°) = − 12.2°C = 261 K.

, where T is in kelvins. V2γ

−1

= V1γ

−1

(T1 / T2 ) and

0.4

(0.50 L)(293 K/823 K) 2.5 = 0.0378 L

(b) ΔU = nCV ΔT = (0.050 mol)(20.76 J/mol ⋅ K)(550°C − 20°C) = 550 J (c) For the compression, Q = 0, so ΔU = Q − W says W = − ΔU = − 550 J. The piston did 550 J of work on the gas. (d) V1 = 0.50 L and V2 = 0.0378 L. T1 = 283 K. Solve for T2 . T1V1γ T2 = T1 (V1/V2 )

0 .4

−1

= (261 K)(0.50 L/0.0378 L)

= T2V2γ

0.4

−1

so

= 733 K = 460°C.

(e) If the initial temperature is too low the final temperature produced by the compression might not be high enough to ignite the fuel. The air could be preheated by an electrical heater before it is compressed.

15.83. Set Up: The initial absolute pressure is 2.39 × 105 Pa and the final absolute pressure is 1.24 × 105 Pa. For a constant pressure process, W = p ΔV . For an adiabatic process, Q = 0 and W = − ΔU = − nCV ΔT . The two processes are shown in the figure below. The gas does positive work when it expands.


15-22

Chapter 15

Solve: Constant pressure process 1 → 2. W = p ΔV = (2.39 × 105 Pa)(1.42 m3 − 0.28 m3 ) = 2.72 × 105 J

Adiabatic process 2 → 3. pV = nRT so T2 =

p2V2 pV C and T3 = 3 3 . W = − V ( p3V3 − p2V2 ) nR nR R

⎛ 20.8 J/mol ⋅ K ⎞ 5 3 5 3 5 W = −⎜ ⎟ ([1.24 × 10 Pa][2.27 m ] − [2.39 × 10 Pa][1.42 m ]) = 1.45 × 10 J. 8 315 J/mol K . ⋅ ⎝ ⎠ The total work done by the gas is 2.72 × 105 J + 1.45 × 105 J = 4.17 × 105 J. Reflect: Each process is an expansion, so the work done during each is positive.

15.84. Set Up: For helium, CV = 12.47 J/mol ⋅ K, C p = 20.78 J/mol ⋅ K and γ = 1.67. ΔU = nCV ΔT for any process of an ideal gas. ΔU = Q − W . Solve: (a) isobaric, Δp = 0 : pV = nRT . If V doubles while p is constant, then T doubles. T1 = 300 K and T2 = 600 K. W = p ΔV = nR ΔT = (4.00 mol)(8.315 J/mol ⋅ K)(300 K) = 9.98 × 103 J.

(b) adiabatic, Q = 0: T1V1γ

−1

= T2V2γ

−1

. γ −1

⎛V ⎞ T2 = T1 ⎜ 1 ⎟ ⎝ V2 ⎠

⎛1⎞ = (300 K) ⎜ ⎟ ⎝2⎠

0.67

= 189 K.

W = − ΔU = − nCV ΔT = − (4.00 mol)(12.47 J/mol ⋅ K)(189 K − 300 K) = 5.54 × 103 J.

(c) The pV diagrams are sketched in the figure below. More work is done in the isobaric process. The area under the curve in the pV diagram is larger for that process.

(d) For the isobaric process, Q = nC p ΔT = (4.00 mol)(20.78 J/mol ⋅ K)(300 K) = 2.49 × 104 J. For the adiabatic pro-

cess, Q = 0. Q is greater for the isobaric process. (e) ΔU = nCV ΔT . For the isobaric process, ΔU = (4.00 mol)(12.47 J/mol ⋅ K)(300 K) = 1.50 × 104 J. For the adia-

batic process, ΔU = − W = − 5.54 × 103 J. The magnitude of the change in temperature is greater for the isobaric process, so the magnitude of the change in internal energy is greater for that process.



15-23

*15.85. Set Up: Use Q = nCV ΔT to calculate the temperature change in the constant volume process and use pV = nRT to calculate the temperature change in the constant pressure process. The work done in the constant volume process is zero and the work done in the constant pressure process is W = pΔV . Use Q = nC p ΔT to calculate the heat

flow in the constant pressure process. ΔU = nCV ΔT , or ΔU = Q − W . For N 2 , CV = 20.76 J/mol ⋅ K and C p = 29.07 J/mol ⋅ K. Solve: (a) For process ab, ΔT =

Q 1.52 × 104 J = = 293 K. Ta = 293 K, so Tb = 586 K. pV = nRT nCV (2.50 mol)(20.76 J/mol ⋅ K)

says T doubles when V doubles and p is constant, so Tc = 2(586 K) = 1172 K = 899D C. (b) For process ab, Wab = 0. For process bc, Wbc = pΔV = nRΔT = (2.50 mol)(8.314 J/mol ⋅ K)(1172 K − 586 K) =

1.22 × 104 J. W = Wab + Wbc = 1.22 × 104 J. (c) For process bc, Q = nC p ΔT = (2.50 mol)(29.07 J/mol ⋅ K)(1172 K − 586 K) = 4.26 × 104 J. (d) ΔU = nCV ΔT = (2.50 mol)(20.76 J/mol ⋅ K)(1172 K − 293 K) = 4.56 × 104 J. Reflect: The total Q is 1.52 × 104 J + 4.26 × 104 J = 5.78 × 104 J.

ΔU = Q − W = 5.78 × 104 J − 1.22 × 104 J = 4.56 × 104 J, which agrees with our results in part (d).

15.86. Set Up: The buoyant force is FB = ρairVg . A lift of 290 kg means mass of hot air in the balloon. m = ρV . mhot = ρ hotV .

ρ hot = ρair − ρ hot =

FB − mhot = 290 kg, where mhot is the g

FB − mhot = 290 kg gives ( ρair − ρ hot )V = 290 kg. g

290 kg 290 kg = 1.23 kg/m3 − = 0.65 kg/m3 . V 500.0 m3

pM pM . ρair = . ρ hotThot = ρairTair and RTair RThot

⎛ρ Thot = Tair ⎜ air ⎝ ρ hot

⎛ 1.23 kg/m3 ⎞ ⎞ = 545 K = 272°C. ⎟ = (288 K) ⎜⎜ 3⎟ ⎟ ⎠ ⎝ 0.65 kg/m ⎠

Solutions to age Problems *15.87. Set Up: The internal energy of an ideal gas is proportional to nRT . The constant of proportionality depends on the number of degrees of freedom that the gas has at the given temperature. Solve: Near room temperature a diatomic gas has 5 degrees of freedom and a monatomic gas has 3 degrees of ⎛5 ⎞ ⎛3 ⎞ freedom. For n moles of each gas the ratio of their internal energies will be ⎜ nRT ⎟ / ⎜ nRT ⎟ = 5/3. The correct ⎝2 ⎠ ⎝2 ⎠ answer is E.

1 nRT . The constant of proportionality 2 depends on the number of degrees of freedom that the gas has at the given temperature. For 9 degrees of freedom the 9 internal energy would be nRT . 2 Solve: Use the work leading to Equation 15.13 to find the molar heat capacity at constant volume: At constant volume, all of the heat added to the gas is used to increase the internal energy of the gas (since there is no work done) 9 9 and we have nCV ΔT = nRΔT . Thus, CV = R and the correct answer is D. 2 2

15.88. Set Up: The internal energy of an ideal gas is proportional to


15-24

Chapter 15

1 R. 2 The constant of proportionality depends on the number of degrees of freedom that the gas has at the given temperature.

*15.89. Set Up: As seen in the previous problem, the specific heat capacity of an ideal gas is proportional to

Solve: Since a monatomic ideal gas has 3 degrees of freedom at all temperatures, we expect its molar heat capacity 3 at constant volume to be R at all temperatures. The correct answer is A. 2 1 R. 2 The constant of proportionality depends on the number of degrees of freedom that the gas has at the given temperature.

15.90. Set Up: As seen in the previous problem, the specific heat capacity of an ideal gas is proportional to

Solve: Since the number of degrees of freedom of a diatomic gas increases from 3 to 7 as the temperature of the gas increases, we expect its molar heat capacity at constant volume to increase in proportion. The correct answer is B.


16

THE SECOND LAW OF THERMODYNAMICS

Answers to Multiple-Choice Problems 1. B 2. B 3. A 4. A, B, C, D 13. C 14. A, D 15. B

5. A

6. A

7. B, D

8. B

9. C

10. B, D

11. A, D

12. B

Solutions to Problems *16.1. Set Up: We have e = W /QH = 0.38. Also, W = QH + QC where W and QH are positive, and QC is negative. In one second the work done by the power plant is W = 750 MJ.

J ⎞ ⎛ 24 h ⎞ ⎛ 3600 s ⎞ ⎛ 13 Solve: The power output of the plant is ⎜ 750 × 106 ⎟ ⎜ ⎟⎜ ⎟ = 6.48 × 10 J/day. In one day the required s 1 day 1 h ⎝ ⎠⎝ ⎠ ⎠⎝ input heat is W 6.48 × 1013 J QH = = = 1.71 × 1014 J. So we have e 0.38 QC = W − QH = 6.48 × 1013 J − 1.71 × 1014 J = −1.1 × 1014 J. Thus, in one day 1.1 × 1014 J of heat is discarded into the

outside air. Reflect: Q is positive when heat enters a system and negative when heat leaves a system. Thus, QC is counted as negative as it leaves the power plant (system 1) and positive as it enters the environment (system 2).

16.2. Set Up: W = QH + QC . W and QH are positive; QC is negative. QH = + 325 J and QC = − 250 J. e=

W Q =1+ C . QH QH

Solve: e = 1 +

−250 J = 0.231 = 23.1,. W = QH + QC = − 250 J + 325 J = + 75 J 325 J

16.3. Set Up: For a heat engine, W = QH − QC . e =

W . QH > 0, QC < 0. QH

Solve: (a) W = 2200 J. QC = 4300 J. QH = W + QC = 6500 J.

2200 J = 0.34 = 34,. 6500 J Reflect: Since the engine operates on a cycle, the net Q equals the net W. But to calculate the efficiency we use the heat energy input, QH . (b) e =


16-1

16-2

Chapter 16

16.4. Set Up: For a heat engine, W = QH − QC . e =

W . QH > 0, QC < 0. QH

Solve: (a) W = 9000 J − 6400 J = 2600 J. (b) e =

W 2600 J = = 0.29 = 29,. QH 9000 J

*16.5. Set Up: For a heat engine, W = QH − QC . e = Solve: (a) e =

W . QH > 0, QC < 0. 1 hp = 746 W. QH

W 3700 J = = 0.230 = 23.0,. QH 1.61 × 104 J

(b) QC = QH − W = 1.61 × 104 J − 3700 J = 1.24 × 104 J. (c) m = (d) P =

1.61 × 104 J 4.60 × 104 J/g

= 0.350 g.

(60.0)(3700 J) = 2.22 × 105 W = 222 kW = 298 hp. 1.00 s

16.6. Set Up: For 1.00 s, W = 180 × 103 J. W = QH − QC . e = Solve: (a) QH =

W . QH > 0, QC < 0. QH

W 180 × 103 J = = 6.43 × 105 J. e 0.280

(b) QC = QH − W = 6.43 × 105 J − 1.80 × 105 J = 4.63 × 105 J.

16.7. Set Up: W = QH − QC . e = Solve: (a) e =

W . QH > 0, QC < 0. QH

W W /t 330 MW = = = 0.25 = 25,. QH QH /t 1300 MW

(b) QC = QH − W so QC /t = QH /t − W/t = 1300 MW − 330 MW = 970 MW. Reflect: The equations for e and W have the same form when written in of power output and rate of heat flow.

16.8. Set Up: ab and cd are constant pressure processes. For a constant pressure process W = p ΔV and Q = n ΔT. pV = nRT gives n ΔT =

⎛ ⎞ p ΔV 7 so Q = ⎜ ⎟ p ΔV . For a diatomic ideal gas C p = 2 R. bc and da are constant R ⎝ R ⎠

volume processes. For a constant volume process, W = 0 and Q = nCV ΔT . pV = nRT gives n ΔT =

V Δp so R

⎛C ⎞ Q = ⎜ V ⎟ V Δp. For a diatomic ideal gas CV = 52 R. 1 atm = 1.013 × 105 Pa and 1 L = 10−3 m3 ⎝ R⎠

Solve: (a) Heat goes into the gas when T increases, and this occurs for ab and da. ⎛ ⎞ 5 −3 3 7 ab Q = ⎜ ⎟ p ΔV = 2 (5.0 atm)(1.013 × 10 Pa/atm)(4.0 × 10 m ) = 7091 J R ⎝ ⎠

()

da

⎛C ⎞ Q = ⎜ V ⎟ V Δp = ⎝ R⎠

( 52 ) (4.0 atm)(1.013 × 105 Pa/atm)(2.0 × 10−3 m3 ) = 2026 J

The total heat that enters is QH = 7091 J + 2026 J = 9120 J. (b) Heat comes out of the gas when T decreases, and this occurs for bc and cd. ⎛C ⎞ bc Q = ⎜ V ⎟ V Δp = 52 (−4.0 atm)(1.013 × 105 Pa/atm)(6.0 × 10−3 m3 ) = − 6078 J ⎝ R⎠

()


The Second Law of Thermodynamics

16-3

⎛ ⎞ 5 −3 3 7 Q=⎜ ⎟ p ΔV = 2 (1.0 atm)(1.013 × 10 Pa/atm)(−4.0 × 10 m ) = − 1418 J ⎝ R ⎠ The total heat that leaves the gas is QC = − 6078 J − 1418 J = − 7500 J.

()

cd

(c) ab W = p ΔV = (5.0 atm)(1.013 × 105 Pa/atm)(4.0 × 10−3 m3 ) = + 2026 J bc W = 0

cd W = p ΔV = (1.0 atm)(1.013 × 105 Pa/atm)(−4.0 × 10−3 m3 ) = − 405 J da W = 0 The total work done in one cycle is W = + 2026 J − 405 J = + 1620 J. Note that QH + QC = + 9120 J + (−7500 J) = 1620 J. W = QH + QC , as it should for a cycle, where ΔU = 0. (d) e =

1620 J W = = 0.178 = 17.8, QH 9120 J

*16.9. Set Up: ca is at constant volume, ab has Q = 0, and bc is at constant pressure. For a constant pressure process W = p ΔV and Q = nC p ΔT . pV = nRT gives n ΔT =

⎛ ⎞ p ΔV so Q = ⎜ ⎟ p ΔV . If γ = 1.40 the gas is R ⎝ R ⎠

diatomic and C p = 72 R. For a constant volume process W = 0 and Q = nCV ΔT . pV = nRT gives n ΔT =

V Δp so R

⎛C ⎞ Q = ⎜ V ⎟ V Δp. For a diatomic ideal gas CV = 52 R. 1 atm = 1.013 × 105 Pa ⎝ R⎠

Solve: (a) Vb = 9.0 × 1023 m3 , pb = 1.5 atm and Va = 2.0 × 1023 m3 . For an adiabatic process paVaγ = pbVbγ . 1.4

γ

⎛ 9.0 × 1023 m3 ⎞ ⎛V ⎞ pa = pb ⎜ b ⎟ = (1.5 atm) ⎜ = 12.3 atm ⎜ 2.0 × 1023 m3 ⎟⎟ ⎝ Va ⎠ ⎝ ⎠ (b) Heat enters the gas in process ca, since T increases. ⎛C ⎞ Q = ⎜ V ⎟ V Δp = 52 (2.0 × 10−3 m3 )(12.3 atm − 1.5 atm)(1.013 × 105 Pa/atm) = 5470 J ⎝ R⎠

()

QH = 5470 J (c) Heat leaves the gas in process bc, since T increases. ⎛ ⎞ 5 −3 3 7 Q=⎜ ⎟ p ΔV = 2 (1.5 atm)(1.013 × 10 Pa/atm)(−7.0 × 10 m ) = − 3723 J ⎝ R ⎠ QC = − 3723 J

()

(d) W = QH + QC = + 5470 J + (−3723 J) = 1747 J

W 1747 J = = 0.319 = 31.9, QH 5470 J Reflect: We did not use the number of moles of the gas. (e) e =

16.10. Set Up: For the Otto cycle, e = 1 −

1 r

γ −1

e= so 1 − e =

QC QH

. R is the compression ratio.

Q W Q =1+ C =1− C QH QH QH

is the fraction that is wasted.

Solve: (a) e = 1 −

1 8.500.67

= 0.762 = 76.2, and 23.8% is the fraction of the gasoline that produces waste heat.


16-4

Chapter 16

(b) For r = 10.0 , e = 1 −

1 10.00.67

= 0.786 = 78.6, and 21.4% is wasted.

*16.11. Set Up and Solve: e = 1 −

1 r

γ −1

.

1 r

γ −1

= 1 − e = 0.350. r 0.40 =

1 and r = 13.8. 0.350

1

. For a monatomic ideal gas γ = 1.67 and for a diatomic ideal gas γ = 1.40. r Solve: The monatomic gas gives a larger e. For a monatomic gas, 1 = 0.741 = 74.1,. e =1− 7.500.67 For a diatomic gas, 1 = 0.553 = 55.3,. e =1− 7.500.40

16.12. Set Up: e = 1 −

γ −1

16.13. Set Up: Apply e = 1 −

1 r

γ −1

. e = 1−

QC QH

. In part (b), QH = 10,000 J. The heat discarded is QC .

1 = 0.594 = 59.4,. 9.500.40 (b) QC = QH (1 − e) = (10,000 J)(1 − 0.594) = 4060 J. Solve: (a) e = 1 −

Reflect: The work output of the engine is W = QH − QC = 10,000 J − 4060 J = 5940 J

16.14. Set Up: For a refrigerator, the coefficient of performance is K =

qQCq . qQCq = 1750 J and qW q = 785 J. qW q

qQH q = qQCq + qW q.

1750 J = 2.23 785 J (b) qQH q = qQCq + qW q = 1750 J + 785 J = 2535 J Solve: (a) K =

*16.15. Set Up: For a refrigerator, qQH q = qQCq + qW q. QC > 0, QH < 0. K = Solve: (a) W =

qQCq . qW q

qQCq 3.40 × 104 J = = 1.62 × 104 J. K 2.10

(b) qQH q = 3.40 × 104 J + 1.62 × 104 J = 5.02 × 104 J. Reflect: More heat is discarded to the high-temperature reservoir than is absorbed from the cold reservoir.

16.16. Set Up: qQH q = qQCq + qW q. Solve: qW q = qQH q − qQCq = 1.44 × 105 J − 9.80 × 104 J = 4.60 × 104 J.

P=

*16.17. Set Up: qQH q = qQCq + qW q. K =

qW q 4.60 × 104 = = 767 W. t 60 s

qQCq . For water, cw = 4190 J/kg ⋅ K and Lf = 3.34 × 105 J/kg. For ice, W

cice = 2010 J/kg ⋅ K. Solve: (a) Q = mcice ΔTice − mLf + mcw ΔTw © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


16-5

Q = (1.80 kg)([2010 J/kg ⋅ K][−5.0 C°] − 3.34 × 105 J/kg + [4190 J/kg ⋅ K][ −25.0 C°]) = − 8.08 × 105 J Q = − 8.08 × 105 J. Q is negative for the water since heat is removed from it. (b) qQCq = 8.08 × 105 J.

W=

qQCq 8.08 × 105 J = = 3.37 × 105 J. K 2.40

(c) qQH q = 8.08 × 105 J + 3.37 × 105 J = 1.14 × 106 J. Reflect: For this device, QC > 0 and QH < 0. More heat is rejected to the room than is removed from the water.

16.18. Set Up: The coefficient of performance is given by K =

qQCq . According to Appendix D, 1Btu/h = 0.293 W qW q

and 1 hp = 746 W. 1 hp (assume 10 1 two significant figures) with an efficiency of 70.0%, which requires an input to the motor of hp/0.700 = 10 W ⎞ ⎛ 0.143 hp = 107 W. The resulting rate of cooling is (1270 Btu/h) ⎜ 0.293 ⎟ = 372 W. Thus, over a period of 1 ⎝ Btu/h ⎠ Solve: According to its specifications, the chiller is driven by a motor that produces an output of

second the input work is W = 107 J and the heat moved is qQCq = 372 J. The coefficient of performance must be K=

qQCq 372 J = = 3.5. qW q 107 J

Reflect: Electric motors are typically rated by their power output. Since the given motor has an output of

1 hp, its 10

input must be somewhat greater due to the fact that the motor is only 70.0% efficient.

*16.19. Set Up: qW q = qQH q − qQCq.

QC T W = − C . QC < 0, QH > 0. e = . QH TH QH

Solve: (a) qW q = qQH q − qQCq = 550 J − 335 J = 215 J. (b)

TC Q Q ⎛ −335 J ⎞ = − C . TC = − TH C = − (620 K) ⎜ ⎟ = 378 J. TH QH QH ⎝ 550 J ⎠

(c) e =

W 215 J = = 0.391 = 39.1,. QH 550 J

16.20. Set Up: The maximum possible efficiencies is the Carnot efficiency, eCarnot = 1 −

TC . In this expression TH

TC and TH must be in kelvins. TC = 8°C = 281 K and TH = 25°C = 298 K. Solve: eCarnot = 1 −

281 K = 0.057 = 5.7,. 298 K

16.21. Set Up: qW q = qQH q − qQCq.

QC T W = − C . QC < 0, QH > 0. e = . QH TH QH

⎛T ⎞ ⎛ 300 K ⎞ 3 Solve: (a) QC = − QH ⎜ C ⎟ = − (6.45 × 103 J) ⎜ ⎟ = − 3.72 × 10 J. ⎝ 520 K ⎠ ⎝ TH ⎠ (b) qW q = qQH q − qQCq = 6.45 × 103 J − 3.72 × 103 J = 2.73 × 103 J (c) e =

W 2.73 × 103 J = = 0.423 = 42.3,. QH 6.45 × 103 J


16-6

Chapter 16

16.22. Set Up: e =

W Q T for any engine. For the Carnot cycle, C = 2 C . TC = 20.0°C + 273.15 K = 293.15 K QH QH TH

W 2.5 × 104 J = = 4.24 × 104 J 0.59 e

Solve: (a) QH =

(b) W = QH + QC so QC = W − QH = 2.5 × 104 J − 4.24 × 104 J = 21.74 × 104 J.

TH = 2TC

⎛ 4.24 × 104 J ⎞ QH = − (293.15 K) ⎜ = 714 K = 441°C. ⎜ 21.74 × 104 J ⎟⎟ QC ⎝ ⎠

Reflect: For a heat engine, W > 0, QH > 0 and QC < 0.

*16.23. Set Up: TC = 273 K, TH = 297 K. For water, Lf = 3.34 × 105 J/kg. QC > 0, QH < 0.

QC T =− C. QH TH

qQH q = qW q + qQCq. Solve: (a) qQCq is the heat removed from the water at 0°C to make ice at 0°C.

qQCq = mLf = (85.0 kg)(3.34 × 105 J/kg) = 2.84 × 107 J. ⎛T ⎞ ⎛ 297 K ⎞ 7 7 QH = − ⎜ H ⎟ QC = − ⎜ ⎟ (2.84 × 10 J) = − 3.09 × 10 J 273 K T ⎝ ⎠ ⎝ C⎠ (b) qW q = qQH q − qQCq = 3.09 × 107 J − 2.84 × 107 J = 2.5 × 106 J.

16.24. Set Up: For a Carnot refrigerator, K = K=

TC . TC = − 10 + 273 = 263 K. TH = 20 + 273 = 293 K. Also, TH − TC

qQCq , where qQCq is the heat that must be removed from the water to cool it to 0°C and freeze it. For water, qW q

c = 4.19 × 103 J/kg ⋅ C° and Lf = 334 × 103 J/kg. qQH q = qW q + qQCq, where qQH q is the heat expelled to the room. Solve: (a) K =

TC 263 K = = 8.77 TH − TC 293 K − 263 K

(b) Q = mcw ΔTw − mLf

Q = (0.375 kg)([4.19 × 103 J/kg ⋅ C°][0°C − 18°C] − 334 × 103 J/kg) = − 1.54 × 105 J 1.54 × 105 J must be removed from the water to convert it to ice at 0° C (c) qW q =

qQCq 1.54 × 105 J = = 1.76 × 104 J K 8.77

(d) qQH q = qW q + qQCq = 1.76 × 104 J + 1.54 × 105 J = 1.72 × 105 J Reflect: More energy is expelled to the room than is removed from the water. K is larger when TH is close to TC ;

less energy must be expelled to move heat energy through a smaller temperature difference.

*16.25. Set Up: For an engine, W and QH are positive and QC is negative. For a refrigerator, QC is positive and W and QH are negative. A heat pump air conditioner takes in heat at a cool place and expels heat into a warm place. A heat pump house heater takes in heat at a cool place and expels heat into a warm place. For both types of heat pumps, W is negative and energy must be supplied to operate the device. bc and da are adiabatic, so the heat flow for these processes is zero. Solve: (a) The cycle is clockwise. More positive work is done during ab and bc than the magnitude of the negative work done in cd and da, so the net work done in the cycle is positive. Heat enters the gas in ab and leaves the gas in cd. (b) The cycle is counterclockwise and the net work done in the cycle is negative. Heat enters the gas in dc and leaves the gas in ba. dc occurs inside the food compartment and ba occurs in the air of the room.



16-7

(c) The cycle is counterclockwise and the net work done in the cycle is negative. Heat enters the gas in dc and leaves the gas in ba. dc occurs inside the house and ba occurs outside. (d) The cycle is counterclockwise and the net work done in the cycle is negative. Heat enters the gas in dc and leaves the gas in ba. dc occurs outside and ba occurs inside the house. Reflect: For a refrigerator the signs of W, QH , and QC are all opposite to what they are for an engine.

16.26. Set Up: ΔS =

Q for each object. For water, Lf = 3.34 × 105 J/kg. T

Solve: (a) The heat flow into the ice is Q = mLf = (0.350 kg)(3.34 × 105 J/kg) = 1.17 × 105 J. The heat flow occurs at T = 273 K, so

ΔS = Q is positive and ΔS is positive.

Q 1.17 × 105 J = = 429 J/K. T 273 K

(b) Q = − 1.17 × 105 J flows out of the heat source, at T = 298 K.

ΔS =

Q 21.17 × 105 J = = − 393 J/K. T 298 K

Q is negative and ΔS is negative. (c) ΔS tot = 429 J/K + (−393 J/K) = + 36 J/K. Reflect: For the total isolated system, ΔS > 0 and the process is irreversible. Q , where T is the Kelvin temperature at which the heat T flow occurs. The heat flows in this problem are irreversible, but since ΔS is path independent the entropy change is

16.27. Set Up: For a reversible isothermal process ΔS =

the same as for a reversible heat flow. The heat flow when the ice melts is Q = mLf , with Lf = 334 × 103 J/kg. Heat flows out of the ocean (Q < 0) and into the air (Q > 0). The heat flow for the ice occurs at T = 0°C = 273.15 K. The heat flow for the ocean occurs at T = 3.50°C = 276.65 K. Solve: Q = mLf = (4.50 kg)(334 × 103 J/kg) = 1.50 × 106 J.

For the ice, ΔS =

Q +1.50 × 106 J = = 5.49 × 103 J/K. T 273.15 K

For the ocean, ΔS =

Q 21.50 × 106 J = = − 5.42 × 103 J/K. 276.65 K T

The net entropy change is 5.49 × 103 J/K + ( −5.42 × 103 J/K) = + 70 J/K. The entropy of the world increases. Q , where qQ q is the heat flow of 125 J each second. This heat flows into the factory air T (Q > 0) at a temperature of 42°C = 315 K. The heat flows out of the furnace (Q < 0) at a temperature of 175°C = 448 K.

16.28. Set Up: ΔS =

Solve: ΔS =

125 J 125 J − = + 0.118 J/K. The operation of the furnace causes the entropy of the factory to 315 K 448 K

increase.

*16.29. Set Up: Apply Qsystem = 0 to calculate the final temperature. The equation Q = mcΔT gives the heat required to warm or cool the water (for water we have c = 4190 J/kg ⋅ K). The temperature of boiling water is T = 100.0J C = 373 K. For an isothermal process we have ΔS =

Q . T

Solve: (a) The heat transfer between 100D C water and 30D C water occurs over a finite temperature difference and the process is irreversible. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16-8

Chapter 16

(b) (270 kg)c(T2 − 30.0D C) + (5.00 kg)c(T2 − 100D C) = 0. T2 = 31.27D C = 304.42 K. (c) As explained in Example 16.8, when the temperature of an object changes, an exact calculation of its entropy change requires calculus; however, we can make a rough estimate by assuming a sequence of nearly isothermal steps. For the water in the tub, we have added (270 kg)(4190 J/kg ⋅ K)(31.27D C − 30.0D C) = 1.44 × 106 J. Since the temp-

erature of the bathwater only increases by 1.27D C we can treat this process as isothermal. Thus, the entropy of the Q 1.44 × 106 J = = 14.75 × 103 J / K. The change in entropy of the boiling T 303 K water is more difficult to estimate due to its large temperature change. Following Example 16.8, we must remove (5.00 kg)(4190 J/kg ⋅ K)(1D C) = 20,950 J of heat for each degree the temperature of the boiling water is reduced. Thus, we can form a chain of nearly isothermal steps starting from 373 K → 372 K and ending with 305 K → 304 K. The approximate change in entropy is 1 1 1 ⎞ ⎛ 1 3 ΔS2 ≈ (−20,950 J) ⎜ + + ... ⎟ = (−20,950 J)(0.2042) = −4.28 × 10 J/K. The total change in entropy ⎝ 373 372 371 305 ⎠ bathwater increases by approximately ΔS1 =

is approximately ΔS = ΔS1 + ΔS2 ≈ + 4.75 × 103 J/K + −4.28 × 103 J/K = 470 J/K. Reflect: ΔSsystem > 0, as it should for an irreversible process.

16.30. Set Up: During the phase transition the gallium is at a constant temperature equal to the melting point of gallium. Your hand is at a constant temperature of 98.6D F = 37.0D C = 310.1 K. Heat qQ q = mLf flows out of your hand and into the gallium. For heat flow at constant temperature we have ΔS =

Q . For gallium, Lf = 8.04 × 104 J/kg and the T

melting point is 29.8D C = 303.0 K. Q 22.01×103 J = 26.48 J/K. = T 310.1 K Q = +6.63 J/K. Heat flows out of your hand, Q is negative, and ΔS is negative. For the gallium we have ΔS = 303.0 K The temperature of the gallium is less than that of your hand and qQ q is the same, so the magnitude of the entropy Solve: qQ q = mLf = (25.0 × 10−3 kg)(8.04 × 104 J/kg) = 2.01 × 103 J. For your hand, ΔS =

change of the gallium is greater than the magnitude of the entropy change of your hand. Reflect: For the gallium, ΔS > 0, so ΔSsystem > 0 and the process is irreversible.

Q . ΔU = Q − W . For an isothermal process of an ideal gas, ΔU = 0 and Q = W . For a T compression, ΔV < 0 and W < 0.

*16.31. Set Up: ΔS =

Solve: Q = W = − 1850 J. ΔS =

21850 J = − 6.31 J/K. 293 K

16.32. Set Up: qQH q leaves the environment at TH = 400°C = 673 K. qQCq enters the environment at TC = 20°C = 293 K. ΔS =

Q T

Solve: For a distance of 1.0 mi the car uses (1.0 mi/35 mi/gal) = 0.0286 gal. For 0.0286 gal, QH = (0.0286 gal)

(1.23 × 108 J/gal) = 3.52 × 106 J. QC = 34 (QH ) = 2.64 × 106 J. 23.52 × 106 J 12.64 × 106 J + = 3780 J/K 673 K 293 K The net entropy change of the environment is positive.

ΔS =



16-9

16.33. Set Up: We are asked to find the entropy change of the person. The person is not an isolated system. In 1.0 s, 0.80(80 J) = 64 J of heat enters the person’s body at 37°C = 310 K. This amount of heat leaves the person at a temperature of 30°C = 303 K. ΔS =

Q . T

Solve: For the person,

164 J 264 J + = − 4.8 × 1023 J/K. 310 K 303 K The entropy of the person can decrease without violating the second law of thermodynamics because the person isn’t an isolated system.

ΔS =

16.34. Set Up: 1 food { calorie = 1000 cal = 4186 J. The heat enters the person’s body at 37°C = 310 K and leaves at a temperature of 30°C = 303 K. ΔS =

Q T

4186 J ⎛ ⎞ Solve: qQ q = (0.80)(2.50 g)(9.3 food { calorie/g) ⎜ = 7.79 × 104 J ⎝ 1 food { calorie ⎟⎠ ΔS =

+7.79 × 104 J −7.79 × 104 J + = − 5.8 J/K. 310 K 303 K

Your body’s entropy decreases. Reflect: The entropy of your body can decrease without violating the second law of thermodynamics because you are not an isolated system.

*16.35. Set Up and Solve: (0.600)(300 W/m 2 ) A = 20.0 × 103 W. A = 111 m 2 . 16.36. Set Up: The power incident on the collector is (200 W/m 2 ) A, where A is the area of the collector. The incident power times the efficiency must equal the power output of 2500 MW. 1 km = 0.6214 mi. Solve: (a) (0.42)(200 W/m 2 ) A = 2500 × 106 W. A = 2.98 × 107 m 2 = 29.8 km 2 = 11.5 mi 2 (b) (0.21)(200 W/m 2 ) A = 2500 × 106 W. A = 5.95 × 107 m 2 = 59.5 km 2 = 23.0 mi 2

*16.37. Set Up: emax = eCarnot = 1 − TC /TH . e =

W W /t W QC QH = . W = QH + QC so = + . For a temperature t t t QH QH /t

change Q = mcΔT . TH = 300.15 K, TC = 279.15 K. For water, ρ = 1000 kg/m3 , so a mass of 1 kg has a volume of 1 L. For water, c = 4190 J/kg ⋅ K. Solve: (a) e = 1 − 279.15 K = 7.0,. 300.15 K Q Q W QH Pout 210 kW (b) = = = 3.0 MW. C = H − = 3.0 MW − 210 kW = 2.8 MW. 0 . 070 t t t t e (c)

m qQC q/t (2.8 × 106 W)(3600 s/h) = = = 6 × 105 kg/h = 6 × 105 L/h. t cΔT (4190 J/kg ⋅ K)(4 K)

Reflect: The efficiency is small since TC and TH don’t differ greatly.

16.38. Set Up: For water, c = 4190 J/kg ⋅ K and ρ = 1.00 × 103 kg/m3 . Solve: The energy collected in 1 h = 3600 s is Q = (0.50)(200 W/m 2 )(30.0 m 2 )(3600 s) = 1.08 × 107 J. The heat goes into the water, so Q = mc ΔT .

m=

Q 1.08 × 107 J = = 57.3 kg. c ΔT (4190 J/kg ⋅ K)(45.0 C°)


16-10

Chapter 16

The volume of water is V =

m

ρ

=

57.3 kg

1.00 × 103 kg/m3 Reflect: This is adequate for a typical household.

= 57.3 × 1023 m3 = 57.3 L.

*16.39. Set Up: W = QC + QH . Since it is a Carnot cycle,

QC T = 2 C . The heat required to melt the ice is QH TH

Q = mLf . For water, Lf = 334 × 103 J/kg. QH > 0, QC < 0. QC = 2mLf . TH = 527D C = 800.15 K. Solve: (a) QH = 1400 J, W = 1300 J. QC = W − QH = 2100 J.

TC = −TH (QC /QH ) = −(800.15 K)[(−100 J)/(400 J)] = +200 K = −73°C (b) The total QC required is − mLf = − (10.0 kg)(334 × 103 J/kg) = −3.34 × 106 J. QC for one cycle is −100 J, so the

−3.34 × 106 J = 3.34 × 104 cycles. −100 J/cycle Reflect: The results depend only on the maximum temperature of the gas, not on the number of moles or the maximum pressure.

number of cycles required is

16.40. Set Up: For an ideal diatomic gas, CV = 52 R, C p = 72 R and γ =

7 = . pV = nRT . ΔU = Q − W . CV 5

Solve: (a) p1 = 1.00 atm.

V1 = V2 = V1 = 8.62 L.

nRT1 (0.350 mol)(0.08206 L ⋅ atm/mol ⋅ K)(300 K) = = 8.62 L. p1 1.00 atm

p1 p2 = so T1 T2

⎛T ⎞ ⎛ 600 K ⎞ p2 = ⎜ 2 ⎟ p1 = ⎜ ⎟ (1.00 atm) = 2.00 atm. T ⎝ 300 K ⎠ ⎝ 1⎠ p3 = 1.00 atm. V3 =

nRT3 (0.350 mol)(0.08206 L ⋅ atm/mol ⋅ K)(492 K) = = 14.1 L. 1.00 atm p3

(b) 1 → 2 : W = 0 since ΔV = 0. Q = ΔU = nCV ΔT = (0.350 mol)

2 → 3 : Q = 0. ΔU = nCV ΔT = (0.350 mol)

( 52 ) (8.315 J/mol ⋅ K)(300 K) = 2183 J.

( 52 ) (8.315 J/mol ⋅ K)(−108 K) = − 786 J.

W = − ΔU = + 786 J.

3 → 1: W = p ΔV = nR ΔT = (0.350 mol)(8.315 J/mol ⋅ K)(−192 K) = − 559 J. Q = nC p ΔT = (0.350 mol)

( 72 ) (8.315 J/mol ⋅ K)(−192 K) = − 1955 J.

ΔU = Q − W = − 1955 J − (−559 J) = − 1396 J.

(c) Wtot = W1 → 2 + W2 → 3 + W3 → 1 = 0 + 786 J + (−559 J) = + 227 J (d) Qtot = Q1 → 2 + Q2 → 3 + Q3 → 1 = 2183 J + 0 + ( −1955 J) = 228 J. Qtot = Wtot , apart from a small difference due to

rounding. W W 227 J (e) e = = = = 0.104 = 10.4,. QH Q1 → 2 2183 J eCarnot = 1 −

TC 300 K =1− = 0.500 = 50.0,. TH 600 K e < eCarnot .



16-11

Reflect: The net ΔU for the cycle is ΔU tot = 2183 J + (−786 J) + ( −1396 J). This is zero, except for a small difference due to rounding. For any cycle, ΔU = 0.

16.41. Set Up: The work done by the engine each cycle is mg Δy , with m = 15.0 kg and Δy = 2.00 m. For a Carnot T QC T = − C . eCarnot = 1 − C . qW q = qQH q − qQCq. QH > 0, QC < 0. QH TH TH Solve: (a) The pV diagram is sketched in the figure below.

engine,

(b) W = mg Δy = (15.0 kg)(9.80 m/s 2 )(2.00 m) = 294 J.

qQCq = qQH q − qW q = 500 J − 294 J = 206 J, and QC = − 206 J.

⎛Q ⎞ ⎛ −206 J ⎞ TC = − TH ⎜ C ⎟ = − (773 K) ⎜ ⎟ = 318 K = 45°C. Q 500 J ⎠ ⎝ ⎝ H⎠ (c) e = 1 −

318 K TC =1− = 0.589 = 58.9,. 773 K TH

(d) qQCq = 206 J. (e) The maximum pressure is for state a. This is also where the volume is a minimum, so Va = 5.00 L = 5.00 × 1023 m3 . Ta = TH = 773 K.

pa =

nRTa (2.00 mol)(8.315 J/mol ⋅ K)(773 K) = = 2.57 × 106 Pa. Va 5.00 × 1023 m3

16.42. Set Up: The least possible qW q and the smallest qQH q is for a Carnot cycle. TC = − 5.0°C = 268 K. TH = 20.0°C = 293 K. For water, cw = 4190 J/kg ⋅ K and Lf = 3.34 × 105 J/kg. qQH q = qW q + qQCq. QH < 0, QC > 0, QC T =− C. QH TH Solve: The heat that must be removed from the water is Q = mc ΔT − mLf = (5.00 kg)([4190 J/kg ⋅ K][ −20.0 C°] − 3.34 × 105 J/kg) = − 2.09 × 106 J. W < 0.

This amount of heat goes into the working substance of the freezer, so QC = 2.09 × 106 J. QH = − QC

QC T = − C so QH TH

TH ⎛ 293 K ⎞ 6 = − (2.09 × 106 J) ⎜ ⎟ = − 2.28 × 10 J. TC ⎝ 268 K ⎠

qQH q = 2.28 × 106 J is expelled into the room. qW q = qQH q − qQCq = 2.28 × 106 J − 2.09 × 106 J = 1.9 − 105 J is the least possible amount of electrical energy that must

be supplied. Reflect: The amount of heat expelled into the room is greater than the amount of heat removed from the water. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16-12

Chapter 16

*16.43. Set Up: The efficiency of the composite engine is e12 =

W1 + W2 , where QH1 is the heat input to the first QH1

engine and W1 and W2 are the work outputs of the two engines. For any heat engine, W = QC + QH , and for a Carnot engine,

Qlow T = 2 low , where Qlow and Qhigh are the heat flows at the two reservoirs that have temperatures Tlow and Qhigh Thigh

Thigh . We know that Qhigh,2 = 2Qlow,1, Tlow,1 = T ′, Thigh,1 = TH , Tlow,2 = TC and Thigh,2 = T ′. Solve: e12 =

Qlow,2 W1 + W2 Qhigh,1 + Qlow,1 + Qhigh,2 + Qlow,2 = . Since Qhigh,2 = −Qlow,1, this reduces to e12 = 1 + . QH1 Qhigh,1 Qhigh,1

Qlow,2 = −Qhigh,2

Tlow,2 Thigh,2

= Qlow,1

⎛T ⎞T ⎛ T′ ⎞T T TC = −Qhigh,1 ⎜ low,1 ⎟ C = −Qhigh,1 ⎜ ⎟ C . This gives e12 = 1 − C . The efficiency ⎜ ⎟ TH T′ ⎝ TH ⎠ T ′ ⎝ Thigh,1 ⎠ T ′

of the composite system is the same as that of the original engine. Reflect: The overall efficiency is independent of the value of the intermediate temperature T ′.

16.44. Set Up: For O2 , CV = 20.85 J/mol ⋅ K and C p = 29.17 J/mol ⋅ K. Solve: (a) p1 = 2.00 atm, V1 = 4.00 L, T1 = 300 K. p2 = 2.00 atm.

V1 V2 = . T1 T2

⎛T ⎞ ⎛ 450 K ⎞ V2 = ⎜ 2 ⎟V1 = ⎜ ⎟ (4.00) = 6.00 L. ⎝ 300 K ⎠ ⎝ T1 ⎠ p p V3 = 6.00 L. 2 = 3 . T2 T3 ⎛T ⎞ ⎛ 250 K ⎞ p3 = ⎜ 3 ⎟ p2 = ⎜ ⎟ (2.00 atm) = 1.11 atm T ⎝ 450 K ⎠ ⎝ 2⎠ V4 = 4.00 L. p3V3 = p4V4 . ⎛V ⎞ ⎛ 6.00 L ⎞ p4 = p3 ⎜ 3 ⎟ = (1.11 atm) ⎜ ⎟ = 1.67 atm. ⎝ 4.00 L ⎠ ⎝ V4 ⎠ These processes are shown in the figure below.

p1V1 (2.00 atm)(4.00 L) = = 0.325 mol RT1 (0.08206 L ⋅ atm/mol ⋅ K)(300 K) process 1 → 2 : W = p ΔV = nR ΔT = (0.325 mol)(8.315 J/mol ⋅ K)(150 K) = 405 J. (b) n =

Q = nC p ΔT = (0.325 mol)(29.17 J/mol ⋅ K)(150 K) = 1422 J. process 2 → 3 : W = 0. Q = nCV ΔT = (0.325 mol)(20.85 J/mol ⋅ K)(−200 K) = − 1355 J.



16-13

process 3 → 4 : ΔU = 0 and ⎛V ⎞ ⎛ 4.00 L ⎞ Q = W = nRT3ln ⎜ 4 ⎟ = (0.325 mol)(8.315 J/mol ⋅ K)(250 K)ln ⎜ ⎟ = − 274 J. V ⎝ 6.00L ⎠ ⎝ 3⎠ process 4 → 1: W = 0. Q = nCV ΔT = (0.325 mol)(20.85 J/mol ⋅ K)(50 K) = 339 J. (c) e =

W 405 J − 274 J = = 0.0744 = 7.44,. QH 1422 J + 339 J eCarnot = 1 −

TC 250 K =1− = 0.444 = 44.4,; TH 450 K

eCarnot is much larger. Reflect: Qtot = + 1422 J + (−1355 J) + ( −274 J) + 339 J = 132 J. This is equal to Wtot , apart from a slight differ-

ence due to rounding. For a cycle, Wtot = Qtot , since ΔU = 0.

*16.45. Set Up: A person with surface area A and surface temperature T = 303 K radiates at a rate H = eσT 4 . The person absorbs heat from the room at a rate H s = Aeσ Ts 4 , where Ts = 293 K is the temperature of the room. In t = 1.0 s, heat Aeσ tT 4 flows into the room and heat Aeσ tTs 4 flows out of the room. The heat flows into and out of the room occur at a temperature of Ts . Solve: For the room, Aeσ tT 4 Aeσ tTs 4 Aeσ t (T 4 − Ts 4 ) ΔS = − = Ts Ts Ts (1.85 m 2 )(1.00)(5.67 × 1028 W/m 2 ⋅ K 4 )(1.0)([303 K]4 − [293 K]4 ) = 0.379 J/K 293 K

ΔS =

16.46. Set Up: e =

W . 1 day = 8.64 × 104 s. For the river water, Q = mcΔT , where the heat that goes into the water QH

is the heat QC rejected by the engine. The density of water is 1000 kg/m3. For an isothermal process we have Q . 18.0° C = 291.1 K. 18.5D C = 291.6 K. T W P 1000 MW so PH = W = = 2.50 × 103 MW. Solve: (a) QH = e e 0.40 ΔS =

(b) The heat input in one day is (2.50 × 109 W)(8.64 × 104 s) = 2.16 × 1014 J. The mass of coal used per day is

2.16 × 1014 J 7

2.65 × 10 J/kg

= 8.15 × 106 kg.

(c) qQH q = qW q + qQCq. qQCq = qQH q − qW q. PC = PH − PW = 2.50 × 103 MW − 1000 MW = 1.50 × 103 MW.

(d) The heat input to the river is 1.50 × 109 J/s. Q = mcΔT and ΔT = 0.5D C gives m =

7.16 × 105 kg. V =

m

ρ

1.50 × 109 J Q = = cΔT (4190 J/kg ⋅ K)(0.5 K)

= 716 m3. The river flow rate must be 716 m3 /s.

(e) To find the entropy change we note that the temperature of the river is nearly constant so we may apply the isothermal approximation discussed in Example 16.8. In one second, 7.16 × 105 kg of water goes from 291.1 K to 291.6 K as

1.50 × 109 J of heat is added. Thus, the entropy change is approximately ΔS ≈

1.50 × 109 J = 5.1 × 106 J/K (rounding to 291.6 K

two place accuracy). Reflect: The entropy of the river increases because heat flows into it. The mass of coal used per second is huge. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16-14

Chapter 16

16.47. Set Up: For a heat engine e =

W T . For a Carnot engine e = 1 − C . Gravitational potential energy is QH TH

U grav = mgh. 1 food { calorie = 1000 cal = 4186 J Solve: (a) e = 1 −

TC 303 K =1− = 0.0226 = 2.26,. This engine has a very low thermal efficiency. TH 310 K

(b) U grav = mgh = (2.50 kg)(9.80 m/s2 )(1.20 m) = 29.4 J. This equals the work output of the engine. e =

W so QH

W 29.4 J = = 1.30 × 103 J e 0.0226 (c) Since 80% of food energy goes into heat, you must eat food with a food energy of 1.30 × 103 J = 1.63 × 103 J. 0.80 QH =

Each candy bar gives (350 food { calorie)(4186 J/food { calorie) = 1.47 × 106 J. The number of candy bars required is

1.63 × 103 J 1.47 × 106 J/candy bar

= 1.11 × 1023 candy bars.

16.48. Set Up: Use Eq. (16.4) to calculate e. For an ideal diatomic gas we have CV = 5R / 2 and C p = CV + R = 7 R /2. The cycle is sketched in the figure below.

Solve: Calculate Q and W for each process.

process 1 → 2 ΔV = 0 implies W = 0 ΔV = 0 implies Q = nCV ΔT = nCV (T2 − T1 ) But pV = nRT and V constant says p1V = nRT1 and p2V = nRT2 . Thus ( p2 − p1 )V = nR (T2 − T1); V Δp = nRΔT (true when V is constant). Then Q = nCV ΔT = nCV (V Δp/nR ) = (CV /R )V Δp = (CV /R)V0 (2 p0 − p0 ) = (CV /R) p0V0 . Q > 0; heat is absorbed by the gas. process 2 → 3 Δp = 0 so W = pΔV = p(V3 − V2 ) = 2 p0 (2V0 − V0 ) = 2 p0V0 (W is positive since V increases.) Δp = 0 implies Q = nC p ΔT = nC p (T2 − T1 )

But pV = nRT and p constant says pV1 = nRT1 and pV2 = nRT2 . Thus p(V2 − V1) = nR(T2 − T1 ); pΔV = nRΔT (true when p is constant).



16-15

Then Q = nC p ΔT = nC p ( pΔV /nR ) = (C p /R ) pΔV = (C p /R)2 p0 (2V0 − V0 ) = (C p /R)2 p0V0 . (Q > 0; heat is absorbed by the gas.) process 3 → 4 ΔV = 0 implies W = 0 ΔV = 0 so Q = nCV ΔT = nCV (V Δp / nR ) = (CV /R )(2V0 )( p0 − 2 p0 ) = 22(CV /R ) p0V0 (Q < 0 so heat is rejected by the gas.) process 4 → 1 Δp = 0 so W = pΔV = p(V1 − V4 ) = p0 (V0 − 2V0 ) = − p0V0 (W is negative since V decreases) Δp = 0 so Q = nC p ΔT = nC p ( pΔV /nR ) = (C p /R ) pΔV = (C p /R ) p0 (V0 − 2V0 ) = −(C p /R) p0V0 (Q < 0 so heat is rejected by the gas.) total work performed by the gas during the cycle: Wtot = W1→ 2 + W2→3 + W3→ 4 + W4→1 = 0 + 2 p0V0 + 0 − p0V0 = p0V0

(Note that Wtot equals the area enclosed by the cycle in the pV-diagram.) total heat absorbed by the gas during the cycle (QH ): Heat is absorbed in processes 1 → 2 and 2 → 3. ⎛ CV + 2C p ⎞ CV p0V0 + 2 p0V0 = ⎜ ⎟ p0V0 R R R ⎝ ⎠ CV + 2(CV + R ) ⎛ 3CV + 2 R ⎞ But C p = CV + R so QH = p0V0 = ⎜ ⎟ p0V0 . R R ⎝ ⎠ QH = Q1→ 2 + Q2→3 =

total heat rejected by the gas during the cycle (QC ): Heat is rejected in processes 3 → 4 and 4 → 1. ⎛ 2CV + C p ⎞ CV p0V0 − p0V0 = 2⎜ ⎟ p0V0 R R R ⎝ ⎠ 2C + (CV + R ) ⎛ 3C + R ⎞ But C p = CV + R so QC = 2 V p0V0 = 2⎜ V ⎟ p0V0 . R R ⎝ ⎠ efficiency QC = Q3→4 + Q4→1 = 22

e=

W p0V0 R R 2 = = = = . QH ([3CV + 2 R ]/ R)( p0V0 ) 3CV + 2 R 3(5R/2) + 2 R 19

e = 0.105 = 10.5, ⎛ 3C + R ⎞ ⎛ 3CV + 2 R ⎞ Reflect: As a check on the calculations we note that QC + QH = 2⎜ V ⎟ p0V0 + ⎜⎝ ⎟⎠ p0V0 = p0V0 = W, ⎝ R ⎠ R

as it should.

*16.49. Set Up: For water, Lf = 3.34 × 105 J/kg. ΔS =

Q . T

Solve: (a) The heat that goes into the ice-water mixture is Q = Lf = (0.160 kg)(3.34 × 105 J/kg) = 5.34 × 104 J. This

is same amount of heat leaves the boiling water, so Q 25.34 × 104 J ΔS = = = − 143 J/K. T 373 K (b) ΔS =

Q 5.34 × 104 J = = + 196 J/K T 273 K


16-16

Chapter 16

(c) For any segment of the rod, the net heat flow is zero, so ΔS = 0. (d) ΔS tot = − 143 J/K + 196 J/K = + 53 J/K.

16.50. Set Up: ab is at constant pressure, bc is at constant volume, and ca is isothermal. For H 2 , CV = 20.42 J/mol ⋅ K and C p = 28.74 J/mol ⋅ K. For a constant pressure process, Q = nC p ΔT and n ΔT =

p ΔV , so R

⎛ ⎞ V Δp ⎛C ⎞ Q=⎜ , so Q = ⎜ V ⎟ V Δp. For an ⎟ p ΔV . For a constant volume process, Q = nCV ΔT and n ΔT = ⎝ R⎠ R R ⎝ ⎠

isothermal process, ΔU = 0 and Q = W = nRT ln (V2 /V1 ). 1 atm = 1.013 × 105 Pa Solve: (a) pV = nRT so heat enters the gas when pV increases. In an isothermal process heat leaves the gas when V

decreases. So, in this cycle heat enters in process ab and leaves in processes bc and ca. (b) QH equals Q for process ab, so

⎛ ⎞ ⎛ 28.74 J/mol ⋅ K ⎞ 5 3 4 QH = ⎜ ⎟ p ΔV = ⎜ ⎟ (0.700 atm)(1.013 × 10 Pa/atm)(0.070 m ) = 1.72 × 10 J ⎝ 8.315 J/mol ⋅ K ⎠ ⎝ R ⎠

For ab, W = p ΔV = (0.700 atm)(1.013 × 105 Pa/atm)(0.070 m3 ) = 4.96 × 103 J. For bc, W = 0. For ca, W = nRTa ln(Va / Vc ). Ta =

paVa (0.700 atm)(1.013 × 105 Pa/atm)(0.0300 m3 ) = = 301 K nR (0.850 mol)(8.315 J/mol ⋅ K)

⎛ 0.0300 m3 ⎞ W = (0.850 mol)(8.315 J/mol ⋅ K)(301 K)ln ⎜ = − 2.56 × 103 J ⎜ 0.100 m3 ⎟⎟ ⎝ ⎠

The total work for the cycle is 4.96 × 103 J + ( −2.56 × 103 J) = 2.40 × 103 J. e=

W 2.40 × 103 J = = 0.140 = 14.0, QH 1.72 × 104 J

Reflect: For a cycle, ΔU = 0 and Qtot = Wtot . The total heat flow must be 2.40 × 103 J, but in calculating e we use

only the heat that flows into the engine. qQCq . If the cycle is run in reverse, W has the same magnitude and opposite qW q sign. Heat enters the refrigerator in processes ac and cb and leaves the refrigerator in process ba. From Problem 16.57, Q for ab is + 1.72 × 104 J, so for ba, Q = − 1.72 × 104 J. This is QH for the refrigerator. In Problem 16.57,

*16.51. Set Up: For a refrigerator, K =

W = 2.40 × 103 J, so here W = − 2.40 × 103 J. For the refrigerator, W < 0, QH < 0 and QC > 0. W = QC + QH Solve: QC = W − QH = − 2.40 × 103 J − (−1.72 × 104 J) = 1.48 × 104 J

K=

1.48 × 104 J 2.40 × 103 J

= 6.17


17

ELECTRIC CHARGE AND ELECTRIC FIELD

Answers to Multiple-Choice Problems 1. D 2. C 15. A

3. C

4. B, C

5. E

6. B

7. C

8. A

9. B

10. B

11. D

12. C

13. B

14. E

Solutions to Problems 17.1. Set Up: Unlike charges attract and like charges repel. In a conductor some of the negative charge is free to move. In an insulator the charge can shift position only slightly. Solve: (a) Aluminum is a conductor and negative charge in the sphere moves away from the rod. The distribution of charge is sketched in Figure (a) below. (b) The charges in the nonconducting sphere displace slightly, with negative charge moving away from the rod. The distribution of charge is sketched in Figure (b) below.

17.2. Set Up: Copper is a conductor, so some of the electrons are free to move. The positive charge on the rod attracts the negative charge in the ball. When the ball is connected to the earth by a conducting wire, charge can flow between the ball and the earth. Solve: (a) Electrons move toward the rod. The distribution of charge is sketched in Figure (a) below. (b) Electrons from the earth are attracted by the region of positive charge on the ball and flow onto the ball, giving it a net negative charge. When the rod is removed, this net charge distributes uniformly over the surface of the ball, as sketched in Figure (b) below.

*17.3. Set Up: For an isolated sphere, the excess charge is uniformly distributed over the surface of the conductor. Unlike charges attract and like charges repel, and in a conductor the excess charge is free to move. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

17-1

17-2

Chapter 17

Solve: (a) The uniform distribution of charge over the surface of each sphere is sketched in Figure (a) below. (b) When the spheres are close to each other, the negative and positive excess charges are drawn toward each other, as shown in Figure (b) below. (c) When the spheres are close to each other, the excess negative charges on each sphere repel, as shown in Figure (c) below.

Reflect: We will learn later in the chapter that the excess charge on a conductor is on the surface of the conductor.

17.4. Set Up: Unlike charges attract and in a conductor the excess charge is free to move. Solve: Negative charge in the earth is pulled to the surface beneath the charged cloud, as sketched in the figure below.

*17.5. Set Up: The charge of one electron is −e = −1.60 × 10−19 C. 1 μ C = 10−6 C; 1 nC = 10−9 C. Solve: (a) N = (b) N =

0 Q 0 2.50 × 10−6 C = = 1.56 × 1013 electrons e 1.60 × 10−19 C

0 Q 0 2.50 × 10−9 C = = 1.56 × 1010 electrons e 1.60 × 10−19 C

17.6. Set Up: The total charge is the number if ions times the charge of each. e = 1.60 × 10−19 C. Solve: N = (5.6 × 1011 /m)(1.5 × 10−2 m) = 8.4 × 109 ions Q = Ne = (8.4 × 109 )(1.60 × 10−19 C) = 1.3 × 10−9 C = 1.3 nC

*17.7. Set Up: Use the mass m of the ring and the atomic mass M of gold to calculate the number of gold atoms. Each atom has 79 protons and an equal number of electrons. N A = 6.02 × 1023 atoms/mol. A proton has charge +e. Solve: The mass of gold is 17.7 g and the atomic weight of gold is 197 g/mol. So the number of atoms is ⎛ 17.7 g ⎞ N A n = (6.02 × 1023 atoms/mol) ⎜ = 5.41 × 1022 atoms. The number of protons is np = (79 protons/atom) ⎝ 197 g mol ⎟⎠ (5.41 × 1022 atoms) = 4.27 × 1024 protons.

Q = (np )(1.60 × 10−19 C/proton) = 6.83 × 105 C. (b) The number of electrons is ne = np = 4.27 × 1024. Reflect: The total amount of positive charge in the ring is very large, but there is an equal amount of negative charge. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Electric Charge and Electric Field

17.8. Set Up: Like charges repel and unlike charges attract. F = k

0 q1 q2 0

r2

17-3

.

Solve: q1 = q2 = +3.00 × 10−6 C. The force is repulsive.

F = (8.99 × 109 N ⋅ m 2 /C2 )

17.9. Set Up: F = k

(3.00 × 10−6 C) 2

0 q1 q2 0

r2

(0.200 m) 2

= 2.02 N

. An electron has charge − e and a proton has charge + e, where e = 1.60 × 10−19 C.

k 0 q1 q2 0 (8.99 × 109 N ⋅ m 2 /C 2 )(1.60 × 10−19 C) 2 = = 1.07 × 10−14 m. Since an electron and a proton 2.00 N F have the same magnitude of charge the distance is the same for two protons as for two electrons.

Solve: r =

17.10. Set Up: F = k

0 q1 q2 0

. Like charges repel and unlike charges attract. r2 Solve: (a) The force is attractive so the unknown charge q2 is positive. Fr 2 (0.200 N)(0.300 m) 2 = = 3.64 × 10−6 C = 3.64 μ C k 0 q1 0 (8.99 × 109 N ⋅ m 2 /C2 )(0.550 × 10−6 C) (b) The unknown charge exerts a downward force of 0.200 N on the other charge. 0 q2 0 =

*17.11. Set Up: The electrical force is given by Coulomb’s law, with k = 8.99 × 109 N ⋅ m 2 /C2 . A proton has charge + e and an electron has charge −e. Solve: (a) F = k

0 q1 q2 0 r2

=

(8.99 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) 2 (1.00 × 10−15 m) 2

= 230 N. Yes, this force is about 52 lbs. 2

⎛ 1.00 × 10−15 m ⎞ (b) The force is smaller than in part (a) by a factor of ⎜ so it is 2.30 × 10−8 N. No, this force is ⎜ 1.00 × 10−10 m ⎟⎟ ⎝ ⎠ very small.

17.12. Set Up: One mole of carbon contains N A = 6.02 × 1023 atoms. Each electron has charge −e = −1.60 × 10−19 C. If charge −q is removed from the sphere, the sphere is left with a charge of + q. Solve: (a) The number of moles is n =

1.00 g = 8.33 × 10−2 mol. The number of atoms is N = nN A = 12.0 g/mol

(8.33 3 10−2 mol)(6.02 3 1023 atoms/mol) = 5.01 3 1022 atoms. Each atom has 6 electrons, so the number of electrons is N e = 6 N = 3.01 × 1023 electrons. Each electron has charge −e so the total negative charge of the electrons is qe = −eN e = −(1.60 × 10−19 C)(3.01 × 1023 ) = −4.82 × 104 C. (b) The sphere would be left with positive charge qsphere = +4.82 × 104 C. The force would have magnitude

F =k

0 qe qsphere 0 2

= (8.99 × 109 N ⋅ m 2 /C2 )

r are opposite sign so the force is attractive.

(4.82 × 104 C)2 (1.50 m) 2

= 9.28 × 1018 N. This is an immense force. The charges

17.13. Set Up: Each electron has a charge of −1.60 × 10−19 C. Solve: (a) The number of excess electrons on your body is Total Excess Charge −55 × 10−6 C = = 3.4 × 1014 Charge of One Electron −1.60 × 10−19 C


17-4

Chapter 17

(b) Assume that your body and the rug form a closed system so that we can apply the principle of charge conservation. Thus, as your body gains an excess charge of −55 μC from the rug, the rug must gain an excess charge of +55 μC from your body. Assuming that the rug is initially electrically neutral, the charge on the rug will now be +55 μC. Reflect: If the rug was not initially electrically neutral we would simply add +55 μC to its initial charge to obtain its final charge.

17.14. Set Up: F = k

0 q1 q2 0

r2

Solve: (a) q1 = q2 = q. F = k (b) q2 = 4q1. F =

k (4q12 )

.

q2 r2

and q = r

F = 3.71 × 10−7 C. q2 = 1.48 × 10−6 C. k

and q1 = 12 r

r2

F 0.220 N = (0.150 m) = 7.42 × 10−7 C, k 8.99 × 109 N ⋅ m 2 /C2

*17.15. Set Up: Use the mass of a sphere and the atomic mass of aluminum to find the number of aluminum atoms in one sphere. Each atom has 13 electrons. Apply Coulomb’s law and calculate the magnitude of charge qq q on each sphere. We have N A = 6.02 × 1023 atoms/mol and qqq = ne′ e, where n′e is the number of electrons removed from one sphere and added to the other. Solve: (a) Assuming that the two spheres are initially neutral, the total number of electrons on each sphere equals the number of protons on each sphere: ⎛ ⎞ 0.0250 kg 24 ne = np = (13)( N A ) ⎜ ⎟ = 7.25 × 10 electrons. . 0 026982 kg / mol ⎝ ⎠ (b) For a force of 1.00 × 104

N to act between the spheres,

F = 1.00 × 104 N = k

q2 r2

.

This gives

qqq = (1.00 × 104 N)(0.800 m) 2 /k = 8.44 × 10−4 C. The number of electrons removed from one sphere and added to the other is n′e = qqq /e = 5.27 × 1015 electrons. (c) ne′ /ne = 7.27 × 10−10. Reflect: When ordinary objects receive a net charge the fractional change in the total number of electrons in the object is very small.

17.16. Set Up: F = k Solve: F = k

q

2

r

2

. 0q0 = r

0 q1 q2 0

r

2

. The charge of an electron is −e = −1.60 × 10−19 C.

F 4.57 × 10−21 N = (0.200 m) = 1.426 × 10−16 C. k 8.99 × 109 N ⋅ m 2 /C2

The number of excess electrons is

17.17. Set Up: Apply F =

0 q 0 1.426 × 10−16 C = = 891. e 1.60 × 10−19 C

k qq1q2q r2

and solve for r. F = 650 N .

(8.99 × 109 N ⋅ m 2 /C2 )(1.0 C) 2 k qq1q2q = = 3.7 × 103 m = 3.7 km 650 N F Reflect: Charged objects typically have net charges much less than 1 C. Solve: r =

17.18. Set Up: A proton has charge +e and an electron has charge −e, with e = 1.60 × 10−19 C. The force between them has magnitude F = k

0 q1 q2 0 r2

=k

e2 r2

and is attractive since the charges have opposite sign. A proton has mass



17-5

G mp = 1.67 × 10−27 kg and an electron has mass 9.11 × 10−31 kg. The acceleration is related to the net force F by G G F = ma . Solve: F = k

proton: ap =

e2 r2

= (8.99 × 109 N ⋅ m 2 /C2 )

(1.60 × 10−19 C)2 (2.0 × 10−10 m) 2

= 5.75 × 10−9 N

F 5.75 × 10−9 N = = 3.4 × 1018 m/s 2 mp 1.67 × 10−27 kg

electron: ae =

F 5.75 × 10−9 N = = 6.3 × 1021 m/s 2 me 9.11 × 10231 kg

The proton has an initial acceleration of 3.4 × 1018 m/s 2 toward the electron and the electron has an initial acceleration of 6.3 × 1021 m/s 2 toward the proton. Note that the force the electron exerts on the proton is equal in magnitude to the force the proton exerts on the electron, but the accelerations produced by this force are different because the masses are different.

*17.19. Set Up: Apply Coulomb’s law. The two forces on q3 must have equal magnitudes and opposite directions. Like charges repel and unlike charges attract. G G qq q q Solve: The force F2 that q2 exerts on q3 has magnitude F2 = k 22 3 and is in the +x direction. The force F1 must r2 be in the − x direction, so q1 must be positive. The condition F1 = F2 gives k

qq1qqq3q r12

=k

qq2qqq3q r22

.

2

2 ⎛r ⎞ ⎛ 2.00 cm ⎞ = 0.750 nC. qq1q = qq2q ⎜ 1 ⎟ = (3.00 nC) ⎜ ⎝ 4.00 cm ⎟⎠ ⎝ r2 ⎠ Reflect: The result for the magnitude of q1 doesn’t depend on the magnitude of q2 .

17.20. Set Up: qe = 21.60 × 10−19 C. qp = 11.60 × 10−19 C. The net force is the vector sum of the forces exerted by each electron. Each force is attractive so is directed toward the electron that exerts it. Solve: Each force has magnitude

F1 = F2 = k

0 q1 q2 0 r2

=k

e2 r2

=

(8.988 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) 2 (1.50 × 10−10 m) 2

= 1.023 × 10−8 N.

The vector force diagram is shown in the figure below.


17-6

Chapter 17

F1x = 1.023 × 10−8 N; F1 y = 0.

F2 x = F2 cos65.0° = 4.32 × 10−9 N; F2 y = F2 sin 65.0° = 9.27 × 10−9 N. Fx = F1x + F2 x = 1.46 × 10−8 N; Fy = F1 y + F2 y = 9.27 × 10−9 N. F = Fx2 + Fy2 = 1.73 × 10−8 N. Fy

9.27 × 10−9 N

= 0.6349 and θ = 32.4°. The net force is 1.73 × 10−8 N and is directed toward a point 1.46 × 10−8 N midway between the two electrons. tan θ =

Fx

=

17.21. Set Up: F = k

0 qq ′0

r2 shown in the figure below.

Solve: F1 = k

F2 = k

. Like charges repel and unlike charges attract. The charges and their forces on q3 are

(1.50 × 10−9 C)(5.00 × 10−9 C)

0 q1 q3 0

= (8.99 × 109 N?m 2 /C2 )

0 q2 q3 0

= (8.99 × 109 N ⋅ m 2 /C2 )

r12

r22

(0.200 m)2

= 1.69 × 10−6 N.

(3.20 × 10−9 C)(5.00 × 10−9 C) (0.400 m)2

= 8.99 × 10−7 N.

G G F1 and F2 are in the same direction so F = F1 + F2 = 2.59 × 10−6 N and the net force is in the 2y direction. Reflect: The forces are vectors and must be added as vectors. We add the forces by adding their magnitudes only because the two force vectors are in the same direction.

17.22. Set Up: F = k

0 qq ′0

r2 shown in the figure below.

Solve: F1 = k

0 q1 q3 0 r12

. Like charges repel and unlike charges attract. The charges and their forces on q3 are

= (8.99 × 109 N ⋅ m 2 /C 2 )

(4.00 × 10−9 C)(0.600 × 10−9 C) (0.200 m) 2

= 5.394 × 10−7 N.



F2 = k

0 q2 q3 0 r22

= (8.99 × 109 N ⋅ m 2 /C2 )

(5.00 × 10−9 C)(0.600 × 10−9 C) (0.300 m) 2

17-7

= 2.997 × 10−7 N.

Fx = F1x + F2 x = 1F1 − F2 = 2.40 × 10−7 N. The net force has magnitude 2.40 × 10−7 N and is in the + x direction.

*17.23. Set Up: F = k

0 qq ′0

. Like charges repel and unlike charges attract. The charges and the forces on the r2 charges q1 and q2 in the dipole are shown in the figure below. Use the coordinates shown. sin θ =

1.50 cm and θ = 48.6°. 2.00 cm

Solve: F1 = F2 = (8.99 × 109 N ⋅ m 2 /C2 )

(5.00 × 10−6 C)(10.0 × 10−6 C) (0.0200 m) 2

= 1.124 × 103 N.

⎛ 1.50 cm ⎞ Fx = F1x + F2 x = 0. Fy = F1 y + F2 y = −2 F1 sin θ = −2(1.124 × 103 N) ⎜ = −1.69 × 103 N. ⎝ 2.00 cm ⎟⎠

The net force has magnitude 1.69 × 103 N and is in the direction from the +5.00 μC to the −5.00 μ C charge. (b) F1x and F2 x each produce clockwise torques and each have a moment arm of 1.50 cm. F1 y and F2 y each have

zero moment arm and produce no torque. τ = 2 F1x (0.0150 m) = 2 F1 cosθ (0.0150 m) = 2(1.124 × 103 N)(cos 48.6°)(0.0150 m) = 22.3 N ⋅ m The torque is clockwise. Reflect: The x components of the two forces are in opposite directions so they cancel and the net force has no x component. But the torques of F1x and F2 x are in the same direction and therefore produce a net torque.

17.24. Set Up: In the O-H-N combination the O− is 0.170 nm from the H + and 0.280 nm from the N − . In the N-H-N combination the N − is 0.190 nm from the H + and 0.300 nm from the other N − . Like charges repel and unlike charges attract. The net force is the vector sum of the individual forces. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

17-8

Chapter 17

0 q1 q2 0

Solve: (a) F = k

r2

=k

e2 r2

O-H-N O − − H1 F = (8.99 × 109 N ⋅ m 2 /C2 ) O − − N − F = (8.99 × 109 N ⋅ m 2 /C2 )

(1.60 × 10−19 C) 2 (0.170 × 10−9 m) 2 (1.60 × 10−19 C) 2 (0.280 × 10

−9

m)

2

= 7.96 × 10−9 N, attractive = 2.94 × 10−9 N, repulsive

N-H-N N − − H + F = (8.99 × 109 N ⋅ m 2 /C2 )

N − − N − F = (8.99 × 109 N ⋅ m 2 /C2 )

(1.60 × 10−19 C) 2 (0.190 × 10−9 m) 2 (1.60 × 10−19 C) 2 (0.300 × 10−9 m)2

= 638 × 10−9 N, attractive = 2.56 × 10−9 N, repulsive

The total attractive force is 1.43 × 10−8 N and the total repulsive force is 5.50 × 10−9 N. The net force is attractive and has magnitude 1.43 × 10−8 N − 5.50 × 10−9 N = 8.80 × 10−9 N. e2

(1.60 × 10−19 C) 2

= 8.22 × 10−8 N r2 (0.0529 × 10−9 m) 2 The bonding force of the electron in the hydrogen atom is a factor of 10 larger than the bonding force of the adeninethymine molecules. (b) F = k

= (8.99 × 109 N ⋅ m 2 /C2 )

*17.25. Set Up: In the O-H-O combination the O− is 0.180 nm from the H1 and 0.290 nm from the other O− . In the N-H-N combination the N − is 0.190 nm from the H1 and 0.300 nm from the other N −. In the O-H-N combination the O− is 0.180 nm from the H1 and 0.290 nm from the other N −. Like charges repel and unlike charges attract. The net force is the vector sum of the individual forces. Solve: F = k

0 q1 q2 0 r

2

=k

e2 r2

. The attractive forces are: O− − H1, 7.10 × 10−9 N; N − − H1, 6.37 × 10−9 N; O− − H + ,

7.10 × 10−9 N. The total attractive force is 2.06 × 10−8 N. The repulsive forces are: O − − O − , 2.74 × 10−9 N; N − − N −, 2.56 × 10−9 N; O − − N − , 2.74 × 10−9 N. The total repulsive force is 8.04 × 10−9 N. The net force is

attractive and has magnitude 1.26 × 10−8 N.

17.26. Set Up: Like charges repel and unlike charges attract. The force increases as the distance between the charges decreases. Solve: The forces on the dipole that is between the slanted dipoles are sketched in Figure (a) below. The forces are attractive because the + and − charges of the two dipoles are closest. The forces are toward the slanted dipoles and so have a net upward component. In Figure (b) below, in adjacent dipoles charges of opposite sign are closer than charges of the same sign so the attractive forces are larger than the repulsive forces and the dipoles attract.



17-9

17.27. Set Up: The central charge will be 1.85 nm from the charge on the left and 1.15 nm from the charge on the right. Like charges repel. The force diagram for the central charge is given in the figure below.

Solve: F1 = k

(2e) 2 r21

= (8.99 × 109 N ⋅ m 2 /C2 ) F2 = k

(2e) 2 r22

4(1.60 × 10−19 C) 2 (1.85 × 10−9 m) 2

= (8.99 × 109 N ⋅ m 2 /C2 )

= 2.690 × 10−10 N

4(1.60 × 10−19 C)2 (1.15 × 10−9 m) 2

= 6.961 × 10−10 N

The net force is 4.27 × 10−10 N, to the left. Reflect: The electrical forces are vectors and must be added as vectors. At the initial position of the central charge the net force on it is zero. If the central charge is displaced in either direction, the net force on it is in the direction that pushes it back toward the equilibrium position.

17.28. Set Up: F = k

0 q1 q2 0 r2

. The new charges are q1′ = 2q1 and q2′ = 2q2 .

Solve: The new force is F ′ = k

0 q1′ q2′ 0 r2

=k

02q1 2q2 0 r2

= 4k

0 q1 q2 0 r2

= 4F .

*17.29. Set Up: In space we assume that the only force accelerating the free proton is the electrical repulsion of the other proton. Coulomb’s law gives the force, and Newton’s second law gives the acceleration: a = Solve: (a) a = (8.99 × 109 N ⋅ m 2 /C2 )

F e2 =k 2. m mr

(1.60 × 10−19 C) 2

= 2.20 × 104 m/s 2 . (1.67 × 10−27 kg)(0.00250 m) 2 (b) The graphs are sketched in the figure below. Reflect: The electrical force of a single stationary proton gives the moving proton an initial acceleration about 20,000 times as great as the acceleration caused by the gravity of the entire earth. As the protons move farther apart, the electrical force gets weaker, so the acceleration decreases. Since the protons continue to repel, the velocity keeps increasing, but at a decreasing rate.

17.30. Set Up: Assume that the charge + Q is located at x = 0, the charge +4Q is located at x = d , and the unknown charge that is required to produce equilibrium, q, is located at a position x = a. Apply F = k

0 q1 q2 0

r2

to each

pair of charges to obtain eqilibrium. Solve: For a charge q to be in equilbrium, it must be placed between the two given positive charges (0 < a < d ) and the magnitude of the force between q and +Q must be equal to the magnitude of the force between q and +4Q:


17-10

Chapter 17

d 4qqqQ as its only root in the required interval =k . Solving for a we obtain (d − a ) = ±4a, which has a = 3 a2 (d − a )2 (0 < a < d ). Furthermore, to conteract the repulsive force between +Q and +4Q the charge q must be negative

k

qqqQ

( q = 2qqq ). The condition that +Q is in equilibrium gives us k

−qQ ( d/ 3)

2

=k

4Q 2 d2

4 . Solving for q we obtain q = 2 Q. 9

d 4 and q = 2 Q. To make sure 3 9 that the problem is well posed, we should check that these conditions also place the charge +4Q is in equilbrium. We 4Q 2 −4qQ k is equal to when the given values for both a and q are substituted. can do this by showing that k d2 (d − a)2

Reflect: We have shown that both q and +Q are in equilibrium provided that a =

G

G

*17.31. Set Up: F = qE . A proton has charge q = + e = +1.60 × 10−19 C. F 20.0 × 10−9 N = = 2.50 N/C. Since the charge is negative, the force and electric field are 0 q 0 8.00 × 10−9 C in opposite directions and the electric field is upward. (b) F = 0 q 0 E = eE = (1.60 × 10−19 C)(2.50 N) = 4.00 × 10−19 N. The charge is positive so the force is in the same Solve: (a) F = 0 q 0 E. E =

direction as the electric field; the force is upward.

G

G

17.32. Set Up: F = qE . A proton has charge q = + e = +1.60 × 10−19 C and mass m = 1.67 × 10227 kg. Solve: (a) The gravity force is downward so the electrical force must be upward. An upward force from a downward electric field requires a negative charge. mg = 0 q 0 E.

0q0 =

mg (1.45 × 10−3 kg)(9.80 m/s 2 ) = = 2.19 × 10−5 C. E 650 N/C

The charge is q = −2.19 × 10−5 C. (b) mg = eE. E =

mg (1.67 × 10−27 kg)(9.80 m/s 2 ) = = 1.02 × 10−7 N/C e 1.60 × 10−19 C

*17.33. Set Up: Use the coordinates shown in the figure below. Since the electric field is uniform, the force is G G G G constant and the acceleration is constant. F = qE and F = ma . υ0 y = 0. For an electron, q = −e = −1.60 × 10−19 C and m = 9.11 × 10−31 kg.

Solve: (a) y = υ0 yt + 12 a y t 2 . a y =

2y t

2

=

2(3.20 × 10−2 m) (1.5 × 10

Fy = ma y = (9.11 × 10 Ey =

Fy 0q0

=

2.59 × 10−16 N 1.60 × 10−19 C

−8

−31

s)

2

= 2.84 × 1014 m/s 2 .

kg)(2.84 × 1014 m/s 2 ) = 2.59 × 10−16 N.

= 1.62 × 103 N/C and E = 1.62 × 103 N/C.



17-11

(b) υ y = υ + a y t = 0 + (2.84 × 1014 m/s 2 )(1.5 × 10−8 s) = 4.26 × 106 m/s Reflect: We could also use the work-energy theorem and set the work done by the force equal to the gain in kinetic energy of the electron.

17.34. Set Up: For a point charge, E = k

0q0 2

G . E is toward a negative charge and away from a positive charge.

r Solve: (a) The field is toward the negative charge so is downward. 3.00 × 10−9 C E = (8.99 × 109 N ⋅ m 2 /C2 ) = 432 N/C. (0.250 m) 2 (b) r =

k 0q0 (8.99 × 109 N ⋅ m 2 /C2 )(3.00 × 10−9 C) = = 1.50 m E 12.0 N/C

17.35. Set Up: For a point charge, E = k Solve: 0 q 0 =

r2

.

Er 2 (6.50 × 103 N/C)(0.100 m) 2 = = 7.23 × 10−9 C k 8.99 × 109 N ⋅ m 2 /C2

17.36. Set Up: For a point charge, E = k Solve: r =

0q0

0q0 r2

.

k 0q0 (8.99 × 109 N ⋅ m 2 /C2 )(5.00 × 10−9 C) = = 3.35 m E 4.00 N/C

*17.37. Set Up: The electric field of a point charge has magnitude E = k

0q0 r2

. A proton has charge

q = e = 1.60 × 10−19 C. Solve: (a) E = (8.99 × 109 N ⋅ m 2 /C2 ) (b) E = (8.99 × 109 N ⋅ m 2 /C2 )

1.60 × 10−19 C (5.0 × 10−15 m) 2

= 5.8 × 1019 N/C

1.60 × 10−19 C

= 5.8 × 109 N/C (5.0 × 10−10 m) 2 Reflect: The electric fields inside atoms are very large.

17.38. Set Up: The proton is stopped by a constant electric field. Thus, the acceleration of the proton is constant and we can use the standard kinematic formulas to find acceleration and time. We know that υ0 x = 4.50 × 106 m/s, υ x = 0 m/s, and Δx = 0.0320 m. Also, we may use Fx = qEx = max to find Ex from ax . Solve: (a) υ x 2 = υ0 x 2 + 2ax Δx. Thus we have 02 = (4.50 × 106 m/s)2 + 2ax (0.0320 m). Solving for ax we obtain a x = −3.16 × 1014 m/s 2 . The electric field is given by

Fx max (1.67 × 10−27 kg)(−3.16 × 1014 m/s 2 ) = = = −3.30 × 106 N/C. q e 1.60 × 10−19 C So the electric field points to the left. υ − υ0 x (b) We can use υ x = υ0 x + axt to find the stopping time, t = x . Using υ x = 0, υ0 x = 4.50 × 106 m/s, and the ax Ex =

value of ax from part (a) we obtain t=

0 − 4.50 × 106 m/s −3.16 × 1014 m/s 2

= 1.42 × 10−8 s = 14.2 ns.


17-12

Chapter 17

(c) The acceleration of the electron will be the same as that found in part (a) for the proton. Thus, the acceleration required for the electron will be

Ex =

max (9.11 × 10−31 kg)(−3.16 × 1014 m/s 2 ) = = +1.80 × 103 N/C, which points to the right. −19 −e −1.60 × 10 C

Reflect: Even a modest electric field, such as the ones in this situation, can produce enormous accelerations for electrons and protons. The electron requires a smaller stopping field due to its smaller mass.

*17.39. Set Up: If the axon is modeled as a point charge, its electric field is E = k

q r2

. The electric field of a point

charge is directed away from the charge if it is positive. Solve: (a) 5.6 × 1011 Na1 ions enter per meter so in a 0.10 mm = 1.0 × 10−4 m section, 5.6 × 107 Na1 ions enter. This number of ions has charge q = (5.6 × 107 )(1.60 × 10−19 C) = 9.0 × 10−12 C. (b) E = k (c) r =

0q0 r

2

= (8.99 × 109 N ⋅ m 2 /C2 )

9.0 × 10−12 C (5.00 × 10−2 m) 2

= 32 N/C, directed away from the axon.

k 0q0 (8.99 × 109 N ⋅ m 2 /C2 )(9.0 × 10−12 C) = = 280 m E 1.0 × 10−6 N/C

0q0 . The net field is r2 the vector sum of the fields produced by each charge. Point A is 0.100 m from q2 and 0.150 m from q1. Point B is G G G 0.100 m from q1 and 0.350 m from q2 . A charge q in an electric field E experiences a force F = qE .

17.40. Set Up: The electric field of a negative charge is directed toward the charge. E = k

Solve: (a) The electric fields due to the charges at point A are shown in Figure (a) above.

E1 = k E2 = k

0 q1 0

= (8.99 × 109 N ⋅ m 2 /C2 )

0 q2 0

= (8.99 × 109 N ⋅ m 2 /C2 )

rA12 rA22

6.25 × 10−9 C (0.150 m) 2 12.5 × 10−9 C (0.100 m) 2

= 2.50 × 103 N/C = 1.124 × 104 N/C

Since the two fields are in opposite directions, we subtract their magnitudes to find the net field. E = E2 − E1 = 8.74 × 103 N/C, to the right. (b) The electric fields at points B are shown in Figure (b) above.

E1 = k E2 = k

0 q1 0

rB12 0 q2 0

= (8.99 × 109 N ⋅ m 2 /C2 )

rB 22

6.25 × 10−9 C

= (8.99 × 109 N ⋅ m 2 /C2 )

(0.100 m) 2 12.5 × 10−9 C (0.350 m)2

= 5.619 × 103 N/C = 9.17 × 102 N/C



17-13

Since the fields are in the same direction, we add their magnitudes to find the net field. E = E1 + E2 = 6.54 × 103 N/C, to the right. (c) At A, E = 8.74 × 103 N/C, to the right. The force on a proton placed at this point would be F = qE = (1.60 × 10−19 C)

(8.74 × 103 N/C) = 1.40 × 10−15 N, to the right.


0q0 2

G . E is directed toward a negative charge and away from a positive

r charge. Let the points a, b, and c be the locations where the field is calculated in parts (a), (b), and (c). The three points and the electric fields produced at those points by each of the two charges are shown in the figure below.

(a) E1 = k

0 q1 0 r12

=

(8.99 × 109 N ⋅ m 2 /C2 )(4.00 × 10−9 C) (0.200 m) 2 E2 = k

0 q2 0 r2 2

=

= 899 N/C

(8.99 × 109 N ⋅ m 2 /C2 )(6.00 × 10−9 C) (0.600 m)2

= 150 N/C

E x = E1x + E2 x = 2899 N/C + ( −150 N/C) = −1049 N/C. The field is 1050 N/C, in the − x direction. (b) E1 = k

0 q1 0 r12

=

(8.99 × 109 N ⋅ m 2 /C2 )(4.00 × 10−9 C)

= 25.0 N/C

(1.20 m) 2 E2 = k

0 q2 0 r2 2

=

(8.99 × 109 N ⋅ m 2 /C2 )(6.00 × 10−9 C) (0.400 m)2

= 337 N/C

E x = E1x + E2 x = 225.0 N/C + 337 N/C = 312 N/C. The field is 312 N/C, in the + x direction. (c) E1 = k

0 q1 0 r12

=

(8.99 × 109 N ⋅ m 2 /C2 )(4.00 × 10−9 C) (0.200 m) 2 E2 = k

0 q2 0 r2 2

=

= 899 N/C

(8.99 × 109 N ⋅ m 2 /C2 )(6.00 × 10−9 C) (1.00 m) 2

= 53.9 N/C

E x = E1x + E2 x = 1899 N/C + ( −53.9 N/C) = 845 N/C. The field is 845 N/C, in the + x direction. Reflect: In each case the two electric fields must be added as vectors.


0q0 2

G . E is toward a negative charge and away from a positive charge. The

r two charges and their fields at each point are shown in Figures (a)-(d) below.


17-14

Chapter 17

G G Solve: (a) E1 = E2 and E1 = E2 are in opposite directions, so the resultant electric field is zero. E x = E y = E = 0. (b) E1 = (8.99 × 109 N ⋅ m 2 /C2 )

E2 = (8.99 × 109 N ⋅ m 2 /C2 )

6.00 × 10−9 C (0.150 m) 2

6.00 × 10−9 C (0.450 m)2

= 2397 N/C

= 266 N/C

E x = E1x + E2 x = E1 + E2 = 2660 N/C. E y = 0. The resultant electric field has magnitude 2660 N/C and is in the + x direction. (c) E1 = (8.99 × 109 N ⋅ m 2 /C2 )

E2 = (8.99 × 109 N ⋅ m 2 /C2 )

6.00 × 10−9 C (0.400 m)2

6.00 × 10−9 C (0.500 m) 2

= 337 N/C. E1x = 0, E1 y = 2337 N/C.

= 216 N/C. θ = 53.1°. E2 x = E2 cosθ = 130 N/C.

E2 y = 2E2 sin θ = 2173 N/C. E x = E1x + E2 x = 0 + 130 N/C = 1130 N/C. E y = E1 y + E2 y = 2337 N/C − 173 N/C = 2510 N/C. tan φ = 0 E=

Ey Ex

E x2

G 0 = 3.92. φ = 75.7° and E makes an angle 360° − 75.7° = 284° counterclockwise from the + x axis.

+ E y2 = 526 N/C.

(d) θ = 53.1°. E1 = E2 = (8.99 × 109 N ⋅ m 2 /C2 )

6.00 × 10−9 C (0.250 m)2

= 863 N/C.

G E x = E1x + E2 x = 0. E y = E1 y + E2 y = 2 E1 sin θ = 2(863 N/C)sin 53.1° = 1380 N/C. E has magnitude 1380 N/C and is in the + y direction.



17-15

G . E is toward a negative charge and away from a positive charge. Let r G G q1 = +0.500 nC and q2 = +8.00 nC. For the net electric field to be zero, E1 and E2 must have equal magnitudes

*17.43. Set Up: For a point charge, E = k

0q0

2

and opposite directions.

Solve: The two charges and the directions of their electric fields in three regions are shown in Figure (a) above. Only in region II are the two electric fields in opposite directions. Consider a point a distance x from q1 so a distance

1.20 m − x from q2 . E1 = E2 gives k

0.500 nC x

2

=k

8.00 nC

(1.20 − x)

2

. 16 x 2 = (1.20 m − x) 2 . 4x = ±(1.20 m − x) and x = 0.24 m

is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from the 0.500 nC charge. (b) Let q2 = −8.00 nC be the negative charge. The two charges and the directions of their electric fields in three G G regions are shown in Figure (b) above. E1 and E2 are in opposite directions in regions I and III. But for the magnitudes of the fields to be equal the point must be closer to the charge q1 that has smaller magnitude, and that occurs only for region I. Consider a point a distance x to the left of q1 so 1.20 m + x from q2 . E1 = E2 gives k

0.500 nC x2

=k

8.00 nC (1.20 + x) 2

. 16 x 2 = (1.20 m + x )2 . 4 x = ± (1.20 m + x) and x = 0.40 m is the positive solution. The

electric field is zero at a point 0.40 m from q1 and 1.60 m from q2 . Reflect: In each case there is only one point along the line connecting the two charges where the net electric field is zero.

17.44. Set Up: The electric field of a negative charge is directed toward the charge. E = k

0q0

. The net field is the r2 vector sum of the fields due to each charge. Label the charges q1 , q2 , and q3 , as shown in Figure (a) below. This

figure also shows additional distances and angles. The electric fields at point P are shown in Figure (b) below. This G G G figure also shows the xy coordinates we will use and the x and y components of the fields E1 , E2 , and E3 .


17-16

Chapter 17

Solve: E1 = E3 = (8.99 × 109 N ⋅ m 2 /C2 )

E2 = (8.99 × 109 N ⋅ m 2 /C2 )

5.00 × 10−6 C

2.00 × 10−6 C (0.0600 m)

2

(0.100 m) 2

= 4.49 × 106 N/C

= 4.99 × 106 N/C

E y = E1 y + E2 y + E3 y = 0 and E x = E1x + E2 x + E3 x = E2 + 2 E1 cos53.1° = 1.04 × 107 N/C E = 1.04 × 107 N/C, toward the −2.00 μC charge. G

G

G

*17.45. Set Up: The force on a charge q that is in an electric field E is F = qE . Solve: (a) The force on + q is F+ = qE , to the right. The force on −q is F− = qE , to the left. The net force on the

dipole is zero. (b) The axis is at the midpoint of the line connecting the charges, and perpendicular to the plane of the figure. The torque is zero when each force, F+ and F− , has zero moment arm. This is the case for θ = 0° and θ = 180°, as

shown in Figure (a) and (b) below.

(c) For θ slightly greater than zero, as shown in Figure (c) above, the torque on the dipole is clockwise and is directed so as to return the dipole to the equilibrium position. For θ slightly less than zero the torque is counterclockwise and again tends to rotate the dipole back to its equilibrium position. The θ = 0° position of Figure (a) above is a stable orientation. For θ slightly less than 180°, as in Figure (d) above, the net torque on the dipole rotates it away from the θ = 180° equilibrium position. The same is true for θ slightly greater than 180°. The θ = 180° position of Figure (b) above is an unstable orientation. (d) The electric field of the dipole is directed from the positive charge and toward the negative charge. Thus, in Figure (a) above the electric field of the dipole is to the left, opposite to the direction of the external field. Reflect: For any orientation of the dipole the net force on the dipole is zero. But only for θ = 0° and θ = 180° is the net torque zero.

G

G G

G

G

17.46. Set Up: Use E in Eq. (17.2) to calculate F , F = ma to calculate a , and a constant acceleration equation to calculate the final velocity. Let +x be east. (a) Solve: Fx = qqqE = (1.602 × 10−19 C)(1.50 N/C) = 2.403 × 10−19 N ax = Fx /m = (2.403 × 10−19 N)/(9.109 × 10−31 kg) = +2.638 × 1011 m/s 2



17-17

υ0 x = +4.50 × 105 m/s, ax = +2.638 × 1011 m/s2 , x − x0 = 0.375 m, υ x = ? υ x2 = υ0 x 2 + 2ax ( x − x0 ) gives υ x = 6.33 × 105 m/s G G Reflect: E is west and q is negative, so F is east and the electron speeds up. (b) Solve:

Fx = − qqqE = − (1.602 × 10−19 C)(1.50 N/C) = −2.403 × 10−19 N

ax = Fx /m = (−2.403 × 10−19 N)/(1.673 × 10−27 kg) = −1.436 × 108 m/s 2

υ0 x = +1.90 × 104 m/s, a x = −1.436 × 108 m/s 2 , x − x0 = 0.375 m, υ x = ?

υ x 2 = υ0 x 2 + 2ax ( x − x0 ) gives υ x = 1.59 × 104 m/s G Reflect: q > 0 so F is west and the proton slows down.

*17.47. Set Up: E = k

0q0 r2

Solve: (a) For r1 = 1.0 cm, E1 = E = k (b) Now r3 = 3r1. E3 = k

0q0 (3r1 ) 2

0q0 r12

. r2 = 2r1 . E2 = k

0q0 (2r1 )2

1 0q0 = k 2 = E /4 4 r1

1 0q0 = k 2 = E /9 9 r1

17.48. Set Up: The analysis is similar to that of Example 17.7c—except that the dipole charges are located on the y-axis and the x-axis is now the axis of symmetry. The electric field genertated by a single point charge has a 0q0 magnitude of E = k 2 : Since the dipole charges are equal in magnitude and equidistant from any given point on the r x-axis, we can conclude that the two dipole charges generate electric fields of the same magnitude at any given point on the x-axis. Solve: In the figure below we show the point P = ( x,0) on the x-axis. At this point the electric field generated by the positive charge points down and to the right ( E+ )—and the electric field generated by the negative charge points down and to the left ( E− ). By symmetry these two electric fields have equal y-components and opposite x-components. We conclude that the total electric field points in the negative y-direction.


17-18

Chapter 17

The magnitude of the total electric field is twice the magnitude of the y-component of the electric field of a single q charge. Thus we have Etotal = 2k 2 cosθ , where θ is measured as shown in the figure above. From the figure we r a qa 2 see that cos θ = and r = x + a 2 . Putting these last three together we obtain Etotal = 2k 2 . r ( x + a 2 )3/2 When x is much larger than a, we have x 2 + a 2 ≈ x 2 . Thus, for large x we have qa qa qa Etotal = 2k 2 ≈ 2k 2 3/2 = 2k 3 . 2 3/2 (x + a ) (x ) x Reflect: Note that the electric field of a point charge decreases in proportion to

in proportion to

1 x3

1 x2

while that of a dipole decreases

.

17.49. Set Up: Field lines are directed away from a positive charge and toward a negative charge. The density of field lines is proportional to the magnitude of the electric field. The field lines represent the resultant field at each point, the net field that is the vector sum of the fields due to each of the three charges. Solve: (a) Since field lines from positive charges and toward negative charges, we can deduce that the top charge is positive, middle is negative, and bottom is positive. (b) The electric field is the smallest on the horizontal line through the middle charge, at two positions on either side where the field lines are least dense. Here the y-components of the field are cancelled between the positive charges and the negative charge cancels the x-component of the field from the two positive charges. Reflect: Far from all three charges the field is the same as the field of a point charge equal to the algebraic sum of the three charges.

17.50. Set Up: The proton has charge + e and the electron has charge −e. The electric field of a point charge is directed away from the charge if it is positive and toward the point charge if it is negative. Use coordinates where the x axis is along line ABC and the + x direction is toward C and where the + y direction is toward the proton. The electric fields due to each charge at A, B, and C are sketched in the figure below.

Solve: (a) At point A the x components of the fields cancel and the net field is in the − y direction. (b) At B both fields are in the − y direction and the net field is in the − y direction. (c) At C, as at A, the x component of the fields cancel and the net field is in the − y direction.



17-19

*17.51. Set Up: The electric field lines point away from the positive charge and into the negative charge. To indicate the relative size of each charge, we should draw roughly four times as many field lines terminating on the −4Q charge as originate from the + Q charge. Solve: The sketch is shown in the figure below.

Reflect: The sketch is similar to that of Figure 17.22b in the textbook—except that in that figure the two charges are equal in magnitude so that the number of field lines terminating on the negative charge is equal to the number of field lines originating from the positive charge.

17.52. Set Up: Electric field points toward negative charge and away from positive charge.

G G G Solve: (a) Figure (a) above shows EQ and E + q at point P. EQ must have the direction shown, to produce a G resultant field in the specified direction. EQ is toward Q, so Q is negative. In order for the horizontal components of

the two fields to cancel, Q and q must have the same magnitude. (b) No. If the lower charge were negative, its field would be in the direction shown in Figure (b) above. The two G G possible directions for the field of the upper charge, when it is positive ( E+ ) or negative ( E− ), are shown. In neither case is the resultant field in the direction shown in the figure in the problem.

*17.53. Set Up: Electric field is directed away from positive charge and toward negative charge. By symmetry, far from the edges of the sheets the field lines are perpendicular to the sheets; there is no reason to prefer to the left or to the right for a component of electric field. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

17-20

Chapter 17

Solve: (a) The fields of each sheet are sketched in Figure (a) above. The solid lines are the field due to the upper positive sheet and the dashed lines are the field due to the lower negative sheet. Between the two sheets the fields are in the same direction and add. Outside the two sheets the fields are in opposite directions and cancel. The net electric field is sketched in Figure (b) above. This result is consistent with the field in the similar case described in Example 17.5. (b) The fields of each sheet are sketched in Figure (c) above. The solid lines are the field due to the upper sheet and the dashed lines are the field due to the lower sheet. Since both sheets are positive, the field of each sheet is directed away from that sheet. Between the two sheets the fields are in opposite directions and cancel. Outside the two sheets the fields are in the same direction and add. The net field is sketched in Figure (d) above. (c) The fields are the same as in part (b), except they are toward each sheet. The net field of the two sheets is sketched in Figure (e) above. Reflect: More advanced treatments show that the electric field of an infinite sheet is uniform and does not depend on the distance from the sheet.

17.54. Set Up: Gauss’s law says Φ E = Solve: (a) Φ E =

Qencl

⑀0

2.50 × 10−6 C 8.854 × 10212 C2 /(N ⋅ m 2 )

. ⑀ 0 = 8.854 × 10−12 C2 /(N ⋅ m 2 ).

= 2.82 × 105 N ⋅ m 2 /C

(b) Qencl = ⑀0ΦE = (8.854 × 10−12 C 2 /(N ⋅ m 2 )) (1.40 N ⋅ m 2 /C) = 1.24 × 10−11 C

17.55. Set Up: Gauss’s law says that the net electric flux Φ E through any closed surface equals Qencl /⑀0 , where Qencl is the net charge enclosed by the surface. Solve: S1 : Qencl = 0 so Φ E = 0

S2 : Qencl = +5.0 μC + 9.0 μC + (−7.0 μC) = +7.0 μC and Φ E = +7.0 μ C/⑀0 = 7.9 × 105 N ⋅ m 2 /C. S3 : Qencl = +9.0 μC + 1.0 μC + (−10.0 μC) = 0 and ΦE = 0. S4 : Qencl = +8.0 μC + (−7.0 μC) = +1.0 μC and Φ E = +1.0 μ C/⑀0 = 1.1 × 105 N ⋅ m 2 /C. S5 : Qencl = 8.0 μ C + (−7.0 μ C) + 5.0 μ C + 9.0 μ C + 1.0 μ C + (−10.0 μC ) = 16.0 μ C and Φ E = +6.0 μ C/⑀0 = 6.8 × 105 N ⋅ m 2 /C.

Reflect: The electric field varies in a complicated way, in both magnitude and direction, for each of the surfaces and direct calculation of the flux as ∑ E⊥ ΔA would be very difficult. Gauss’s law gives us a very simple way of

calculating the net flux.

17.56. Set Up: The cube is a closed surface so we can apply Gauss’s law to it: Φ E =

Qencl

⑀0

. By symmetry the flux

is the same through each of the 6 faces of the cube.



Solve: (a) The total flux through the cube is Φ E =

Qencl

⑀0

=

8.00 × 10−9 C 8.854 × 10−12 C2 /(N ⋅ m 2 )

17-21

= 904 N ⋅ m 2 /C.

Φ E 904 N ⋅ m 2 /C = = 151 N ⋅ m 2 /C. 6 6

(b) The flux through each face is

*17.57. Set Up: Example 17.10 shows that E = 0 inside a uniform spherical shell and that E = k

0q0 r2

outside the

shell. Solve: (a) E = 0 (b) r = 0.060 m and E = (8.99 × 109 N ⋅ m 2 /C2 ) (c) r = 0.110 m and E = (8.99 × 109 N ⋅ m 2 /C2 )

15.0 × 10−6 C (0.060 m) 2

= 3.75 × 107 N/C

15.0 × 10−6 C

= 1.11 × 107 N/C (0.110 m) 2 Reflect: Outside the shell the electric field is the same as if all the charge were concentrated at the center of the shell. But inside the shell the field is not the same as for a point charge at the center of the shell; inside the shell the electric field is zero.

17.58. Set Up: The electric field required to produce a spark 6 in. long is 6 times as strong as the field needed to produce a spark 1 in. long. By Gauss’s law, q = ⑀0 EA and the electric field is the same as for a point-charge, E=

1 q . 4π ⑀0 r 2

Solve: (a) The electric field for 6-in. sparks is E = 6 × 2.00 × 104 N/C = 1.20 × 105 N/C

The charge to produce this field is q = ⑀0 EA = ⑀0 E (4π r 2 ) = (8.85 × 10−12 C 2 /N ⋅ m 2 )(1.20 × 105 N/C)(4π )(0.15 m) 2 = 3.00 × 10−7 C. (b) Using Coulomb’s law gives E = (9.00 × 109 N ⋅ m 2 /C2 )

3.00 × 10−7 C

(0.150 m) 2 Reflect: It takes only about 0.3 μC to produce a field this strong.

= 1.20 × 105 N/C.

17.59. Set Up: The method of Example 17.10 shows that the electric field outside the sphere is the same as for a point charge of the same charge located at the center of the sphere. The charge of an electron has magnitude e = 1.60 × 10−19 C. Solve: (a) E = k

0q0 r2

. For r = R = 0.150 m, E = 1150 N/C so 0q0 =


Er 2 (1150 N/C)(0.150 m)2 = = 2.88 × 10−9 C. k 8.99 × 109 N ⋅ m 2 /C2 2.88 × 10−9 C

1.60 × 10−19 C/electron

(b) r = R + 0.100 m = 0.250 m. E = k

0q0 r2

= 1.80 × 1010 electrons.

= (8.99 × 109 N ⋅ m 2 /C2 )

2.88 × 10−9 C (0.250 m) 2

= 414 N/C.

17.60. Set Up: Apply Gauss’s law. In each case consider a small Gaussian surface in the region of interest. Solve: (a) Pick any point in a region where the electric field is uniform. Construct a small cubical Gaussian surface surrounding that point so that the uniform electric field is perpendicular to two opposite faces and parallel to the other four faces of the cube. The total flux through the cube must be zero since the uniform field enters one face and exits the opposite face of the cube. According to Gauss’s law the net charge within the cube must be zero. Since we may make this cube as small as we wish, the net charge at the given point must be zero.


17-22

Chapter 17

(b) The fact that there is no charge in a region of space does NOT mean that the electric field is uniform. Consider a closed volume close to, but not including, a point charge. The field diverges there, but there is no charge in that region. Reflect: The electric field within a region can depend on charges located outside the region. But the flux through a closed surface depends only on the net charge contained within that surface.

*17.61. Set Up: The charge distribution has spherical symmetry, so the electric field, if nonzero, is radial and depends only on the distance from the center of the shell. Solve: (a) Apply Gauss’s law to a sphere of radius r < a and concentric with the shell. The electric field, if nonzero, is constant over the Gaussian surface and perpendicular to it, so ΦE = E (4π r 2 ). But no charge is enclosed by the Gaussian surface, so Qencl = 0. Gauss’s law gives E (4π r 2 ) = 0 and E = 0. (b) Apply Gauss’s law to a sphere of radius r > b. ΦE = E (4π r 2 ). Qencl = Q. Gauss’s law gives E (4 π r 2 ) =

E=

Q 4π ⑀0r 2

Q

⑀0

and

.

(c) The thick shell can be constructed from a series of concentric thin shells. E = 0 inside each of these thin shells, so E = 0 inside the thick shell. Reflect: In using Gauss’s law it is very helpful to select a Gaussian surface of appropriate symmetry, so it is simple to express the flux in of the electric field at the surface.

17.62. Set Up: Section 17.9 shows that the net charge of a charged conductor is entirely on its outer surface. Solve: (a) q = 0 on the surface of the inner cavity of the conducting car. (b) All the net charge is on the outer surface, −850 μC.

*17.63. Set Up: E = 0 everywhere within the conductor. Any net charge must be on the inner and outer surfaces of the conductor.

Solve: (a) and (b) Apply Gauss’s law to a surface that is within the conductor, just outside the cavity, as shown in the figure above. E = 0 everywhere on the Gaussian surface so ΦE = 0 for that surface. Gauss’s law then says that

Qencl for this surface is zero. The conductor is neutral, so if the outer surface has charge −12 μC the inner surface must have charge +12 μC. To make Qencl = 0 there must be −12 μC within the hole. Reflect: The charge in the hole creates the charge separation in the conductor. It pulls +12 μC to the inner surface and that leaves −12 μC on the outer surface.

17.64. Set Up: E = 0 everywhere within the conductor. Any net charge must be on the inner and outer surfaces of the conductor. Solve: Apply Gauss’s law to a surface that is within the conductor, just outside the cavity, as shown in the figure below. E = 0 everywhere on the Gaussian surface so ΦE = 0 for that surface.



17-23

(a) Qencl = 0 for the Gaussian surface means that q = 0 on the inner surface, so the +16 μC of net charge is on the

outer surface of the conductor. (b) Qencl = 0 for the Gaussian surface means that there must be +11 nC on the inner surface of the conductor. Since the total net charge of the conductor is +16 nC, if there is +11 nC on the inner surface there must be +5 nC on the outer surface.

17.65. Set Up: F = k

0 qq′0

. Like charges repel and unlike charges attract. Charges q1 and q2 and the forces they r2 G G exert on q3 at the origin are sketched in Figure (a) and (b) below. For the net force on q3 to be zero, F1 and F2

from q1 and q2 must be equal in magnitude and opposite in direction.

G G G Solve: (a) Since F1 , F2 , and Fnet are all in the + x direction, F = F1 + F2 . This gives 4.00 × 10−6 N = k

0 q1 q3 0 r12

+k

0 q2 q3 0 r2 2

.

⎛ 4.50 × 10−9 C 2.50 × 10−9 C ⎞ ⎜ ⎟ = + q 3 2 ⎟ ⎜ [ 0.200 m ]2 8.99 × 109 N ⋅ m 2 /C2 . 0 300 m [ ] ⎝ ⎠ 4.00 × 10−6 N

and q3 = 3.17 × 10−9 C = 3.17 nC. G G G G G (b) Both F1 and F2 are in the + x direction, so Fnet = F1 + F2 is in the + x direction. G G (c) The forces F1 and F2 on q3 in each of the three regions are sketched in Figure (b) above. Only in regions I (to G G the left of q2 ) and III (to the right of q1 ) are F1 and F2 in opposite directions. But since 0 q2 0 < 0 q1 0 , q3 must be closer to q2 than to q1 in order for F1 = F2 , and this is the case only in region I. Let q3 be a distance d to the left of q2 , so it is a distance d + 0.500 m from q1 . F1 = F2 gives k

4.50 nC (d + 0.500 m)

2

=k

2.50 nC

d2

. 1.80d 2 = (d + 0.500 m) 2 .

1.80d = ± (d + 0.500 m). The positive solution is d = 1.46 m. This point is at x = −0.300 m − 1.46 m = −1.76 m. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

17-24

Chapter 17

Reflect: At the point found in part (c) the electric field is zero. The force on any charge placed at this point will be zero.

G

G

17.66. Set Up: Use constant acceleration equations to calculate the upward acceleration a, and then apply F = qE to calculate the electric field. Let +y be upward. An electron has charge q = − e. Solve: (a) υ0 y = 0 and a y = a, so y − y0 = υ0 yt + 12 a yt 2 gives y − y0 = 12 at 2 . Then a =

2( y − y0 ) t

2

=

2(4.50 m) (3.00 × 10−6 s) 2

=

1.00 × 1012 m/s 2 . E=

F ma (9.11 × 10−31 kg) (1.00 × 1012 m/s 2 ) = = = 5.69 N/C q q 1.60 × 10−19 C

The force is up, so the electric field must be downward since the electron has negative charge. (b) The electron’s acceleration is ~ 1011 g , so gravity must be negligibly small compared to the electrical force. Reflect: Since the electric field is uniform, the force it exerts is constant and the electron moves with constant acceleration.

*17.67. Set Up: F = k

0 qq ′ 0

. Like charges repel and unlike charges attract. The three charges and the forces on q3 r2 are shown in the figure below.

Solve: (a) F1 = k

0 q1q3 0 r12

= (8.99 × 109 N ⋅ m 2 /C2 )

(5.00 × 10−9 C)(6.00 × 10−9 C) (0.0500 m)2

= 1.079 × 10−4 C.

θ = 36.9°. F1x = + F1 cos θ = 8.63 × 10−5 N. F1 y = + F1 sin θ = 6.48 × 10−5 N. F2 = k

0 q2 q3 0 r22

= (8.99 × 109 N ⋅ m 2 /C2 )

(2.00 × 10−9 C)(6.00 × 10−9 C) (0.0300 m)2

= 1.20 × 10−4 C.

F2 x = 0, F2 y = − F2 = −1.20 × 10−4 N. Fx = F1x + F2 x = 8.63 × 10−5 N. Fy = F1 y + F2 y = 6.48 × 10−5 N + (−1.20 × 10−4 N) = −5.52 × 10−5 N. (b) F = Fx 2 + Fy 2 = 1.02 × 10−4 N. tan φ = 0

Fy Fx

0 = 0.640. φ = 32.6°, below the + x axis.

Reflect: The individual forces on q3 are computed from Coulomb’s law and then added as vectors, using

components. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


17.68. Set Up: F = k

17-25

0 q1 q2 0

. Like charges repel and unlike charges attract. The positions of the three charges are r2 sketched in Figure (a) below and each force on q3 is shown. The distance between q1 and q3 is 5.00 cm.

Solve: (a) F1 = k

0 q1q3 0 r12

= (8.99 × 109 N ⋅ m 2 /C2 )

(3.00 × 10−9 C)(5.00 × 10−9 C) (5.00 × 10−2 m) 2

= 5.394 × 10−5 N

F1x = − F1 cos θ = − (5.394 × 10−5 N)(0.600) = −3.236 × 10−5 N; F1 y = 2F1 sin θ = − (5.394 × 10−5 N)(0.800) = −4.315 × 10−5 N F2 = k

0 q2 q3 0 r22

= (8.99 × 109 N ⋅ m 2 /C2 )

(2.00 × 10−9 C)(5.00 × 10−9 C) (3.00 × 10−2 m)2

= 9.989 × 10−5 N

F2 x = 9.989 × 10−5 N; F2 y = 0

Fx = F1x + F2 x = 9.989 × 10−5 N + (−3.236 × 10−5 N) = 6.75 × 10−5 N; Fy = F1 y + F2 y = −4.32 × 10−5 N G Fy (b) F and its components are shown in Figure (b) above. F = Fx 2 + Fy 2 = 8.01 × 10−5 N. tan θ = 0 0 = 0.640 and Fx G θ = 32.6°. F is 327° counterclockwise from the + x axis.


17-26

Chapter 17

0 qq ′0

*17.69. Set Up: F = k

r2

. Like charges repel and unlike charges attract.

Solve: (a) The charges and the forces on the −1.00 μC charge are shown in Figure (a) above. The distance r14 G G between q1 and q4 is r = 12 2(0.200 m) = 0.1414 m. F2 and F3 are equal in magnitude and opposite in direction, G G G G so F2 + F3 = 0 and Fnet = F1 .

()

0 q1 q4 0

F1 = k

r14

2

= (8.99 × 109 N ⋅ m 2 /C2 )

(3.00 × 10−9 C)(1.00 × 10−6 C) (0.1414 m) 2

= 1.35 × 10−3 N.

The resultant force has magnitude 1.35 × 10−3 N and is directed away from the vacant corner. (b) The charges and the forces on the −1.00 μC charge are shown in Figure (b) above. The distance r14 between q1

and q4 is

2(0.200 m) = 0.2828 m. Use coordinates as shown in the figure. F2 x = − F3 x and F1x = 0, so Fx = 0.

Fy = F1 y + F2 y + F3 y = F1 + 2 F2 cos 45°. F1 = k F2 = k

0 q1 q4 0 r14

2

0 q2 q4 0 r24 2

= (8.99 × 109 N ⋅ m 2 /C2 )

(3.00 × 10−9 C)(1.00 × 10−6 C)

= (8.99 × 109 N ⋅ m 2 /C2 )

(0.2828 m) 2 (3.00 × 10−9 C)(1.00 × 10−6 C) (0.200 m)2

= 3.372 × 10−4 N. = 6.742 × 10−4 N

Fy = 3.372 × 10−4 N + 2(6.742 × 10−4 N)cos 45° = 1.29 × 10−3 N. The resultant force has magnitude 1.29 × 10−3 N and is directed toward the center of the square.

17.70. Set Up: The force on the negatively charged electron is opposite to the direction of the field. Since the field is uniform, the force and acceleration are constant. Use coordinates with + y upward and + x to the right. The electron is at x = +0.0200 m when y = +0.00500 m. a x = 0. υ0 x = +5.00 × 106 m/s, υ0 y = 0. For an electron,

q = −e = −1.60 × 10−19 C and m = 9.11 × 10−31 kg. x = υ yt + 12 ax t 2 . t= y = υ0 yt + 12 a y t 2 gives a y = Fy = ma y and E y = −

Fy e

=−

2y

t2

=

ma y e

x

υ0 x

=

0.0200 m 6

5.00 × 10 m/s

2(0.00500 m) (4.00 × 10−9 s) 2

=−

= 4.00 × 10−9 s.

= 6.25 × 1014 m/s 2 .

(9.11 × 10−31 kg)(6.25 × 1014 m/s 2 ) 1.60 × 10

−19

C

= −3.56 × 103 N/C.

3

The electric field has magnitude 3.56 × 10 N/C.



17-27

*17.71. Set Up: The ball is in equilibrium, so for it ∑ Fx = 0 and ∑ Fy = 0. The force diagram for the ball is given

G G G in the figure below. FE is the force exerted by the electric field. F = qE . Since the electric field is horizontal, FE is

horizontal. Use the coordinates shown in the figure. The tension in the string has been replaced by its x and y components.

Solve: ∑ Fy = 0 gives Ty − mg = 0. T cos θ − mg = 0 and T =

mg . cos θ

∑ Fx = 0 gives FE − Tx = 0. FE − T sin θ = 0. ⎛ mg ⎞ −3 2 −2 FE = ⎜ ⎟ sin θ = mg tan θ = (12.3 × 10 kg)(9.80 m/s )tan17.4° = 3.78 × 10 N ⎝ cosθ ⎠ FE 3.78 × 10−2 N = = 3.41 × 104 N/C 0 q 0 1.11 × 10−6 C G G q is negative and FE is to the right, so E is to the left in the figure. FE = 0 q 0 E so E =

Reflect: The larger the electric field E the greater the angle the string makes with the wall.

17.72. Set Up: E = k

0q0 2

G G . E is toward a negative charge and away from a positive charge. At the origin, E1 due

r to the 25.00 nC charge is in the + x direction, toward the charge. Solve: (a) E1 = (8.99 × 109 N ⋅ m 2 /C2 )

(5.00 × 10−9 C) (1.20 m) 2

= 31.2 N/C. E1x = +31.2 N/C.

E x = E1x + E2 x . E x = +45.0 N/C, so E2 x = Ex − E1x = +45.0 N/C − 31.2 N/C = 13.8 N/C.

G 0Q 0 E r 2 (13.8 N/C)(0.600 m)2 E is away from Q so Q is positive. E2 = k 2 gives 0 Q 0 = 2 = = 5.53 × 10−10 C. k r 8.99 × 109 N ⋅ m 2 /C2 (b) E x = −45.0 N/C, so E2 x = E x − E1x = −45.0 N/C − 31.2 N/C = −76.2 N/C.

G E r 2 (76.2 N/C)(0.600 m) 2 E is toward Q so Q is negative. 0 Q 0 = 2 = = 3.05 × 10−9 C. 9 2 2 k 8.99 × 10 N ⋅ m /C

17.73. Set Up: The weight of the raindrop is w = mg and is downward. We can calculate the mass of the raindrop G G 4 from the known density of water: m = ρ V , where ρ = 103 kg/m3 and V = π r 3. The electric force is F = qE , 3 where E = 150 N/C.


17-28

Chapter 17

Solve: (a) To balance the weight of the raindrop the electric force must be upward. Since the electric field is downward the net charge on the raindrop must be negative. For equilibrium we must have w = mg = 0 q 0 E . Thus

0q |=

mg ⎛ 4 3 ⎞ 4 = ⎜ π r ρ g ⎟ /E = π (1.0 × 10−5 m)3 (103 kg/m3 )(9.80 m/s 2 )/(150 N/C) = 2.7 × 10−13 C. ⎝ ⎠ E 3 3

0 q 0 2.7 × 10−13 C = = 1.7 × 106. e 1.60 × 10−19 C Reflect: Although this may appear to be a large number in absolute , the excess number of electrons represents only about 10−7, of the total number of electrons in the raindrop.


17.74. Set Up: It is rather difficult to calculate the flux directly from Eq. (17.5) since the magnitude and direction of the electric field varies across the faces of the cube. A much easier approach is to use Gauss’s law to calculate the total flux through the cube. Let the cube be the Gaussian surface. The charge enclosed is the point charge. 9.60 × 10−6 C Solve: (a) ΦE = Qencl /⑀0 = = 1.084 × 106 N ⋅ m 2 /C. 8.854 × 10−12 C2 /N ⋅ m 2 By symmetry the flux is the same through each of the six faces, so the flux through one face is 1 (1.084 × 106 N ⋅ m 2 /C) = 1.81 × 105 N ⋅ m 2 /C. 6 (b) Reflect: In part (a) the size of the cube did not enter into the calculations. The flux through one face depends only on the amount of charge at the center of the cube. So the answer to (a) would not change if the size of the cube were changed.

*17.75. Set Up: The force on the electron is upward in the figure so the electric field must be downward. To produce a net electric field that is downward, it must be that q1 is positive, q2 is negative, and 0 q1 0 = 0 q2 0. The field due to q1 and q2 at the location of the electron are sketched in the figure below. The electron is r = 3.61 cm from each charge. tan θ =

3.00 cm and θ = 56.3°. 2.00 cm

Solve: The net electric field is E = 2 E1 y = 2k

q1 =

q1 r

2

cos θ . F = eE =

2keq1 r

2

cos θ . F = ma so

2keq1 r2

cos θ = ma and

r 2ma (3.61 × 10−2 m) 2 (9.11 × 10−31 kg)(8.25 × 1018 m/s 2 ) = = 6.14 × 1026 C. 2ke cos θ 2(8.99 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C)cos56.3°

q2 = −6.14 × 10−6 C. Reflect: The force that an electric field exerts on a negative charge is opposite to the direction of the electric field. We could also do this problem by considering the force that each charge exerts on the electron.



17-29

17.76. Set Up: When the forces balance, a = 0 and the molecule moves with constant velocity. q Kυ Solve: (a) F = FD so qE = KRv and

= R E Eq Eq ⎛ ⎞ ⎛ ET ⎞ q T =⎜ (b) υ = and is constant. x = υ t = ⎜ ⎟ ⎝ KR ⎠ ⎝ K ⎟⎠ R KR

ET ⎛ ET ⎞ q ⎛ q⎞ ⎛ q⎞ ⎛ q⎞ ⎛ q⎞ (c) x = ⎜ , where is constant. ⎜ ⎟ = 2 ⎜ ⎟ and ⎜ ⎟ = 3⎜ ⎟ ⎝ K ⎟⎠ R ⎝ R⎠ 2 ⎝ R ⎠1 ⎝ R⎠3 ⎝ R ⎠1 K

⎛ ET ⎞ ⎛ q ⎞ ⎛ ET ⎞ ⎛ q ⎞ ⎛ ET ⎞ ⎛ q ⎞ ⎛ ET ⎞ ⎛ q ⎞ x2 = ⎜ = 2⎜ = 2 x1; x3 = ⎜ = 3⎜ = 3 x1 ⎝ K ⎟⎠ ⎜⎝ R ⎟⎠ 2 ⎝ K ⎟⎠ ⎜⎝ R ⎟⎠ 1 ⎝ K ⎟⎠ ⎜⎝ R ⎟⎠ 3 ⎝ K ⎟⎠ ⎜⎝ R ⎟⎠ 1

*17.77. Set Up: The force between the proton and electron is given by FE = k have a =

υ2 r

q1q2 r2

=k

e2 r2

. For a circular orbit we

.

Solve: Applying F = ma gives k

υ=

e2 r2

= me

υ2 r

. Solving for υ we obtain

ke2 (8.99 × 109 N ⋅ m 2 /C2 )(1.602 × 10−19 C) 2 = = 2.2 × 106 m/s. −31 −11 me r (9.109 × 10 kg)(5.29 × 10 m)

Reflect: Although this speed is fast, it is only about 1% the speed of light.

Solutions to age Problems 17.78. Set Up: Positive charges will have a tendancy to move in the direction of the applied electric field and negative charges will have a tendancy to move in a direction that is opposite the applied electric field. Solve: Since we do not know the sign of the charge on the molecules, we cannot determine the direction of the applied electric field. The correct answer is D.

*17.79. Set Up: Molecules are in a constant state of agitation and tend to diffuse randomly in the absence of a perturbing force. Although an external electric field can alter the distribution of the proteins, the total number of protein molecules will remain fixed. Solve: Once the external electric field is removed, the random motion of the proteins will cause them to distribute themselves randomly within the membrane. Since the total number of proteins must remain fixed, the final density of proteins must be intermediate between the minimum value shown at 0D and the maximum value shown at 180D. The correct answer is C.


18

ELECTRIC POTENTIAL AND CAPACITANCE

Answers to Multiple-Choice Problems 1. A, C 2. C 3. A 14. A, C 15. C

4. A, D

5. C

6. B, C

7. B, D

8. C

9. B

10. C

11. D

12. A

13. A

Solutions to Problems *18.1. Set Up: Since the charge is positive the force on it is in the same direction as the electric field. Since the field is uniform the force is constant and W = Fs cos φ . G G Solve: (a) F is upward and s is to the right, so φ = 90° and W = 0. G G (b) F is upward and s is upward, so φ = 0° W = Fs = qEs = (28.0 × 10−9 C)(4.00 × 104 N/C)(0.670 m) = 7.50 × 10−4 J. G G (c) F is upward and s is at 45.0° below the horizontal, so φ = 135.0°. W = Fs cos φ = qEs cos φ = (28.0 × 10−9 C)(4.00 × 104 N/C)(2.60 m)cos135.0° = −2.06 × 10−3 J. Reflect: The work is positive when the displacement has a component in the direction of the force and it is negative when the displacement has a component opposite to the direction of the force. When the displacement is perpendicular to the force the work done is zero.

18.2. Set Up: (a) The proton has charge + e and mass 1.67 × 10−27 kg. Let point a be at the negative plate and point b be at the positive plate. The electric field is directed from the positive plate toward the negative plate. The force on the positively charged proton is in this direction. The proton moves in a direction opposite to the electric force, so the work done by the electric force is negative. Solve: Wa →b = − qEd = −eEd = −(1.60 × 10−19 C)(2.80 × 106 N/C)(0.100 m) = −4.48 × 10−14 J

18.3. Set Up: U = k Solve: r =

q1 q2 r

kq1 q2 (8.99 × 109 N ⋅ m 2 / C2 )(−7.20 × 10−6 C)(12.30 × 10−6 C) = = 0.372 m U −0.400 J

18.4. Set Up: Wa →b = U a − U b . U = k Solve: U a = k

q1q2 . ra = 0.150 m, rb = (0.250 m)2 + (0.250 m) 2 = 0.354 m. r

q1 q2 (8.99 × 109 N ⋅ m 2 / C2 )(2.40 × 10−6 C)(−4.30 × 10−6 C) = = −0.619 J. ra 0.150 m


18-1

18-2

Chapter 18

Ub = k

q1 q2 (8.99 × 109 N ⋅ m 2 /C2 )(2.40 × 10−6 C)(−4.30 × 10−6 C) = = −0.262 J. rb 0.354 m Wa →b = U a − U b = −0.619 J − (−0.262 J ) = −0.357 J.

qq′ . Let q1 = r +3.00 nC and q2 = +2.00 nC. Let q3 be the electron. The potential energy of the electron is the sum of U13 and U 23 .

*18.5. Set Up: An electron has charge q = −e = −1.60 × 10−19 C. For a pair of point charges, U = k Solve: (a) r1 = r2 = 0.250 m.

⎛ [3.00 × 10−9 C][−1.60 × 10−19 C] ⎞ q1 q3 −17 = (8.99 × 109 N ⋅ m 2 /C2 ) ⎜ J. ⎟⎟ = −1.73 × 10 ⎜ . r1 0 250 m ⎝ ⎠ ⎛ [2.00 × 10−9 C][−1.60 × 10−19 C] ⎞ q q −17 U 23 = k 2 3 = (8.99 × 109 N ⋅ m 2 /C2 ) ⎜ J. ⎟⎟ = −1.15 × 10 ⎜ r2 0.250 m ⎝ ⎠

U13 = k

U = U13 + U 23 = −2.88 × 10−17 J (b) r1 = 0.100 m and r2 = 0.400 m.

U13 = k

⎛ [3.00 × 10−9 C][−1.60 × 10−19 C] ⎞ q1 q3 −17 = (8.99 × 109 N ⋅ m 2 / C2 ) ⎜ J. ⎟⎟ = −4.32 × 10 ⎜ . r1 0 100 m ⎝ ⎠

U 23 = k

⎛ [2.00 × 10−9 C][−1.60 × 10−19 C] ⎞ q2 q3 −18 = (8.99 × 109 N ⋅ m 2 / C2 ) ⎜ J. ⎟⎟ = −7.19 × 10 ⎜ . r2 0 400 m ⎝ ⎠ U = U13 + U 23 = −5.04 × 10−17 J.

Reflect: The potential energy is negative since the charge of the electron has opposite sign from the charge of each of the other two particles. The magnitude of the potential energy increases when the electron moves toward the larger charge.

18.6. Set Up: For a pair of point charges the electrical potential energy is U = k

qq ′ . In the O-H-N combination r

the O− is 0.170 nm from the H + and 0.280 nm from the N − . In the N-H-N combination the N − is 0.190 nm from the H + and 0.300 nm from the other N2. U is positive for like charges and negative for unlike charges. Solve: (a) O-H-N (1.60 × 10−19 C) 2 O− -H + : U = −(8.99 × 109 N ⋅ m 2 /C2 ) = −1.35 × 10−18 J 0.170 × 10−9 m

O− -N − : U = (8.99 × 109 N ⋅ m 2 /C2 )

(1.60 × 10−19 C)2 0.280 × 10−9 m

= +8.22 × 10−19 J

N-H-N N − -H + : U = −(8.99 × 109 N ⋅ m 2 /C2 ) N − -N − : U = (8.99 × 109 N ⋅ m 2 /C2 )

(1.60 × 10−19 C) 2 0.190 × 10−9 m

(1.60 × 10−19 C) 2 −9

= −1.21 × 10−18 J = +7.67 × 10−19 J

0.300 × 10 m The total potential energy is U tot = −1.35 × 10−18 J + 8.22 × 10−19 J − 1.21 × 10−18 J + 7.67 × 10−19 J = −9.71 × 10−19 J. (b) In the hydrogen atom the electron is 0.0529 nm from the proton. (1.60 × 10−19 C) 2 U = − (8.99 × 109 N ⋅ m 2 /C2 ) = −4.35 × 10−18 J. 0.0529 × 10−9 m The magnitude of the potential energy in the hydrogen atom is about a factor of 4 larger than what it is for the adenine-thymine bond. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Electric Potential and Capacitance

18.7. Set Up: For a pair of point charges the electrical potential energy is U = k

18-3

qq′ . In the O-H-O combination the r

O− is 0.180 nm from the H + and 0.290 nm from the other O− . In the N-H-N combination the N − is 0.190 nm from the H + and 0.300 nm from the other N − . U is positive for like charges and negative for unlike charges. In the O-H-N combination the O− is 0.180 nm from the H + and 0.290 nm from the N − . U is positive for like charges and negative for unlike charges. Solve: O-H-O O− -H + , U = −1.28 × 10−18 J; O− -O − , U = +7.93 × 10−19 J

N-H-N N − -H + , U = −1.21 × 10−18 J; N − -N − , U = +7.67 × 10−19 J O-H-N O − -H + , U = −1.28 × 10−18 J; O− -N − , U = +7.93 × 10−19 J The total potential energy is −3.77 × 10−18 J + 2.35 × 10−18 J = −1.42 × 10−18 J. Reflect: For pairs of opposite sign the potential energy is negative and for pairs of the same sign the potential energy is positive. The net electrical potential energy is the algebraic sum of the potential energy of each pair.

18.8. Set Up: The initial potential energy is converted to kinetic energy after the charges have moved far apart. qq′ to each pair of charges and add to get the total electrical potential energy. In part (b) the distance r between the 4.00 μ C and 3.00 μ C charges is 18.0 cm.

Apply U = k

⎛q q qq q q ⎞ ⎛ 8.99 × 109 N ⋅ m 2 /C2 ⎞ −9 −9 Solve: (a) U = k ⎜ 1 2 + 1 3 + 2 3 ⎟ = ⎜ ⎟⎟ ([24.00 × 10 C][−2.00 × 10 C] + ⎜ . 0 100 m r r r 13 23 ⎠ ⎝ ⎝ 12 ⎠

[−2.00 × 10−9 C][23.00 × 1029 C] + [−3.00 × 10−9 C][24.00 × 1029 C]) = 2.34 × 10−6 J ⎛ [4.00 × 10−6 C][2.00 × 10−6 C] ⎛q q qq q q ⎞ (b) U = k ⎜ 1 2 + 1 3 + 2 3 ⎟ = (8.99 × 109 N ⋅ m 2 /C2 ) ⎜ + ⎜ 0.100 m r13 r23 ⎠ ⎝ r12 ⎝

[2.00 × 10−6 C][3.00 × 10−6 C] [4.00 × 10−6 C][3.00 × 10−6 C] ⎞ + ⎟⎟ = 1.68 J 0.150 m 0.180 m ⎠ The maximum total kinetic energy will be 1.68 J. The kinetic energy increases as the electrical potential energy decreases, as the charges move apart. The maximum kinetic energy is achieved only after a long time.

*18.9. Set Up: Call the three charges 1, 2, and 3. U = U12 + U13 + U 23. U12 = U 23 = U13 because the charges are equal and each pair of charges has the same separation, 0.500 m. Solve: U =

3kq 2 3k (1.2 × 10−6 C) 2 = = 0.078 J. 0.500 m 0.500 m

Reflect: When the three charges are brought in from infinity to the corners of the triangle, the repulsive electrical forces between each pair of charges do negative work and electrical potential energy is stored. qq′ 18.10. Set Up: U = k . U 2 = 3U1 . r1 = R r ⎛U ⎞ ⎛U ⎞ Solve: Ur = kqq′, which is constant. So, U1r1 = U 2 r2 . r2 = ⎜ 1 ⎟ r1 = ⎜ 1 ⎟ R = R / 3 ⎝ U2 ⎠ ⎝ 3U1 ⎠

*18.11. Set Up: For a pair of oppositely charged parallel metal plates, Vab = Ed . F = 0 q 0 E. Solve: (a) E =

Vab 360 V = = 8000 V/m = 8000 N/C d 45.0 × 10−3 m

(b) F = 0 q 0 E = (2.40 × 10−9 C)(8000 N/C) = 1.92 × 10−5 N


18-4

Chapter 18

18.12. Set Up: For a pair of oppositely charged metal parallel metal plates, Vab = Ed . Solve: d =

Vab 4.75 × 103 V = = 1.58 × 10−3 m = 1.58 mm E 3.00 × 106 V/m

*18.13. Set Up: For two oppositely charged parallel plates, Vab = Ed , where Vab is the potential difference between the two plates, E is the uniform electric field between the plates, and d is the separation of the plates. An G G G electric field E exerts a force F = 0 q 0 E on a charge placed in the field. a = F/m. An electron has charge −e and mass 9.11 × 10−31 kg. Solve: (a) E =

Vab 25 V = = 1.25 × 104 V/m d 2.0 × 1023 m

(b) F = 0 q 0 E = eE . a =

F eE (1.60 × 10−19 C)(1.25 × 104 V/m) = = = 2.20 × 1015 m/s 2 −31 m m 9.11 × 10 kg

Reflect: The electric field is the same at all points between the plates (away from the edges) so the acceleration would be the same at all points between the plates as it is for a point midway between the plates.

18.14. Set Up: For two oppositely charged sheets of charge, Vab = Ed . The positively charged sheet is the one at higher potential.

Vab 70 × 10−3 V = = 9.3 × 106 V/m. The electric field is directed inward, toward the interior of the d 7.5 × 10−9 m G axon, since the outer surface of the membrane has positive charge and E points away from positive charge and toward negative charge. Section 18.9 explores the effects of a material other than air between the plates. (b) The outer surface has positive charge so it is at higher potential than the inner surface. Solve: (a) E =

18.15. Set Up: Vab = Ed for parallel plates. Solve: d =

Vab 1.5 V = = 1.5 × 106 m = 1.5 × 103 km E 1.0 × 10−6 V/m

18.16. Set Up: The work-energy theorem says Wa →b = Kb − K a .

Wa→b = Va − Vb . Point a is the starting and point q

b is the ending point. Since the field is uniform, Wa →b = Fs cosφ = E 0 q 0 s cosφ . The field is to the left so the force on the positive charge is to the left. The particle moves to the left so φ = 0° and the work Wa →b is positive. Solve: (a) Wa →b = Kb − K a = 1.50 × 10−6 J − 0 = 1.50 × 10−6 J (b) Va − Vb =

Wa →b 1.50 × 10−6 J = = 357 V. Point a is at higher potential than point b. q 4.20 × 10−9 C

(c) E 0 q 0 s = Wa →b , so E =

Wa →b Va − Vb 357 V = = = 5.95 × 103 V/m. 0q0 s s 6.00 × 1022 m

*18.17. Set Up: From Example 18.4, with Vb = 0 and x rather than y as the distance from the negative plate, Vx = Ex. Solve: E =

Vx 5.0 V = = 25 V/m x 0.200 m

Reflect: Vx = Ex says the potential increases linearly with x and the graph of Vx versus x should be a straight line, in agreement with the figure in the problem.



18-5

G

18.18. Set Up: (a) The direction of E is always from high potential to low potential so point b is at higher potential. (b) Apply Example 18.4 to relate Vb − Va to E. Solve: According to Example 18.4 we can calculate the magnitude of the electric field, E, to be V − Va +240 V E= b = = 800 V/m xb − xa 0.90 m − 0.60 m (c) Wb→ a = q(Vb − Va ) = ( −0.200 × 10−6 C)(+240 V) = −4.80 × 10−5 J. Reflect: The electric force does negative work on a negative charge when the negative charge moves from high potential (point b) to low potential (point a).

18.19. Set Up: For a point charge V = Solve: (a) r =

kq . r

kq (8.99 × 109 N ⋅ m 2 /C2 )(2.50 × 10−11 C) = = 2.50 × 10−3 m = 2.50 mm V 90.0 V

⎛V ⎞ ⎛ 90.0 V ⎞ (b) Vr = kq = constant so V1 r1 = V2 r2 . r2 = r1 ⎜ 1 ⎟ = (2.50 mm) ⎜ ⎟ = 7.50 mm. ⎝ 30.0 V ⎠ ⎝ V2 ⎠

18.20. Set Up: Apply conservation of energy, Eq. (18.3). Use Eq. (18.10) to express U in of V. (a) Solve: K1 + qV1 = K 2 + qV2

q (V1 − V2 ) = K 2 − K1;

q = −1.602 × 10−19 C

K1 = 12 meυ 1 2 4.099 × 10−18 J;

K 2 = 12 meυ2 2 = 2.915 × 10217 J

K 2 − K1 = −156 V q Reflect: The electron gains kinetic energy when it moves to higher potential. (b) Solve: Now K1 = 2.915 × 10−17 J, K 2 = 0 V1 − V2 =

K 2 − K1 = +182 V q Reflect: The electron loses kinetic energy when it moves to lower potential. V1 − V2 =

*18.21. Set Up: Apply conservation of energy to points A and B. Solve: We know that K A + U A = K B + U B and U = qV , so K A + qVA = K B + qVB . Thus, K B = K A + q (VA − VB ) = 0.00250 J + ( −5.00 × 10−6 C)(200 V − 800 V) = 0.00550 J.

Which gives

υ B = 2 K B /m = 7.42 m/s. Reflect: It is faster at B; a negative charge gains speed when it moves to higher potential.

18.22. Set Up: For a single point charge V =

kq . The total potential is the sum of the potentials due to the two r

point charges. WB → A = q (VB − VA ). Solve: (a) VA = (b) VB =

kq1 kq2 8.99 × 109 N ⋅ m 2 /C2 + = (2.40 × 10−9 C + [ −6.50 × 10−9 C]) = −737 V r1 A r2 A 0.050 m

⎛ 2.40 × 10−9 C −6.50 × 10−9 C ⎞ kq1 kq2 + = (8.99 × 109 N ⋅ m 2 /C2 ) ⎜ + ⎟⎟ = −704 V ⎜ 0.080 m r1B r2 B 0.060 m ⎝ ⎠

(c) WB → A = q (VB − VA ) = (2.50 × 10−9 C)( −704 V − (−737 V )) = 8.2 × 10−8 J © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

18-6

Chapter 18

*18.23. Set Up: U = k

qQ . Conservation of energy says U a + K a = U b + Kb . K = 12 mυ 2 . r

Solve: (a) U = (8.99 × 109 N ⋅ m 2 /C2 )

(1.20 × 10−6 C)(4.60 × 10−6 C) = 0.198 J 0.250 m

(b) (i) U a = +0.198 J, K a = 0. rb = 0.500 m so U b = 12 U a = 0.099 J.

K b = U a + K a − U b = 0.198 J − 0.099 J = 0.099 J. υb =

2 Kb 2(0.099 J) = = 26.6 m/s. m 2.80 × 10−4 kg

(ii) rb = 5.00 m so U b = 0.0099 J. Kb = 0.198 J − 0.0099 J = 0.188 J. υb = 36.7 m/s. (iii) rb = 50.0 m so U b = 0.00099 J. Kb = 0.198 J − 0.00099 J = 0.197 J. υb = 37.5 m/s. Reflect: As the charge q moves away from Q the repulsive force does positive work on q and its kinetic energy increases.

18.24. Set Up: Let a be when they are 0.750 nm apart and b when they are very far apart. A proton has charge + e and mass 1.67 × 10−27 kg. As they move apart the protons have equal kinetic energies and speeds. Solve: (a) They have maximum speed when they are far apart and all their initial electrical potential energy has been converted to kinetic energy. K a + U a = Kb + U b . K a = 0 and U b = 0, so

Kb = U a = k

e2 (1.60 × 10−19 C) 2 = (8.99 × 109 N ⋅ m 2 /C2 ) = 3.07 × 10−19 J. ra 0.750 × 10−9 m

Kb = 12 mυb 2 + 12 mυb 2 , so Kb = mυb 2 and υb =

Kb 3.07 × 10−19 J = = 1.36 × 104 m/s m 1.67 × 10−27 kg

(b) Their acceleration is largest when the force between them is largest and this occurs at r = 0.750 nm, when they are closest.

F =k

2

⎛ 1.60 × 10−19 C ⎞ = (8.99 × 109 N ⋅ m 2 /C2 ) ⎜ = 4.09 × 10−10 N 2 ⎜ 0.750 × 10−9 m ⎟⎟ r ⎝ ⎠

e2

a=

F 4.09 × 10−10 N = = 2.45 × 1017 m/s 2 m 1.67 × 10−27 kg

18.25. Set Up: Apply K a + U a = Kb + U b , with K a = 0 and υb = υ . Solve: (a) Kb = U a − U b = q (Va − Vb ) = −e(Va − Vb ). An electron gains energy when it moves to higher potential, so

Vb − Va = V and Kb = eV . Kb = 12 mυ 2 so

υ= (b) υ =

2(1.60 × 10−19 C)(95 V) 9.11 × 10−31 kg

2 Kb 2eV = . m m

= 5.8 × 106 m/s

Reflect: A positive charge gains kinetic energy when it moves from high potential to low potential but a negative charge gains kinetic energy when it moves from low potential to high potential.

18.26. Set Up: From Problem 18.25, υ = 2eV/m for a particle with charge of magnitude e. The speed of light is c = 3.00 × 108 m/s. An electron has mass 9.11 × 10−31 kg and a proton has mass 1.67 × 10−27 kg.

Solve: (a) υ = (0.010)c = 3.00 × 106 m/s. V =

mv 2 (9.11 × 10−31 kg)(3.00 × 106 m/s) 2 = = 26 V 2e 2(1.60 × 10−19 C)



18-7

(b) From Problem 18.25, the kinetic energy the particle gains is K = eV so V = 26 V gives electrons and protons the same kinetic energy. But the protons must be accelerated by a potential decrease whereas the electrons are accelerated by a potential increase. (c) υ =

2eV 2(1.60 × 10−19 C)(26 V) = = 7.0 × 104 m/s = (0.024%)c m 1.67 × 10−27 kg q r

*18.27. Set Up: Treat the gold nucleus as a point charge so that V = k . According to conservation of energy we have K a + U a = Kb + U b , where U = qV . Solve: Assume that the alpha particle is at rest before it is accelerated and that it momentarily stops when it arrives at its closest approach to the surface of the gold nucleus. Thus we have K a = Kb = 0, which implies that U a = U b . Since U = qV we conclude that the accelerating voltage must be equal to the voltage at its point of closest approach to the

surface of the gold nucleus: q 79(1.60 × 10−19 C) = (8.99 × 109 N ⋅ m 2 /C2 ) = 4.2 × 106 V. r (7.3 × 10−15 m + 2.0 × 10−14 m) Reflect: Although the alpha particle has kinetic energy as it approaches the gold nucleus this is irrelevant to our solution since energy is conserved for the whole process. Va = Vb = k

18.28. Set Up: Example 18.4 shows that V = Ey, where y is the distance from the negative plate. V = 0 at the negative plate. For y = 6.0 cm, V = 12.0 V. The potential gradient is

ΔV . Δy

12.0 V V 6.0 V = 2.0 V/cm. y = = = 3.0 cm; V = 6.0 V when y = 3.0 cm. The +6.0 V equipo6.0 cm E 2.0 V/cm tential surface is a flat sheet parallel to the plates and midway between them. V 2.0 V (b) y = = = 1.0 cm. The +2.0 V equipotential surface is a flat sheet parallel to the plates and 1.0 cm E 2.0 V/cm from the negative plate and 5.0 cm from the positive plate. ΔV 12.0 V (c) = = 200 V/m Δy 6.0 × 10−2 m

Solve: (a) E =

*18.29. Set Up: Example 18.4 shows that V = Ey, where y is the distance from the negative plate. V = 0 at the negative plate. For y = 25 cm, V = 50.0 V. 50.0 V = 2.0 V/cm. So V = (2.0 V/cm) y. 25 cm V = +10.0 V for y = 5 cm, V = +20.0 V for y = 10 cm, V = +30.0 V for y = 15 cm, V = +40.0 V for y = 20 cm and V = +50.0 V for y = 25 cm. The equipotential surfaces are drawn in the figure below. Solve: (a) E =


18-8

Chapter 18

(b) Yes, they are separated by 5 cm. (c) The equipotential surfaces are flat sheets parallel to the plates. Reflect: The electric field lines are straight lines perpendicular to the plates, so are perpendicular to the equipotential surfaces, as they must be. The electric field is uniform so the equipotential lines of constant potential difference are equally spaced.

18.30. Set Up: For a point charge, V =

kq . r

kq (8.99 × 109 N ⋅ m 2 / C2 )(5.00 × 10−12 C) 4.50 V ⋅ cm = = V = 1.00 V at r = 4.50 cm; V = 2.00 V V V V at r = 2.25 cm; V = 3.00 V at r = 1.50 cm; V = 4.00 V at r = 1.12 cm and V = 5.00 V at r = 0.90 cm. The equipotential surfaces are sketched in the figure below. Solve: (a) r =

(b) These equipotential surfaces are not equally spaced in distance. The V = 1.00 V and V = 2.00 V surfaces are farther apart than the V = 4.00 V and V = 5.00 V surfaces. The equipotential surfaces are closer together where E is larger. (c) The equipotential surfaces are concentric spheres with the point charge at their common center.

q r

*18.31. Set Up: For a uniformly charged sphere V = k . Solve: (a) At the surface of the sphere we have q (8.99 × 109 N ⋅ m 2 / C2 )(1.50 × 10−6 C) V =k = = 26.970 kV = 27.0 kV. 0.500 m r q kq (b) Solving V = k for r we obtain r = . This allows us to find the distance between any two equipotential r V kq kq kq(V1 − V2 ) − = . surfaces V1 and V2 : r2 − r1 = V2 V1 V1V2



18-9

We desire to space the equipotentials in 500 V increments outside the sphere. Thus, the first equipotential is at 26.970 kV − 500 V = 26.470 kV and the second equipotential is at 26.970 kV − 2(500 V) = 25.970 kV. The required spacing of the first two equipotentials is kq(V1 − V2 ) (8.99 × 109 N ⋅ m 2 /C2 )(1.50 × 10−6 C)(500 V) = 0.0098 m = 0.98 cm. r2 − r1 = = V1V2 (26.470 × 103 V)(25.970 × 103 V) The 20th equipotential is at 26.970 kV − 20(500 V) = 16.970 kV and the 21st equipotential is at 26.970 kV − 21(500 V) = 16.470 kV. Thus, the required spacing is r2 − r1 =

kq (V1 − V2 ) (8.99 × 109 N ⋅ m 2 /C2 )(1.50 × 10−6 C)(500 V) = = 0.024 m = 2.4 cm. V1V2 (16.970 × 103 V)(16.470 × 103 V)

(c) The increased spacing of the equipotentials is due to the fact that the electric field is weaker at greater distances from the charged sphere.

18.32. Set Up: The electric field lines are perpendicular to the equipotential surfaces. The electric field lines are closer together where E is larger, and where E is larger the equipotential surfaces for equal potential differences are closer together. The projection of the equipotential surfaces into the plane of the figure are lines that satisfy these two requirements. Solve: For the electric field given in the problem it is not possible to draw equipotential lines that satisfy both requirements. The electric field shown in the problem cannot be produced by an electrostatic distribution of charges.

18.33. Set Up and Solve: The electric field lines and intersections of the equipotential surfaces with the plane of the drawing are shown in Figure 18.11b in the textbook. Reflect: The figure shows that the electric field lines are always perpendicular to the equipotential surfaces. Also, the electric field lines are closer together where E is larger, and where E is larger the equipotential surfaces for equal potential differences are closer together. 18.34. Set Up: The mass of a drop is m = ρV , where V = 43 π r 3 (the volume of a sphere). ρ = 0.820 g/cm3 = 820 kg/m3 . The drops have r = 5.00 × 10−7 m. The electric field and electric force magnitudes are related by F = 0 q 0 E , where q is the net charge of the drop. Solve: (a) F = w, so 0 q 0 E = mg and E =

mg . 0 q 0 = 5e = 8.00 × 10−19 C. 0q0

m = ρV = ρ 43 π r 3 = (820 kg/m3 ) E=

( 43 )π (5.00 × 10−7 m)3 = 4.29 × 10−16 kg

(4.29 × 10−16 kg)(9.80 m/s 2 ) 8.00 × 10−19 C

= 5.26 × 103 V/m

(b) V = Ed = (5.26 × 103 V/m)(2.25 × 10−2 m) = 118 V (c) E =

0q0 =

V 73.8 V = = 3.28 × 103 V/m. d 2.25 × 1022 m

mg (4.29 × 10−16 kg)(9.80 m/s 2 ) = = 1.28 × 10−18 C. 3 E 3.28 × 10 V/m

The number of electrons is N =

0 q 0 1.28 × 10−18 C = = 8. e 1.60 × 10−19 C

*18.35. Set Up: 1 eV = 1.60 × 10219 J. K = 12 mυ 2 . c = 3.00 × 108 m/s. An electron has mass 9.11 × 10−31 kg and a proton has mass 1.67 × 10−27 kg. Solve: (a) K = 1.00 eV = 1.60 × 10−19 J. υ =

2K m


18-10

Chapter 18

2(1.60 × 10−19 J)

electron: υe = proton: υ p =

9.11 × 10−31 kg 2(1.60 × 10−19 J) 1.67 × 10

−27

(b) K = 1.00 keV = 1.60 × 10

kg −16

= 5.93 × 105 m/s

= 1.38 × 104 m/s.

mp υe = = 1836 υp me

J

7

electron: υe = 1.87 × 10 m/s; proton: υp = 4.38 × 105 m/s (c) υe = υ p = (0.0100)c = 3.00 × 106 m/s

K e = 12 meυe 2 = 12 (9.11 × 10−31 kg)(3.00 × 106 m/s)2 = 4.10 × 10−18 J = 0.0256 keV

K p = 12 mp υp 2 = 12 (1.67 × 10−27 kg)(3.00 × 106 m/s)2 = 7.52 × 10−15 J = 47.0 keV The proton energy is larger by a factor of mp /me . Reflect: When we use K = 12 mυ 2 we should express all quantities in SI units.

18.36. Set Up: C = Solve: (a) C =

Q Vab

1.25 × 10−6 C = 1.11 × 10−7 F = 0.111 μ F 11.3 V

(b) Q = CVab = (7.28 × 10−6 F)(25.0 V) = 1.82 × 10−4 C = 182 μ C

18.37. Set Up: C =

⑀0 A d

, C=

Q and V = Ed . V

A 0.00122 m 2 Solve: (a) C = ⑀0 = ⑀0 = 3.29 pF 0.00328 m d (b) V =

Q 4.35 × 10−8 C = = 13.2 kV C 3.29 × 10−12 F

V 13.2 × 103 V = = 4.02 × 106 V/m 0.00328 m d Reflect: The electric field is uniform between the plates, at points that aren’t close to the edges. (c) E =

18.38. Set Up: For a parallel-plate capacitor, Vab = Ed , E =

Q

⑀0 A

and C =

Q . Vab

Solve: (a) Vab = Ed = (4.00 × 106 V/m)(2.50 × 10−3 m) = 1.00 × 104 V (b) A =

Q 80.0 × 10−9 C = = 2.26 × 10−3 m 2 = 22.6 cm 2 E⑀0 ( 4.00 × 106 V/m )(8.854 × 10−12 C2 /(N ⋅ m 2 ) )

(c) C =

Q 80.0 × 10−9 C = = 8.00 × 10−12 F = 8.00 pF Vab 1.00 × 104 V

*18.39. Set Up: For a parallel-plate capacitor C = Solve: (a) Vab = (b) A =

Cd

⑀0

=

Q Q ⑀ A . C = 0 . Vab = Ed . The surface charge density is σ = . A Vab d

Q 0.200 × 10−6 C = = 400 V C 500.0 × 10−12 F

(500.0 × 10−12 F)(0.600 × 10−3 m) 8.854 × 10−12 C2 /(N ⋅ m 2 )

= 0.0339 m 2



(c) E =

18-11

Vab 400 V = = 6.67 × 105 V/m d 0.600 × 1023 m

Q 0.200 × 10−6 C = = 5.90 × 10−6 C/m 2 A 0.0339 m 2 Reflect: If the plates are square, each side is about 18 cm in length. We could also calculate σ from E =σ /⑀0 . (d) σ =

18.40. Set Up: C = ⑀0 Solve: C =

A when there is air between the plates. d

(8.854 × 10−12 F/m)(3.0 × 10−2 m) 2 5.0 × 10−3 m

*18.41. Set Up: C =

= 1.59 × 10−12 F = 1.59 pF

Q ⑀ A . C= 0 . d Vab

Solve: (a) Q = CVab = (10.0 × 10−6 F)(12.0 V) = 1.20 × 10−4 C = 120 μ C (b) When d is doubled C is halved, so Q is halved. Q = 60 μ C. (c) If r is doubled, A increases by a factor of 4. C increases by a factor of 4 and Q increases by a factor of 4. Q = 480 μ C. Reflect: When the plates are moved apart, less charge on the plates is required to produce the same potential difference. With the separation of the plates constant, the electric field must remain constant to produce the same potential difference. The electric field depends on the surface charge density, σ . To produce the same σ , more charge is required when the area increases.

18.42. Set Up: C =

Q ⑀ A . C= 0 . Vab d

Solve: (a) 12.0 V

Q and Q is constant, so V doubles. V = 24.0 V. C (ii) When r is doubled, A increases by a factor of 4. V decreases by a factor of 4 and V = 3.0 V. (b) (i) When d is doubled, C is halved. Vab =

A d

*18.43. Set Up: For parallel disks we have C = ⑀0 , where A = π r 2 . The voltage between the charged plates is given by V =

Q Qd = . C ⑀0 A

Solve: (a) C = (8.854 × 10−12 F/m)

π (0.0750 m) 2 (0.00350 m)

= 44.7 pF

(b) Once the battery is disconnected, the charge remains fixed. Use the equation V =

voltage for the two different plate separations:

Qd and look at the ratio of the ⑀0 A

V 3.50 cm = = 10.0. Thus the voltage is increased by a 6.00 volts 3.50 mm

factor of 10 to a final value of 600 volts. Reflect: The work you do in pulling the oppositely charged plates apart provides the energy needed to increase the voltage between the plates.

18.44. Set Up: C = Solve: (a) d =

Q ⑀ A . Vab = Ed . C = 0 . Vab d

Vab 1.00 × 102 V = = 1.00 × 10−2 m = 1.00 cm. E 1.00 × 104 N/C


18-12

Chapter 18

A= A

A = π r 2 so r =

π

Cd

⑀0

=

(5.00 × 10−12 F)(1.00 × 10−2 m)

= 5.65 × 10−3 m 2 .

8.854 × 10−12 C2 /(N ⋅ m 2 )

= 4.24 × 10−2 m = 4.24 cm.

(b) Q = CVab = (5.00 × 10−12 F)(1.00 × 102 V) = 5.00 × 10−10 C = 500 pC

18.45. Set Up: Use C = ⑀0 A

d

⑀0π r 2

and solve for d. Estimate r = 1.0 cm. A = π r 2 .

=

⑀0π (0.010 m) 2

= 2.8 mm. C d 1.00 × 10−12 F Reflect: The separation between the pennies is nearly a factor of 10 smaller than the diameter of a penny, so it is a reasonable approximation to treat them as infinite sheets. Solve: C =

so d =

⑀0 A

A d

18.46. Set Up: Since the capacitor is disconnected from the battery, Q does not change. C = ⑀0 . C = Solve: (a) C = ⑀0

Q . V

A A A C . Cnew = ⑀0 = ⑀0 = . 2d 2 d d new

(b) Q = Q0 (c) V0 =

Q0 Q Q Q . Vnew = 0 = 0 = 2 0 = 2V0 C Cnew C /2 C

*18.47. Set Up: The capacitors between b and c are in parallel. This combination is in series with the 15 pF capacitor. C1 = 15 pF, C2 = 9.0 pF and C3 = 11 pF. Solve: (a) For capacitors in parallel, Ceq = C1 + C2 + g so C23 = C2 + C3 = 20 pF (b) C1 = 15 pF is in series with C23 = 20 pF. For capacitors in series,

1 1 1 = + + g so Ceq C1 C2

1 1 1 CC (15 pF)(20 pF) = + and C123 = 1 23 = = 8.6 pF. C123 C1 C23 C1 + C23 15 pF + 20 pF Reflect: For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors. For capacitors in series the equivalent capacitance is smaller than any of the individual capacitors.

18.48. Set Up: For capacitors in series the voltage across the combination equals the sum of the voltages in the individual capacitors. For capacitors in parallel the voltage across the combination is the same as the voltage across each individual capacitor. Solve: (a) Connect the capacitors in series so their voltages will add. (b) V = V1 + V2 + V3 + g= NV1, where N is the number of capacitors in the series combination, since the capacitors are identical.

N=

V 500 V = = 5000. V1 0.10 V

*18.49. Set Up: For capacitors in parallel the voltages are the same and the charges add. For capacitors in series, the charges are the same and the voltages add. C = Q / V . C1 and C2 are in parallel and C3 is in series with the parallel combination of C1 and C2 . Solve: (a) C1 and C2 are in parallel and so have the same potential across them: V1 = V2 =

Q2 40.0 × 10−6 C = = C2 3.00 × 10−6 F

13.33 V. Therefore we have Q1 = V1C1 = (13.33 V)(6.00 × 10−6 F) = 80.0 × 10−6 C. Since C3 is in series with the



18-13

parallel combination of C1 and C2 , its charge must be equal to their combined charge:

Q3 = 40.0 × 10−6 C + 80.0 × 10−6 C = 120.0 × 10−6 C. (b) The total capacitance is found from

1 1 1 1 1 and Ctot = 3.21 μF. = + = + Ctot C12 C3 9.00 × 10−6 F 5.00 × 10−6 F

Vab = Reflect: V3 =

Qtot 120.0 × 10−6 C = = 37.4 V. Ctot 3.21 × 10−6 F

Q3 120.0 × 10−6 C = = 24.0 V. Vab = V1 + V3. C3 5.00 × 10−6 F

18.50. Set Up: For capacitors in series,

1 1 1 = + + g For N equivalent capacitors in series, Ceq = C /N . Ceq C1 C2

For capacitors in parallel, Ceq = C1 + C2 + g For N equivalent capacitors in parallel, Ceq = NC . Solve: There are many ways to achieve the required equivalent capacitance. In each case one simple solution is shown in the figure below.

18.51. Set Up: Use the rules for V for capacitors in series and parallel: for capacitors in parallel the voltages are the same and for capacitors in series the voltages add. Solve: V1 = Q1 /C1 = (150 μC)/(3.00 μF) = 50 V. Since the capacitors C1 and C2 are in parallel we know that V2 = V1 = 50 V. Thus we have V3 = 120 V − V1 = 70 V. Reflect: Now that we know the voltages, we could also calculate Q for the other two capacitors.

18.52. Set Up: In series, the charges on the capacitors are the same and the sum of the potential differences across the capacitors is the applied potential difference. C = Solve: (a) V1 + V2 = 48.0 V.

Q . Let C1 = 4.00 μ F and C2 = 6.00 μ F. Q1 = Q2 = Q. Vab

Q Q + = 48.0 V. C1 C2

⎛ [ 4.00 × 10−6 μ F][6.00 × 10−6 μ F] ⎞ ⎛ CC ⎞ Q = ⎜ 1 2 ⎟ (48.0 V) = ⎜ (48.0 V) = 1.15 × 10−4 C. ⎜ 4.00 × 10−6 μ F + 6.00 × 10−6 μ F ⎟⎟ ⎝ C1 + C2 ⎠ ⎝ ⎠ The charge on each capacitor is 115 μ C.


18-14

Chapter 18

Q 1.15 × 10−4 C Q 1.15 × 10−4 C = = . . V = = = 19.2 V. 28 8 V 2 C1 4.00 × 10−6 F C2 6.00 × 10−6 F The potential difference across the 4.00 μ F capacitor is 28.8 V and the potential difference across the 6.00 μ F capacitor is 19.2 V. (b) V1 =

*18.53. Set Up: C =

Q . For two capacitors in parallel, Ceq = C1 + C2 . For two capacitors in series, V 1 1 1 CC = + and Ceq = 1 2 . Ceq C1 C2 C1 + C2

For capacitors in parallel, the voltages are the same and the charges add. For capacitors in series, the charges are the same and the voltages add. Let C1 = 3.00 μ F, C2 = 5.00 μ F and C3 = 6.00 μ F.

Solve: (a) The equivalent capacitance for C1 and C2 in parallel is C12 = C1 + C2 = 8.00 μ F. This gives the circuit

shown in Figure (a) above. In that circuit the equivalent capacitance is C=

C12C3 (8.00 μ F)(6.00 μ F) = = 3.43 μ F. C12 + C3 8.00 μ F + 6.00 μ F

This gives the circuit shown in Figure (b) above. In Figure (b), Q = CV = (3.43 × 10−6 F)(24.0 V) = 8.23 × 10−5 C. In Figure (a) each capacitor therefore has charge 8.23 × 10−5 C. The potential differences are

V3 =

Q3 8.23 × 10−5 C Q12 8.23 × 10−5 C and = = . V = = = 10.3 V. 13 7 V 12 C3 6.00 × 10−6 F C12 8.00 × 10−6 F

Note that V3 + V12 = 24.0 V. Then in the original circuit, V1 = V2 = V12 = 10.3 V.

Q1 = V1C1 = (10.3 V)(3.00 × 10−6 F) = 3.09 × 10−5 C. Q2 = V2C2 = (10.3 V)(5.00 × 10−6 F) = 5.15 × 10−5 C.

Q1 = 30.9 μ C, Q2 = 51.5 μ C and Q3 = 82.3 μ C. Note that Q1 + Q2 = Q3 . (b) V1 = 10.3 V, V2 = 10.3 V and V3 = 13.7 V Reflect: Note that Q1 + Q2 = Q3 , V1 = V2 and V1 + V3 = 24.0 V

18.54. Set Up: C =

Q . For two capacitors in parallel, Ceq = C1 + C2 . For two capacitors in series, V 1 1 1 CC = + and Ceq = 1 2 . Ceq C1 C2 C1 + C2

For capacitors in parallel, the voltages are the same and the charges add. For capacitors in series, the charges are the same and the voltages add. Label the capacitors as shown in Figure (a) below.



18-15

Solve: The equivalent capacitance for C1 and C2 in series is

C12 =

C1C2 C = . C1 + C2 2

This gives the circuit shown in Figure (b) above. The equivalent capacitance for C12 and C3 in parallel is

C123 = C12 + C3 =

3C . 2

This gives the circuit shown in Figure (c) above. The equivalent capacitance for C123 and C4 in series is

C123C4 (3C /2)(C ) = = 0.600C = 2.40 × 10−6 F. C123 + C4 3C /2 + C This gives the circuit shown in Figure (d) above. In Figure (d) above, Q = CeqVab = (2.40 × 10−6 F)(28.0 V) = 6.72 × 10−5 C. This is the charge on each capacitor in Ceq =

Figure (c) above. Q4 = 67.2 μ C and

V4 =

67.2 μC Q4 67.2 μ C = 11.2 V. = = 16.8 V. V123 = (3/2)(4.00 μF) C4 4.00 μ F

Note that V4 + V123 = Vab . Then in Figure (b) above, V3 = V12 = 11.2 V. Q3 = C3V3 = (4.00 μ F)(11.2 V) = 44.8 μ C.

Q12 = C12V12 = 12 (4.00 μ F)(11.2 V) = 22.4 μ C. Q1 = Q2 = 22.4 μ C. V1 =

Q1 22.4 μ C Q 22.4 μ C = = 5.6 V. V2 = 2 = = 5.6 V. C1 4.00 μ F C2 4.00 μ F

In summary, Q1 = 22.4 μ C, Q2 = 22.4 μC, Q3 = 44.8 μ C, Q4 = 67.2 μ C, V1 = 5.6 V, V2 = 5.6 V, V3 = 11.2 V, and

V4 = 16.8 V.

18.55. Set Up: For capacitors in series the voltages add and the charges are the same;

1 1 1 = + + g For Ceq C1 C2

Q . V Solve: (a) The equivalent capacitance of the 5.0 μF and 8.0 μ F capacitors in parallel is 13.0 μ F. When these two

capacitors in parallel the voltages are the same and the charges add; Ceq = C1 + C2 + g C =

capacitors are replaced by their equivalent we get the network sketched in the figure below. The equivalent capacitance of these three capacitors in series is 3.47 μ F. (b) Q tot = CtotV = (3.47 μ F)(50.0 V) = 174 μ C © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

18-16

Chapter 18

(c) Qtot is the same as Q for each of the capacitors in the series combination shown in the figure below, so Q for each of the capacitors is 174 μ C.

Reflect: The voltages across each capacitor in the figure above are V10 =

Q tot Q = 17.4 V, V13 = tot = 13.4 V and C10 C13

Qtot = 19.3 V. V10 + V13 + V9 = 17.4 V + 13.4 V + 19.3 V = 50.1 V. The sum of the voltages equals the applied C9 voltage, apart from a small difference due to rounding. V9 =


1 1 1 = + + g For Ceq C1 C2

Q . V Solve: (a) The equivalent capacitance of the 18.0 nF, 30.0 nF, and 10.0 nF capacitors in series is 5.29 nF. When these capacitors are replaced by their equivalent we get the network sketched in the figure below. The equivalent capacitance of these three capacitors in parallel is 19.3 nF, and this is the equivalent capacitance of the original network.

capacitors in parallel the voltages are the same and the charges add; Ceq = C1 + C2 + g C =

(b) Qtot = CeqV = (19.3 nF)(25 V) = 482 nC (c) The potential across each capacitor in parallel network of the figure above is 25 V.

Q6.5 = C6.5V6.5 = (6.5 nF)(25 V) = 162 nC. (d) 25 V

*18.57. Set Up: The charge across a capacitor charged to a voltage V is Q = CV . The energy stored in a charged 1 1 1 Q2 . For capacitors in parallel we have Ceq = C1 + C2 and for capacitors in series capacitor is U = CV 2 = QV = 2 2 2 C CC we have Ceq = 1 2 . C1 + C2 Solve: (a) For the capacitors in parallel we have Ceq = C1 + C2 = 2.5 μF + 5.0 μF = 7.5 μF. The charge provided by

the battery is Q = CV = (7.5 μ F)(12 V) = 90 μC.



(b) For the capacitors in series we have Ceq =

18-17

C1C2 (2.5 μF)(5.0 μ F) = = 1.67 μF. The charge provided by the C1 + C2 2.5 μ F + 5.0 μF

battery is Q = CV = (1.67 μ F)(12 V) = 20 μC. 1 1 (c) For the capacitors in parallel we have U = QV = (90 μC)(12 V) = 5.4 × 10−4 J. For the capacitors in series we 2 2 1 1 −4 have U = QV = (20 μC)(12 V) = 1.2 × 10 J. 2 2

Reflect: Since we also know the charge on each equivalent capacitor we could also use either U =

1 Q2 or 2 Ceq

1 U = CeqV 2 to calculate the energy supplied by the battery. 2

18.58. Set Up: The energy density is u = 12 ⑀0 E 2 . Vab = Ed . E =

Vab 400 V = = 8.00 × 104 V/m. d 5.00 × 10−3 m

u = 12 ⑀0 E 2 = 12 (8.854 × 10−12 C 2 /(N ⋅ m 2 ))(8.00 × 104 V/m) 2 = 0.0283 J/m3 .

18.59. Set Up: C =

Q . U = 12 CV 2 . V

Solve: (a) Q = CV = (5.0 μF)(1.5 V) = 7.5 μC. U = 12 CV 2 = 12 (5.0 μF)(1.5 V) 2 = 5.62 μ J (b) U = 12 CV 2 = 12 C (Q /C )2 = Q 2 /2C . Q = 2CU = 2(5.0 × 10−6 F)(1.0 J) = 3.2 × 10−3 C.

V=

18.60. Set Up: C = Solve: (a) V =

Q 3.2 × 10−3 C = = 640 V. C 5.0 × 10−6 F

Q V

2 Q ⎛ Q⎞ Q so U = 12 QV = 12 Q ⎜ ⎟ = ⎝ C ⎠ 2C C

(b) Q = CV so U = 12 (CV )V = 12 CV 2

*18.61. Set Up: For a parallel plate capacitor we have C =

⑀0 A d

. The stored energy can be expressed either as

Q2 2C

2

CV , whichever is more convenient for the calculation. Since d is halved, C doubles. 2 Solve: (a) If the separation distance is halved while the charge is kept fixed, then the capacitance increases and the stored energy, which was 8.38 J, decreases since U = Q 2 /2C . Therefore the new energy is 4.19 J. or as

(b) If the voltage is kept fixed while the separation is decreased by one half, then the doubling of the capacitance leads to a doubling of the stored energy to 16.8 J, using U = CV 2 /2, when V is held constant throughout. Reflect: When the capacitor is disconnected, the stored energy decreases because of the positive work done by the attractive force between the plates. When the capacitor remains connected to the battery, Q = CV tells us that the charge on the plates increases. The increased stored energy comes from the battery when it puts more charge onto the plates.

18.62. Set Up: The energy stored in a capacitor is

Q2 ⑀ A . The charge of an electron is 1.60 × 10−19 C. C = 0 . 2C d


18-18

Chapter 18

Solve: (a) Q = 2C (25.0 J) = 2(5.00 × 10−9 F)(25.0 J) = 5.00 × 10−4 C. The number of electrons with this

5.00 × 10−4 C

= 3.12 × 1015 electrons. 1.60 × 10−19 C/electron (b) To double the stored energy, halve the capacitance. To do this, either double the plate separation or halve the plate area. magnitude of charge is

*18.63. Set Up: The two capacitors are in series. For capacitors in series the voltages add and the charges are the same.

1 1 1 Q = + + gC = . U = 12 CV 2 . Ceq C1 C2 V

Solve: (a)

CC 1 1 1 (150 nF)(120 nF) = 66.7 nF. so Ceq = 1 2 = = + C1 + C2 150 nF + 120 nF Ceq C1 C2

Q = CV = (66.7 nF)(36 V) = 2.4 × 10−6 C = 2.4 μC (b) Q = 2.4 μC for each capacitor. (c) U = 12 CeqV 2 = 12 (66.7 × 10−9 F)(36 V)2 = 43.2 μJ (d) We know C and Q for each capacitor so rewrite U in of these quantities. U = 12 CV 2 = 12 C (Q /C ) 2 = Q 2 /2C

150 nF: U =

(2.4 × 10−6 C) 2

= 19.2 μJ; 120 nF: U =

(2.4 × 10−6 C) 2

2(150 × 10−9 F) 2(120 × 10−9 F) Note that 19.2 μ J + 24.0 μJ = 43.2 μJ, the total stored energy calculated in part (c).

= 24.0 μJ

Q 2.4 × 10−6 C Q 2.4 × 10−6 C = = 16 V; 120 nF: V = = = 20 V 9 − C 150 × 10 F C 120 × 10−9 F Note that these two voltages sum to 36 V, the voltage applied across the network. Reflect: Since Q is the same the capacitor with smaller C stores more energy (U = Q 2 /2C ) and has a larger voltage (V = Q /C ). 150 nF: V =

18.64. Set Up: For capacitors in parallel, the voltages are the same and the charges add. Ceq = C1 + C2 . C =

Q . V

U = 12 CV 2 . Solve: (a) Ceq = C1 + C2 = 35 nF + 75 nF = 110 nF. Qtot = CeqV = (110 × 10−9 F)(220 V) = 24.2 μC (b) V = 220 V for each capacitor.

35 nF: Q35 = C35V = (35 × 10−9 F)(220 V) = 7.7 μC; 75 nF: Q75 = C75V = (75 × 10−9 F)(220 V) = 16.5 μC.

Note that Q35 + Q75 = Qtot . (c) U tot = 12 CeqV 2 = 12 (110 × 10−9 F)(220 V) 2 = 2.66 mJ (d) 35 nF: U 35 = 12 C35V 2 = 12 (35 × 10−9 F)(220 V)2 = 0.85 mJ;

75 nF: U 75 = 12 C75V 2 = 12 (75 × 10−9 F)(220 V) 2 = 1.81 mJ. Since V is the same the capacitor with larger C stores more energy. (e) 220 V for each capacitor.

*18.65. Set Up: C =

Q . Let C1 = 20.0 μF and C2 = 10.0 μF. The energy stored in a capacitor is Vab 1 QVab 2

= 12 CVab 2 =

2 1 Q . 2V ab



18-19

Solve: (a) The initial charge on the 20.0 μF capacitor is

Q = C1 (800 V) = (20.0 × 10−6 F)(800 V) = 0.0160 C. (b) In the final circuit, charge Q is distributed between the two capacitors and Q1 + Q2 = Q. The final circuit contains

only the two capacitors, so the voltage across each is the same, V1 = V2 . V=

Q Q Q C so V1 = V2 gives 1 = 2 . Q1 = 1 Q2 = 2Q2 . C C2 C1 C2

Using this in Q1 + Q2 = 0.0160 C gives 3Q2 = 0.0160 C and Q2 = 5.33 × 10−3 C. Q = 2Q2 = 1.066 × 10−2 C. V1 =

Q1 1.066 × 10−2 C Q2 5.33 × 10−3 C V = = 533 V . = = = 533 V. 2 C1 20.0 × 10−6 F C2 10.0 × 10−6 F

The potential differences across the capacitors are the same, as they should be. (c) Energy = 12 C1V 2 + 12 C2V 2 = 12 (C1 + C2 )V 2 = 12 (20.0 × 10−6 F + 10.0 × 10−6 F)(533 V) 2 = 4.26 J. (d) The 20.0 μF capacitor initially has energy = 12 C1V 2 = 12 (20.0 × 10−6 F)(800 V) 2 = 6.40 J. The decrease in stored

energy that occurs when the capacitors are connected is 6.40 J − 4.26 J = 2.14 J. Reflect: The decrease in stored energy is because of conversion of electrical energy to other forms during the motion of the charge when it becomes distributed between the two capacitors. Thermal energy is generated by the current in the wires and energy is emitted in electromagnetic waves.


1 1 1 = + + ". For Ceq C1 C2

Q . U = 12 CV 2 . V for the network by replacing each series or parallel combination by its equivalent. The

capacitors in parallel the voltages are the same and the charges add; Ceq = C1 + C2 + " C = Solve: (a) Find Ceq

successive simplified circuits are shown in Figure (a)-(c) below. U tot = 12 CeqV 2 = 12 (2.19 × 10−6 F)(12.0 V) 2 = 1.58 × 10−4 J = 158 μJ

(b) From Figure (c) above, Qtot = CeqV = (2.19 × 10−6 F)(12.0 V) = 2.63 × 10−5 C. From Figure (b) above, Qtot = 2.63 × 10−5 C. V4.8 =

Q4.8 2.63 × 10−5 C = = 5.48 V. C4.8 4.80 × 10−6 F

U 4.8 = 12 CV 2 = 12 (4.80 × 10−6 F)(5.48 V) 2 = 7.21 × 10−5 J = 72.1 μJ This one capacitor stores nearly half the total stored energy. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

18-20

Chapter 18

18.67. Set Up: Apply C = Q /V . C =

⑀0 A d

. The work done to double the separation equals the change in the stored

Q2 1 . energy. U = CV 2 = 2 2C Solve: (a) V = Q /C = (2.55 μC)/(920 × 10−12 F) = 2770 V (b) The equation C =

⑀ 0A

says that the capacitance is halved when the separation is doubled. Since voltage is inversely d proportional to capacitance when Q is fixed (V = Q/C), the voltage doubles to 5540 V when the separation is halved. Q 2 (2.55 × 10−6 C) 2 = = 3.53 × 10−3 J. When the separation is doubled while Q stays the same, the (c) U = 2C 2(920 × 10212 F) capacitance halves, and the energy stored doubles. So the amount of work done to move the plates equals the difference in energy stored in the capacitor, which is 3.53 × 10−3 J. Reflect: The oppositely charged plates attract each other—this means that positive work must be done by an external force to pull the plates apart.

18.68. Set Up: With the dielectric present we have C = KC0 . We know that V = Ed and C = Q /V hold with or without the presence of the dielectric. Solve: (a) V = Ed = (3.00 × 104 V/m)(1.50 × 10−3 m) = 45.0 V. Q = C0V = (5.00 × 10−12 F)(45.0 V) = 2.25 × 10−10 C.

(b) With the dielectric, C = KC0 = (2.70)(5.00 pF) = 13.5 pF. V is still 45.0 V, so Q = CV = (13.5 × 10−12 F)(45.0 V) =

6.08 × 10−10 C. Reflect: The presence of the dielectric increases the amount of charge that can be stored for a given potential difference and electric field between the plates. Q increases by a factor of K. A d equation applies whether or not a dielectric is present. (8.85 × 10−12 F/m)(1.0 × 10−4 m 2 ) Solve: (a) C = (10) = 1.18 μF per cm 2 . 7.5 × 10−9 m

*18.69. Set Up: C = KC0 = K ⑀0 . A = 1.0 cm 2 = 1.0 × 10−4 m 2 . V = Ed for a parallel plate capacitor; this

V 85 mV = = 1.13 × 106 V/m. Kd (10)(7.5 × 10−9 m) Reflect: The dielectric material increases the capacitance and decreases the electric field that corresponds to a given potential difference.

(b) E =

18.70. Set Up: The capacitance with the dielectric present is C = KC0 =

K ⑀0 A . Vab = Ed . The dielectric strength d

is the maximum allowed field between the plates. V 4.00 × 103 V = 2.00 × 10−4 m. Solve: d = ab = 6 E 20.0 × 10 V/m A=

Cd (1.50 × 10−9 F)(2.00 × 10−4 m) = = 0.0106 m 2 . K ⑀0 (3.20)(8.854 × 10−12 C2 /(N ⋅ m 2 ))

18.71. Set Up: C = KC0 . U = 12 CV 2 . Solve: (a) With the dielectric, C = (3.75)(12.5 μF) = 46.9 μF.

before: U = 12 C0V 2 = 12 (12.5 × 10−6 F)(24.0 V) 2 = 3.60 mJ



18-21

after: U = 12 CV 2 = 12 (46.9 × 10−6 F)(24.0 V) 2 = 13.5 mJ (b) ΔU = 13.5 mJ − 3.6 mJ = 9.9 mJ. The energy increased. Reflect: The power supply must put additional charge on the plates to maintain the same potential difference when the dielectric is inserted. U = 12 QV , so the stored energy increases.

18.72. Set Up: The capacitance with the dielectric present is C = KC0 =

K ⑀0 A . Vab = Ed . The dielectric strength d

is the maximum allowed field between the plates. Solve: (a) A =

Cd (0.200 × 10−6 F)(0.0800 × 10−3 m) = = 0.723 m 2 K ⑀0 (2.50)(8.854 × 10−12 C2 /(N ⋅ m 2 ))

(b) Vab = 12 (50.0 × 106 V/m)(0.0800 × 10−3 m) = 2.00 × 103 V

*18.73. Set Up: We know that C = Q /V , C = KC0 , and V = Ed . Table 18.1 gives the dielectric constant K = 3.1 for Mylar. Solve: (a) ΔQ = Q − Q0 = ( K − 1)Q0 = ( K − 1)C0V0 = (2.1)(2.5 × 10−7 F)(12 V) = 6.3 × 10−6 C. (b) σ i = σ (1 − 1/K ) so Qi = Q(1 − 1/K ) = (9.3 × 10−6 C)(1 − 1/3.1) = 6.3 × 10−6 C. (c) The addition of the Mylar doesn’t affect the electric field since the induced charge cancels the additional charge drawn to the plates. Reflect: E = V /d and V is constant so E doesn’t change when the dielectric is inserted.

18.74. Set Up: The potential at the surface of a uniformly charged sphere is V =

kQ . For a sphere we have R

4 V = π R3. When the raindrops merge, the total charge and volume is conserved. 3 Solve: (a) V =

kQ k (−1.20 × 10−12 C) = = −16.6 V. R 6.50 × 10−4 m

(b) The volume doubles, so the radius increases by the cube root of two: Rnew = 3 2 R = 8.19 × 10−4 m and the new

kQnew k (−2.40 × 10−12 C) = = −26.4 V. Rnew 8.19 × 10−4 m Reflect: The charge doubles but the radius also increases and the potential at the surface increases by only a factor of 2 = 22/3. 1/3 2 charge is Qnew = 2Q = − 2.40 × 10−12 C. The new potential is Vnew =

*18.75. Set Up: For a point charge, E =

k 0q0 r2

and V =

kq . The electric field is directed toward a negative charge r

and away from a positive charge. Solve: (a) V > 0 so q > 0.

2 4.98 V V kq/r ⎛ kq ⎞ ⎛ r ⎞ = 0.415 m. = = ⎜ ⎟⎟ = r. r = ⎜ ⎟ 2 ⎜ 12.0 V/m E k 0 q 0 /r ⎝ r ⎠ ⎝ kq ⎠

rV (0.415 m)(4.98 V) = = 2.30 × 10−10 C k 8.99 × 109 N ⋅ m 2 /C2 (c) q > 0, so the electric field is directed away from the charge. (b) q =

Reflect: The ratio of V to E due to a point charge increases as the distance r from the charge increases, because E falls off as 1/r 2 and V falls off as 1/r.


18-22

Chapter 18

18.76. Set Up: At each point (a and b), the potential is the sum of the potentials due to both spheres. The voltmeter reads the difference between these two potentials. The spheres behave as point-charges since the meter is connected to the surface of each one. Solve: (a) Call a the point on the surface of one sphere and b the point on the surface of the other sphere, call r the radius of each sphere, and call d the center-to-center distance between the spheres. The potential difference Vab between points a and b is then Vb − Va = Vab =

1 ⎡ −q q −q ⎞ ⎤ 2q ⎛ 1 1⎞ ⎛q + −⎜ + − ⎟. ⎟ = ⎜ 4π ⑀0 ⎢⎣ r d − r ⎝ r d − r ⎠ ⎥⎦ 4π ⑀0 ⎝ d − r r ⎠

Substituting the numbers gives 1 1 ⎞ ⎛ Vb − Va = 2(175 µC)(9.00 × 109 N ⋅ m 2 /C2 ) ⎜ − = −8.40 × 106 V ⎝ 0.750 m 0.250 m ⎟⎠

The meter reads 8.40 MV. (b) Since Vb − Va is negative, Va > Vb , so point a is at the higher potential. Reflect: An easy way to see that the potential at a is higher than the potential at b is that it would take positive work to move a positive test charge from b to a since this charge would be attracted by the negative sphere and repelled by the positive sphere.

*18.77. Set Up: With air between the layers, E0 =

Q

⑀0 A

=

σ and V0 = E0 d . The energy density in the electric ⑀0

field is u = 12 ⑀0 E 2 . The volume of a shell of thickness t and average radius R is 4π R 2t . The volume of a solid sphere of radius R is

4 π R3. 3

Solve: (a) E0 =

With the dielectric present, E =

E0 V and V = 0 . K K

σ 0.50 × 10−3 C/m 2 = = 5.6 × 107 V/m ⑀0 8.854 × 10−12 C2 /(N ⋅ m 2 )

(b) V0 = E0d = (5.6 × 107 V/m)(5.0 × 10−9 m) = 0.28 V. The outer wall of the cell is at higher potential, since it has

positive charge. (c) For the cell, Vcell = 43 π R3 and 1/3

⎛ 3V ⎞ R = ⎜ cell ⎟ ⎝ 4π ⎠

1/3

⎛ 3(10−16 m3 ) ⎞ =⎜ ⎟⎟ ⎜ 4π ⎝ ⎠

= 2.9 × 10−6 m.

The volume of the cell wall is Vwall = 4π R 2t = 4π (2.9 × 10−6 m) 2 (5.0 × 10−9 m) = 5.3 × 10−19 m3 .

u0 = 12 ⑀0 E0 2 =

1 (8.854 × 10−12 2

C2 /(N ⋅ m 2 ))(5.6 × 107 V/m)2 = 1.39 × 104 V/m3 .

The total electric field in the cell wall is (1.39 × 104 V/m3 )(5.3 × 10−19 m3 ) = 7 × 10−15 V. (d) E =

E0 5.6 × 107 V/m V 0.28 V = 0.052 V. = = 1.0 × 107 V/m and V = 0 = K 5.4 K 5.4

18.78. Set Up: Let a be the initial situation, where the alpha particle is very far from the gold nucleus and has kinetic energy K = 10.0 MeV = 1.60 × 10−12 J. At a the gold nucleus has zero kinetic energy. Let b be at the distance of closest approach, when the distance between the two particles is rb . Conservation of energy says K a + U a = Kb + U b . U = k

q1q2 . The alpha particle has charge q1 = +2e and the gold r

nucleus has charge q2 = +79e. Solve: K a = 1.60 × 10−12 J and Kb = 0, since at the distance of closest approach the alpha particle has momentarily

come to rest. U a = 0, since ra is very large. Ub = k

(+2e)( +79e) . rb



18-23

Conservation of energy gives K a = U b . rb =

(2)(79)e2k (2)(79)(1.60 × 10−19 C)2 (8.99 × 109 N ⋅ m 2 /C2 ) = = 2.27 × 10−14 m Ka 1.60 × 10−12 J

18.79. Set Up: The electric force between the electron and proton is attractive and has magnitude F = circular motion the acceleration is arad = υ 2 /r. U = − k Solve: (a) (b) K =

mυ 2 ke 2 = 2 and υ = r r

ke2 r2

. For

e2 . We know that e = 1.60 × 10−19 C and 1 eV = 1.60 × 10−19 J. r

ke2 . mr

1 2 1 ke2 1 mυ = =− U 2 2 r 2

1 1 ke 2 1 k (1.60 × 10−19 C) 2 (c) E = K + U = U = − =− = −2.17 × 10−18 J = −13.6 eV. 2 2 r 2 5.29 × 10−11 m Reflect: The total energy is negative, so the electron is bound to the proton. Work must be done on the electron to take it far from the proton.

18.80. Set Up: The proton has charge qp = + e and mass mp = 1.67 × 10−27 kg. The alpha particle has charge qα = +4e and mass mα = 4mp = 6.68 × 10−27 kg. We can apply both conservation of energy and conservation of F 0q q 0 , where F = k 1 22 . m r Solve: acceleration: The maximum force and hence the maximum acceleration occurs just after they are released, when r = 0.225 nm.

linear momentum to the system. a =

F = (8.99 × 109 N ⋅ m 2 /C2 ) ap =

(2)(1.60 × 10−19 C)2 (0.225 × 10−9 m) 2

= 9.09 × 10−9 N

F 9.09 × 10−9 N F 9.09 × 10−9 N 18 2 = = 5 . 44 × 10 m/s ; a = = = 1.36 × 1018 m/s 2 α mp 1.67 × 10−27 kg mα 6.68 × 10−27 kg

The acceleration of the proton is larger by a factor of mα /mp . speed: Conservation of energy says U i + Ki = U f + K f . Ki = 0 and U f = 0, so K f = U i . Ui = k

qq ′ (2)(1.60 × 10−19 C)2 = (8.99 × 109 N ⋅ m 2 /C2 ) = 2.05 × 10−18 J, r 0.225 × 10−9 m

so the total kinetic energy of the two particles when they are far apart is K f = 2.05 × 10−18 J. Conservation of linear momentum says how this energy is divided between the proton and alpha particle. pi = pf . 0 = mpυp − mαυα and

⎛ mp ⎞ ⎟υp . ⎝ mα ⎠

υα = ⎜

2

K f = 12 mpυp 2 +

υp =

1 m υ 2= 1 m υ 2 2 α α 2 p p

+

1m 2 α

⎛ mp ⎞ 2 ⎜ ⎟ υp = ⎝ mα ⎠

1 m υ 2 ⎛1 + 2 p p ⎜

⎝

mp ⎞ ⎟ mα ⎠

−18

2Kf 2(2.05 × 10 J) = = 4.43 × 104 m/s mp (1 + (mp /mα )) (1.67 × 10−27 kg) 1 + 14

(

)

⎛ mp ⎞ 4 4 1 ⎟υp = 4 (4.43 × 10 m/s) = 1.11 × 10 m/s ⎝ mα ⎠ The maximum acceleration occurs just after they are released. The maximum speed occurs after a long time.

υα = ⎜


18-24

Chapter 18

*18.81. Set Up: C = Solve: (a) C =

⑀0 A d

=

⑀0 A d

Q Q . Vab = Ed and E = . The stored energy is Vab ⑀0 A

(8.854 × 10−12 C2 /(N ⋅ m 2 ))(0.200 m)2 0.800 × 10

(b) Q = CVab = (4.43 × 10 (c) E =

. C=

−11

−2

m

1 QVab 2

= 12 CVab 2 =

1 2

Q2 . C

= 4.43 × 10−11 F

F)(120 V) = 5.32 × 10−9 C

Vab 120 V = = 1.50 × 104 V/m d 0.800 × 10−2 m

(d) Energy = 12 QV = 12 (5.32 × 10−9 C)(120 V) = 3.19 × 10−7 J (e) Since the battery is disconnected, the charge Q on the capacitor stays constant. ⑀ A (a) C = 0 so C = 12 (4.43 × 10−11 F) = 2.22 × 10−11 F. d

(b) The charge can’t change, so Q = 5.32 × 10−9 C. (c) E =

Q

⑀0 A

. Since Q doesn’t change, E doesn’t change and E = 1.50 × 104 V/m.

(d) Energy =

Q2 . 2C

Q doesn’t change and C changes by a factor of

1 , 2

so the stored energy doubles and becomes 6.38 × 10−7 J.

Reflect: Since the stored energy increases, work must be done by the force that pulls the plates apart.

18.82. Set Up: Since the battery remains connected, V remains constant and Q changes. ⑀0 A (8.854 × 10−12 C2 /(N ⋅ m 2 ))(0.200 m) 2 −11 Solve: (a) C =

d

=

1.60 × 10−2 m

= 2.21 × 10

F

(b) Q = CVab = (2.21 × 10−11 F)(120 V) = 2.65 × 10−9 C (c) E =

Vab 120 V = = 7.50 × 103 V/m d 1.60 × 10−2 m

(d) Energy = 12 CV 2 = 12 (2.21 × 10−11 F)(120 V) 2 = 1.59 × 10−7 J

*18.83. Set Up: C = Solve: (a) C = (5.00)

Q K ⑀0 A . Energy = 12 QVab = 12 CVab 2 = . C= Vab d

(8.854 × 10−12 C2 /(N ⋅ m 2 ))(16.0 × 10−4 m 2 ) 0.200 × 10−2 m

1 2

Q2 . C

= 3.54 × 10−11 F

(b) Q = CVab = (3.54 × 10−11 F)(300 V) = 1.06 × 10−8 C (c) Energy = 12 CV 2 = 12 (3.54 × 10−11 F)(300 V) 2 = 1.59 × 10−6 J

18.84. Set Up: P = E /t , where E is the total light energy output. The energy stored in the capacitor is U = 12 CV 2 . Only 95% of the stored energy is available: E = 0.95U Solve: (a) The power output is 600 W, and 95% of the original energy is converted, so E = Pt = (2.70 × 105 W)(1.48 × 10−3 s) = 400 J. U = 400 J = 421 J. 0.95 2 U 2(421 J) (b) U = 12 CV 2 so C = 2 = = 0.054 F. V (125 V) 2 Reflect: For a given V, the stored energy increases linearly with C.



*18.85. Set Up: For capacitors in series, the equivalent resistance Ceq is given by

18-25

1 1 1 = + + g. For Ceq C1 C2

capacitors in parallel, the equivalent capacitance Ceq is given by Ceq = C1 + C2 + g

Solve: (a) Using the rules for combining capacitors in series and in parallel gives the sequence of equivalent networks shown in the figure above. The equivalent capacitance of the network is 2.3 μ F. (b) In Figure (d) above the three capacitors in series have the same capacitance, so the voltage across each is (420 V)/3 = 140 V. Q1 = C1V1 = (6.9 μF)(140 V) = 966 μC. The voltage across C2 in Figure (c) above is 140 V, so

Q2 = C2V2 = (4.6 μF)(140 V) = 644 μC. Reflect: We could continue to analyze the networks in the figure above and find V and Q for each capacitor in the network.

18.86. Set Up: The capacitor is equivalent to two capacitors in parallel, as shown in the figure below. Each of these two capacitors have plates that are 12.0 cm by 6.0 cm. For a parallel-plate capacitor with dielectric filling the volume A between the plates, C = K ⑀0 . d


18-26

Chapter 18

Solve: (a) C = C1 + C2 . C2 = ⑀0

A (8.854 × 10−12 F/m)(0.120 m)(0.060 m) = = 1.42 × 10−11 F d 4.50 × 10−3 m

C1 = KC2 = (3.40)(1.42 × 10−11 F) = 4.83 × 10−11 F. C = C1 + C2 = 6.25 × 10−11 F = 62.5 pF.

(b) U = 12 CV 2 = 12 (6.25 × 10−11 F)(18.0 V) 2 = 1.01 × 1 0−8 J (c) Now C1 = C2 and C = 2(1.42 × 10−11 F) = 2.84 × 10−11 F.

U = 12 CV 2 = 12 (2.84 × 10−11 F)(18.0 V)2 = 4.60 × 10−9 J. The plexiglass increases the capacitance and that increases the energy stored for the same voltage across the capacitor.

18.87. Set Up: For a single capacitor with dielectric of dielectric constant K completely filling the volume between K ⑀0 A its plates, C =

. The capacitor is sketched in Figure (a) below. It is equivalent to two capacitors in series as d shown in Figure (b) below. For two capacitors C1 and C2 in series the equivalent capacitance C is given by 1 1 1 = + . C C1 C2

Solve: (a)

1 1 1 C1 + C2 K ⑀ A 2K ⑀ A K ⑀ A 2 K 2⑀ 0 A = + = . C1 = 1 0 = 1 0 and C2 = 2 0 = . C C1 C2 C1C2 d /2 d d /2 d

⎛ 2 K1⑀0 A ⎞⎛ 2 K 2⑀0 A ⎞ ⎜ ⎟⎜ ⎟ 2⑀ A ⎛ K K ⎞ d ⎠⎝ d ⎠= 0 1 2 C=⎝ ⎜ ⎟ 2 K1⑀0 A 2 K 2⑀0 A d ⎝ K1 + K 2 ⎠ + d d 2⑀0 A K 2 K ⑀0 A = , which is the correct expression. d 2K d 2⑀ A ⎛ 1 ⎞ ⑀ A (c) For K1 = K 2 = 1.0, C → 0 ⎜ ⎟ = 0 , which is the correct expression. d ⎝2⎠ d Reflect: Our result applies only when each dielectric fills half the space between the plates. When K1 ≠ K 2 the (b) For K1 = K 2 , C →

electric field has a different value in each dielectric.



18-27

Solutions to age Problems 18.88. Set Up: We are told that the specific capacitance of the egg membrane is roughly 1 μF/cm 2 . Although it is not defined in the text we may reasonably guess that specific capacitance is equal to C /A, where A is the surface area Q C of the membrane. In this context, the equation Q = CV may be written as = V. Also, we are told that one mole A A of Na + or K + ions have a net charge of 105 C. Solve: The voltage difference between the inside and the outside of the egg swings from −70 mV to +30 mV. Thus, the change in voltage difference is ΔV = +30 mV − ( −70 mV) = 100 mV. The amount of charge moved per ΔQ C = ΔV = (10−6 F/cm 2 )(0.100 V) = 10−7 C/cm 2 . Using the conversion factor A A ⎛ 10−7 C ⎞ ⎛ 1 mole ⎞ = 10−12 mole/cm 2 of Na + given for the Faraday number, this amount of charge corresponds to ⎜ ⎜ cm 2 ⎟⎟ ⎜⎝ 105 C ⎟⎠ ⎝ ⎠ ions. The correct answer is C.

unit area of membrane must be

*18.89. Set Up: Assume that the egg cell is spherical with a radius of 100 μm = 10−2 cm. The volume of the cell 4 is then V = π r 3 and its surface area is A = 4π r 2 . We are told that the initial concentration of Na + ions is 3 30 mmoles/liter within the cell. Solve: From the previous problem we know that 10−12 mole/cm 2 of Na + ions move into the egg during fertilization. Thus, the change in concentration is 4π r 2 (10−12 mole/cm 2 ) 3(10−12 mole/cm 2 ) Δc = = = 3 × 10−10 mole/cm3. −2 4 3 10 cm πr 3

The fractional change in concentration is

Δc (3 × 10−10 mole/cm3 )(103 mmoles/mole)(103 cm3 /liter) = = 10−5. c 30 mmoles/liter

The correct answer is B.

18.90. Set Up: A single Ca +2 ion carries a charge of +2e, which is twice as great as that of a single Na + ion. Solve: The amount of charge needed to produce the given change in Vm is fixed. Thus, we only need half as

many calcium ions as sodium ions since each calcium ion carries twice as much charge as a sodium ion. The correct answer is A. 1 2

*18.91. Set Up: The change in electrical energy stored in the cell can be calculated from U = CV 2 . The specific capacitance of the cell is 1 μF/cm 2 . Assume that the egg cell is spherical with a diameter of 100 μm = 10−2 cm as before. Solve: The capacitance of the cell is 4π r 2 (1 μ F/cm 2 ) = 4π (10−2 cm) 2 (1 μ F/cm 2 ) = 0.00126 μ F = 1.26 nF. The change in electrical energy needed to return the cell from +30 mV back to −70 mV is 1 1 ΔU = C (Vf2 − Vi2 ) = (1.26 × 10−9 F)[(−70 × 10−3 V) 2 − (+30 × 10−3 V) 2 ] = 2.5 × 10−12 J ≈ 3 pJ. 2 2 The correct answer is D.


CURRENT, RESISTANCE , AND DIRECT-CURRENT CIRCUITS

19

Answers to Multiple-Choice Problems 1. C 2. B 3. D 14. A 15. E

4. B, D

5. A

6. A

7. C

8. A

9. C

10. C

11. A

12. C

13. C

Solutions to Problems 19.1. Set Up: 1A = 1 C/s. An electron has charge of magnitude e = 1.60 × 10−19 C. The current tells us how much charge flows in a given time. We can find the number of electrons that correspond to a certain amount of charge. 1 ⎛ ⎞ 18 Solve: (a) 1.50 A = (1.50 C/s) ⎜ ⎟ = 9.38 × 10 electrons/s −19 C/electron ⎠ ⎝ 1.60 × 10 (b) ΔQ = I Δt = (1.50 A)(300 s) = 450 C ΔQ 7.50 C = = 5.00 s I 1.50 A Reflect: The amount of charge associated with typical currents is quite large. In Chapter 17 we found that the force between objects that have net charge 1.0 C is immense. The same amount of charge enters and leaves any part of the circuit each second and each circuit element remains neutral. (c) Δt =

19.2. Set Up: I =

ΔQ Δt

Solve: ΔQ = I Δt = (25,000 A)(40 × 10−6 s) = 1.0 C

*19.3. Set Up: The number of ions that enter gives the charge that enters the axon in the specified time. I =

ΔQ . Δt

Solve: ΔQ = (5.6 × 1011 ions)(1.60 × 10−19 C/ion) = 9.0 × 10−8 C

I=

ΔQ 9.0 × 10−8 C = = 9.0 μA Δt 10 × 10−3 s

19.4. Set Up: Negative charge moving from A to B is equivalent to an equal magnitude of positive charge going from B to A. I =

ΔQ . The current direction is the direction of flow of positive charge. Δt


19-1

19-2

Chapter 19

Solve: The total positive charge moving from B to A is

ΔQ = (5.11 × 1018 + 2[3.24 × 1018 ])(1.60 × 10−19 C) = 1.85 C. I =

ΔQ 1.85 C = = 62 mA. Positive charge flows from Δt 30 s

B to A so the current is in this direction.

*19.5. Set Up: The volume of the cylindrical wire is given by V = π r 2 L, where r = (2.05 mm)/2 = 1.025 mm and L = 25.0 cm. The current ing a given point in the wire is I =

ΔQ . We assume that the density of free electrons in Δt

copper is 8.5 × 1028 electrons/m3. We assume that all of the electrons travel with the drift velocity υd (see section 19.1). Solve: (a) The number of free electrons in the wire is π (1.025 × 10−3 m)2 (0.250 m)(8.5 × 1028 electrons/m3 ) = 7.0 ×1022. (b) Imagine that you are standing near the end of the wire where the current exits. In a time equal to t =

L

υd

=

0.25 m

υd

all of the free electrons in this section of wire will exit past you. Thus, ΔQ = (7.0 × 1022 electrons)(1.60 × 10−19 C/electron) = 1.1× 104 C in a time t. The resulting current is I = 1.55 A =

ΔQ 1.1 × 104 C = . Solving for the drift velocity we obtain Δt (0.25 m)/υd

υd = (1.55 A)(0.25 m)/(1.1 × 104 C) = 3.5 × 10−5 m/s = 0.035 mm/s. Reflect: The charges inside of a conductor that can move are known as free charges. The current in a conductor is carried by its free charges.

19.6. Set Up: V = IR. For copper, Table 19.1 gives that ρ = 1.72 × 10−8 Ω ⋅ m and for silver, ρ = 1.47 × 10−8 Ω ⋅ m. Solve: (a) R =

ρL A

(1.72 × 10−8 Ω ⋅ m)(2.00 m)

=

π (0.814 × 10−3 m)2

= 1.65 × 10−2 Ω.

V = (12.5 × 10−3 A)(1.65 × 10−2 Ω) = 2.06 × 10−4 V. (b) V =

I ρ L V IL V V . = = constant, so s = c . A ρ A ρs ρc ⎛ 1.47 × 10−8 Ω ⋅ m ⎞ ⎛ρ ⎞ Vs = Vc ⎜ s ⎟ = (2.06 × 10−4 ) ⎜ = 1.76 × 10−4 V. ⎜ 1.72 × 10−8 Ω ⋅ m ⎟⎟ ρ ⎝ c⎠ ⎝ ⎠

19.7. Set Up: R = Solve: L =

ρL A

=

ρL π (d /2) 2

. From Table 19.1, copper has ρ = 1.72 × 10−8 Ω ⋅ m.

π ( d /2) 2 R π (1.628 × 10−3 m) 2 (1.00 Ω) = = 121 m ρ 4(1.72 × 10−8 Ω ⋅ m)

19.8. Set Up: The resistivity of the wire can be used to identify what material it is made of. R = ρL/A and the radius of the wire is half its diameter. Solve: Solve for ρ and substitute the numerical values. π ([0.00205 m]/2) 2 (0.0290 Ω) ρ = AR/L = π ( D/2)2 R/L = = 1.47 × 10−8 Ω ⋅ m 6.50 m Reflect: This result is the same as the resistivity of silver, which suggests that the material is silver.

*19.9. Set Up: R =

ρL A

. The length of the wire in the spring is the circumference π d of each coil times the

number of coils. Solve: L = (75)π d = (75)π (3.50 × 10−2 m) = 8.25 m. A = π r 2 = π d 2 /4 = π (3.25 × 10−3 m)2 /4 = 8.30 × 10−6 m 2 . © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Current, Resistance, and Direct-Current Circuits

19-3

RA (1.74 Ω)(8.30 × 10−6 m 2 ) = = 1.75 × 10−6 Ω ⋅ m. L 8.25 m Reflect: The value of ρ we calculated is about a factor of 100 times larger than ρ for copper. The metal of the

ρ=

spring is not a very good conductor.

19.10. Set Up: R = Solve:

ρ d

2

=

ρL A

=

ρL π d 2 /4

. For aluminum, ρal = 2.63 × 10−8 Ω ⋅ m. For copper, ρc = 1.72 × 10−8 Ω ⋅ m.

Rπ ρ ρ = constant, so al2 = c2 . 4L d al dc d c = d al

19.11. Set Up: R = Solve: (a) R = (b) R =

ρal L A

ρL A

ρc 1.72 × 10−8 Ω ⋅ m = (3.26 mm) = 2.64 mm. ρal 2.63 × 10−8 Ω ⋅ m

. For aluminum, ρal = 2.63 × 10−8 Ω ⋅ m. For copper, ρc = 1.72 × 10−8 Ω ⋅ m.

(2.63 × 10−8 Ω ⋅ m)(3.80 m)

=

= 2.00 × 10−4 Ω

(1.00 × 10−2 m)(5.00 × 10−2 m)

ρc L Rπ r 2 (2.00 × 10−4 Ω)π (0.750 × 10−3 m) 2 L = = = 0.0205 m = 2.05 cm . ρc πr2 1.72 × 10−8 Ω

19.12. Set Up: R = Solve: Rnew

ρL

ρL

=

. Lnew = 3L; d new = 2d

π ( d /2) 2 ρ Lnew ρ3 L 3 ρL = = = = 3R 2 2 4 π (d /2)2 4 π ( d new /2) π (2d /2) A

*19.13. Set Up: R =

ρL A

. Lnew = 3L. The volume of the wire remains the same when it is stretched.

Solve: Volume = LA so LA = Lnew Anew . Anew =

Rnew =

L Lnew

ρ Lnew

=

A=

A . 3

ρ(3L)

=9

ρL

= 9 R. Anew A/3 A Reflect: When the length increases the resistance increases, and when the area decreases the resistance increases.

19.14. Set Up: RT = R0[1 + α (T − T0 )]. T = 0.0°C. T = 11.5°C and R0 = 100.00 Ω. Solve: RT = (100.00 Ω)[1 + (0.00040 (C°) −1)(0.0°C − 11.5°C)] = 99.54 Ω

*19.15. Set Up: First use Ohm’s law to find the resistance at 20.0°C; then calculate the resistivity from the resistance. Finally use the dependence of resistance on temperature to calculate the temperature coefficient of resistance. Ohm’s law is R = V/I, R = ρL/A, R = R0[1 + α (T – T0)], and the radius is one-half the diameter. Solve: (a) At 20.0°C, R = V/I = (15.0 V)/(18.5 A) = 0.811 Ω. Using R = ρL/A and solving for ρ gives ρ = RA/L = Rπ(D/2)2/L = (0.811 Ω)π[(0.00500 m)/2]2/(1.50 m) = 1.06 × 10−5 Ω ⋅ m. (b) At 92.0°C, R = V/I = (15.0 V)/(17.2 A) = 0.872 Ω. Using R = R0[1 + α (T – T0)] with T0 taken as 20.0°C, we have 0.872 Ω = (0.811 Ω)[1 + α (92.0°C – 20.0°C)]. This gives α = 0.00105 (°C) −1 Reflect: The results are typical of ordinary metals.

19.16. Set Up: RT = R0[1 + α (T − T0 )] Solve: T − T0 =

( RT /R0 ) − 1

α

=

(215.8 Ω /217.3 Ω) − 1 −0.00050 (°C) −1

= 13.8 C°. T = 13.8 C° + 4.0°C = 17.8°C.


19-4

Chapter 19

19.17. Set Up: Ohm’s law says V = IR, where R is a constant. Solve: (a) Ohm’s law says a graph of V versus I is a straight line. That is the case here, so this object obeys Ohm’s law. (b) V = IR says R is the slope of a graph of V versus I. (c) R =

0.60 V − 0.30 V 4.0 × 10−3 A − 2.0 × 10−3 A

= 150 Ω

Reflect: We can always define R = V /I , but only when R is independent of I does the object obey Ohm’s law.

19.18. Set Up: Ohm’s law says R =

Vab is a constant. I

Solve: (a) The graph is given in the figure below

(b) The graph of Vab versus I is a straight line so NichromeTM obeys Ohm’s law. (c) R is the slope of the graph in part (a). R =

15.52 V − 1.94 V = 3.88 Ω. 4.00 A − 0.50 A

*19.19. Set Up: Apply RT = R0 (1 + α (T − T0 )). Since V = IR and V is the same,

RT I = 20 . For tungsten, R20 IT

α = 4.5 × 10−3 (C°) −1. Solve: The ratio of the current at 20°C to that at the higher temperature is (0.860 A)/(0.220 A) = 3.909. RT = 1 + α (T − T0 ) = 3.909, where T0 = 20°C. R20 R /R − 1 3.909 − 1 T = T0 + T 20 = 20°C + = 666°C. α 4.5 × 10−3 (C°) −1 Reflect: As the temperature increases, the resistance increases and for constant applied voltage the current decreases. The resistance increases by nearly a factor of four.

19.20. Set Up: Since the battery has negligible internal resistance, E = IR. E 1.50 V = = 83.3 Ω I 18.0 × 10−3 A E 12.0 V (b) I = = = 0.144 A R 83.3 Ω (c) E = IR = (0.453 A)(83.3 Ω) = 37.7 V Solve: (a) R =



19-5

*19.21. Set Up: V = IR, where I is the current through the battery and V is the potential difference applied to the person. Solve: V = IR = (5.0 × 10−3 A)(1000 Ω) = 5.0 V. This is well within the range of common household voltages.

19.22. Set Up: Current is measured in A, potential difference in V. Problem 19.21 says the resistance of a wet human is 1 k Ω. V = IR. Solve: (a) Volts are not current; he probably meant a potential difference of 12 V. V 6V (b) I = = = 16 mA. Problem 19.21 says currents above 10 mA are dangerous, so this current is dangerous. R 1 kΩ

19.23. Set Up: Use R =

ρL A

, where A = π r 2 and r = 0.050 m, to calculate R and then apply V = IR. For copper

we have ρ = 1.72 × 10−8 Ω ⋅ m. Solve: R =

ρL A

=

(1.72 × 10−8 Ω ⋅ m)(100 × 103 m)

π (0.050 m)2

= 0.219 Ω. V = IR = (125 A)(0.219 Ω) = 27.4 V.

Reflect: Even though the resistance of the cable is relatively low, there is a significant voltage drop due to the high current ing through the cable.

ρL

19.24. Set Up: R = Solve: (a) R =

4ρ L

πd 2

A

=

ρL π ( d /2) 2

. V = IR. Table 19.1 gives ρ = 2.44 × 10−8 Ω ⋅ m for gold.

4(2.44 × 10−8 Ω ⋅ m)(6.40 m)

=

π (0.840 × 10−3 m)2

= 0.282 Ω

(b) V = IR = (1.15 A)(0.282 Ω) = 0.324 V

*19.25. Set Up: R =

ρL A

. V = IR so I =

Solve: (a) Lnew = 2 L. I new = (b) d new = 2d . I new =

V VA V π (d /2) 2 V π d 2 = = = . ρL 4ρ L R ρL

Vπd 2 Vπd 2 = = 1I 4ρ Lnew 4ρ (2 L) 2

V π d new 2 V π (2d ) 2 = = 4I 4ρ L 4ρ L

V π d new 2 V π (2d ) 2 = = 2I 4ρ Lnew 4ρ(2 L) Reflect: I increases when R decreases. R decreases when L decreases or d increases. (c) Lnew = 2 L; d new = 2d . I new =

V I = 10.5 Ω + Rinternal .

19.26. Set Up: The total resistance of the circuit can be calculated from Ohm’s law: Rtotal = . Since the internal resistance of the battery is in series with the resistance of the light bulb we have Rtotal

V 6.00 V = = 17.1 Ω. Thus, the internal resistance of I 0.350 A = Rtotal − 10.5 Ω = 17.1 Ω − 10.5 Ω = 6.6 Ω.

Solve: From the known current and voltage we have Rtotal =

the battery is Rinternal

Reflect: The internal resistance of the battery is a significant fraction of the total resistance and limits the maximum 6.00 V ≈ 0.9 A. current that this battery can provide to I max = 6.6 Ω

*19.27. Set Up: When the switch is open there is no current and the terminal voltage of the battery equals its emf, E. When the switch is closed, current I flows and the terminal voltage V of the battery is V = E − Ir. The current I is the same at all points of the circuit.


19-6

Chapter 19

Solve: E = 3.08 V. V = E − Ir , with V = 2.97 V and I = 1.65 A, so r =

V − E 3.08 V − 2.97 V = = 0.067 Ω. I 1.65 A

V 2.97 V = = 1.80 Ω. I 1.65 A Reflect: When current flows through the battery, the terminal voltage is less than the emf because of the voltage across the internal resistance. V − IR = 0 and R =

19.28. Set Up: When the switch is open, there is no current. The voltmeter reads the voltage across the circuit element it is connected across. The emf of the battery is 12.0 V. Solve: (a) I = 0, so the terminal voltage of the battery equals the emf. The voltmeter reads 12.0 V. (b) I = 0, so the voltage V = IR across the resistor is zero. The voltmeter reads zero. (c) The sum of the potential changes around the circuit must be zero, so if there is a potential rise of 12.0 V across the battery, there must be a potential drop of 12.0 V across the switch. The voltmeter reads 12.0 V. E 12.0 V = = 2.40 A. (a) The terminal voltage of the battery is (d) E − Ir − IR = 0 and I = r + R 4.70 Ω + 0.30 Ω E − Ir = 12.0 V − (2.40 A)(0.30 Ω) = 11.3 V. (b) The voltage across the resistor is IR = (2.40 A)(4.70 Ω) = 11.3 V. (c) The meter is now a path of zero resistance and the voltage across it is zero.

*19.29. Set Up: The terminal voltage of the battery is given by Vab = E − Ir . The terminal voltage also equals the voltage across the resistor, so Vab = IR. We have two unknowns (E , r ) so we need two equations. Solve: (a) With the 1500 M Ω resistor, I =

Vab 2.50 V = = 1.67 × 10−9 A. R 1500 × 106 Ω

Vab = E − Ir gives 2.50 V = E − (1.67 × 10−9 A)r . Vab 1.75 V = = 0.350 A. R 5.00 Ω Vab = E − Ir gives 1.75 V = E − (0.350 A)r . Subtracting the second equation from the first gives 0.75 V = (0.350 A)r and r = 2.14 Ω. E = 1.75 V + (0.350 A) r = 2.50 V

With the 5.00 Ω resistor, I =

(b) Now R is 7.00 Ω . I =

E 2.50 V = = 0.274 A R + r 7.00 Ω + 2.14 Ω Vab = E − Ir = 2.50 V − (0.274 A)(2.14 Ω) = 1.91 V. Or,

Vab = IR = 1.91 V. Reflect: The smaller R is, the larger the current and the smaller the terminal voltage. With the 1500 M Ω resistor the current is very small and the terminal voltage differs only very little from the emf of the battery.

19.30. Set Up: For a battery with internal resistance r we have Vab = E − Ir. We know that the terminal voltage of the battery drops from 12.0 V to 7.0 V when the battery provides a current of 150 A. E − Vab 12.0 V − 7.0 V Solve: Solving for the internal resistance, r, we obtain r = = = 0.033 Ω. I 150 A 7.0 V = 0.047 Ω. 150 A The internal resistance of the battery plus the resistance of the starter circuit adds to the total resistance, which is 12.0 V = 0.0800 Ω. 150 A

Reflect: The resistance of the starter motor circuit (including its cables and connectors) must be

19.31. Set Up: When current es through a battery in the direction from the − terminal toward the + terminal, the terminal voltage Vab of the battery is Vab = E − Ir .



19-7

E − Vab 24.0 V − 21.2 V = = 0.700 Ω. I 4.00 A V 21.2 V = 5.30 Ω. (b) Vab − IR = 0 so R = ab = I 4.00 A 24.0 V Reflect: The total resistance in the circuit is R + r = 6.00 Ω. I = = 4.00 A, which agrees with the value 6.00 Ω specified in the problem. Solve: (a) Vab = E − Ir gives r =

19.32. Set Up: The voltmeter reads the potential difference Vab between the terminals of the battery. For the open circuit we have I = 0. The circuit is sketched in Figure (a) below.

Solve: Vab = E = 3.08 V Set Up: When the switch is closed the circuit appears as in Figure (b) below.

3.08 V − 2.97 V E − 2.97 V = 0.067 Ω. . Thus, we have r = 1.65 A I V 2.97 V = IR so R = ab = = 1.80 Ω. I 1.65 A

Solve: Vab = E − Ir = 2.97 V, so r =

Finally, we have Vab

Reflect: When current flows through the battery there is a voltage drop across its internal resistance and its terminal voltage V is less than its emf.

*19.33. Set Up: For a resistor, P = VI and V = IR. P 327 W = = 21.8 A V 15.0 V V 15.0 V = 0.688 Ω (b) R = = I 21.8 A Solve: (a) I =

19.34. Set Up: For a resistor, P = I 2 R = Solve: (a) I = (b) P =

V2 = VI and V = IR. R

P 5.0 W = = 0.0183 A. V = IR = (0.0183 A)(15 × 103 Ω) = 275 V. R 15 × 103 Ω

V2 (120 V) 2 = = 1.6 W R 9.0 × 103 Ω


19-8

Chapter 19

*19.35. Set Up: The voltmeter reads the terminal voltage of the battery, which is the potential difference across the appliance. The terminal voltage is less than 15.0 V because some potential is lost across the internal resistance of the battery. The equation P = V 2 /R gives the power dissipated by the appliance. The equation Vab = E − Ir relates the terminal voltage of the battery to its internal resistance. (a) Solve: P = (11.3 V) 2 /(75.0 Ω) = 1.70 W (b) The current in the circuit can be calculated from the terminal voltage of the battery and the resistance of the 11.3 V E − Vab attached appliance: I = = 0.151 A. The internal resistance of the battery, r, is given by r = = 75.0 Ω I 15.0 V − 11.3 V 24.5 Ω. 0.151 A Reflect: The full 15.0 V of the battery would be available only when no current (or a very small current) is flowing in the circuit. This would be the case if the appliance had a resistance much greater than 24.5 Ω.

19.36. Set Up: Power is energy per unit time. The power delivered by a voltage source is P = Vab I . Solve: (a) P = (25 V)(12 A) = 300 W. (b) Energy = Pt = (300 W)(3.0 × 10−3 s) = 0.90 J

19.37. Set Up: P = Solve: (a) R =

V2 R

V 2 (120 V) 2 = = 240 Ω P 60 W

(b) P increases when R decreases, so the 100 W bulb has less resistance. V 2 (120 V) 2 R= = = 144 Ω P 100 W Reflect: P = I 2R. The bulb with smaller R draws more current and this more than compensates for smaller R in I 2R.

19.38. Set Up: V = IR and P = Solve: (a) I = (b) P =

V2 . R

V 120 V = = 0.12 A. Yes, this is very dangerous. R 1.0 × 103 Ω

V2 (120 V) 2 = = 14.4 W. We can also calculate P as I 2 R or VI. These expressions also give 14.4 W. R 1.0 × 103 Ω

*19.39. Set Up: P = VI and energy is the product of power and time. Solve: P = (500 V)(80 × 10−3 A) = 40 W.

Energy = Pt = (40 W)(10 × 10−3 s) = 0.40 J. Reflect: The energy delivered depends not only on the voltage and current but also on the length of the pulse.

19.40. Set Up: P = I 2 R = Solve: (a) P =

V2 = VI . V = IR. The heater consumes 540 W when V = 120 V. Energy = Pt . R

V2 V 2 (120 V) 2 so R = = = 26.7 Ω R P 540 W

(b) P = VI so I =

P 540 W = = 4.50 A V 120 V



19-9

(c) Energy = (0.540 kW)(1 h) = 0.540 kWh. The cost is (0.540 kWh)(7.2 cents/kWh) = 3.9 cents. (d) Assuming that R remains 26.7 Ω, P =

V 2 (110 V)2 = = 453 W. P is smaller by a factor of (110/120)2 . R 26.7 Ω

19.41. Set Up: The power delivered by the battery can be calculated in two different ways: P =

Energy or P = VI . Time

5.25 h = (5.25 h)(3600 s/h) = 1.89 × 104 s. Solve: The average power delivered by the battery is P =

Energy 3.15 × 104 J = = 1.6667 W. Thus, the current must Time 1.89 × 104 s

P 1.6667 W = = 0.450 A. V 3.70 V Reflect: The energy stored in the battery can be expressed in joules or watt-hours: using the equation Energy = Pt ,

be I =

we can express the stored energy as either 3.15 × 104 J or (1.6667 W)(5.25 h) = 8.75 W ⋅ h.

19.42. Set Up: For an emf, P = VI . For a resistor, P = I 2 R.

Solve: (a) 12.0 V − I (1.0 Ω + 5.0 Ω) = 0 and I = 2.0 A. The rate at which chemical energy is being converted to

electrical energy within the battery is Pε = VI = (12.0 V)(2.0 A) = 24.0 W. (b) Pr = I 2 r = (2.0 A) 2 (1.0 Ω) = 4.0 W (c) PR = I 2 R = (2.0 A) 2 (5.0 Ω) = 20.0 W

Note that Pε = Pr + PR .

*19.43. Set Up: P = I 2 R =

V2 = VI . V = IR. The heater consumes 540 W when V = 120 V. Energy = Pt . R

V2 V 2 (120 V) 2 so R = = = 26.7 Ω R P 540 W P 540 W = 4.50 A (b) P = VI so I = = V 120 V Solve: (a) P =

V 2 (110 V)2 = = 453 W. P is smaller by a factor of (110/120) 2 . R 26.7 Ω Reflect: (d) With the lower line voltage the current will decrease and the operating temperature will decrease. R will be less than 26.7 Ω and the power consumed will be greater than the value calculated in part (c). (c) Assuming that R remains 26.7 Ω, P =

19.44. Set Up: V = IR. P = I 2 R. The total resistance is the resistance of the person plus the internal resistance of the power supply. V 14 × 103 V = = 1.17 A Solve: (a) I = Rtot 10 × 103 Ω + 2000 Ω (b) P = I 2 R = (1.17 A) 2 (10 × 103 Ω) = 1.37 × 104 J = 13.7 kJ (c) Rtot =

V 14 × 103 V = = 14 × 106 Ω. The resistance of the power supply would need to be I 1.00 × 10−3 A 14 × 106 Ω − 10 × 103 Ω = 14 × 106 Ω = 14 MΩ.

ρL

. V = IR. P = I 2 R. A (5.0 Ω ⋅ m)(1.6 m) Solve: (a) R = = 1.0 × 103 Ω π (0.050 m) 2

19.45. Set Up: R =


19-10

Chapter 19

(b) V = IR = (100 × 10−3 A)(1.0 × 103 Ω) = 100 V (c) P = I 2 R = (100 × 10−3 A)2 (1.0 × 103 Ω) = 10 W

19.46. Set Up: For resistors in parallel,

1 1 1 = + + g For resistors in series, Req = R1 + R2 + g These rules Req R1 R2

may have to be applied in several ways. Solve: (a) All four resistors are in parallel.

1 1 1 1 1 = + + + so Req = 2.9 Ω. Req 25 Ω 12 Ω 5.0 Ω 45 Ω

(b) The 75 Ω and 25 Ω resistors are in parallel.

1 1 1 = + so their equivalent is Rp = 18.8 Ω. The 9.0 Ω, Rp 75 Ω 25 Ω

18.8 Ω, and 18 Ω resistors are in series, so Req = 9.0 Ω + 18.8 Ω + 18 Ω = 45.8 Ω. (c) The 13 Ω and 15 Ω resistors are in series, so their equivalent is Rs = 13 Ω + 15 Ω = 28 Ω. The 32 Ω, 28 Ω, and

14 Ω resistors are in parallel, so resistors are in parallel so

1 1 1 1 and their equivalent is Rp1 = 7.2 Ω. The 72 Ω and 45 Ω = + + Rp1 32 Ω 28 Ω 14 Ω

1 1 1 and their equivalent is Rp2 = 27.7 Ω. This gives a network with 19 Ω, = + Rp2 72 Ω 45 Ω

7.2 Ω, 27.7 Ω and 24 Ω in series. The equivalent resistance is Req = 19 Ω + 7.2 Ω + 27.7 Ω + 24 Ω = 77.9 Ω. (d) The 12 Ω, 13 Ω and 14 Ω resistors are in series and their equivalent is Rs = 12 Ω + 13 Ω + 14 Ω = 39 Ω. Rs and

the 22 Ω resistor are in parallel and their equivalent Rp is given by

1 1 1 , so Rp = 14.1 Ω. Then Rp = + Rp 39 Ω 22 Ω

and the 11 Ω resistor are in series and Req = 14.1 Ω + 11 Ω = 25.1 Ω.

*19.47. Set Up: The resistors can be placed in any combination of series and parallel. The equivalent resistance of two resistors in series is Req = R1 + R2 , which is larger than either resistance by itself. However, the equivalent

1 , which is always less than the smaller of the two 1 1 + R1 R2 resistances. Thus, to obtain the maximum equivalent resistance the three resistors should all be in series and to obtain the minimum equivalent resistance they should all be in parallel. Solve: (a) For three resistors in series we have Req = R1 + R2 + R3 = 36 Ω + 47 Ω + 51 Ω = 134 Ω. resistance of two resistors in parallel is given by Req =

1 = 15 Ω. 1 1 1 + + 36 Ω 47 Ω 51 Ω Reflect: In addition to these two extreme cases, we can combine the three resistors in a variety of other combinations: for example, we can form a series combination of a single resistor and a pair of resistors that are connected in parallel. Can you find all possible combinations of these three resistors? (b) For three resistors in parallel we have Req =

19.48. Set Up: R =

ρL

so the resistance is proportional to the length of the bar. The bars can be represented by the A circuit shown in the figure below.



Solve: The two resistors in parallel have an equivalent resistance

equivalent resistance R =

19-11

( R /3) R = . Then the three resistors in series have 2 6

2 R R 2 R 9 R 3R + + = = . 3 6 3 6 2

*19.49. Set Up: For a parallel connection the full line voltage appears across each resistor. The equivalent resistance is

1 1 1 RR = + and Req = 1 2 . Req R1 R2 R1 + R2

(40.0 Ω)(90.0 Ω) = 27.7 Ω 40.0 Ω + 90.0 Ω V 120 V = = 4.33 A (b) I = Req 27.7 Ω Solve: (a) Req =

V 120 V V 120 V = = 3.00 A. For the 90.0 Ω resistor, I 2 = = = 1.33 A. R1 40.0 Ω R2 90.0 Ω Reflect: If either resistor is removed, the voltage and current for the other resistor is unchanged. (c) For the 40.0 Ω resistor, I1 =

19.50. Set Up: Let R1 = 1.60 Ω, R2 = 2.40 Ω, R3 = 4.80 Ω. For a parallel network the full battery voltage appears across each resistor. The equivalent resistance Req is given by Solve: (a) (b) I1 = (c) I =

V2 1 1 1 1 = + + . P= . Req R1 R2 R3 R

1 1 1 1 = + + and Req = 0.80 Ω. Req 1.60 Ω 2.40 Ω 4.80 Ω

V 28.0 V V 28.0 V V 28.0 V = = 17.5 A. I 2 = = = 11.7 A. I 3 = = = 5.8 A. R1 1.60 Ω R2 2.40 Ω R3 4.80 Ω

V 28.0 V = = 35.0 A. This is also equal to I1 + I 2 + I 3 . Req 0.80 Ω

(d) The voltage is 28.0 V across each resistor. V 2 (28.0 V) 2 V 2 (28.0 V) 2 V 2 (28.0 V)2 = = 490 W. P2 = = = 327 W. P3 = = = 163 W. (e) P1 = R1 1.60 Ω R2 2.40 Ω R3 4.80 Ω

(f) Since P =

V2 and V is the same for all three resistors, the resistor with the smallest R dissipates the greatest R

power.

19.51. Set Up: Let R1 = 1.60 Ω, R2 = 2.40 Ω, R3 = 4.80 Ω. For a series network, the current is the same in each resistor and the sum of voltages for each resistor equals the battery voltage. The equivalent resistance is Req = R1 + R2 + R3 . P = I 2 R. Solve: (a) Req = 1.60 Ω + 2.40 Ω + 4.80 Ω = 8.80 Ω (b) I =

V 28.0 V = = 3.18 A Req 8.80 Ω

(c) I = 3.18 A, the same as for each resistor. (d) V1 = IR1 = (3.18 A)(1.60 Ω) = 5.09 V. V2 = IR2 = (3.18 A)(2.40 Ω) = 7.63 V.

V3 = IR3 = (3.18 A)(4.80 Ω) = 15.3 V. Note that V1 + V2 + V3 = 28.0 V. (e) P1 = I 2 R1 = (3.18 A)2 (1.60 Ω) = 16.2 W. P2 = I 2 R2 = (3.18 A)2 (2.40 Ω) = 24.3 W. P3 = I 2 R3 = (3.18 A) 2 (4.80 Ω) = 48.5 W.


19-12

Chapter 19

(f) Since P = I 2 R and the current is the same for each resistor, the resistor with the greatest R dissipates the greatest power. Reflect: When resistors are connected in parallel, the resistor with the smallest R dissipates the greatest power.

19.52. Set Up: For resistors in parallel, the voltages are the same and the currents add.

1 1 1 = + so Req R1 R2

R1R2 . For resistors in series, the currents are the same and the voltages add. Req = R1 + R2 . R1 + R2 Solve: The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits shown in 60.0 V = 12.0 A. This is the current through each of the resistors the figure below. Req = 5.00 Ω. In Figure (c), I = 5.00 Ω in Figure (b) below. Req =

V12 = IR12 = (12.0 A)(2.00 Ω) = 24.0 V. V34 = IR34 = (12.0 A)(3.00 Ω) = 36.0 V. Note that V12 + V34 = 60.0 V. V12 is the voltage across R1 and across R2 , so I1 = voltage across R3 and across R4 , so I 3 =

V12 24.0 V V 24.0 V = = 8.00 A and I 2 = 12 = = 4.00 A. V34 is the R1 3.00 Ω R2 6.00 Ω

V34 36.0 V V 36.0 V = = 3.00 A and I 4 = 34 = = 9.00 A. R3 12.0 Ω R4 4.00 Ω

*19.53. Set Up: For resistors in parallel, the voltages are the same and the currents add.

1 1 1 = + so Req R1 R2

R1R2 . For resistors in series, the currents are the same and the voltages add. Req = R1 + R2 . R1 + R2 Solve: The rules for combining resistors in series and parallel lead to the sequence of equivalent circuits shown in 48.0 V = 16.0 A. In Figure (b), the voltage across each resistor is the figure below. Req = 3.00 Ω. In Figure (c), I = 3.00 Ω 48.0 V 48.0 V 48.0 V, so I12 = = 12.0 A and I 34 = = 4.00 A. Note that I12 + I 34 = 16.0 A. Then in Figure (a), 4.00 Ω 12.0 Ω I1 = I 2 = 12.0 A and I 3 = I 4 = 4.00 A. Req =



19-13

19.54. Set Up: We can use the power rating in P = V 2 /R to find the resistance of a bulb. Then we can find the voltage across each bulb when they are connected in series and in parallel. V 2 (120 V) 2 = = 144 Ω Solve: R = P 100 W V 120 V = = 0.417 A. The voltage across each bulb is V1 = IR = 60 V. (a) Req = 2 R = 288 Ω. I = Req 288 Ω V 2 (60 V) 2 = = 25 W R 144 Ω (b) In parallel, the voltage across each bulb is 120 V and the power dissipated by each is 100 W. P=

*19.55. Set Up: We need to reduce the given resistance without removing it from the circuit. We can do this by soldering a second resistor in parallel with the incorrect one. The equivalent resistance for resistors in parallel is 1 1 1 = + . Req R1 R2 Solve: We wish for the equivalent resistance to equal 36.5 kΩ. Thus, the unknown resistance is given by

1 = R1

Req R2 1 1 (36.5 kΩ)(69.8 kΩ) = = 76.5 kΩ. − . This simplifies to R1 = Req R2 R2 − Req (69.8 kΩ − 36.5 kΩ) Reflect: There would probably be no easy way to solve this problem if we had soldered a resistor that was smaller than the desired value into the circuit.

19.56. Set Up: For two resistor connected in series we have Req = R1 + R2 and for two resistors in parallel we have R1R2 V2 . The power dissipated by a resistor is P = . We know that R2 = 68 kΩ. R1 + R2 R Solve: We wish for the power dissipated by parallel combination of resistors to be 4 times that of the series comV2 V 2 ( R1 + R2 ) 4V 2 = = . Simplify this expression (cancel the common voltage bination. Thus, we have R1R2 R1R2 R1 + R2 R1 + R2 Req =

term and cross-multiply) to obtain:

R1 2 − 2R1R2 + R2 2 = ( R1 − R2 )2 = 0.

Thus, the desired resistance is

R1 = R2 = 68 kΩ. Reflect: When two identical resistors are connected in series the equivalent resistance is double that of a single resistor—and when we connect two identical resistors in parallel the equivalent resistance is half that of V2 a single resistor. Thus, the power dissipation, , increases by a factor of four when changing from a series to a Req

parallel combination of a pair of identical resistors. *19.57. Set Up: Assume the unknown currents have the directions shown in the figure below. We have used the junction rule to write the current through the 10.0 V battery as I1 + I 2 . There are two unknowns, I1 and I 2 , so we will need two equations. Three possible circuit loops are shown in the figure.


19-14

Chapter 19

Solve: (a) Apply the loop rule to loop (1), going around the loop in the direction shown: + 10.0 V − (30.0 Ω) I1 = 0

and I1 = 0.333 A. (b) Apply the loop rule to loop (3): +10.0 V − (20.0 Ω) I 2 − 5.00 V = 0 and I 2 = 0.250 A. (c) I1 + I 2 = 0.333 A + 0.250 A = 0.583 A Reflect: For loop (2) we get +5.00 V + I 2 (20.0 Ω) − I1(30.0 Ω) = 5.00 V + (0.250 A)(20.0 Ω) − (0.333 A)(30.0 Ω) = 5.00 V + 5.00 V − 10.0 V = 0, so that with the currents we have calculated the loop rule is satisfied for this third loop.

19.58. Set Up: The circuit diagram is given in the figure below. The junction rule has been used to find the magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown currents.

Solve: The loop rule applied to loop (1) gives: +20.0 V − (1.00 A)(1.00 Ω) + (1.00 A)(4.00 Ω) + (1.00 A)(1.00 Ω) − E1 − (1.00 A)(6.00 Ω) = 0 E1 = 20.0 V − 1.00 V + 4.00 V + 1.00 V − 6.00 V = 18.0 V.

The loop rule applied to loop (2) gives: +20.0 V − (1.00 A)(1.00 Ω) − (2.00 A)(1.00 Ω) − E 2 − (2.00 A)(2.00 Ω) − (1.00 A)(6.00 Ω) = 0 E 2 = 20.0 V − 1.00 V − 2.00 V − 4.00 V − 6.00 V = 7.0 V.

19.59. Set Up: The circuit is sketched in the figure below. When we go around loop (1) in the direction shown there is a potential rise across the 200.0 V battery, so there must be a drop across R and the current in R must be in the direction shown in the figure. Similar analysis of loops (2) and (3) tell us that currents I 2 and I 5 must be in the directions shown. The junction rule has been used to label the currents in all the other branches of the circuit.



Solve: (a) Apply the loop rule to loop (1): +200.0 V − I1R = 0. R = (b) Loop (2): +160.0 V − I 2 (40.0 Ω) = 0. I 2 =

Loop (3): +160.0 V − I 5 (20.0 Ω) = 0. I 5 =

19-15

+200.0 V +200.0 V = = 20.0 Ω I1 10.0 A

160.0 V = 4.00 A 40.0 Ω

160.0 V = 8.00 A 20.0 Ω

A2 reads I 2 = 4.00 A. A3 reads I 2 + I 5 = 12.0 A. A4 reads I1 + I 2 = 14.0 A. A5 reads I 5 = 8.00 A. Reflect: The sum of potential changes around loop (4) is +200.0 V − I1R + I 2 (40.0 Ω) − I 5 (20.0 Ω) = 200.0 V − (10.0 A)(20.0 Ω) + (4.00 A)(40.0 Ω) − (8.00 A)(20.0 Ω) = 200.0 V − 200.0 V − 160.0 V = 0.

The loop rule is satisfied for loop (4) and this is a good check of our calculations.

19.60. Set Up: The circuit diagram is given in the figure below. The junction rule has been used to find the magnitude and direction of the current in the upper branch of the circuit. There are no remaining unknown currents.

Solve: (a) The junction rule gives that the current in R is 2.00 A to the left. (b) The loop rule applied to loop (1) gives: − (2.00 A) R + 28.0 V − (6.00 A)(3.00 Ω) = 0.

R=

28.0 V − 18.0 V = 5.00 Ω. 2.00 A

(c) The loop rule applied to loop (2) gives:

−(4.00 A)(6.00 Ω) + E − (6.00 A)(3.00 Ω) = 0. E = 24.0 V + 18.0 V = 42.0 V.

*19.61. Set Up: We can use the power consumption in the 5.00 Ω resistor to find the current through it. The circuit, unknown currents, and a circuit loop are all shown in the figure below.


19-16

Chapter 19

Solve: P = I 2 R so I1 =

I2 =

20.0 W = 2.00 A. The loop rule for loop (1) gives 5.00 Ω +15.0 V − I 2 (7.00 Ω) − (2.00 A)(5.00 Ω) = 0.

15.0 V − 10.0 V = 0.714 A. The ammeter reads 0.714 A. 7.00 Ω

19.62. Set Up: P = I 2 R so the power consumption of the 6.0 Ω resistor allows us to calculate the current through it. Unknown currents I1 , I 2 and I 3 are shown in the figure below. The junction rule says that I1 = I 2 + I 3 . In the figure below the two 20.0 Ω resistors in parallel have been replaced by their equivalent (10.0 Ω ).

Solve: (a) P = I 2 R gives I1 =

P = R

24 J/s = 2.0 A. The loop rule applied to loop (1) gives: 6.0 Ω

− (2.0 A)(3.0 Ω) − (2.0 A)(6.0 Ω) + 25 V − I 2 (10.0 Ω + 19.0 Ω + 1.0 Ω) = 0. I2 =

25 V − 18 V = 0.233 A. 30.0 Ω



19-17

(b) I 3 = I1 − I 2 = 2.0 A − 0.233 A = 1.77 A. The loop rule applied to loop (2) gives: −(2.0 A)(3.0 Ω + 6.0 Ω) + 25 V − (1.77 A)(17 Ω) − E − (1.77 A)(13 Ω) = 0.

E = 25 V − 18 V − 53.1 V = −46.1 V. The emf is 46.1 V and the polarity of the battery is opposite to what is shown in the figure in the problem; the + terminal is adjacent to the 13 Ω resistor.

*19.63. Set Up: The time constant is τ = RC . Solve: C =

τ

=

R

2.00 s = 4.00 × 10−3 F = 4.00 mF 500.0 Ω

19.64. Set Up: We know that q = Q0e2t /τ , where τ = RC. The fractional charge remaining at any time is thus, q = e2t / τ . Q0 Solve: The time constant for this circuit is τ = RC = (1.5 × 105 Ω)(6.0 × 10−6 F) = 0.90 s. At t = 1.8 s, which is 2τ ,

the fractional charge remaining is

q = e2t / τ = e− (1.8 s) / (0.90 s) = e−2 = 0.14 or 14%. Q0

Reflect: This result can be applied to any RC circuit: After two time constants 14% of the initial charge remains on any capacitor discharging through a simple RC circuit.

*19.65. Set Up: The time constant is RC = (0.895 × 106 Ω) (12.4 × 10−6 F) = 11.1 s. Solve: (a) At t = 0 s: q = CE (1 − e−t/RC ) = 0.

At t = 5 s: q = C E (1 − e−t/RC ) = (12.4 × 10−6 F)(60.0 V)(1 − e− (5.0 s)/(11.1 s) ) = 2.70 × 10−4 C. At t = 10 s: q = CE (1 − e−t/RC ) = (12.4 × 10−6 F)(60.0 V)(1 − e− (10.0 s)/(11.1 s) ) = 4.42 × 10−4 C. At t = 20 s: q = CE (1 − e−t/RC ) = (12.4 × 10−6 F)(60.0 V)(1 − e−(20.0 s)/(11.1 s) ) = 6.21 × 10−4 C. At t = 100 s: q = C E (1 − e−t/RC ) = (12.4 × 10−6 F)(60.0 V)(1 − e− (100 s)/(11.1 s) ) = 7.44 × 10−4 C. (b) The current at time t is given by: i =

At t = 0 s: i = At t = 5 s: i =

60.0 V 8.95 × 105 Ω 60.0 V 8.95 × 105 Ω

At t = 10 s: i = At t = 20 s: i =

e−0 / 11.1 = 6.70 × 10−5 A. e−5 / 11.1 = 4.27 × 10−5 A.

60.0 V 8.95 × 105 Ω 60.0 V 8.95 × 105 Ω

At t = 100 s: i =

E −t/RC e . R

e−10 / 11.1 = 2.72 × 10−5 A. e−20 / 11.1 = 1.11 × 10−5 A.

60.0 V 8.95 × 105 Ω

e−100 / 11.1 = 8.20 × 10−9 A.

(c) The graphs of q (t ) and i (t ) are given in Figures (a) and (b) below. Reflect: The charge on the capacitor increases in time as the current decreases.


19-18

Chapter 19

q C

19.66. Set Up: The loop rule gives E = iR + . Just after the circuit is completed, q = 0. After a long time the capacitor is fully charged and i = 0.

Solve: (a) q = 0 so the voltage across the capacitor is zero. (b) q = 0 so iR = E; the voltage drop across the resistor is 500 V. (c) q = 0 (d) i =

E 500 V = = 0.111 A R 4500 Ω

q = E and the voltage across the capacitor is 500 V. C (b) i = 0 so the voltage across the resistor is zero.

(e) (a) i = 0 so

(c) q = EC = (500 V)(6.00 × 10−6 F) = 3.00 × 10−3 C = 3.00 mC. (d) i = 0

19.67. Set Up: The capacitor discharges exponentially through the voltmeter. Since the potential difference across the capacitor is directly proportional to the charge on the plates, the voltage across the plates decreases exponentially with the same time constant as the charge. The reading of the voltmeter obeys the equation V = V0e− t / RC , where RC is the time constant. Solve: (a) Solving for C and evaluating the result when t = 4.00 s gives t 4.00 s C= = = 8.49 × 10−7 F R ln(V /V0 ) ⎛ 12.0 V ⎞ 6 (3.40 × 10 Ω)ln ⎜ ⎝ 3.00 V ⎟⎠ (b) τ = RC = (3.40 × 106 Ω)(8.49 × 10−7 F) = 2.89 s Reflect: In most laboratory circuits, time constants are much shorter than this one.

19.68. Set Up: The time constant is τ = RC , so q = Qfinal (1 − e−t / RC ) and i = I 0e−t /RC . Solve: (a) q = 0.95Qfinal gives 0.95 = 1 − e−t /RC . e−t /τ = 0.05. −

t

τ

= ln (0.05).

t

τ

= 3.0. Three time constants are required.

(b) e− t /τ = 0.05 so i = 0.05 I 0 .



19-19

*19.69. Set Up: q = C /V . After a long time the current has dropped to zero, there is no potential drop across the resistor and the full battery voltage is applied to the capacitor network. Just after the switch is closed the charge and voltage across each capacitor is zero and the battery voltage equals the voltage drop across the resistor. The time constant is τ = RC , where C is the equivalent capacitance of the network. i = I 0e − t /RC . Solve: (a) The 20.0 pF, 30.0 pF, and 40.0 pF capacitors are in series and their equivalent capacitance Cs is given by

1 1 1 1 = + + . Cs 20.0 pF 30.0 pF 40.0 pF Cs = 9.23 μF. After a long time the voltage across the 10.0 pF capacitor is 50.0 V. The charge on this capacitor is q10 = C10V10 = (10.0 pF)(50.0 V) = 500 pC. The voltage across Cs is 50.0 V and q5 = C5V5 = (9.23 pF)(50.0 V) = 461 pC. This is the charge on each of the capacitors in series, so q20 = q30 = q40 = 461 pC. (b) V10 = 50.0 V. V20 =

q20 461 pC q 461 pC = = 23.1 V. V30 = 30 = = 15.4 V. C20 20.0 pF C30 30.0 pF V40 =

q40 461 pC = = 11.5 V. C40 40.0 pF

Note that V20 + V30 + V40 = 50.0 V, as it should. (c) I =

E 50.0 V = = 2.50 A R 20.0 Ω

(d) The 10.0 pF capacitor and Cs are in parallel so their equivalent is Ceq = 10.0 pF + 9.23 pF = 19.2 pF.

τ = RC = (20.0 Ω)(19.2 pF) = 384 ps Reflect: The charges on the capacitors start at zero and increase to their final values. The current starts at its maximum value and then decays to zero.

19.70. Set Up: The time constant for the circuit is τ = RC . While charging, q = Qfinal (1 − e−t /RC ) and i = I 0e−t /RC . In the discharging circuit both the charge and current decrease to 1/e of their initial values in time t = τ . Solve: τ = RC = (125 Ω)(1.50 μF) = 188 μs (a) q = Qfinal (1 − e − t /RC ). q =

Qfinal 1 1 gives = 1 − e− t /RC . e − t /RC = 1 − and e e e t = − RC (ln[1 − 1/e]) = − (188 μs)( −0.458) = +86.1 μs

(b) i = I 0e − t /RC = I 0e − (86.1 μs)/(188 μs) = 0.633I 0 (c) t = τ = 188 μs

19.71. Set Up: Equations (19.17) give i and q as functions of time during charging. During discharging, both i and q decay exponentially, as in Eq. (19.19) for i. Solve: (a) The graphs are the same as in Figure 19.33a and b in the textbook and are given in Figure (a) below.


19-20

Chapter 19

(b) The graphs of i and q versus t for a discharging capacitor are sketched in Figure (b) below.

19.72. Set Up: There is a single current path so the current is the same at all points in the circuit. Assume the current is counterclockwise. Solve: (a) Apply the loop rule, traveling around the circuit in the direction of the current: +16.0 V − I (1.6 Ω + 5.0 Ω + 1.4 Ω + 9.0 Ω) − 8.0 V = 0. 16.0 V − 8.0 V = 0.471 A. 17.0 Ω Our calculated I is positive so I is counterclockwise, as we assumed. (b) Vb + 16.0 V − I (1.6 Ω) = Va . Vab = 16.0 V − (0.471 A)(1.6 Ω) = 15.2 V I=

*19.73. Set Up: The ohmmeter reads the equivalent resistance between points a and b. Replace series and parallel combinations by their equivalent. For resistors in parallel we have

1 1 1 = + . For resistors in series we have Req R1 R2

Req = R1 + R2 . Solve: Circuit (a): The 75.0 Ω and 40.0 Ω resistors are in parallel and have equivalent resistance 26.09 Ω. The 25.0 Ω and 50.0 Ω resistors are in parallel and have an equivalent resistance of 16.67 Ω. The equivalent network is 1 1 1 given in Figure (a) below. , so Req = 18.7 Ω. = + Req 100.0 Ω 23.05 Ω



19-21

Circuit (b): The 30.0 Ω and 45.0 Ω resistors are in parallel and have equivalent resistance 18.0 Ω. The equivalent network is given in Figure (b) below.

1 1 1 , so Req = 7.5 Ω. = + Req 10.0 Ω 30.3 Ω

Reflect: In circuit (a) the resistance along one path between a and b is 100.0 Ω, but that is not the equivalent

resistance between these points. A similar comment can be made about circuit (b).

19.74. Set Up: We know that the power used by the refrigerator is given by P = VI , and the corresponding energy used by the refrigerator when operated for a time t is E = Pt. Solve: When the refrigerator is operating the power used is P = VI = (120 V)(3.5 A) = 420 W, which is 0.42 kW. In a 30-day month there are (30 days)(24 h/day) = 720 h. Since the refrigerator operates 50% of the time, the refrigerator will be in use for 360 h over the 30-day period. Thus, the amount of energy used during the month is E = Pt = (0.42 kW)(360 h) = 150 kW ⋅ h. The total cost to run the refrigerator for the month is (150 kW ⋅ h) ($0.12/kW ⋅ h) = $18. Reflect: Since most household appliances operate at 120 V, the cost of operation of a given appliance is roughly proportional to its current draw and the percent of time it is in use. For example, an appliance that uses 7.0 A of 7.0 A 5, ⋅ ⋅ $18 = $3.60 per month to current and operates 5% of the time during the month would cost roughly 3.5 A 50, operate.

*19.75. Set Up: RT = R0 (1 + α [T − T0 ]). R =

V . P = VI . When the temperature increases the resistance increases I

and the current decreases. Solve: (a)

V V = (1 + α [T − T0 ]). I 0 = IT (1 + α [T − T0 ]). IT I 0 T − T0 =

(b)

I 0 − IT 1.35 A − 1.23 A = = 217 C°. α IT (1.23 A)(4.5 × 10−4 (C°) −1 ) T = 20°C + 217 C° = 237°C

(i) P = VI = (120 V)(1.35 A) = 162 W (ii) P = (120 V)(1.23 A) = 148 W


19-22

Chapter 19

19.76. Set Up: R = Solve:

ρL A

so each piece has resistance R /3. For resistors in parallel,

1 ⎛ 1 ⎞ 9 = 3⎜ = . Req = R /9. This can also be obtained as decreasing L by ⎝ R /3 ⎠⎟ R Req

overall decrease of

1 1 1 1 = + + . Req R1 R2 R3 1 3

and increasing A by 3, for an

1 . 9

*19.77. Set Up: The circuit is sketched in the figure below. rtotal is the combined internal resistance of both batteries. The power delivered to the bulb is I 2 R. Energy = Pt .

Solve: (a) rtotal = 0. The loop rule gives 1.5 V + 1.5 V − I (17 Ω) = 0. I = 0.1765 A. P = I 2 R = (0.1765 A) 2 (17 Ω) = 0.530 W.

This is also (3.0 V)(0.1765 A). (b) Energy = (0.530 W)(5.0 h)(3600 s/h) = 9540 J P 0.265 W = = 0.125 A. R 17 Ω 3.0 V − (0.125 A)(17 Ω) = 7.0 Ω. The loop rule gives 1.5 V + 1.5 V − IR − Irtotal . rtotal = 0.125 A

(c) P =

0.530 W = 0.265 W. P = I 2 R so I = 2

Reflect: When the power to the bulb has decreased to half its initial value, the total internal resistance of the two batteries is nearly half the resistance of the bulb. Compared to a single battery, using two identical batteries in series doubles the emf but also doubles the total internal resistance.

19.78. Set Up: The power dissipated in a resistor is I 2 R. The power for an emf is EI . When the current es through an emf from the − to the + terminal, chemical energy is converted to electrical energy. When the current es through the emf from the + to the − terminal, electrical energy is converted to chemical energy. Solve: (a) Assume the current in the circuit is counterclockwise. The loop rule gives: 12.0 V − 8.0 V − I (1.0 Ω + 8.0 Ω + 1.0 Ω) = 0. I = 0.40 A. (b) P = I 2 R + I 2r1 + I 2r2 = I 2 ( R + r1 + r2 ) = (0.40 A) 2 (8.0 Ω + 1.0 Ω + 1.0 Ω) = 1.6 W (c) In E1 chemical energy is being converted to electrical energy, at the rate (12.0 V)(0.40 A) = 4.8 W.

19.79. Set Up: Electrical energy is deposited in his body at the rate P = I 2 R. Heat energy Q produces a temperature change ΔT according to Q = mcΔT , where c = 4190 J/kg ⋅ C°.

Solve: (a) P = I 2 R = (25,000 A) 2 (1.0 kΩ) = 6.25 × 1011 W. The energy deposited is Pt = (6.15 × 1011 W)(40 × 10−6 s) = 2.5 × 107 J.

Find ΔT when Q = 2.5 × 107 J.

ΔT =

Q 2.5 × 107 J = = 80 C° mc (75 kg)(4190 J/kg ⋅ C°)



19-23

(b) An increase of only 63 C° brings the water in the body to the boiling point; part of the person’s body will be vaporized. Reflect: Even this approximate calculation shows that being hit by lightning is very dangerous.

19.80. Set Up: V = IR. R =

ρL

. A Solve: (a) The tube of seawater has resistance ρ L (0.13 Ω ⋅ m)(0.20 m) R= = = 260 Ω. A 1.0 × 10−4 m 2

V 3.0 V = = 11.5 mA R 260 Ω (b) The object and the remaining tube of seawater are in series. The 10 cm tube of seawater has resistance 130. Ω. The object is 10 cm long, has the same cross-sectional area and half the resistance of seawater, so its resistance is 65. Ω. The total equivalent resistance of the tube is 130 Ω + 65 Ω = 195 Ω. The current now is V 3.0 V = 15.4 mA. I= = R 195 Ω The increase in current due to the proximity of the object is 15.4 mA − 11.5 mA = 3.9 mA. I=

*19.81. Set Up: The network is sketched in the figure below. If the current in R1 is I, it is I /2 for R2 and R3 . P = I 2 R. Since the current is greatest in R1 , that resistor dissipates the most power.

Solve: P1 = 32.0 W so I =

P 32.0 W = = 4.00 A. The total power dissipated therefore is R 2.00 Ω

(4.00 A) 2 (2.00 Ω) + (2.00 A) 2 (2.00 Ω) + (2.00 A) 2 (2.00 Ω) = 48.0 W.

Reflect: R1 is dissipating its maximum of 32.0 W but R2 and R3 are dissipating only 8.0 W each.

19.82. Set Up: Assume that the dielectric has a cross-sectional area A and a plate separation d . Use R=ρ

V VA L d to calculate the resistance of the dielectric and I = = to calculate the leakage current. The =ρ R ρd A A

capacitance is given by C = K

A Q Qd ⑀0 and the voltage between the plates of the capacitor is given by V = = . d C KA⑀ 0

⎛ Qd ⎞ A Q Solve: Putting the expression for I and V together we obtain I = ⎜ = , which is the desired result. ⎟ ⑀ ρ ⑀0ρ KA d K 0⎠ ⎝ Reflect: Note that we can minimize the leakage current by making the term K ρ large.

*19.83. Set Up: Since they are in parallel, the voltage across each device is 120 V. P = VI . The total current drawn from the outlet is the sum of the individual currents. P 1800 W 1400 W 75 W Solve: (a) Toaster: I = = = 15.0 A. Frypan: I = = 11.7 A. Lamp: I = = 0.625 A. V 120 V 120 V 120 V (b) I = 15.0 A + 11.7 A + 0.625 A = 27.3 A. This is greater than 20 A and the circuit breaker will blow. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

19-24

Chapter 19

19.84. Set Up: The resistance of each half of a wire is 0.500 Ω. The combination can be represented by the resistor network shown in the figure below.

Solve: The two resistors in parallel have an equivalent resistance of 0.250 Ω. Then the three in series have an equivalent resistance of 0.500 Ω + 0.250 Ω + 0.500 Ω = 1.25 Ω.

19.85. Set Up: Let R be the resistance of each resistor. Ptot = Solve: When the resistors are in series, Req = 3R and Ps =

V2 . Req

V2 . When the resistors are in parallel, Req = R /3. 3R

V2 V2 =3 = 9 Ps = 9(27 W) = 243 W. R /3 R Reflect: In parallel, the voltage across each resistor is the full applied voltage V. In series, the voltage across each resistor is V /3 and each resistor dissipates less power. Pp =

19.86. Set Up: For resistors in series, the currents are the same and the voltages add. Req = R1 + R2 + … For resistors in parallel, the currents add and the voltages are the same.

1 1 1 = + +… Req R1 R2

Solve: (a) Replacing series and parallel combinations of resistors by their equivalents gives the equivalent networks as shown in the figure below. The equivalent resistance of the network is 8.0 Ω.



19-25

(b) The voltage across the 8.0 Ω resistor in Figure (a) above is (2.4 A)(8.0 Ω) = 19.2 V. This is the voltage across the 4.0 Ω resistor in Figure (b) above. The current through the 4.0 Ω resistor is 4.8 A. This is also the current through the 20.0 Ω resistor and the voltage across that resistor is 96.0 V. Therefore, the voltage between points x and y is 19.2 V + 96.0 V = 115.2 V. This voltage is divided equally between the 6.0 Ω resistors in Figure (b), so the voltage between points x and y is (115.2 V)/2 = 57.6 V. The voltmeter will read 57.6 V.

*19.87. Set Up: The current delivered by the power plant can be calculated from its power and the known voltage: I=

P . The power dissipated by the wires can be calculated from the current and resistance of the wires: P = I 2 R. V

Solve: (a) With a voltage of 220 V the current is I =

150 × 103 W = 681.8 A. With this current the power 220 V

dissipated by the wire is P = I 2 R = (681.8 A) 2 (0.25 Ω) = 116 × 103 W. Thus, the percentage of the total power 116 kW ⋅ 100, = 77,. 150 kW (b) We can repeat the steps above—or notice that 22 kV is 100 times the previous voltage. Since current is inversely proportional to voltage and the power dissipated is proportional to current squared the power dissipated will now be

dissipated by the wire is

2

⎛ 1 ⎞ ⎜⎝ ⎟ ⋅ 77, = 0.0077,. 100 ⎠ Reflect: Power is most efficiently transferred at relatively low current and high voltage.

19.88. Set Up: Because of the polarity of each emf, the current in the 7.00 Ω resistor must be in the direction shown in the figure below. Let I be the current in the 24.0 V battery.

Solve: The loop rule applied to loop (1) gives: +24.0 V − (1.80 A)(7.00 Ω) − I (3.00 Ω) = 0. I = 3.80 A. The

junction rule then says that the current in the middle branch is 2.00 A, as shown in Figure (b) above. The loop rule applied to loop (2) gives: +E − (1.80 A)(7.00 Ω) + (2.00 A)(2.00 Ω) = 0 and E = 8.6 V.

*19.89. Set Up: There is no current in the middle branch because there is not a complete conducting path for that branch. There is only a single current in the circuit, as shown in the figure below. To find Vab = Va − Vb , start at point b and travel to point a.


19-26

Chapter 19

Solve: 12.0 V − 8.0 V − I (2.00 Ω + 2.00 Ω + 1.00 Ω + 2.00 Ω + 1.00 Ω + 1.00 Ω) = 0. I = 0.444 A. Going from a to b

through the 12.0 V battery gives Va + I (2.00 Ω) + I (1.00 Ω) − 12.0 V + I (1.00 Ω) + 10.0 V = Vb . Va − Vb = 12.0 V − 10.0 V − (0.444 A)(4.00 Ω) = +0.22 V.

Or, going from a to b through the 8.0 V battery gives Va − I (2.00 Ω) − I (1.00 Ω) − 8.0 V − I (2.00 Ω) + 10.0 V = Vb and Va − Vb = −2.0 V + (0.444 A)(5.00 Ω) = +0.22 V.

The voltmeter reads 0.22 V. Va − Vb is positive, so point a is at higher potential. Reflect: Since there is no current in the middle branch, the two resistances in that branch could be removed without affecting the potential difference Vab . The 10.0 V emf doesn’t affect the current but does affect Vab

19.90. Set Up: iR =

q . The charge is reduced to 1/e of its maximum value when t = τ = RC . C

Solve: (a) At t = 0, q = iRC = (0.250 A)(4600 Ω)(0.800 × 10−9 F) = 9.20 × 10−7 C = 0.920 μC. (b) t = τ = (4600 Ω)(0.800 × 10−9 F) = 3.68 × 10−6 s = 3.68 μs.

q . At t = 0, q = 0. τ = RC . C E 400.0 V Solve: (a) At t = 0, E = iR and R = = = 5.00 × 105 Ω. i 0.800 × 10−3 A

19.91. Set Up: E = iR +

(b) τ = RC so C =

τ R

=

6.00 s 5.00 × 105 Ω

= 1.20 × 10−5 F.

19.92. Set Up: As the resistance R varies, the current in the circuit also varies, which causes the potential drop across the internal resistance of the battery to vary. The largest current will occur when R = 0, and the smallest current will occur when R → ∞. The largest terminal voltage will occur when the current is zero ( R → ∞ ) and the smallest terminal voltage will be when the current is a maximum ( R = 0 ). Solve: (a) As R → ∞, I → 0, so Vab → E = 15.0 V, which is the largest reading of the voltmeter. When R = 0, the current is largest at (15.0 V)/(4.00 Ω) = 3.75 A, so the smallest terminal voltage is Vab = E − rI = 15.0 V − (4.00 Ω)(3.75 A) = 0. (b) From part (a), the maximum current is 3.75 A when R = 0, and the minimum current is 0.00 A when R → ∞. (c) The graphs are sketched in the figure below



19-27

Reflect: Increasing the resistance R increases the terminal voltage, but at the same time it decreases the current in the circuit.

Solutions to age Problems *19.93. Set Up: We know that resistance can be calculated from the dimensions of the conductor according to L R = ρ . We are told that ρ = 100 Ω ⋅ cm = 1 Ω ⋅ m. Conductance is defined to be the reciprocal of resistance and so it A 1 A is given by = , and it is measured in Ω −1, which is known as a siemens (S) (so we have S = Ω −1 ). R ρL Solve: Based on our model the resistance of the channel is R = ρ

conductance is

L (1 Ω ⋅ m)(0.3 × 10−9 ) = ≈ 109 Ω. Thus, the A π (0.3 × 10−9 ) 2

1 1 = 9 = 1 nS. The correct answer is B. R 10 Ω

V , so we can find the current in a single channel from the applied R voltage, which we know is 50 mV, and the resistance of the channel. The density of channels can then be found from

19.94. Set Up: According to Ohm’s law I =

the known current density, which is 5 mA/cm 2 . Resistance is the reciprocal of the conductance so we have 1 1 R= = = 1011 Ω. 10 pS 10−11 S Solve: The current is

5 mA/cm 2 5 × 10−10 mA/channel

I=

V 50 mV = = 5 × 10−10 mA R 1011 Ω

per channel. To produce

5 mA/cm 2

we need

= 1010 channels/cm 2 . Note that this result is clearly larger than the first three answers so we 2

convert our answer using 1 μm = 10

−4

10

cm. Thus, we have 10

channels ⎛ 10−4 cm ⎞ ⋅⎜ = 100 channels/μ m 2 . Thus, the ⎜ 1 μ m ⎟⎟ cm 2 ⎝ ⎠

correct answer is D.

*19.95. Set Up: The time constant for a single channel is τ = RC , where R is the resistance of the channel and C is its capacitance. Assuming the results of the previous problem we have that the channel density is 1010 channels/cm 2 and the resistance of a single channel is 1011 Ω. We are told that the specific capacitance is 1 μF/cm 2 . Solve: The capacitance of a single channel is

1 μF/cm 2 1010 channels/cm 2

= 10−10 μ F. Thus, the time constant of the channel

must be τ = RC = (1011 Ω)(10−10 μ F) = 10 μs. The correct answer is B.


20

MAGNETIC FIELD AND MAGNETIC FORCES

Answers to Multiple-Choice Problems 1. B, C 2. C 3. C 13. A 14. B 15. A

4. B

5. D

6. A

7. C

8. C

9. B

10. B, C

11. C

12. B

Solutions to Problems G *20.1. Set Up: The force F on the particle is in the direction of the deflection of the particle. The directions of G G G G G υ , B and F are shown in the figure below. Apply the right-hand rule to the directions of υ and B. See if your G thumb is in the direction of F , or opposite to that direction. Use F = 0 q 0 vB sin φ with φ = 90° to calculate F.

G G G Solve: (a) When you apply the right-hand rule to υ and B, your thumb points east. F is in this direction, so the charge is positive. (b) F = 0 q 0υ B sin φ = (8.50 × 10−6 C)(4.75 × 103 m/s)(1.25 T)sin 90° = 0.0505 N G

G

G

20.2. Set Up: The force F on the ion is downward, in the direction of the deflection. The directions of υ and F G G G are shown in Figure (a) below. Use the right-hand rule to relate the directions of υ , B, and F . The charge is positive. Use F = 0 q 0υ B sin φ with φ = 90° to calculate B. e = 1.60 × 10−19 C.


20-1

20-2

Chapter 20

G G G G Solve: B is perpendicular to both υ and F , so B is either toward the top of the page or toward the bottom of the G page. If B is toward the top of the page, as shown in Figure (b) above, then when you apply the right-hand rule your G G thumb points into the page. This is the specified direction of F , so B is directed as shown in Figure (b). F = 0 q 0υ B sin φ gives

B=

6.94 × 10−15 N F = = 0.850 T 0 q 0υ sin φ 6(1.60 × 10−19 C)(8.50 × 103 m/s)sin 90° G

G

*20.3. Set Up: The directions of υ and B are shown in the figure below. A proton has charge e = +1.60 × 10−19 C. F = 0 q 0υ B sin φ . In part (a), φ = 55.0°. An electron has charge −e.

G Solve: (a) The right-hand rule says F is into the page in the figure above. F = 0 q 0υ B sin φ = (1.60 × 10−19 C)(3.60 × 103 m/s)(0.750 T)sin 55.0° = 3.54 × 10−16 N. G G (b) F is maximum when φ = 90°, when υ is perpendicular to B. Fmax = 0 q υ 0 B = 4.32 × 10−16 N. F is minimum when G G φ = 0° or 180°, when υ is either parallel or antiparallel to B. Fmin = 0. (c) 0 q 0 is the same for an electron and a proton, so F = 3.45 × 10−16 N, the same as for a proton. Since the proton and

electron have charges of opposite sign, the forces on them are in opposite directions. The force on the electron is directed out of the page in the figure above. G G Reflect: Only the component of υ perpendicular to B contributes to the magnetic force. Therefore, this force is G zero when the charged particle moves along the direction of B and the force is maximum when the particle moves in G a direction perpendicular to B. When the sign of the charge changes, the force reverses direction. And, even though the force magnitude is the same for the electron and proton, the effect of the force (the acceleration) for the electron would be much greater, because of its smaller mass.

20.4. Set Up: The gravity force is downward so the force from the magnetic field must be upward. The charge’s velocity and the forces are shown in the figure below. Since the charge is negative, the magnetic force is opposite to G the right-hand rule direction. The minimum magnetic field is when the field is perpendicular to υ. The force is also G G perpendicular to B, so B is either eastward or westward.


Magnetic Field and Magnetic Forces

20-3

G G Solve: If B is eastward, the right-hand rule direction is into the page and FB is out of the page, as required. G mg (0.195 × 10−3 kg)(9.80 m/s 2 ) = = 1.91 T. Therefore, B is eastward. mg = 0 q 0υ B sin φ . φ = 90° and B = υ 0 q 0 (4.00 × 104 m/s)(2.50 × 10−8 C)

20.5. Set Up: The charge is positive, so the magnetic force is in the right-hand rule direction. The figure below G G G G shows υ and B. F = ma relates the force and acceleration.

G G G Solve: (a) The right-hand rule applied to υ and B gives that FB is in the +z direction. FB = 0 q 0υ B sin φ = (3.50 × 10−8 C)(2.00 × 105 m/s)(0.80 T)sin 90° = 5.60 × 10−3 N

(b) a =

G F 5.60 × 10−3 N = = 1.12 m/s 2 . a is in the +z direction. − 3 m 5.00 × 10 kg

G G Reflect: The force and acceleration are perpendicular to the plane formed by the directions of υ and B.

G

20.6. Set Up: Use F = 0 q 0υ B sin φ to relate the force F to the angle φ between the direction of υ and the direction G of B. Since 0 q 0 , υ , and B are unknown and constant, set up a ratio that involves just F and φ .

Solve:

F F1 F = 0 q 0υ B, which is constant. Therefore, = 2 . sin φ sin φ1 sin φ2 ⎛ sin φ2 ⎞ ⎛ sin 30.0° ⎞ F2 = F1 ⎜ ⎟ = F⎜ ⎟ = 1.93F . ⎝ sin15.0° ⎠ ⎝ sin φ1 ⎠

20.7. Set Up: Use F = 0 q 0υ B sin φ to calculate the force and then use a = F/m to calculate the acceleration. The

G G charge q of the nucleus is 4e. The directions of υ and B are shown in the figure below, where the plane of the G page is horizontal and upward is directed out of the page. The acceleration has the same direction as F .

Solve: F = 0 q 0υ B sin φ = 4(1.60 × 10−19 C)(1.35 × 103 m/s)(1.12 T)sin 90° = 9.68 × 10−16 N.

a=

F 9.68 × 10−16 N = = 6.45 × 1010 m/s 2 . m 1.50 × 10−26 kg

G The right-hand rule gives that F is north, so the acceleration is north.


20-4

Chapter 20

20.8. Set Up: F = 0 q 0υ B sin φ. Since the particle has negative charge, the force is opposite to the right-hand rule (rhr) direction. G G Solve: (a) The directions of υ and B are shown in Figure (a) below. The angle φ is 130°. F = (2.50 × 10−8 C)(40.0 × 103 m/s)(2.00 T)sin130° = 1.53 × 10−3 N. The rhr direction is in the + z direction so the force is in the − z direction.

G G (b) The directions of υ and B are shown in Figure (b) above. φ = 90°. F = 0 q 0υ B sin φ = (2.50 × 10 −8 C)(40.0 × 103 m/s)(2.00 T)sin 90° = 2.00 × 10−3 N. G G The right-hand rule direction and the direction of F are also shown in Figure (b); F is 140° counterclockwise from the + x axis, so is 50° counterclockwise from the + y axis.

G

G

*20.9. Set Up: Take υ and B as shown in the textbook. The direction of the magnetic force at each point is given by the right-hand rule. We may use F = q υ Bsinφ to find the magnitude of the force and a = F /m to find the acceleration of the charge. According to the figure φ = 45D. G Solve: According to the right-hand rule, F is in the − z direction. The acceleration of the charge is given by a = qqqυ Bsinφ /m = (1.22 × 10−8 C)(3.00 × 104 m/s)(1.25 T)sin 45D /(1.81 × 10−3 kg) = 1.79 × 10−1 m/s 2 . Reflect: For a given velocity and magnetic field strength, the maximum acceleration of the charge occurs when the magnetic field is perpendicular to the velocity of the charge. In this case, the maximum acceleration would be 0.253 m/s 2 . On the other hand, if the magnetic field were parallel to the velocity of the charge, the charge would have an acceleration of zero.

20.10. Set Up: For the alpha particles to emerge from the plates undeflected, the magnetic force on them must exactly cancel the electric force. The battery produces an electric field between the plates, which acts on the alpha 1 particles. First use energy conservation to find the speed of the alpha particles as they enter the plates: qV = mυ 2 . 2 The electric field between the plates due to the battery is E = Vb /d . For the alpha particles not to be deflected, the magnetic force must cancel the electric force, so qυ B = qE , giving B = E /υ. Solve: Solve for the speed of the alpha particles just as they enter the region between the plates. Their charge is 2e.

υα =

2(2e)V 4(1.60 × 10−19 C)(1750V) = = 4.11 × 105 m/s −27 m 6.64 × 10 kg

The electric field between the plates, produced by the battery, is E = Vb /d = (150 V)/(0.00820 m) = 18,300 V/m The magnetic force must cancel the electric force: B = E /υα = (18,300 V/m)/(4.11 × 105 m/s) = 0.0445 T



20-5

The magnetic field is perpendicular to the electric field. If the charges are moving to the right and the electric field points upward, the magnetic field is out of the page. Reflect: The sign of the charge of the alpha particle does not enter the problem, so negative charges of the same magnitude would also not be deflected.

G

*20.11. Set Up: The relation between υ, B, and E is given by Eq. (20.3). E points from the positive plate to the negative plate, so is directed upward in the figure. The electric field is related to the potential difference V between the plates by V = Ed , where d is the separation between the plates. E 3.33 × 103 V/m V 150 V = 1.02 T. = = 3.33 × 103 V/m. B = = − 2 v 3.25 × 103 m/s d 4.50 × 10 m (b) The forces and fields are shown in the figure below. Assume a positive charge; the same field directions also G G G work for a negative charge. E is upward so the electric force FE is upward. In order for FB to be downward, to G G oppose FE , B must be out of the page, as shown. Solve: (a) E =

Reflect: Ions with speed υ = E /B are undeflected no matter what is the sign or magnitude of their charge. If v is not equal to E /B, then either FB or FE is larger than the other, there is a net force on the charge and the charge does

not travel in a straight line.

G

20.12. Set Up: Apply Newton’s second law and F = q vB sin φ , where φ is the angle between the direction of v

G and the direction of the magnetic field, B. Solve: (a) The smallest possible acceleration is zero, when the motion is parallel to the magnetic field. The greatest acceleration is when the velocity and magnetic field are at right angles: qvB (1.6 × 10−19 C)(2.50 × 106 m/s)(7.4 × 10−2 T) a= = = 3.25 × 1016 m/s 2 . m (9.11 × 10−31 kg) qvB sin φ , then sinφ = 0.25 and φ = 14.5°. m Reflect: The force and acceleration decrease as the angle φ approaches zero. (b) If a = 14 (3.25 × 1016 m/s 2 ) =

*20.13. Set Up: Eq. (20.4) says R =

mυ . A proton has mass m = 1.67 × 10−27 kg and charge 0q0 B

q = + e = 1.60 × 10−19 C. Solve: υ =

R 0 q 0 B (0.0613 m)(1.60 × 10−19 C)(0.250 T) = = 1.47 × 106 m/s. m 1.67 × 10−27 kg

20.14. Set Up: Since the particle moves perpendicular to the uniform magnetic field, the radius of its path is G mυ G R= . The magnetic force is perpendicular to both υ and B. The alpha particle has charge 0q0 B

q = 12e = 3.20 × 10−19 C.


20-6

Chapter 20

(6.64 × 10−27 kg)(35.6 × 103 m/s)

= 6.73 × 10−4 m = 0.673 mm. (3.20 × 10−19 C)(1.10 T) The alpha particle moves in a circular arc of diameter 2 R = 1.35 mm. (b) For a very short time interval the displacement of the particle is in the direction of the velocity. The magnetic force is always perpendicular to this direction so it does no work. The work-energy theorem therefore says that the kinetic energy of the particle, and hence its speed, is constant. (c) The acceleration is Solve: (a) R =

a=

FB 0 q 0υ B sin φ (3.20 × 10−19 C)(35.6 × 103 m/s)(1.10 T)sin 90° = = = 1.88 × 1012 m/s 2 . m m 6.64 × 10−27 kg

We can also use a =

υ2 R

and the result of part (a) to calculate (35.6 × 103 m/s)2

= 1.88 × 1012 m/s 2 , 6.73 × 10−4 m G G the same result. The acceleration is perpendicular to υ and B and so is horizontal, toward the center of curvature of the particle’s path. G G G (d) The unbalanced force ( FB ) is perpendicular to υ , so it changes the direction of v but not its magnitude, which a=

is the speed.

20.15. Set Up: A proton and a deuteron each have charge + e = 1.60 × 10−19 C, but they have different masses. The mυ −27

mass of a proton is 1.67 × 10 Solve: (a) υ =

kg. The radius of the circular arc that the particle moves in is R =

0q0 B

.

R 0 q 0 B (0.556 m)(1.60 × 10−19 C)(0.500 T) = = 1.33 × 107 m/s. m 3.34 × 10−27 kg

The speed is not changed by the magnetic field so the deuteron has this speed both just before it enters the field and just after it leaves the field. R υ R R = = constant and 1 = 2 . (b) Only the mass changes. m1 m 2 m 0q0 B

⎛ 1.67 × 10−27 kg ⎞ ⎛m ⎞ R2 = R1 ⎜ 2 ⎟ = (55.6 cm) ⎜ ⎟ = 27.7 cm. −27 ⎜ kg ⎠⎟ ⎝ m1 ⎠ ⎝ 3.34 × 10 Reflect: The magnetic force does no work on the moving charge so the motion in a magnetic field is at constant speed. The radius of the path decreases as m decreases.

20.16. Set Up: The magnetic force on the beam bends it through a quarter circle. The distance that particles in the beam travel is s = Rq, and the radius of the quarter circle is R = mv/qB. Solve: Solving for R gives R = s/q = s/(p/2) = (1.18 cm)/(π/2) = 0.751 cm. Solving for the magnetic field: B = mv /qR = (1.67 × 10−27 kg)(1200 ms)/[(1.60 × 10−19 C)(0.00751 m)] = 1.67 × 10−3 T Reflect: This field is about 10 times stronger than the Earth’s magnetic field, but much weaker than many laboratory fields. mυ *20.17. Set Up: R = . 0q0 B Solve: RB =

⎛B ⎞ ⎛ B ⎞ mυ = constant. R1 B1 = R2 B2 and R2 = R1 ⎜ 1 ⎟ = R ⎜ 1 ⎟ = R/3. 0q0 B ⎝ 2⎠ ⎝ 3B1 ⎠

Reflect: The radius of the circular path decreases when B increases.



20.18. Set Up: For motion in an arc of a circle, a =

20-7

υ2

and the net force is radially inward, toward the center of the R circle. The direction of the force is shown in the figure below. Since the electron has negative charge, the force on it is opposite to the right-hand rule direction. The magnetic force does no work and the electron moves at constant speed. The distance from A to B along the path is π R, where R = 5.00 cm. An electron has charge q = − e = −1.60 × 10−19 C and mass 9.11 × 10−31 kg. A proton has charge q = + e and mass 1.67 × 10−27 kg.

G G Solve: (a) To produce a force in the direction shown, the magnetic field must be directed into the page. F = ma

gives 0 q 0υ B sin φ = m

υ2 R

. φ = 90° and B=

(b) t =

πR

υ

=

π (5.00 × 10−2 m) 6

1.41 × 10 m/s

mυ (9.11 × 10−31 kg)(1.41 × 106 m/s) = = 1.61 × 10−4 T 0 q 0 R (1.60 × 10219 C)(5.00 × 10−2 m)

= 1.11 × 10−7 s

G (c) Since the proton has positive charge, B must be directed out of the page to produce a force in the required direction. mυ (1.67 × 10−27 kg)(1.41 × 106 m/s) B= = = 0.294 T. 0 q 0 R (1.60 × 10219 C)(5.00 × 10−2 m)

*20.19. Set Up: In the acceleration process the electric potential energy decrease, 0 qV 0 , equals the increase in kinetic energy,

1 mυ 2 . 2

Use this to find the speed υ of the protons after they have been accelerated. In the magnetic

field,

R=

mυ . 0q0 B

Protons have charge 1e and mass 1.67 × 10−27 kg. Electrons have charge −e and mass 9.11 × 10−31 kg. Solve: (a) 0 qV 0 = 12 mυ 2 . υ =

20 qV 0 2(1.60 × 10219 C)(0.745 × 103 V) mυ = = 3.78 × 105 m/s. R = and m 0q0 B 1.67 × 10−27 kg

B= (b) B =

mυ (1.67 × 10227 kg)(3.78 × 105 m/s) = = 4.51 × 10−3 T. 0q0 R (1.60 × 10−19 C)(0.875 m)

Bp Be υ B mυ mυ = . . In = = constant and only m changes. mp me m 0q0 R 0q0 R 0q0 R

⎛m ⎞ ⎛ 9.11 × 10−31 kg ⎞ Be = ⎜ e ⎟ Bp = ⎜ (4.51 × 10−3 T) = 2.46 × 10−6 T. ⎜ 1.67 × 10−27 kg ⎟⎟ ⎜ mp ⎟ ⎝ ⎠ ⎝ ⎠ Reflect: For lighter particles a smaller force and hence smaller B is needed to produce the same path. To produce electrons of the same speed as the protons a smaller accelerating voltage would be required.


20-8

Chapter 20

20.20. Set Up: The magnitude of the magnetic force on the bullet is given by F = q υ B sin φ and the acceleration of the bullet is given by a = F /m. We know that the magnetic field is perpendicular to the velocity of the bullet so that

φ = 90D. Solve: (a) F = q υ B sin φ = (1.65 × 10−6 C)(425 m/s)(5.50 × 10−4 T)sin 90D = 3.86 × 10−7 N. (b) Assuming that only the magnetic force acts on the bullet, we have a = F /m = (3.86 × 10−7 N)/(3.25 × 10−3 kg) =

1.19 × 10−4 m/s 2 . Reflect: The acceleration calculated in part (b) ignores the gravitational force acting on the bullet, which would produce an acceleration that is nearly 105 times larger.

20.21. Set Up: An electron has charge q = −e = −1.60 × 10−19 C and mass 9.11 × 10−31 kg. When it moves in a υ2

. When a particle of charge −e is accelerated through a potential R difference of magnitude V, it gains kinetic energy eV. Solve: 12 mυ 2 = eV and

circular path of radius R, its acceleration is

υ=

2eV 2(1.60 × 10−19 C)(2.00 × 103 V) = = 2.65 × 107 m/s. −31 m 9.11 × 10 kg

G υ2 G F = ma gives 0 q υ 0 B sin φ = m . φ = 90° and R B=

mυ (9.11 × 10−31 kg)(2.65 × 107 m/s) = = 8.38 × 10−4 T. 0q0 R (1.60 × 10−19 C)(0.180 m)

20.22. Set Up: For no deflection the magnetic and electric forces must be equal in magnitude and opposite in direction. v = E /B for no deflection. With only the magnetic force, | q | vB = mv 2 /R Solve: (a) v = E /B = (1.56 × 104 V/m)/(4.62 × 10−3 T) = 3.38 × 106 m/s. G G G (b) The directions of the three vectors v, E, and B are sketched in the figure below. (c) R =

mv (9.11 × 10−31 kg)(3.38 × 106 m/s) = = 4.17 × 10−3 m. | q | B (1.60 × 10−19 C)(4.62 × 10−3 T)

2π m 2π R 2π (4.17 × 10−3 m) = = = 7.74 × 10−9 s. |q|B v (3.38 × 106 m/s) Reflect: For the field directions shown in the figure below, the electric force is toward the top of the page and the magnetic force is toward the bottom of the page. T=

*20.23. Set Up: In a velocity selector, E = υ B (Section 20.2). For motion in a circular arc in a magnetic field B′, mυ R=

0 q 0 B′

(Section 20.3). The ions have charge + e.

Solve: (a) υ =

E 155 V/m = = 4.92 × 103 m/s B 0.0315 T



(b) m =

R 0 q 0 B′

υ

=

(0.175 m)(1.60 × 10−19 C)(0.0175 T) 4.92 × 103 m/s

20-9

= 9.96 × 10−26 kg

Reflect: Ions with larger m / 0 q 0 move in a path of larger radius.

20.24. Set Up: R =

mυ . After completing one semicircle the separation between the ions is the difference in the 0q0 B

diameters of their paths, or 2( R13 − R12 ). A singly ionized ion has charge +e. Solve: (a) B =

mυ (1.99 × 10−26 kg)(8.50 × 103 m/s) = = 8.46 × 10−3 T. 0q0 R (1.60 × 10−19 C)(0.125 m)

(b) The only difference between the two isotopes is their masses.

R R R υ = = constant and 12 = 13 . m12 m13 m 0q0 R

⎛ 2.16 × 10−26 kg ⎞ ⎛m ⎞ R13 = R12 ⎜ 13 ⎟ = (12.5 cm) ⎜ = 13.6 cm. ⎜ 1.99 × 10−26 kg ⎟⎟ ⎝ m12 ⎠ ⎝ ⎠ The diameter is 27.2 cm. (c) The separation is 2( R13 − R12 ) = 2(13.6 cm − 12.5 cm) = 2.2 cm. This distance can be easily observed.

*20.25. Set Up: 0 q 0 , υ , and B are the same as in Problem 20.24. For m = 1.99 × 10−26 kg (12 C), R12 = 12.5 cm. The separation of the isotopes at the detector is 2( R15 − R14 ). Solve: R =

R R mυ R υ . = = constant. 14 = 12 and m14 m12 0q0 B m 0q0 B

⎛ 2.32 × 10−26 kg ⎞ ⎛m ⎞ R14 = R12 ⎜ 14 ⎟ = (12.5 cm) ⎜ = 14.6 cm. ⎜ 1.99 × 10−26 kg ⎟⎟ ⎝ m12 ⎠ ⎝ ⎠ ⎛ 2.49 × 10−26 kg ⎞ ⎛m ⎞ R15 = R12 ⎜ 15 ⎟ = (12.5 cm) ⎜ = 15.6 cm. ⎜ 1.99 × 10−26 kg ⎟⎟ ⎝ m12 ⎠ ⎝ ⎠ The separation of the isotopes at the detector is 2( R15 − R14 ) = 2(15.6 cm − 14.6 cm) = 2.0 cm. Reflect: The separation is large enough to be easily detectable. Since the diameter of the ion path is large, about 30 cm, the uniform magnetic field within the instrument must extend over a large area.

G

20.26. Set Up: F = IlB sin φ . The direction of F is given by applying the right-hand rule to the directions of

G I and B. G Solve: (a) The current and field directions are shown in Figure (a) below. The right-hand rule gives that F is directed to the south, as shown. φ = 90° and F = (1.20A)(1.00 × 10−2 m)(0.588 T) = 7.06 × 10−3 N.


20-10

Chapter 20

G (b) The right-hand rule gives that F is directed to the west, as shown in Figure (b) above φ = 90° and F = 7.06 × 10−3 N, the same as in part (a).

G (c) The current and field directions are shown in Figure (c) above. The right-hand rule gives that F is 60.0° north of west. φ = 90° so F = 7.06 × 10−3 N, the same as in part (a).

20.27. Set Up: F = IlB sin φ . The current and field are perpendicular, so φ = 90°. 1 gauss = 10−4 T. Solve: F = (50,000 A)(1 m)(0.5 × 10−4 T)sin 90° = 2.5 N

20.28. Set Up: F = IlB sin φ . Since the field is perpendicular to the rod it is perpendicular to the current and φ = 90°. Solve: I =

F 0.13 N = = 9.7A lB (0.200 m)(0.067 T)

G

*20.29. Set Up: F = IlB sin φ . The direction of F is determined by applying the right-hand rule to the directions of G I and B. 1 gauss = 10−4 T.

G Solve: (a) The directions of I and B are sketched in Figure (a) below. φ = 90° and

F = (1.5 A)(2.5 m)(0.55 × 10−4 T) = 2.1 × 10−4 N. G The right-hand rule says that F is directed out of the page, so is upward.

G G (b) The directions of I and B are sketched in Figure (b) above. φ = 90° and F = 2.1 × 10−4 N. F is directed east to west. G (c) B and the direction of the current are antiparallel. φ = 180° and F = 0. (d) The magnetic force of 2.1 × 10−4 N is not large enough to cause significant effects. G Reflect: The magnetic force is a maximum when the directions of I and B are perpendicular and it is zero when the current and magnetic field are either parallel or antiparallel.

20.30. Set Up: The magnetic force is F = ILB sin φ . The poles of the magnet have a diameter D = 3.50 cm. Solve: From the figure below we see that the length of the wire between the poles is L = D / sin 65D. Thus, the F 0.0514 N = = 1.05 A. current flowing through the wire is I = LB sin φ ⎛ 0.0350 m ⎞ D (1 40 T)sin 65 . ⎜⎝ ⎟ sin 65D ⎠



20-11

Reflect: Note that the answer does not depend on the angle between the wire and the magnetic field. This is due to the fact that a longer wire is required to span the gap as the angle between the magnetic field and wire is increased. Thus, F = ILB sin φ = I ( D / sin φ ) B sin φ = IDB is independent of φ .

G

G

20.31. Set Up: F = IlB sin φ . The direction of F is given by the right-hand rule applied to the directions of I and B. Solve: (a) segment da: φ = 180° and F = 0 segment ab: φ = 90° and F = IlB = (8.0 A)(0.200 m)(0.750 T) = 1.2 N. G G G The directions of I, B, and F are shown in Figure (a) below. F is directed into the page. segment bc: φ = 0° and G G G F = 0 segment cd: φ = 90° and F = 1.2 N. The directions of I, B, and F are shown in Figure (b) below. F is directed out of the page.

G G (b) Fab and Fcd are equal in magnitude and opposite in direction. Their vector sum is zero and the net force on the

entire circuit is zero. Reflect: The net force on any current loop in a uniform magnetic field is zero.

20.32. Set Up: The direction of the magnetic force on each current segment in the field is shown in the figure G G below. By symmetry, Fa = Fb . Fa and Fb are in opposite directions so their vector sum is zero. The net force equals Fc . F = IlB sin φ . For Fc , φ = 90° and l = 0.450 m.


20-12

Chapter 20

Solve: Fc = IlB = (6.00 A)(0.450 m)(0.666 T) = 1.80 N. The net force is 1.80 N, directed to the left.

*20.33. Set Up: Label the three segments in the field as a, b, and c. Let x be the length of segment a. Segment b has length 0.300 m and segment c has length 0.600 cm − x. Figure (a) below shows the direction of the force on each segment. For each segment, φ = 90°. The total force on the wire is the vector sum of the forces on each segment.

G G Solve: Fa = IlB = (4.50 A) x (0.240 T). Fc = (4.50 A)(0.600 m − x)(0.240 T). Since Fa and Fc are in the same

direction their vector sum has magnitude Fac = Fa + Fc = (4.50 A)(0.600 m)(0.240 T) = 0.648 N and is directed toward the bottom of the page in Figure (a). Fb = (4.50 A)(0.300 m)(0.240 T) = 0.324 N and is directed to the right. G G The vector addition diagram for Fac and Fb is given in Figure (b) above. F = Fac 2 + Fb 2 = (0.648 N) 2 + (0.324 N) 2 = 0.724 N. tan θ =

Fac 0.648 N and = Fb 0.324 N

θ = 63.4°. The net force has magnitude 0.724 N and its direction is specified by θ = 63.4° in Figure (b). Reflect: All three current segments are perpendicular to the magnetic field, so φ = 90° for each in the force equation. The direction of the force on a segment depends on the direction of the current for that segment.

20.34. Set Up: F = IlB sin φ . The direction of the magnetic force is given by the right-hand rule applied to the

G directions of I and B. The torque due to a force equals the force times the moment arm, the perpendicular distance between the axis and the line of action of the force. Solve: (a) The direction of the magnetic force on each segment of the circuit is shown in the figure below. For segments bc and da the current is parallel or antiparallel to the field and the force on these segments is zero.



20-13

G G (b) Fab acts at the hinge and therefore produces no torque. Fcd tends to rotate the loop about the hinge so it does produce a torque about this axis. Fcd = IlB sin φ = (5.00 A)(0.200 m)(1.20 T)sin 90° = 1.20 N. (c) τ = Fl = (1.20 N)(0.350 m) = 0.420 N ⋅ m. The torque is directed so as to rotate side cd out of the plane of the page in the figure above.

*20.35. Set Up: τ = IAB sin φ. Since the plane of the loop is parallel to the field, the field is perpendicular to the normal to the loop and φ = 90°. The magnetic moment of the loop is μ = IA. Solve: (a) τ = IAB = (6.2 A)(0.050 m)(0.080 m)(0.19 T) = 4.7 × 10−3 N ⋅ m (b) μ = IA = (6.2 A)(0.050 m)(0.080 m) = 0.025 A ⋅ m 2 Reflect: The torque is a maximum when the field is in the plane of the loop and φ = 90°.

20.36. Set Up: τ = IAB sin φ . The maximum torque is for φ = 90°, where φ is the angle between the field and the normal to the plane of the loop. The coil has area A = π r 2 = π (4.3 × 10−2 m) 2 = 5.81 × 10−3 m 2 . Solve: (a) τ = (15)(2.7 A)(5.81 × 10−3 m 2 )(0.56 T) = 0.132 N ⋅ m 2 . The plane of the coil is parallel to the field. (b) τ = 0.71τ max means sin φ = 0.71 and φ = 45°. The plane of the coil makes an angle of 45° with the field

direction.

*20.37. Set Up: τ = IAB sin φ . The coil as viewed along the axis of rotation is shown in Figure (a) below for its original position and in Figure (b) after it has rotated 30.0°.

G G G G Solve: (a) The forces on each side of the coil are shown in Figure (a). F1 + F2 = 0 and F3 + F4 = 0. The net force on the coil is zero. φ = 0° and sin φ = 0, so τ = 0. The forces on the coil produce no torque. (b) The net force is still zero. φ = 30.0° and the net torque is τ = (1)(1.40 A)(0.220 m)(0.350 m)(1.50 T)sin 30.0° = 0.0808 N ⋅ m. The net torque is clockwise in Figure (b) above and is directed so as to increase the angle φ . © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

20-14

Chapter 20

Reflect: For any current loop in a uniform magnetic field the net force on the loop is zero. The torque on the loop depends on the orientation of the plane of the loop relative to the magnetic field direction.

20.38. Set Up: The torque on a solenoid is τ = NIAB sin φ . The angle φ is the angle between the axis of the solenoid and the direction of the field. Solve: (a) τ = NIAB sin φ = (165)(1.20 A)(6.75 × 10−4 m 2 )(1.12 T)sin 90° = 0.150 N ⋅ m (b) φ = 0° and τ = 0. (c) φ = 35.0° and τ = (0.150 N ⋅ m)sin 35.0° = 0.0860 N ⋅ m

20.39. Set Up: We know that τ = NIAB sin φ , where A = π r 2 . The direction of the force on each section of the loop is given by the right-hand rule. Solve: (a) τ = NI (π r 2 ) B sinφ = 50(2.50 A)π (0.100 m)2 (0.450 T)sin 45D = 1.25 N ⋅ m. (b) Imagine that you are standing to the south and looking down on the coil. The current in the north side of the coil is to your right. Since the magnetic field points down and away from you, the right-hand rule shows that the magnetic force on the north section of the coil must be up and away from you. A similar analysis shows that the magnetic force on the south side of the coil must be down and toward you. Thus, the net torque on the coil will tend to raise the north side of the coil. It will also tend to lower the south side of the coil, which cannot happen since the coil is sitting on the tabletop. Reflect: The net magnetic force on the coil is zero but the net magnetic torque on the coil is not zero.

20.40. Set Up: For a long, straight wire, B = Solve: (a) I = (b) B =

2π rB

μ0

=

μ0 I . μ0 = 4π × 10−7 T ⋅ m/A. 2π r

(2π )(0.040 m)(5.50 × 10−4 T) 4π × 10−7 T ⋅ m/A

(4π × 10−7 T ⋅ m/A)(110 A) 2π (8.00 × 10−2 m)

= 110 A

= 2.75 × 10−4 T. B is proportional to 1/r so the field is half what it is in part (a).

*20.41. Set Up: For a long straight wire, B =

μ0 I . The magnetic field of the earth is of the order of 1024 T. 2π r

μ0 = 4π × 10−7 T ⋅ m/A. Solve: B =

(4π × 10−7 T ⋅ m/A)(10 A) = 4.0 × 10−5 T. This field is the same order of magnitude as the earth’s field. 2π (0.050 m)

20.42. Set Up: The long current-carrying wire produces a magnetic field. The magnetic field due to a long wire is μ I B= 0 . 2π r Solve: First solve for the current, then substitute the numbers using the above equation. (a) Solving for the current gives I = 2π rB /μ0 = 2π (0.0200 m)(1.00 × 10−4 T)/(4π × 10−7 T ⋅ m/A) = 10.0 A (b) The earth’s horizontal field points northward, so at all points directly above the wire the field of the wire would point northward. (c) At all points directly east of the wire, its field would point northward. Reflect: Even though the Earth’s magnetic field is rather weak, it requires a fairly large current to cancel this field.

20.43. Set Up: 1 gauss = 10−4 T. For a long straight wire, B = Solve: I =

2π rB

μ0

=

2π (0.050 m)(1.0 × 10−10 T) 4π × 10

−7

T ⋅ m/A

μ0 I . μ0 = 4π × 10−7 T ⋅ m/A. 2π r

= 2.5 × 10−5 A = 25 μ A.



20.44. Set Up: B = Solve: B =

20-15

μ0 I . μ0 = 4π × 10−7 T ⋅ m / A. 2π r

(4π × 10−7 T ⋅ m/A)(12 × 10−3 A) = 1.2 × 10−7 T = 0.12 μT 2π (0.020 m)

*20.45. Set Up: For a long, thin, straight wire we have B =

μ0 I . 2π r

μ0 I (4π × 10−7 T ⋅ m/A)(0.65 A) = = 4.3 × 10−5 T. 2π r 2π (0.0030 m) μ I Reflect: B = 0 applies precisely only to an infinitely long straight wire. 2π r Solve: B =

20.46. Set Up: B = Solve: Br =

μ0 I . B1 = B when r1 = R. Calculate r2 when B2 = B /3. 2π r

⎛B ⎞ μ0 I ⎛ B ⎞ = constant and B1r1 = B2r2 . r2 = r1 ⎜ 1 ⎟ = R ⎜ ⎟ = 3R. B 2π ⎝ B/3 ⎠ ⎝ 2⎠

*20.47. Set Up: B = Solve: (a) Br =

μ0 I . B1 = 8.0 μ T when r1 = 2.0 cm. 2π r

⎛ r1 ⎞ μ0 I ⎛ 2.0 cm ⎞ = constant and B1r1 = B2r2 . B2 = B1 ⎜ ⎟ = (8.0 μ T) ⎜ ⎟ = 4.0 μ T. 2π ⎝ 4.0 cm ⎠ ⎝ r2 ⎠

⎛B ⎞ ⎛ 8.0 μ T ⎞ (b) r2 = r1 ⎜ 1 ⎟ = (2.0 cm) ⎜ ⎟ = 16.0 cm. ⎝ 1.0 μ T ⎠ ⎝ B2 ⎠ (c)

⎛I ⎞ B B B μ0 ⎛ 2I ⎞ = = constant and 1 = 2 . B2 = B1 ⎜ 2 ⎟ = (8.0 μ T) ⎜ ⎟ = 16.0 μ T I1 I 2 I I 2π r ⎝ I ⎠ ⎝ 1⎠

Reflect: B is proportional to I and to 1/r. When I is doubled, B is doubled. When r is doubled, B is halved. μ I 20.48. Set Up: B = 0 . μ0 = 4π × 10−7 T ⋅ m/A. 2π r Solve: Assume that the lines are 10 m off the ground. At this distance (4π × 10−7 T ⋅ m/A)(100 A) B= = 2 × 10−6 T. 2π (10 m) This is about 4% of the earth’s magnetic field. The magnetic field is too small to be of concern.

20.49. Set Up: For a long, straight wire B =

μ0 I . If the currents are in the same direction, their fields add. If a pair 2π r

of wires carry current in opposite directions, their fields are in opposite directions and cancel. Solve: (a) μ I (4π × 10−7 T ⋅ m/A)(0.300 A) B=6 0 =6 = 1.44 × 10−7 T 2π r 2π (2.50 m) (b) The fields of two pairs cancel and the net field is due to the remaining two wires. μ I B = 2 0 = 13 (1.44 × 10−7 T) = 4.80 × 1028 T 2π r Reflect: In a typical household appliance cord there are two closely spaced conductors carrying current in opposite directions and the currents in the cord produce very little net magnetic field.


20-16

Chapter 20

20.50. Set Up: The net magnetic field at each point is the vector sum of the fields due to each wire. For each wire G μ I B = 0 and the direction of B is determined by the right-hand rule described in Section 20.7. Let the wire with 2π r 12.0 A be wire 1 and the wire with 10.0 A be wire 2. G μ I (4π × 10−7 T ⋅ m/A)(12.0 A) = 1.6 × 10−5 T. The direction of B1 is out of the page. Solve: (a) Point Q: B1 = 0 1 = 2π r1 2π (0.15 m) B2 =

μ0 I 2 (4π × 10−7 T ⋅ m/A)(10.0 A) = = 2.5 × 10−5 T. 2π r2 2π (0.080 m)

G G G The direction of B2 is out of the page. Since B1 and B2 are in the same direction, B = B1 + B2 = 4.1 × 10−5 T and G B is directed out of the page. Point P: B1 = 1.6 × 10−5 T, directed into the page. B2 = 2.5 × 10−5 T, directed into the page. B = B1 + B2 = 4.1 × 10−5 T G and B is directed into the page. G G (b) B1 is the same as in part (a), out of the page at Q and into the page at P. The direction of B2 is reversed from what it was in (a) so is into the page at Q and out of the page at P. G G G Point Q: B1 and B2 are in opposite directions so B = B2 − B1 = 2.5 × 10−5 T − 1.6 × 10−5 T = 9.0 × 10−6 T and B is directed into the page. G G G Point P: B1 and B2 are in opposite directions so B = B2 − B1 = 9.0 × 10−6 T and B is directed out of the page.

*20.51. Set Up: B =

G μ0 I . The direction of B is given by the right-hand rule in Section 20.7. Call the wires a and 2π r

b, as indicated in the figure below. The magnetic fields of each wire at points P1 and P2 are also shown in the figure.

Solve: (a) At P1, Ba = Bb and the two fields are in opposite directions, so the net field is zero. G μ I μ I G (b) Ba = 0 . Bb = 0 . Ba and Bb are in the same direction so 2π ra 2π rb

μ0 I ⎛ 1 1 ⎞ (4π × 10−7 T ⋅ m/A)(4.00 A) ⎡ 1 1 ⎤ −6 + ⎜ + ⎟= ⎢ ⎥ = 6.67 × 10 T 2π ⎝ ra rb ⎠ 2π ⎣ 0.300 m 0.200 m ⎦ G B has magnitude 6.67 μ T and is directed toward the top of the page. B = Ba + Bb =

Reflect: At points directly to the left of both wires the net field is directed toward the bottom of the page.

20.52. Set Up: B =

G μ0 I . The direction of B is given by the right-hand rule in Section 20.7. 2π r

Solve: (a) The magnetic fields of the two wires are in opposite directions at points between the wires, as shown in Figure (a) below. Consider a point that is a distance x from wire 1 and hence 0.400 m − x from wire 2. B1 = B2 gives

μ0 I1 μ0 I 2 25.0 A 75.0 A = . = . 3.00 x = 0.400 m − x and x = 0.100 m. (0.400 m − x) 2π x 2π (0.400 m − x) x © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


20-17

The net field is zero at a point between the wires, 10.0 cm from the wire carrying 25.0 A and 30.0 cm from the wire carrying 75.0 A.

(b) The magnetic fields of the two wires are in opposite directions at points along the extension of the line connecting the wires and either to the right or to the left of both wires. But for the fields to cancel, they must have equal magnitudes and this happens only at points closer to the wire carrying the smaller current. Therefore, consider a point P that is a distance x to the left of wire 1 and hence 0.400 m + x to the left of wire 2, as shown in Figure (b) above.

μ0 I1 μ0 I 2 25.0 A 75.0 A = . = . (0.400 m + x) 2π x 2π (0.400 m + x) x 3.00 x = 0.400 m + x and x = 0.200 m. The net field is zero at a point 20.0 cm from the wire carrying 25.0 A and 60.0 cm from the wire carrying 75.0 A.

*20.53. Set Up: F =

μ0 lII ′ . μ0 = 4π × 10−7 N/A 2 . Parallel conductors carrying currents in opposite directions 2π r

repel each other. Parallel conductors carrying currents in the same direction attract each other. (4π × 10−7 N/A 2 )(15 m)(25 A)(75 A) Solve: F = = 0.0161 N. Since the currents are in the same direction the force 2π (0.35 m) is attractive. Reflect: The currents are large but the force per meter on each wire is very small.

20.54. Set Up: Label wires 1 and 2. Find the direction of the magnetic field of one wire at the location of the other. Then use the right-hand rule to find the direction of the force on the second wire due to this magnetic field. Solve: Figure (a) below shows the magnetic fields at the wires for the case where the currents are both to the left. The directions of the forces on the wires exerted by these fields are also shown. The wires attract.

Figure (b) above shows the magnetic fields at the wires for the case where one current is to the left and the other is to the right. The directions of the forces on the wires exerted by these fields are also shown. The wires repel.

*20.55. Set Up: Measurements on a typical extension cord give that the centers of the two wires are separated by μ lII ′ about 4 mm. F = 0 . 4π × 10−7 N/A 2 . 2π r


20-18

Chapter 20

Solve: F =

(4π × 10−7 N/A 2 )(2.0 m)(5.0 A) 2 2π (4.0 × 10−3 m)

= 2.5 × 10−3 N. Since the currents are in opposite directions the wires

repel.

20.56. Set Up: The attractive force between two parallel wires is F =

μ0 LI1 I 2 . The power input of the motor is 2π r

65 hp. We know that 1 hp = 746 W. The relation between power, voltage, and current is P = VI . Solve: (a) We can find the current from I =

P (65 hp)(746 W/hp) = = 80.8 A. V 600 V

(4π × 10−7 T ⋅ m/A)(80.8 A) 2 = 2.4 × 10−3 N/m. 2π (0.55 m) Reflect: If the current from the cables is in the same direction, the force will be attractive; however, if the current runs in opposite directions the force will be repulsive. (b) The attractive force between the wires per unit length is F /L =

20.57. Set Up: At the center of a circular loop B =

μ0 I

G . 4π × 10−7 T ⋅ m/A. The direction of B can be found by

2R the right-hand rule: with the thumb of your right hand in the direction of the current your fingers curl around the wire in the direction of the magnetic field. 2 RB 2(0.11 m)(0.50 × 10−4 T) Solve: (a) I = = = 8.8 A. μ0 4π × 10−7 T ⋅ m/A (b) The plane of the loop should be perpendicular to the earth’s field and the current should be directed as shown in the figure below.

Reflect: A large current is required. Note that B =

μ0 I

only applies at the center of the circular loop. The current 2R produces magnetic field at other points but only at the center of the loop do we have a simple expression for the magnitude of the field.

20.58. Set Up: The magnetic field at the center of N circular loops is B = Solve: B =

(4π × 10−7 T ⋅ m/A)(600)(0.500 A) 2(2.00 × 10−2 m)

N μ0 I . 2R

= 9.42 × 10−3 T

*20.59. Set Up: The magnetic field at the center of N circular loops is B = Solve: N =

N μ0 I . 2R

2 RB 2(6.00 × 10−2 m)(6.39 × 10−4 T) = = 24.4. Therefore, 24 turns are required. μ0 I (4π × 10−7 T ⋅ m/A)(2.50 A)



20.60. Set Up: B = Solve: I =

2 RB

μ0

=

μ0 I 2R

20-19

. R = 8 cm. 1 gauss = 1 × 10−4 T.

2(8 × 1022 m)(3.0 × 10−12 T) 4π × 10−7 T ⋅ m/A

= 3.8 × 10−7 A.

20.61. Set Up: Apply Eq. (20.13), where the radius of the coil is R = 0.0240 m. 2aB 2(0.0240 m) (0.0580 T) μ NI Solve: B = 0 , so I = = = 2.77 A. μ0 N (4π × 1027 T ⋅ m/A) (800) 2R Reflect: Despite the large number of turns, a sizable current is necessary to produce the given magnetic field.

20.62. Set Up: Let wire 1 be the inner wire with diameter 20.0 cm and let wire 2 be the outer wire with diameter

G G 30.0 cm. To produce zero net field, the fields B1 and B2 of the two wires must have equal magnitudes and opposite G μ I directions. At the center of a wire loop we have B = 0 . The direction of B is given by the right-hand rule applied 2R to the current direction. ⎛R ⎞ μ I μ I μ0 I1 μ0 I 2 ⎛ 15.0 cm ⎞ Solve: B1 = 0 , B2 = 0 . B1 = B2 gives = and I 2 = ⎜ 2 ⎟ I1 = ⎜ ⎟ (12.0 A) = 18.0 A. The 2 R1 R 2 R2 2 R1 2 R2 ⎝ 10.0 cm ⎠ ⎝ 1⎠ G G directions of I1 and of its field are shown in the figure below. Since B1 is directed into the page we need B2 to be directed out of the page, and so I 2 is counterclockwise.

*20.63. Set Up: The magnetic field at the center of a circular loop is B =

μ0 I 2R

. By symmetry each segment of the

loop that has length Δl contributes equally to the field, so the field at the center of a semicircle is loop. Since the straight sections produce no field at P, the field at P is B = Solve: B =

μ0 I 4R

μ0 I 4R

1 2

that of a full

.

G G . The direction of B is given by the right-hand rule: B is directed into the page.

Reflect: For a quarter-circle section of wire the magnetic field at its center of curvature is B =

20.64. Set Up: B = μ0 nI . n =

μ0 I 8R

.

N 750 coils = = 5000 coils/m. L 0.150 m

Solve: B = (4π × 10−7 T ⋅ m/A)(5000 coils/m)(7.00 A) = 0.0440 T.


20-20

Chapter 20

20.65. Set Up: At the center of a long solenoid B = μ0 nI = μ0 Solve: I =

BL

μ0 N

=

(0.150 T)(1.40 m) (4π × 10−7 T ⋅ m/A)(4000)

= 41.8 A

20.66. Set Up: At the center of a long solenoid B = μ0 Solve: (a) N =

BL

μ0 I

=

N I. L

(0.0279 T)(0.400 m) (4π × 10−7 T ⋅ m/A)(12.0 A)

N I . The total length of wire is N (2π r ). L

= 740 turns

(b) The total length of wire is N (2π r ) = (740)(2π )(1.40 × 1022 m) = 65.1 m.

*20.67. Set Up: The magnetic field at the center of a circular loop is Bloop = of a solenoid is Bsolenoid = μ0 nI , where n = Solve: (a) Bloop =

μ0 I 2R

=

μ0 I 2R

. The magnetic field at the center

N is the number of turns per meter. L

(4π × 10−7 T ⋅ m/A)(2.00 A) N 1000 = 200 m −1. = 2.51 × 10−5 T. (b) n = = 2(0.050 m) L 5.00 m

Bsolenoid = μ0 nI = (4π × 10−7 T ⋅ m/A)(200 m −1)(2.00 A) = 5.03 × 10−4 T. Bsolenoid = 20 Bloop . The field at the center

of a circular loop depends on the radius of the loop. The field at the center of a solenoid depends on the length of the solenoid, not on its radius. Reflect: Eq. (20.14) for the field at the center of a solenoid is only correct for a very long solenoid, one whose length L is much greater than its radius R. We cannot consider the limit that L gets small and expect the expression for the solenoid to go over to the expression for N circular loops.

20.68. Set Up: At the center of a solenoid, Bsolenoid = μ0 nI = μ0 Bwire =

N I . A distance R from a long straight wire, L

μ0 I . 2π R

⎛ 450 ⎞ (1.75 A) = 2.83 × 10−3 T. Solve: (a) Bsolenoid = (4π × 10−7 T ⋅ m/A) ⎜ ⎝ 0.35 m ⎟⎠

(4π × 10−7 T ⋅ m/A)(1.75 A)

= 3.50 × 10−5 T. 2π (1.0 × 10−2 m) The magnetic field due to the wire is much less than the field at the center of the solenoid. For the solenoid, the fields of all the wires add to give a much larger field.

(b) Bwire =

*20.69. Set Up: The magnetic field at the center of N circular loops is B =

μ0 NI 2R

. The magnetic field of the earth

is given to be 0.55 G = 5.5 × 10−5 T. The number of turns can be determined from the length of the wire and the diameter of the coil: N = L /(πD ) Solve: The number of turns is N = L /(π D) =

2(0.22 m)(5.5 × 10−5 T) 18(4π × 10−7 T ⋅ m/A)

2RB 25 m = 18. The required current is I = = μ0 N π (0.44 m)

= 1.1 A.

μ0 NI

N , is determined by the ratio . If a given length of wire 2R R is used to construct the coil, we know that halving R would double the number of turns (N) that we can make — and so the B at the center of the coil would be quadrupled.

Reflect: For a fixed current, the magnetic field, B =



20-21

20.70. Set Up: Figure 20.42 in the text shows that outside a toroidal solenoid there is no magnetic field. According μ NI to Eq. (20.15), inside the toroidal solenoid the magnetic field is given by B = 0 . The torus extends from 2π r r1 = 15.0 cm to r2 = 18.0 cm. Solve: (a) r = 0.12 m, which is outside the torus, so B = 0. μ NI μ (250)(8.50 A) = 2.66 × 10−3 T. (b) r = 0.16 m, so B = 0 = 0 2π r 2π (0.160 m) (c) r = 0.20 m, which is outside the torus, so B = 0. Reflect: The magnetic field inside the torus is proportional to 1/r , so it varies somewhat over the cross-section of the torus.

*20.71. Set Up: The Biot and Savart law is ΔB =

μ0 I Δl sin θ

4π

r2

G . The wire and the vector r from each short

segment to point P are shown in the figure below. For segment A, tan θ A = For segment C, θC = 90° and rC = 5.00 cm.

5.00 cm and θ A = 35.5°. rA = 8.60 cm. 7.00 cm

⎛ 4π × 10−7 T ⋅ m / A ⎞ (10.0 A)(2.00 × 10−3 m)sin 35.5° = 1.57 × 10−7 T. Solve: (a) ΔB = ⎜ ⎟⎟ −2 2 ⎜ 4 π (8 60 10 m) . × ⎝ ⎠ G The right-hand rule says that ΔB is directed out of the page. ⎛ 4π × 10−7 T ⋅ m/A ⎞ (10.0 A)(2.00 × 10−3 m)sin 90° (b) ΔB = ⎜ = 8.00 × 10−7 T. ⎟ 2 −2 4 π (5 00 10 m) . × ⎝ ⎠ G The right-hand rule says that ΔB is directed out of the page. Reflect: Both segments produce magnetic field at P that is in the same direction. At P the field due to segment C is larger than that due to segment A.

μ0 I Δl sin θ G . The wire and the vector r from the short segment 2 4π r to point P are shown in the figure below. r = 0.150 m and θ = 90°.

20.72. Set Up: The Biot and Savart law is ΔB =

⎛ 4π × 10−7 T ⋅ m/A ⎞ (5.00 A)(1.50 × 10−2 m)sin 90° Solve: ΔB = ⎜ = 3.33 × 10−7 T. ⎟ 4π (0.150 m) 2 ⎝ ⎠ G The right-hand rule says that ΔB is directed into the page.


20-22

Chapter 20

20.73. Set Up: Assume that the tightly bound wires have negligible diameters. By symmetry the field lines are circles centered on the wires and the magnitude of the field depends only on the distance from the wires. Ampere’s law says ∑ BiΔs = μ0 I encl . Apply this law to a circular path with radius r = 10 m and with the wires at its center. G Solve: As we move along the chosen circular path B has a constant magnitude and is tangent to the path. If we face southward as we look at the circular path, the net current points toward us and (according to the right-hand rule) the magnetic field circulates in a counterclockwise direction around the circular path. Thus, if we traverse the circular path in the counterclockwise direction we have ∑ B||Δs = B ∑ Δs = B(2π r ), with Iencl = 17.5 A + 11.3 A − 23.0 A = 5.8 A. Ampere’s law therefore gives B(2π r) = μ0 Iencl and B =

μ0 I encl (4π × 10−7 T ⋅ m/A)(5.8 A) = = 1.16 × 10−7 = 0.17 μT. 2π r 2π (10.0 m)

Reflect: We could also obtain this result by using the superposition principle since each current I produces a known μ I magnetic field equal to B = 0 . As we face southward, northbound currents create counterclockwise magnetic field 2π r and southbound currents create clockwise fields. Thus, the net magnetic field has a magnitude equal to μ |I −I | B = 0 north south , with the direction of circulation determined by the larger current. 2π r

20.74. Set Up: By symmetry the magnetic field lines of a cylindrical wire are similar to that of a long, straight conductor—and this is true at points both inside and outside of the conductor. Thus, applying Ampere’s law we μ I obtain ∑ BiΔs = B ∑ Δs = B(2π r ) = μ0 I encl . This reduces to B = 0 encl at points both inside and outside the conductor. 2π r Solve: Outside the wire, where r > R, we have I encl = I , where I is the current carried by the wire. Thus, we have µ0 I at points outside the wire. Inside the wire, where r < R, only a fraction of the total current es 2π r through our path: since the current is uniformly distributed over the cross-sectional area we have Boutside =

2

I encl = I

2

π r2 μ I μ I⎛r⎞ μ Ir ⎛r⎞ = I ⎜ ⎟ . Thus, we have Binside = 0 encl = 0 ⎜ ⎟ = 0 2 at points inside the wire. Note that the 2 2π r 2π r ⎝ R ⎠ πR 2π R ⎝R⎠

magnitude of Boutside decreases as r increases while the magnitude of Binside increases as r increases. Thus, the maximum value of B occurs at the surface where r = R, and we have Bmax = field has half its maximum value when,

μ0 Ir

2π R 2

μ0 I . For points inside the wire, the 2π R

1⎛ μ I ⎞ = ⎜ 0 ⎟ , which gives r = R/2. Outside the wire, we have 2 ⎝ 2π R ⎠

μ0 I 1 ⎛ μ0 I ⎞ = ⎜ ⎟ , which gives r = 2R. 2π r 2 ⎝ 2π R ⎠ Reflect: The magnetic field increases from r = 0 to r = R , and decreases thereafter. Thus, the magnetic field has its maximum value at points on the surface of the wire where r = R. Inside of the wire, the field has half the maximum value at points where r = R / 2, while outside the wire the field has half its maximum value at points where r = 2R.

*20.75. Set Up: μ = K m μ0 , with μ0 = 4π × 10−7 T ⋅ m/A. The magnetic field in the material is a factor of K m greater than it is in vacuum. Solve: (a) μ = (1.00026)(4π × 10−7 T ⋅ m/A) = 1.257 × 10−6 T ⋅ m/A (b) Binside = K m Bexternal = (1.00026)(1.3500 T) = 1.3504 T Reflect: The magnetic field inside the paramagnetic material is larger than the external field, but only slightly larger.

20.76. Set Up: K m = Binside /Boutside . μ = K m μ0 Solve: (a) K =

1.5023 T = 1.0015. 1.5000 T

(b) μ = K m μ0 = (1.0015)(4π × 10−7 T ⋅ m/A) = 1.259 × 10−6 T ⋅ m/A © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


20-23

*20.77. Set Up: Use υ y 2 = υ0 y 2 + 2a y y to calculate υ y . With + y downward, υ0 y = 0, a y = +9.80 m / s 2 and y = 125 m. The direction of the force is given by the right-hand rule and the magnitude is given by F = 0 q 0υ B sin φ .

The charge of the ball is −(4.00 × 108 )e. G G Solve: The directions of υ and B are shown in the figure below, where the downward direction is into the page. The right-hand rule (rhr) direction is north, but since the charge is negative the force is opposite the rhr direction and is to the south. At the bottom of the shaft the speed of the ball is

υ = 2 g (125 m) = 2(9.80 m / s 2 )(125 m) = 49.5 m / s. 0 q 0 = (4.00 × 108 )(1.60 × 10−19 C) = 6.40 × 10−11 C.

φ = 90°, so F = 0 q 0υ B = (6.40 × 10−11 C)(49.5 m/s)(0.250 T) = 7.92 × 10−10 N.

Reflect: The magnetic force is much less than the gravity force on the ball.

G

G

20.78. Set Up: The magnetic force FB must be upward and equal to mg. The direction of FB is determined by the direction of I in the circuit. FB = IlB sin φ , with φ = 90°. I =

V , where V is the battery voltage. R

G Solve: (a) The forces are shown in the figure below. The current I in the bar must be to the right to produce FB upward. To produce current in this direction, point a must be the positive terminal of the battery.

(b) FB = mg . IlB = mg . m =

IlB VlB (175 V)(0.600 m)(1.50 T) = = = 3.21 kg. g Rg (5.00 Ω)(9.80 m/s 2 )

20.79. Set Up: The maximum current I is the current that results in an upward magnetic force FB that equals the weight mg of the bar. I =

V . F = IlB sin φ , with φ = 90°. R

Solve: FB = mg . IlB = mg .

VlB mgR (0.750 kg)(9.80 m/s 2 )(25.0 Ω) = mg . V = = = 817 V. R (0.500 m)(0.450 T) lB

Reflect: The current required to lift the bar is quite large, about 33 A. For a bar of less mass, less current would be required to lift the bar off the s and break the circuit.


20-24

Chapter 20

20.80. Set Up: Choose the y-axis parallel to the straight portion of the wire with the positive y-axis at the top of the page as usual. By symmetry the y-component of force acting on the upper half of the semicircle cancels the y-component of force acting on the lower half of the semicircle. Thus, the net force on the entire wire is parallel to the positive x-axis. Solve: In the figure below a small section of the semicircular portion of the wire is shown. This section has a length ΔL and is short enough to be treated as a straight wire. Since the magnetic field is perpendicular to the plane of the wire we have φ = 90D ; thus, the magnitude of the force on our section is F = IBΔL sin 90D = IBΔL. The x-component of this force is Fx = IBΔL cosθ = IBΔy, where θ is the angle between the force and the positive x-axis (as shown in the figure below). The x-component of net force acting on the entire wire is ∑ Fx = ∑ IBΔy = IB ∑ Δy = (3.40 A)(2.20 T)(3.00 m) = 22.4 N. Reflect: Notice that the curve has no effect on the net force. In other words, the net force is the same as if the wire were simply a straight wire 3.00 m long oriented along the y-axis.

*20.81. Set Up: The ion has charge q = + e. After being accelerated through a potential difference V the ion has kinetic energy qV. The acceleration in the circular path is υ 2 /R. Solve: K = qV = + eV .

υ=

1 mυ 2 2

2eV = m

= eV and

2(1.60 × 10−19 C)(220 V) 1.16 × 10−26 kg

= 7.79 × 104 m/s. FB = 0 q 0υ B sin φ . φ = 90°.

G υ2 G F = ma gives 0 q υ 0 B=m . R

R=

mυ (1.16 × 10−26 kg)(7.79 × 104 m/s) = = 7.81 × 10−3 m = 7.81 mm. 0q0 B (1.60 × 10−19 C)(0.723 T)

20.82. Set Up: The magnetic field of a long straight wire is given by B = a circular loop is B =

μ0 I

μ0 I 2πr

. The magnetic field at the center of

. Let 1 refer to the wire and let 2 refer to the loop. The center of the loop is a distance 2R G r = 3.25 cm from the wire. By the right-hand rule, at the center of the loop the magnetic field B1 due to the wire is G directed into the page and the magnetic field B2 due to the loop is directed out of the page. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Solve: B1 =

20-25

μ0 I1 (4π × 10−7 T ⋅ m/A)(10.0 A) = = 6.15 × 10−5 T. 2π r 2π (3.25 × 10−2 m) B2 =

μ0 I 2 2R

=

(4π × 1027 T ⋅ m/A)(12.0 A) 2(3.25 × 10−2 m)

= 2.32 × 10−4 T.

G G G G Since B1 and B2 are in opposite directions, B = B2 − B1 = 1.70 × 10−4 T. B2 is larger than B1 , so the net field is in G the direction of B2 , out of the page. G

*20.83. Set Up: The direction of the current is shown in the figure below. The direction of B is given by the rightμ I hand rule. B = 0 . 2π r

Solve: (a) Below the line the magnetic field is toward the east. (4π × 10−7 T ⋅ m/A)(800 A) B= = 2.91 × 10−5 T. 2π (5.50 m) (b) The field due to the current is the same order of magnitude as the earth’s field so it is problem.

20.84. Set Up: Calculate the magnetic field of the wire at the location of the electron. Then calculate the force this field exerts on the electron. The electron has negative charge so the force on it is opposite to the right-hand rule direction. Solve: The wire and electron are shown in Figure (a) below. At the electron, the field of the wire has magnitude μ I (4π × 10−7 T ⋅ m/A)(2.50 A) B1 = 0 = = 1.11 × 10−5 T 2π r 2π (4.50 × 10−2 m) and is directed out of the page. By the right-hand rule the force on the electron is to the right. The force has magnitude F = 0 q 0 vB sin φ = (1.60 × 10−19 C)(6.00 × 104 m/s)(1.11 × 10−5 T)sin 90° = 1.07 × 10−19 N.

(b) The wire and electron are shown in Figure (b) above. The magnitude of the force is the same as in part (a), 1.07 × 10219 N. The force is directed away from the wire.

20.85. Set Up: B =

G μ0 I . The direction of B is given by the right-hand rule. 2π r


20-26

Chapter 20

Solve: (a) The currents are the same so points where the two fields are equal in magnitude are equidistant from the two wires. The net field is zero along the dashed line shown in Figure (a) below.

(b) For the magnitudes of the two fields to be the same at a point, the point must be 3 times closer to the wire with the smaller current. The net field is zero along the dashed line shown in Figure (b) above. (c) As in (a), the points are equidistant from both wires. The net field is zero along the dashed line shown in Figure (c) above. Reflect: The lines of zero net field consist of points at which the fields of the two wires have opposite directions and equal magnitudes.

20.86. Set Up: Since the magnetic force is always perpendicular to the velocity of the object, a charged particle mυ

moving perpendicular to the field travels in an arc of a circle with radius R =

. 0q0 B Solve: (a) The direction of curvature of the path tells us the direction of the force on the particle. If this is the same G G as the right-hand direction for υ and B, then the particle has positive charge it is opposite to this direction then the particle has negative charge. (b) At any point the momentum of the electron is given by mυ = R 0 q 0 B. A measurement of the radius of curvature, R,

of the path gives the momentum of the particle. If the mass is known then the speed is momentum divided by mass. mυ (c) The electron is slowing down and by R = this causes the radius of curvature of the path to decrease. The 0q0 B electron loses kinetic energy as it moves through the liquid. (d) A neutron has no charge. The magnetic field exerts no force on it, so it moves in a straight line.

*20.87. Set Up: The current in the bar is horizontal toward the bottom of the figure, so the magnetic force on it is vertically upwards (out of the page). The net force on the bar is equal to the magnetic force minus the gravitational force, so Newton’s second law gives the acceleration. The bar is in parallel with the 10.0-Ω resistor, so we must use circuit analysis to find the initial current through it. First find the current. The equivalent resistance across the battery is 30.0 Ω, so the total current is 4.00 A, half of which goes through the bar. Applying Newton’s second law to the bar gives ∑ F = ma = FB − mg = iLB − mg .



20-27

Solve: Solving for the acceleration gives iLB − mg (2.0A)(1.50 m)(1.60 T) − 3.00 N a= = = 5.88 m/s 2 . m (3.00 N/9.80 m/s 2 )

The direction is upward. Reflect: Once the bar is free of the conducting wires, its acceleration will become 9.8 m/s 2 downward since only gravity will be acting on it.

20.88. Set Up: The wires repel each other since they carry currents in opposite directions, so the wires will move away from each other until the magnetic force is just balanced by the force due to the spring. Call x the distance the μ I 2L . springs each stretch. The force of the spring is kx and the magnetic force on each wire is Fmag = 0 2π x Solve: On each wire, Fspr = Fmag , and there are two spring forces on each wire. Therefore 2kx =

μ0 I 2 L , which 2π x

μ0 I 2 L . 2π k Reflect: Since μ0 /2π is small, x will probably be much less than the length of the wires.

gives x =

20.89. Set Up: R =

mυ 2π R . The time T is given by T = . υ 0q0 B

2π ⎛ mυ ⎞ 2π m . ⎜ ⎟= υ ⎝ 0q0 B ⎠ 0q0 B (b) The time for one revolution is the same for both. An ion moving in a larger circle is traveling faster. (c) To complete one revolution in the same time, an ion moving in a larger circle must travel faster. We can also see mυ . this from R = 0q0 B Solve: (a) T =

(d) From R =

mυ 0 q 0 BR we get υ = . m 0q0 B

Reflect: Since the orbital period depends only on B and on the nature of the ion (m/ 0 q 0 ) and not on the speed (and

radius of the path) of the ion, the rate at which the electric field is pulsed does not have to change as the ions are accelerated.

20.90. Set Up: 1 eV = 1.60 × 10−19 J. The energy of the ions and their speed are related by K = 12 mυ 2 . The mass of a proton is 1.67 × 10−27 kg. From Problem 20.93, the speed υ of the particle and the magnetic field B are related qBR . The radius R is related to the circumference C by C = 2π R. m ⎛ 1.60 × 10−19 J ⎞ −13 J. K = 12 mυ 2 gives Solve: (a) K = 1.25 MeV = (1.25 × 106 eV) ⎜ ⎟⎟ = 2.00 × 10 ⎜ 1 eV ⎝ ⎠

by υ =

υ= (b) υ =

2K 2(2.00 × 10−13 J) = = 1.55 × 107 m / s m 1.67 × 10−27 kg

qBR mυ C 6.4 × 103 m = = 1.02 × 103 m. gives B = . R= m 2π 2π qR

B=

(1.67 × 10−27 kg)(1.55 × 107 m / s) (1.60 × 10219 C)(1.02 × 103 m)

= 1.59 × 10−4 T.


20-28

Chapter 20

*20.91. Set Up: Turning the charged loop creates a current, and the external magnetic field exerts a torque on that current. The current is I = q /T = q /(1/f ) = qf = q (ω /2π ) = qω /2π . The torque is τ = μ B sin φ . Solve: In this case, φ = 90° and μ = AB, giving τ = IAB. Combining the results for the torque and current and

⎛ qω ⎞ 2 2 1 using A = π r 2 gives τ = ⎜ ⎟ π r B = 2 qω r B ⎝ 2π ⎠ Reflect: Any moving charge creates a current—so turning the loop creates a current, which in turn causes a magnetic field.

20.92. Set Up: Label the three segments in the magnetic field 1, 2, and 3, as shown in Figure (a) below. The force on a current carrying conductor is F = IlB sin φ , where φ is the angle between the direction of the current and the direction of the magnetic field. The direction of the force on each segment is given by the right-hand rule and is G G G shown in the figure. The sum of F1 and F3 is the same as the force F13 on a wire 0.307 m long. Section 2 has length 0.800 m. The current in each segment is perpendicular to the magnetic field, so φ = 90°.

Solve: F13 = IlB sin φ = (6.50 A)(0.307 m)(0.280 T)sin 90° = 0.559 N.

F2 = IlB sin φ = (6.50 A)(0.800 m)(0.280 T)sin 90° = 1.46 N. The forces and a coordinate system are shown in G Figure (b) above. F2 has been resolved into its x and y components. Fx = F2 x + F13 x = 2F2 cos60.0° = −(1.46 N)(cos60.0°) = −0.730 N.

Fy = F2 y + F13 y = F2 sin 60.0° + F13 = +(1.46 N)(sin 60.0°) + 0.559 N = +1.83 N. G Fx , Fy and the resultant total force F are shown in Figure (c). The resultant force has magnitude 1.97 N and is at 68.3° clockwise from the left-hand straight segment.

Solutions to age Problems *20.93. Set Up: The torque on a magnetic dipole is given by τ = μ B sin φ . The magnetic moment of the proton is 1.4 × 10−26 J/T = 1.4 × 10−26 N ⋅ m/T. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


20-29

Solve: Using the given values we have τ = μ B sin φ = (1.4 × 10−26 N ⋅ m/T)(2 T)sin 90D = 2.8 × 10−26 N ⋅ m. Thus, the

correct answer is C.

20.94. Set Up: To be a candidate for MRI, the net magnetic moment of the nucleus must not be zero. Solve: Since the protons and neutrons will tend to pair with opposite orientations to their magnetic moments, only nuclei with an odd number of nucleons (protons and neutrons) are candidates for MRI. The only one of the listed isotopes with an odd number of nucleons is 31 P15 . The correct answer is D.


21

ELECTROMAGNETIC INDUCTION

Answers to Multiple-Choice Problems 1. A 14. C

2. B 3. B 15. A

4. B

5. B

6. A

7. C

8. A

9. C

10. D

11. C

12. A

13. D

Solutions to Problems 21.1. Set Up: Φ B = BA cos φ . The normal to the loop is in the z direction. The loop has area A = π r 2 = π (6.50 × 10−2 m) 2 = 1.33 × 10−2 m 2 .

Solve: (a) φ = 0°. Φ B = BA = (0.230 T)(1.33 × 10−2 m 2 ) = 3.06 × 10−3 Wb. (b) φ = 53.1°. Φ B = BA cos53.1° = (3.06 × 103 Wb)cos53.1° = 1.84 × 10−3 Wb. (c) φ = 90°. ΦB = 0. Reflect: The flux is a maximum when the magnetic field is perpendicular to the plane of the loop and the flux is zero when the magnetic field is parallel to the plane of the loop.

G

21.2. Set Up: ΦB = BA cos φ , where φ is the angle between B and the normal to the surface. Solve: (a) The normal to the surface is parallel to the x axis. φ = 90° and Φ B = 0. (b) The normal to the surface is parallel to the z axis and φ = 0°. Φ B = (0.128 T)(0.300 m) 2 = 0.0115 Wb. The flux

is into the enclosed volume.

G (c) The normal to the surface and the direction of B are shown in the figure below. 40.0 cm tan φ = and φ = 53.1°. 30.0 cm Φ B = BA cos φ = (0.128 T)(0.300 m)(0.500 m)cos53.1° = 0.0115 Wb. The flux is out of the shaded volume. (d) The net flux is zero, since the magnitude of the flux into the volume equals the magnitude of the flux out of the volume.


21-1

21-2

Chapter 21

*21.3. Set Up: The total flux through the bottle is zero because it is a closed surface. The total flux through the bottle is the flux through the plastic plus the flux through the open cap, so the sum of these must be zero. Φ plastic + Φ cap = 0. Φ plastic = −Φ cap = − B A cos ϕ = − B (π r 2 )cos ϕ Solve: Substituting the numbers gives Φ plastic = −(1.75 T )π (0.0125 m) 2 cos 25D = −7.8 × 10−4 Wb Reflect: It would be impossible to calculate the flux through the plastic directly because of the complicated shape of the bottle, but with a little thought we can find this flux through a simple indirect calculation.

21.4. Set Up: E =

ΔΦB . Φ B = BA cos φ . φ = 0°, A is constant and B is changing. Δt

ΔB = (0.0900 m 2 )(0.190 T/s) = 0.0171 V. Δt E 0.0171 V = 0.0285 A. (b) I = = R 0.600 Ω

Solve: (a) E = A

*21.5. Set Up: E = N Solve: E = N

ΔΦB . ΦB = BA cos φ . φ = 0°, A is constant and B is changing. Δt

ΔΦB ΔB E NA ΔB/Δt = NA .I= = . Δt Δt R R

ΔB RI (40.0 Ω)(0.150 A) = = = 10.6 T/s. NA (200)π (3.00 × 10−2 m) 2 Δt

21.6. Set Up: E =

ΔΦB . ΦB = BA cos φ . φi = 0°. φf = 90°. ΦB is the flux through each turn of the coil. Δt

Solve: (a) ΦB,i = BA cos0° = (6.0 × 10−5 T)(12 × 10−4 m 2 )(1) = 7.2 × 10−8 Wb. The total flux through the coil is

N ΦB,i = (200)(7.2 × 10−8 Wb) = 1.44 × 10−5 Wb. ΦB ,f = BA cos90° = 0. (b) E =

N Φi − N Φf 1.44 × 10−5 Wb = = 3.6 × 10−4 V = 0.36 mV. Δt 0.040 s

21.7. Set Up: E =

G ΔΦB . ΦB = BA cos φ . φ is the angle between the normal to the loop and B , so Δt

φi = 90.0° − 37.0° = 53.0° and φf = 0°. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Electromagnetic Induction

21-3

NBA cos φf − cos φi (80)(1.10 T)(0.250 m)(0.400 m) = 0 cos0° − cos53.0°0 = 58.4 V. Δt 0.0600 s Reflect: The flux changes because the orientation of the coil relative to the magnetic field changes. Solve: E =

21.8. Set Up: E = Solve: E =

ΔΦB . ΦB = BA cos φ . Bi = μ0 nI and Bf = 0. φ = 0°. Δt

2μ0 nIA 2(4π × 10−7 T ⋅ m/A)(40 × 102 turns/m)(0.250 A)(6.00 × 10−4 m 2 ) = = 3.02 × 10−5 V = 30.2 μ V Δt 0.0500 s

*21.9. Set Up: Since the field is uniform, ΦB = BA cosφ . Since the plane of the circuit is perpendicular to the field, ΔB ΔB ΔΦB and = 0.25 T/s. The current in . A is constant and B is changing, so E = A Δt Δt Δt the circuit is given by Ohm’s law, E = IR. E 0.045 V Solve: E = (0.300 m)(0.600 m)(0.25 T/s) = 0.045 V. I = = = 3.0 mA. R 15 Ω Reflect: The induced emf depends on the rate of change of the flux so it depends on the rate at which B is changing. It does not depend on the initial magnitude of the field.

φ = 0° and cos φ = 1. E =

21.10. Set Up: Apply Faraday’s law. Since the loop initially rests on the table and the magnetic field is perpendicular to the table we may assume that the initial magnetic flux is simply ΦB = EAcosφ = EA. The final flux is zero since the loop is removed from the field. Solve: (a) E =

ΔΦB 0 − (1.5 T)π (0.120 m) 2 = = 34 V. Δt 2.0 × 10−3 s

(b) According to Lenz’s law the induced current will attempt to maintain the upward magnetic field in the loop; thus, the induced current is counterclockwise. Reflect: The shorter the removal time, the larger the average induced emf.

*21.11. Set Up: The maximum emf is produced when the opposite sides of the coil are moving perpendicular to the field with speed υ = ω ( L /2). The maximum emf induced in each of the two sides on one turn is BLυ . The maximum total emf is 2NBLυ = NBL2ω . Solve: (a) E max = NBL2ω . ω =

E max

=

(20.0 × 10−3 V)

= 3.70 rad/s. NBL (15)(0.0250 T)(0.120 m)2 (b) The emf changes polarity as the coil rotates and the average emf is zero. Reflect: The maximum emf is proportional to the angular velocity of the coil. As ω increases, the flux through the coil changes more rapidly. 2

21.12. Set Up: The field of the induced current is directed to oppose the change in flux in the primary circuit. Solve: (a) The magnetic field in A is to the left and is increasing. The flux is increasing so the field due to the induced current in B is to the right. To produce magnetic field to the right, the induced current flows through R from right to left. (b) The magnetic field in A is to the right and is decreasing. The flux is decreasing so the field due to the induced current in B is to the right. To produce magnetic field to the right the induced current flows through R from right to left. (c) The magnetic field in A is to the right and is increasing. The flux is increasing so the field due to the induced current in B is to the left. To produce magnetic field to the left the induced current flows through R from left to right. (d) The magnetic field of A is constant so the flux through B is constant and there is no induced emf and no induced current.

*21.13. Set Up: The field of the induced current is directed to oppose the change in flux.


21-4

Chapter 21

Solve: (a) The field is into the page and is increasing so the flux is increasing. The field of the induced current is out of the page. To produce field out of the page the induced current is counterclockwise. (b) The field is into the page and is decreasing so the flux is decreasing. The field of the induced current is into the page. To produce field into the page the induced current is clockwise. (c) The field is constant so the flux is constant and there is no induced emf and no induced current.

21.14. Set Up: The field of the induced current is directed to oppose the change in flux in coil B. When S is closed the magnetic field of coil A is directed to the right. The field of A is stronger at points closer to A. The field of A is proportional to the current in A. Solve: (a) The flux due to field directed to the right decreases (goes to zero) when S is opened. The field due to the induced current is to the right inside of coil B. To produce field in this direction, the induced current es through the resistor from a to b. (b) The flux through coil B due to the field of A increases, so the field due to the induced current is to the left inside of coil B. To produce field in this direction, the induced current es through the resistor from b to a. (c) Decreasing R causes the current in coil A to increase, so the field of A increases. The flux through coil B due to the field of A increases, so the field due to the induced current is to the left inside of coil B. To produce field in this direction, the induced current es through the resistor from b to a.

21.15. Set Up: Use the right-hand rule to determine the direction of the magnetic field in the solenoid and use Lenz’s law to determine the direction of the induced current in the ring. Solve: Based on the figure we can deduce that the magnetic field of the solenoid points toward the ring. As the solenoid moves upward, the magnitude of the flux in the ring increases. As seen from the solenoid, the induced current must then move counterclockwise to oppose the increase in flux. If the solenoid stops moving toward the ring and its current decreases, the magnetic flux in the ring will begin to decrease. To oppose this, the induced current in the ring will now flow in the clockwise direction as seen from the solenoid. Reflect: When the flux in the ring is increasing, the currents in the ring and solenoid are in opposite directions—this creates a repulsive magnetic force as expected by Lenz’s law. Similarly, when the flux in the ring is decreasing, the currents in the ring and solenoid are in the same direction, which produces an attractive magnetic force.

21.16. Set Up: The magnetic force on a + charge in the moving bar is given by the right-hand rule. The direction of the current is in the direction of this force. Lenz’s law says the field of the induced current opposes the change in flux through the circuit. G G Solve: (a) The directions of υ and B are shown in the figure below. The magnetic force is toward the top of the bar, so the induced current is from b to a through the bar and counterclockwise in the circuit.

(b) The magnetic field is into the page and the flux is increasing as the area of the circuit increases. The magnetic field of the induced current therefore is out of the page inside the circuit. To produce a magnetic field in this direction the induced current must be counterclockwise.



21-5

21.17. Set Up: Use the right-hand rule to find the direction of the magnetic field due to the long wire at the location of each loop. Lenz’s law says that the magnetic field of the induced current is directed to oppose the change in flux through the circuit. Solve: The magnetic field of the long wire is directed out of the page at C and into the page at A. When the current increases, the magnetic field increases. Therefore, the magnetic field of the induced current in loop C is directed into the page inside the loop, to oppose the increase in flux out of the page due to the current in the long wire. To produce magnetic field in this direction, the induced current in C is clockwise. The magnetic field of the induced current in loop A is directed out of the page inside the loop, to oppose the increase in flux into the page due to the current in the long wire. To produce magnetic field in this direction the induced current in A is counterclockwise. Reflect: If the direction of I were reversed the directions of the induced currents would reverse. If I were decreasing rather than increasing, the directions of the induced currents would reverse.

21.18. Set Up: The magnetic field of the bar magnet is directed into the S pole so is toward the top of the page in the figure in the problem. The loop and magnetic field of the bar magnet, as viewed from below the loop, are shown in the figure below. Lenz’s law says that the magnetic field of the induced current is directed to oppose the change in flux through the circuit.

Solve: (a) The strength of the magnetic field decreases with distance from the coil. When the coil is dropped it moves away from the bar magnet and the flux through the coil decreases and to oppose this decrease the flux due to the induced current is into the page. To produce magnetic field inside the coil that is into the page, the induced current must be clockwise. (b) When the magnet is dropped, it moves toward the coil and the flux through the coil increases. To oppose this increase, the flux due to the induced current is out of the page. To produce magnetic field inside the coil that is out of the page the induced current must be counterclockwise. (c) Both objects fall at the same rate. The distance between them doesn’t change. The flux through the coil doesn’t change and there is no current induced in the coil.

*21.19. Set Up: By Lenz’s law, the induced current flows to oppose the flux change that caused it. Solve: The magnetic field is outward through the round coil and is decreasing, so the magnetic field due to the induced current must also point outward to oppose this decrease. Therefore the induced current is counterclockwise. Reflect: Careful! Lenz’s law does not say that the induced current flows to oppose the magnetic flux. Instead it says that the current flows to oppose the change in flux.

21.20. Set Up: For a bar magnet the magnetic field is directed into the south pole and out of the north pole. The loop and the current induced in it, as viewed by the observer, are shown in the figure below. Lenz’s law says that the magnetic field of the induced current is directed to oppose the change in flux through the loop.


21-6

Chapter 21

Solve: The magnetic field of the induced current is directed out of the page inside the loop. The magnitude of the flux decreases when the bar magnet moves away from the loop. Since the induced flux tends to oppose this decrease, the flux due to the bar magnet is directed out of the page. The magnetic field of the bar magnet is away from pole C, so pole C is a north pole and pole A is a south pole.

*21.21. Set Up: The bar and magnetic field are shown in the figure below. Vab = υ BL. To determine which end is at higher potential consider the direction of the magnetic force on a positive charge in the moving rod. Accumulation of positive charge produces higher potential.

Solve: (a) Vab = υ BL = (11.5 m/s)(1.22 T)(0.150 m) = 2.10 V. (b) The magnetic force on a positive charge in the bar is eastward. Positive charge accumulates at end a of the bar and that end (the east end) is at higher potential. (c) If the bar is moving east to west the force is to the north. The force is perpendicular to the bar and there is no charge accumulation at its ends; the potential difference between its ends is zero. Reflect: If the bar moves parallel to the magnetic field there is no magnetic force and no potential difference is produced.

21.22. Set Up: The rod moves perpendicular to the magnetic field, so Eq. (21.6) applies. To determine which end is at higher potential, consider the direction of the magnetic force on a positive charge in the moving rod. The end where positive charge accumulates is the higher potential end. V 0.450 V = 0.833 T. (a) Vab = υ BL so B = ab = υ L (4.50 m/s)(0.120 m) (b) The rod is sketched in Figure (a) below. The magnetic force on a positive charge in the moving rod is shown. The G G right-hand rule applied to υ and B says this force is toward end b and this end is at higher potential. (c) The rod is sketched in Figure (b) below. The magnetic force now is perpendicular to the rod. There is no charge accumulation at its ends and no induced emf.



21-7

*21.23. Set Up: There will be a motional emf between the underbody and roof of your car due to the motion of your car through the earth’s magnetic field. Follow the derivation of Equation 21.7—except that the velocity of the car is not perpendicular to the magnetic field. Thus, the magnetic force on charges within the body of your car is given by FB = q υ B sinφ , which leads to E = EL = υ BL sinφ . The speed of the car is 95 km/h = (95 km/h) (103 m/km)(1 h/3600 s) = 26.39 m/s.

Solve: E = υ BL sin φ = (26.39 m/s)(5.5 × 10−5 T)(1.5 m)sin35D = 1.248 × 10−3 V = 1.2 mV. Using the right-hand rule

we conclude that an electron in the body of the car will experience a downward force (while the car moves in the given direction). This implies that the roof of the car will be at a higher potential than the underbody. Reflect: The maximum motional emf occurs when the velocity of the car is perpendicular to the direction of the magnetic field. For example, if the car were moving due east, the induced emf would be E = υ BL sin φ = (26.39 m/s) (5.5 × 10−5 T)(1.5 m)sin90D = 2.2 mV.

21.24. Set Up: We must modify the derivation of the equation E = υ BL, which assumes that the velocity of the bar is perpendicular to the axis of the bar (leading to a magnetic force that is parallel to the axis of the bar). When the velocity of the bar is not perpendicular to the bar, it produces a magnetic force that is no longer aligned with the bar. The component of the magnetic force that is parallel to the bar is qυ B sin θ , where θ is the angle between the axis of the bar and its velocity. Assuming that an equal electric force balances the parallel component of the magnetic force we obtain a motional emf given by E = υ BL sin θ . Solve: (a) E = υ BL sin θ = (2.50 m/s)(1.20 T)(1.41 m)(sin 37.0D ) = 2.55 V, with a at the higher potential because

positive charges are pushed toward that end. (b) Same as (a) except θ = 53.0°, giving 3.38 V, with a at the higher potential. (c) Zero, since the velocity is parallel to the magnetic field. (d) The bar must move perpendicular to its length, for which the emf is 4.23 V. For Vb > Va , it must move upward and to the left (toward the second quadrant) perpendicular to its length. Reflect: The orientation of the bar affects the potential induced across its ends.

21.25. Set Up: E = υ BL. Use Lenz’s law to determine the direction of the induced current. The force Fext required to maintain constant speed is equal and opposite to the force FI that the magnetic field exerts on the rod because of the current in the rod. Solve: (a) E = υ BL = (7.50 m/s)(0.800 T)(0.500 m) = 3.00 V G (b) B is into the page. The flux increases as the bar moves to the right, so the magnetic field of the induced current is out of the page inside the circuit. To produce magnetic field in this direction the induced current must be counterclockwise, so from b to a in the rod. G E 3.00 V = 2.00 A. FI = ILB sin φ = (2.00 A)(0.500 m)(0.800 T)sin 90° = 0.800 N. FI is to the left. To (c) I = = R 1.50 Ω keep the bar moving to the right at constant speed an external force with magnitude Fext = 0.800 N and directed to the right must be applied to the bar.


21-8

Chapter 21

Reflect: The force on the rod due to the induced current is directed to oppose the motion of the rod. This agrees with Lenz’s law.

21.26. Set Up and Solve: (a) Each slab of flowing blood has maximum width d and is moving perpendicular to the field with speed υ. E = υ BL becomes E = υ Bd . (b) B =

1.0 × 10−3 V E = = 1.3 T υ d (0.15 m/s)(5.0 × 10−3 m)

(c) The blood vessel has cross sectional area A = π d 2 /4. The volume of blood that flows past a cross section of the

vessel in time t is π (d 2 /4)υt . The volume flow rate is volume/time = R = π d 2υ /4.

υ=

π d 2 ⎛ E ⎞ π dE E so R = . ⎜ ⎟= Bd 4 ⎝ Bd ⎠ 4 B

*21.27. Set Up: Let solenoid 1 have 100 turns and solenoid 2 have 500 turns. M =

N 2ΦB 2 . The field within the i1

μ0 N1 i1 . 2π r μ NiA Solve: ΦB 2 = B1 A = 0 1 1 . 2π r μ N N A (4π × 10−7 Wb/m ⋅ A)(100)(500)(4.00 × 10−4 m2 ) = 4.00 × 10−5 H. M= 0 1 2 = 2π r 2π (0.100 m) Reflect: If we interchange the roles of coils 1 and 2 we calculate the same value of M.

turns of solenoid 1 is B1 =

21.28. Set Up: Example 21.8 derives that for this configuration of coils, M =

μ0 N1 N 2 A l

where l is the length of

coil 1. Solve: M =

(4π × 10−7 Wb/m ⋅ A)(800)(50)π (0.200 × 10−2 m)2 = 6.32 × 10−6 H = 6.32 μ H 0.100 m

21.29. Set Up: E1 = M

Δi2 Δt

and E 2 = M

N Φ Δi1 . M = 2 B 2 , where ΦB 2 is the flux through one turn of the i1 Δt

second coil. Solve: (a) M = (b) ΦB 2 =

E2 0 Δi1/Δt 0

=

1.65 × 10−3 V = 6.82 × 10−3 H = 6.82 mH 0.242 A/s

Mi1 (6.82 × 10−3 H)(1.20 A) = = 3.27 × 10−4 Wb N2 25

Δi2 = (6.82 × 10−3 H)(0.360 A/s) = 2.46 × 10−3 V = 2.46 mV Δt Reflect: We can express M either in of the total flux through one coil produced by a current in the other coil, or in of the emf induced in one coil by a changing current in the other coil. (c) E1 = M

21.30. Set Up: The emf E 2 in solenoid 2 produced by changing current i1 in solenoid 1 is given by E 2 = M The mutual inductance of two such solenoids is derived in Example 21.8 to be M =

μ0 AN1 N 2 l

Δi1 . Δt

, where A and l are

the cross-sectional area and length of solenoid 1. μ0 = 4π × 10−7 T ⋅ m/A.



Solve: (a) M = (b) E 2 =

21-9

(4π × 10−7 T ⋅ m/A)π (6.00 × 10−4 m) 2 (6750)(15) = 2.88 × 10−7 H = 0.288 μ H 0.500 m

Δi1 = (2.88 × 10−7 H)(37.5 A/s) = 1.08 × 10−5 V Δt

*21.31. Set Up: Apply Equation 21.11. Solve: (a) M =

N 2ΦB 2 400(0.0320 Wb) = = 1.96 H 6.52 A i1

N1ΦB1 Mi (1.96 H)(2.54 A) so ΦB1 = 2 = = 7.11 × 10−3 Wb 700 i2 N1 Reflect: M relates the current in one coil to the flux through the other coil. Equation 21.11 shows that M is the same for a pair of coils, no matter which one has the current and which one has the flux. (b) M =

21.32. Set Up: The inductance L is related to the flux ΦB through one turn by N ΦB = Li. The voltage across the coil is related to the rate at which the current in it is changing by E = L

Δi . Δt

Li (4.5 × 10−3 H)(11.5 A) = = 4.1 × 10−4 Wb 125 N Δi E 1.16 V (b) = = = 258 A/s Δt L 4.5 × 10−3 H Solve: (a) ΦB =

*21.33. Set Up: E = L Solve: L =

Δi Δt

0.0160 V E = = 0.250 H Δi/Δt 0.0640 A/s

21.34. Set Up: E = L

Δi Δt

Solve: E = (0.260 H)(18.0 × 10−3 A/s) = 4.68 × 10−3 V = 4.68 mV

21.35. Set Up: Example 21.10 shows that the inductance of a toroidal solenoid is L = the coil is related to the rate at which the current in it is changing by E = L Solve: (a) N =

μ0 N 2 A . The voltage across 2π r

Δi . Δt

2π rL 2π (0.0600 m)(2.50 × 10−3 H) = = 1940 turns μ0 A (4π × 10−7 T ⋅ m/A)(2.00 × 10−4 m 2 )

Δi E 2.00 V = = = 800 A/s Δt L 2.50 × 10−3 H Reflect: The inductance is determined solely by how the coil is constructed. The induced emf depends on the rate at which the current through the coil is changing. (b)

21.36. Set Up: L =

N ΦB N . The magnetic field inside a long solenoid is uniform and equal to B = μ0 i. i l

Solve: ΦB = BA. L =

N ΦB μ0 N 2 A = . i l


21-10

Chapter 21

Reflect: The self-inductance depends only on the geometrical parameters of the coil and is independent of the current that flows in the coil.

*21.37. Set Up: Combine the two expressions for L: L = N ΦB / i and L = E/(Δi/Δt ) , where ΦB is the average flux through one turn of the solenoid. (12.6 × 10−3 V)(1.40 A) = 238 turns. (0.00285 Wb)(0.0260 A/s) Reflect: The induced emf depends on the time rate of change of the total flux through the solenoid. Solve: Solving for N we have N = Ei/[Φ B (Δi/Δt )] =

21.38. Set Up: A transformer transforms voltages according to circuit of resistance R is Reff =

R ( N 2 /N1 ) 2

V2 N 2 = . The effective resistance of a secondary V1 N1

. N 2 = 275 and V1 = 25.0 V.

Solve: (a) V2 = V1( N 2 /N1) = (25.0 V)(834/275) = 75.8 V (b) Reff =

R ( N 2 /N1)

2

=

125 Ω (834/275) 2

= 13.6 Ω

*21.39. Set Up: A transformer transforms voltages according to circuit of resistance R is Reff =

R ( N 2 /N1)

V2 N 2 = . The effective resistance of a secondary V1 N1

. Resistance R is related to P and V by P = 2

V2 . Conservation of energy R

requires P1 = P2 so V1 I1 = V2 I 2 . Solve: (a) V1 = 240 V and we want V2 = 120 V, so use a step-down transformer with N 2 /N1 = 12 . (b) P = VI , so I =

P 1600 W = = 6.67 A. 240 V V

(c) The resistance R of the blower is R = Reff =

9.00 Ω (1/2) 2

V 2 (120 V) 2 = = 9.00 Ω. The effective resistance of the blower is 1600 W P

= 36.0 Ω.

Reflect: I 2 V2 = (13.3 A)(120 V) = 1600 W. Energy is provided to the primary at the same rate that it is consumed in

the secondary. Step-down transformers step up resistance and the current in the primary is less than the current in the secondary.

21.40. Set Up: For an ideal transformer, the input power is equal to the output power. We know that the relation between power, voltage, and current is given by P = VI . Ohm’s law, V = IR, gives the relation between current and resistance on both the input and output of the transformer. P 15 W Solve: (a) Assuming that the output power is 15 W, we have I = = = 2.5 A. V 6.0 V V 6.0 V = 2.4 Ω. (b) Since we know both the voltage of the secondary circuit and its current, we have R = = I 2.5 A P 15 W (c) Assuming that the input power is 15 W, we have I = = = 0.068 A. Thus, the equivalent resistance of the V 220 V V 220 V input circuit is R = = = 3200 Ω. I 0.068 A E 2 6.0 V N 2 = = = 0.027. Reflect: We could also solve this problem by finding the ratio of the number of turns: E1 220 V N1



Thus, we can find the equivalent resistance of the input by using Equation 21.17: R1 =

*21.41. Set Up: V1 = 120 V. V2 = 13,000 V. P = VI and P1 = P2 . Solve: (a)

2.4 Ω (0.27) 2

21-11

= 3300 Ω.

N1 V1 = . N 2 V2

N 2 V2 13,000 V = = = 108 120 V N1 V1

(b) P = V2 I 2 = (13,000 V)(8.50 × 10−3 A) = 110 W

P 110 W = = 0.917 A V1 120 V Reflect: Since the power supplied to the primary must equal the power delivered by the secondary, in a step-up transformer the current in the primary is greater than the current in the secondary. (c) I1 =

21.42. Set Up: Example 21.10 shows that the inductance of a toroidal solenoid is L =

μ0 N 2 A . The energy stored 2π r

in an inductor is U = 12 LI 2 . ⎛ μ N2A ⎞ 2 4π rU 4π(0.150 m)(0.390 J) = = 2850. Solve: U = 12 ⎜ 0 I and N = 2 −7 ⎜ 2π r ⎟⎟ μ0 AI (4π × 10 Wb/m ⋅ A)(5.00 × 10−4 m 2 )(12.0 A)2 ⎝ ⎠ Reflect: The stored energy is proportional to the square of the number of turns even though the magnetic field within the solenoid is directly proportional to the number of turns. We will see in Section 21.10 that the magnetic field energy depends on the square of the magnetic field.

21.43. Set Up: U = 12 LI 2 Solve: (a) U = 12 (10.2 × 10−3 H)(1.15 A) 2 = 6.74 mJ (b) I =

2U 2(1.0 J) = = 14 A. This is not a reasonable current in a circuit; it is too large. L 10.2 × 10−3 H

21.44. Set Up: U = 12 LI 2 . The self-inductance of a solenoid is found in Problem 21.37 to be L =

μ0 AN 2 l

. The

length l of the solenoid is the number of turns divided by the turns per unit length. 2U 2(10.0 J) Solve: (a) L = 2 = = 8.89 H I (1.50 A) 2 (b) L =

μ0 AN 2 l

. If α is the number of turns per unit length, then N = α l and L = μ0 Aα 2l . For this coil

α = 10 coils/mm = 10 × 103 coils/m. l=

L

=

8.89 H

= 56.3 m. (4π × 10 T ⋅ m/A)π (0.0200 m) 2 (10 × 103 coils/m) 2 This is not a practical length for laboratory use. Reflect: The number of turns is N = (56.3 m)(10 × 103 coils/m) = 5.63 × 105 turns. The length of wire in the

μ0 Aα

2

−7

solenoid is the circumference C of one turn times the number of turns. C = π d = π (4.00 × 10−2 m) = 0.126 m. The length of wire is (0.126 m)(5.63 × 105 ) = 7.1 × 104 m = 71 km. This length of wire will have a large resistance and I 2 R electrical energy loses will be very large.

*21.45. Set Up: (a) The magnetic field inside a solenoid is B = μ0nI . © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

21-12

Chapter 21

Solve: B =

(4π × 10−7 T ⋅ m/A)(400)(80.0 A) = 0.161 T 0.250 m

(b) Set Up: The energy density in a magnetic field is u =

B2 . 2 μ0

(0.161T)2

= 1.03 × 104 J/m3 2(4π × 10−7 T ⋅ m/A) (c) Set Up: The total stored energy is U = uV. Solve: U = uV = u (lA) = (1.03 × 104 J/m3 )(0.250 m)(0.500 × 10−4 m 2 ) = 0.129 J Solve: u =

(d) Set Up: The energy stored in an inductor is U = 12 LI 2 . Solve: Solving for L and putting in the numbers gives 2U 2(0.129 J) L= 2 = = 4.02 × 10−5 H I (80.0 A) 2 Reflect: An inductor stores its energy in the magnetic field inside of it.

21.46. Set Up: Energy = Pt. U = 12 LI 2 . Solve: (a) Energy = (200 W)(24 h)(3600 s/h) = 1.73 × 107 J (b) L =

2U I2

=

2(1.73 × 107 J) (80.0 A)2

*21.47. Set Up: U = 12 LI 2 . 1/ 2

⎛U ⎞ Solve: I 2 = I1 ⎜ 2 ⎟ ⎝ U1 ⎠

= 5.41 × 103 H U I2

= 12 L = constant and

U1

I1 2

=

U2

I2 2

.

1/ 2

⎛ 9.0 mJ ⎞ = I⎜ ⎟ ⎝ 3.0 mJ ⎠

= I 3.

21.48. Set Up: The time constant is τ = L /R. The current as a function of time is i =

E (1 − e−t/τ ). The energy R

stored in the inductor is U = 12 LI 2 . Solve: (a) τ =

L 11.0 × 10−3 H = = 7.33 × 10−5 s = 73.3 μs R 150.0 Ω

(b) The maximum current is when t → ∞ and is equal to I = (c) i = (0.0800 A)(1 − e−73.3 μs/73.3 μs ) = 0.0506 A

E 12.0 V = = 0.0800 A. R 150.0 Ω

(d) U max = 12 LI 2 = 12 (11.0 × 10−3 H)(0.0800 A)2 = 3.52 × 10−5 J = 35.2 μJ

Δi Δi = 0. L = υ L , the voltage across the Δt Δt E E − ( R/L )t inductor. The current as a function of time is given by i = (1 − e at t → ∞. ). At t = 0, i = 0. i = imax = R R Δi E 6.00 V Solve: (a) At t = 0, i = 0 and = = = 2.40 A/s. Δt L 2.50 H Δi Δi (b) At t = 0, = 2.40 A/s and υ L = L = (2.50 H)(2.40 A/s) = 6.00 V. Initially, i = 0, the voltage across the Δt Δt resistor is zero, and the full battery emf appears across the inductor.

*21.49. Set Up: The loop rule applied to the circuit gives E − iR − L


Electromagnetic Induction (c) The time constant is τ =

21-13

L 2.50 H = = 0.313 s. When t = 0.313 s, R 8.00 Ω

E 6.00 V (1 − e−1 ) = (1 − e−1 ) = 0.474 A. R 8.00 Ω E 6.00 V (d) When t → ∞, i = = = 0.750 A. R 8.00 Ω Reflect: Initially i = 0 and the full battery voltage is across the inductor. After a long time, the full battery voltage is across the resistor. i=

21.50. Set Up: After S1 has been closed a long time, the current has reached a value

250.0 V = 0.500 A. When 500.0 Ω

S1 is opened and S 2 is closed, a current decay R-L circuit is produced and i = I 0e2( R /L )t . Solve: (a) I 0 = 0.500 A

L 300.0 × 10−3 H = = 6.00 × 10−4 s = 0.600 ms R 500.0 Ω (c) When t → ∞, i → 0. (b) τ =

21.51. Set Up: At all times υ1 + υ2 = 25.0 V. The voltage across the resistor depends on the current through it and the voltage across the inductor depends on the rate at which the current through it is changing. Immediately after closing the switch the current thorough the inductor is zero. After a long time the current is no longer changing. Solve: (a) i = 0 so υ1 = 0 and υ 2 = 25.0 V. The ammeter reading is A = 0. (b) After a long time, υ 2 = 0 and υ1 = 25.0 V. υ1 = iR and i =

υ1

=

25.0 V = 1.67 A. The ammeter reading is 15.0 Ω

R A = 1.67 A. (c) None of the answers in (a) and (b) depend on L so none of them would change. Reflect: The inductance L of the circuit affects the rate at which current reaches its final value. But after a long time the inductor doesn’t affect the circuit and the final current does not depend on L.

E (1 − e−t/τ ), with τ = L /R. The energy stored in the inductor is U = 12 Li 2 . The R maximum current occurs after a long time and is equal to E/R.

21.52. Set Up: We know that i =

Solve: (a) We know that imax = E/R; thus, i = imax /2 when (1 − e − t /τ ) = 12 , which implies that e − t /τ = 12 . Solving for

t we obtain −t /τ = ln

( 12 ) , which gives t = L lnR 2 = (ln 2)(150.25.0×Ω10

−3

H)

= 17.3 μs

(b) We know that U = 12 U max when i = imax / 2. This gives us 1 − e −t /τ = 1/ 2, so we have e −t /τ = 1 − 1/ 2 = 0.2929.

Solving for t we obtain t = − L ln (0.2929)/R = 30.7 μs. Reflect: τ = L /R = 2.50 × 10−5 s = 25.0 μs. The time in part (a) is 0.692τ and the time in part (b) is 1.23τ .

*21.53. Set Up: The current as a function of time is given by i =

E (1 − e−( R/L )t ). The energy stored in the inductor R

1 times its maximum value. 2 1 1 ⎛ R⎞ E 1E Solve: (a) The maximum current is . i = gives = 1 − e− ( R /L )t and e− ( R /L )t = . 2⎜ ⎟ t = ln 2 2 ⎝ L⎠ 2R R is U = 12 Li 2 . U reaches

1 2

its maximum value when i is

t =2

L ln

( 12 ) .

( 12 ) = 2 (1.50 × 10−3 H)ln ( 12 ) = 1.39 μs.

R

0.750 × 103 Ω


21-14

(b)

Chapter 21

L ln (1 − 1/ 2) (1.50 × 10−3 H)ln (1 − 1/ 2) 1 = 1 − e− ( R /L )t and t = − =2 = 2.46 μs. R 2 0.750 × 103 Ω

21.54. Set Up: The resonant angular frequency of an L-C circuit is ω = Solve: ω =

1 (5.25 × 10

−3

H)(12.0 × 10−6 F)

1 . LC

= 3980 rad/s

*21.55. Set Up: The energy stored in a capacitor is U C = 12 Cυ 2 . The energy stored in an inductor is U L = 12 Li 2 . Energy conservation requires that the total stored energy be constant. The current is a maximum when the charge on the capacitor is zero. Solve: (a) Initially υ = 16.0 V and i = 0. U L = 0 and U C = 12 Cυ 2 = 12 (5.00 × 10−6 F)(16.0 V) 2 = 6.40 × 10−4 J. The total energy stored is 0.640 mJ. (b) The current is maximum when q = 0 and U C = 0. U C + U L = 6.40 × 10−4 J so U L = 6.40 × 10−4 J. 1 Li 2 2 max

= 6.40 × 10−4 J and imax =

2(6.40 × 10−4 J) 3.75 × 10−3 H

= 0.584 A.

21.56. Set Up: The energy stored in a capacitor is U C = 12 Cυ 2 . The energy stored in an inductor is U L = 12 Li 2 . Energy conservation requires that the total stored energy be constant. The current is a maximum when the charge on the capacitor is zero. Q 175 μC = 11.7 V. 12 Cυmax2 = 12 Limax 2 . Solve: (a) υ max = = C 15.0 μF imax = υmax

C 15.0 × 10−6 F = (11.7 V) = 0.641 A. L 5.00 × 10−3 H

At this instant q = 0. (b) υ max = 11.7 V. At this instant i = 0. (c) U max = 12 Limax 2 = 1.03 mJ. This is when i = 0.641 A.

21.57. Set Up: This is an R-L circuit and i (t ) is given by Equation 21.25. When t → ∞, i → if = V /R. Solve: (a) R =

V 12.0 V = = 1860 Ω. if 6.45 × 10−3 A

− Rt −(1860 Ω)(7.25 × 10−4 s) Rt = = 0.963 H. = −ln(1 − i /if ) and L = ln(1 − i /if ) ln(1 − (4.86/6.45)) L Reflect: The current after a long time depends only on R and is independent of L. The value of R /L determines how rapidly the final value of i is reached. (b) i = if (1 − e− ( R /L )t ) so

21.58. Set Up: When the loop is entering or leaving the region of magnetic field the flux through it is changing and there is an induced emf. The magnitude of this induced emf is E = BLυ , as in Example 21.4. The length L is 0.750 m. When the loop is totally within the field the flux through the loop is not changing and there is no induced emf. The E and direction given by Lenz’s law. induced current has magnitude I = R



21-15

E BLυ (1.25 T)(0.750 m)(3.0 m/s) = = = 0.225 A. The flux through the loop is directed out of the 12.5 Ω R R page and is increasing, so the magnetic field of the induced current is into the page inside the loop and the induced current is clockwise. (b) The flux is not changing so E and I are zero. E (c) I = = 0.225 A. The flux through the loop is directed out of the page and is decreasing, so the magnetic field of R the induced current is out of the page inside the loop and the induced current is counterclockwise. (d) Let clockwise currents be positive. At t = 0 the loop is entering the field. It is totally in the field at time ta and Solve: (a) I =

beginning to move out of the field at time tb . The graph of the induced current as a function of time is sketched in the figure below.

*21.59. Set Up: At t = 0 the angle φ between the normal to the loop and the magnetic field is zero. φ = ω t . ΦB = BA cos φ. From the discussion of generators in Section 21.3, the rate of change of ΦB is −ω BA sin ω t . The induced emf is ω AB sin ω t . Solve: (a) ΦB = BA cos ω t . The graph of ΦB versus t is sketched in Figure (a) below. (b) The graph of −ω BA sin ω t versus t is sketched in Figure (b) below. (c) The graph of E = ω AB sin ω t versus t is sketched in Figure (c) below. (d) Doubling ω halves the period of E and doubles its amplitude. The graph of E = 2ω AB sin (2ω t ) versus t is sketched in Figure (d) below.


21-16

Chapter 21

ΔΦB to calculate the average emf. Apply Lenz’s law to Δt calculate the direction of the induced current. ΦB = BA. The flux changes because the area of the loop changes.

21.60. Set Up: Apply Faraday’s law in the form Eav = − N

ΔΦB ΔA π r2 π (0.0650/2 m) 2 =B =B = (0.950 T) = 0.0126 V. Δt Δt Δt 0.250 s (b) Since the magnetic field is directed into the page and the magnitude of the flux through the loop is decreasing, the induced current must produce a field that goes into the page to oppose this change. Therefore the current flows from point a through the resistor to point b (which is from point b through the coil to point a). Reflect: According to Faraday’s law there will be an induced emf whenever there is a change in the magnetic flux. This can occur due to a change in B, A, or to a change in the orientation between B and A. Solve: (a) Eav =

*21.61. Set Up: Use U C = 12 CVC 2 (energy stored in a capacitor) to solve for C. Then use Equation 21.30 and ω = 2π f to solve for the L that gives the desired current oscillation frequency. Solve: VC = 12.0 V; U C = 12 CVC 2 so C = 2U C /VC 2 = 2(0.0160 J)/(12.0 V)2 = 222 μ F 1

1

. Setting f = 3500 Hz gives L = 9.31 μ H 2π LC (2π f )2 C Reflect: Note that f is in Hz and ω is in rad/s; we must be careful not to confuse the two.

We know that f =

so L =

21.62. Set Up: Closing S2 and simultaneously opening S1 produces an L-C circuit with initial current through the inductor of 3.50 A. When the current is a maximum the charge q on the capacitor is zero and when the charge q is a maximum the current is zero. Conservation of energy says that the maximum energy 12 Limax 2 stored in the inductor equals the maximum energy

1 2

qmax 2 stored in the capacitor. C



Solve: (a)

1 Li max2 2

=

1 2

21-17

qmax 2 . C

qmax = ( LC )imax = (2.0 × 10−3 H)(5.0 × 10−6 F)(3.50A) = 3.50 × 10−4 C = 0.350 mC. (b) When q is maximum, i = 0. Δi . E is limited by the loop rule, so the rate of change of the Δt current is limited and the current in the inductor can’t jump suddenly from one value to another. Solve: (a) The current in the inductor is zero when the switch is open and is still zero immediately after the switch is closed. The current through the 30.0 Ω resistor equals the current through the inductor and therefore is zero. The voltage across the 20.0 Ω resistor must equal the battery voltage, so the current through this resistor is 60.0 V = 3.00 A. 20.0 Ω (b) After a long time the current is no longer changing and the voltage across the inductor is zero. The voltage across each resistor is 60.0 V, so the current through the 20.0 Ω resistor is 3.00 A and the current through the 30.0 Ω resistor is 2.00 A. (c) The current through the inductor just after the switch is opened is the same as it was just before the switch was opened, 2.00 A. With the switch open there is only one current loop and the current is 2.00 A through each resistor.

*21.63. Set Up: The emf in the inductor is E = L

Solutions to age Problems 21.64. Set Up: We will apply Lenz’s law to determine the direction of the induced current. As viewed from above, the figure shows the current pulse in the coil is in the counterclockwise direction. Also, the figure shows that direction-1 for the induced current is clockwise and direction-2 is counterclockwise. Solve: As the current pulse increases, it produces an increaward magnetic field. To oppose this the induced current must flow clockwise (direction-1). As the current pulse decreases its upward magnetic field decreases and the induced current must flow counterclockwise (direction-2) to oppose this. The correct answer is C. ΔΦB . Δt Solve: Since the magnetic flux increases with increasing area, the induced emf will be greatest on the periphery of the dotted line. The correct answer is B.

21.65. Set Up: We will apply Faraday’s law: E =


22

ALTERNATING CURRENT

Answers to Multiple-Choice Problems 1. D 2. D 3. A 14. D 15. C

4. E

5. A

6. D

7. B

8. A

9. D

10. C

11. C

12. B, C, D

13. C

Solutions to Problems 22.1. Set Up: I rms = I/ 2 . I is the current amplitude, the maximum value of the current. 1.50 A = 1.06 A. 2 Reflect: The current amplitude is larger than the root-mean square current. Solve: I = 1.50A gives I rms =

22.2. Set Up: The specified value is the root-mean-square current; I rms = 0.34 A. i = I cos ω t and I rms = I/ 2 . (a) I rms = 0.34 A (b) I = 2 I rms = 2 (0.34 A) = 0.48 A. (c) Since the current is positive half of the time and negative half of the time, its average value is zero. (d) Since I rms is the square root of the average of i 2 , the average square of the current is I rms 2 = (0.34 A) 2 = 0.12 A 2 .

*22.3. Set Up: The reactance of capacitors and inductors depends on the angular frequency at which they are operated, as well as their capacitance or inductance. The reactances are X C = 1/ω C and X L = ω L.

1 LC 1 1 = = 7560 rad/s LC (5.00 mH)(3.50 μ F)

Solve: (a) Equating the reactances gives ω L = (b) Using the numerical values we get ω =

1

ωC

⇒ω =

XC = XL = ω L = (7560 rad/s)(5.00 mH) = 37.8 Ω Reflect: At other angular frequencies, the two reactances could be very different.

22.4. Set Up: The reactance of an inductor is X L = ω L = 2π fL. The reactance of a capacitor is XC =

1 1 = . ωC 2π fC

The frequency f is in Hz. (a) At 60.0 Hz, X L = 2π (60.0 Hz)(0.450 H) = 170 Ω. X L is proportional to f so at 600 Hz, X L = 1700 Ω.


22-1

22-2

Chapter 22

(b) At 60.0 Hz, X C =

1 2π (60.0 Hz)(2.50 × 10−6 F)

(c) X L = X C says 2π fL =

= 1.06 × 103 Ω. X C is proportional to 1/f , so at 600 Hz, XC = 106 Ω.

1 1 1 and f = = = 150 Hz. 2π fC 2π LC 2π (0.450 H)(2.50 × 10−6 F)

*22.5. Set Up: We have VL = I ω L, where ω is the angular frequency, in rad/s and f =

ω 2π

is the frequency in Hz.

VL (12.0 V) = = 1.63 × 106 Hz. 2ω IL 2π (2.60 × 10−3 A)(4.50 × 10−4 H) Reflect: When f is increased, I decreases. Solve: VL = I ω L so f =

22.6. Set Up: The current amplitude is I =

V 1 . XC = . XC ωC

Solve: (a) I = V ωC = (60.0 V)(100 rad/s)(2.20 × 10−6 F) = 0.0132 A. I is proportional to ω . (b) I = (10)(0.0132 A) = 0.132 A (c) I = (10)(0.132 A) = 1.32 A

22.7. Set Up: The current amplitude is I =

V 1 . . XC = XC ωC

Solve: (a) I = V ωC = (25.0 V)(1000 rad/s)(0.0100 × 10−6 F) = 2.50 × 10−4 A = 0.250 mA. I is proportional to C. (b) I = (100)(0.250 mA) = 25.0 mA (c) I = (100)(25.0 mA) = 2.50 A

22.8. Set Up: The current amplitude is I = Solve: (a) I =

V

ωL

=

V . X L = ω L. XL

25.0 V = 2.50 A. I is proportional to 1/L. (1000 rad/s)(0.0100 H)

2.50 A = 25.0 mA 100 25.0 mA = 0.250 mA (c) I = 100

(b) I =

22.9. Set Up: I rms = I/ 2 and Vrms = V/ 2 . Solve: (a) The voltage amplitude V is the maximum voltage so V = 12.0 V.

I = 2 I rms = (b) Vrms

2(7.50 mA) = 10.6 mA.

V 12.0 V = = = 8.48 V 2 2

22.10. Set Up: For an R-L-C circuit we have Z = R 2 + ( X L − X C ) 2 , where X L = ω L, X C = ω = 2π f .

1 , R = 65 Ω, and ωC

Solve: (a) We must find Z at the frequency ω = 2π (300 Hz) = 600π rad/s. We have X L = ω L = (600π rad/s)

(35 × 10−3 H) = 66 Ω, and X C =

1

ωC

=

1 (600π rad/s)(8.0 × 10−6 F)

= 66 Ω. Thus we have

Z = (65 Ω) 2 + (66 Ω − 66 Ω) 2 = 65 Ω.


Alternating Current

22-3

(b) We repeat all of the above calculations for ω = 2π (30.0 × 103 Hz) = 6π × 104 rad/s. We have X L = (6π × 104 rad/s)

(35 × 10−3 H) = 6.6 × 103 Ω, and X C =

1 (6π × 104 rad/s)(8.0 × 10−6 F)

= 0.66 Ω. Thus we have

Z = (65 Ω) 2 + (6.6 × 103 Ω − 0.66 Ω) 2 = 6,600 Ω. Reflect: Near 300 Hz the capacitive reactance is equal to the inductive reactance so the circuit behaves as though it were a simple 65 Ω resistance. At low frequencies (much less than 300 Hz) the capacitive reactance dominates and at high frequencies (much higher than 300 Hz) the inductive reactance dominates.

*22.11. Set Up: Use Equation 22.18: V 2 = VR2 + (VL − VC ) 2 Solve: V = (30.0 V) 2 + (50.0 V − 90.0 V)2 = 50.0 V Reflect: Note that the voltage amplitudes do not simply add to give 170.0 V for the source voltage.

22.12. Set Up: The impedance of an R-L circuit is given by Z = R 2 + X L2 = R 2 + ω 2 L2 , where ω = 2π f . Compare the impedances Z1 and Z 2 that correspond to angular frequencies ω1 and ω2 , respectively: set Z 2 = 2 Z1 and solve for ω2 in of ω1. Solve: We have

ω2 =

3R 2 + 4ω12 L2

is f 2 =

2

L

R 2 + ω22 L2 = 2 R 2 + ω12 L2 , so that R 2 + ω22 L2 = 4( R 2 + ω12 L2 ). Solving for ω2 we obtain

=

3(1500 Ω)2 + 4(2π )2 (120 Hz)2 (350 × 10−3 H)2 (350 × 10

−3

H)

2

= 7.57 × 103 rad/s. Thus, the desired frequency

3

ω2 7.57 × 10 rad/s = = 1200 Hz. 2π 2π

Reflect: You can also solve this problem by calculating a numerical value for the impedance of the circuit at 120 Hz, and then solve for the frequency at which the circuit’s impedance has twice that value.

1 V X − XC . I = . tan φ = L . R ωC Z Solve: (a) 1000 rad/s: X L = ω L = (1000 rad/s)(0.900 H) = 900 Ω. *22.13. Set Up: Z = R 2 + ( X L − X C ) 2 . X L = ω L. X C =

XC =

1 1 = = 500 Ω. ωC (1000 rad/s)(2.00 × 10−6 F)

Z = (200 Ω) 2 + (900 Ω − 500 Ω) 2 = 447 Ω. 750 rad/s: X L = ω L = (750 rad/s)(0.900 H) = 675 Ω. XC =

1

ωC

=

1 (750 rad/s)(2.00 × 10−6 F)

= 667 Ω.

Z = (200 Ω) 2 + (675 Ω − 667 Ω) 2 = 200 Ω. 500 rad/s: X L = ω L = (500 rad/s)(0.900 H) = 450 Ω. XC =

1 1 = = 1000 Ω. ωC (500 rad/s)(2.00 × 10−6 F)

Z = (200 Ω) 2 + (450 Ω − 1000 Ω)2 = 585 Ω. V . Z decreases and then increases. I increases as ω varies from 1000 rad/s to around 750 rad/s and then I Z decreases as ω continues to decrease. X − X C 900 Ω − 500 Ω (c) tan φ = L = = 2.00 and φ = +63.4°. R 200 Ω (b) I =


22-4

Chapter 22

(d) The phasor diagram is sketched in the figure below

22.14. Set Up: Z = R 2 + ( X L − X C ) 2 . X L = ω L = 2π fL. X C =

1 1 X − XC = . tan φ = L . R ωC 2π fC

Solve: (a) X L = 2π fL = (2π )(500 Hz)(0.100 H) = 314 Ω.

XC =

1 1 = = 637 Ω. 2π fC (2π )(500 Hz)(0.500 × 10−6 F)

X L − X C 314 Ω − 637 Ω = = −1.62 and φ = −58.3°. The source 200 Ω R voltage lags the current. The phasor diagram is sketched in Figure (a) below.

Z = (200 Ω) 2 + (314 Ω − 637 Ω) 2 = 380 Ω. tan φ =

(b) X L = 2π fL = (2π )(1000 Hz)(0.100 H) = 628 Ω.

XC =

1 1 = = 318 Ω. 2π fC (2π )(1000 Hz)(0.500 × 10−6 F)

X L − X C 628 Ω − 318 Ω = = +1.55 and φ = +57.2°. The source 200 Ω R voltage leads the current. The phasor diagram is sketched in Figure (b) above.

Z = (200 Ω) 2 + (628 Ω − 318 Ω) 2 = 369 Ω. tan φ =


Alternating Current

22-5

Reflect: When f increases, X L increases and X C decreases. When X C > X L the phase angle is negative and when

X L > X C the phase angle is positive.

22.15. Set Up: 20.0 W is the average power P. For a pure resistance, P = Vrms I rms = I rms 2 R. Solve: (a) The average power is one-half the maximum power, so the maximum instantaneous power is 40.0 W. 20.0 W P (b) I rms = = = 0.167 A Vrms 120 V (c) R =

P I rms2

=

20.0 W (0.167 A)2

= 720 Ω

Vrms R . cos φ = . Z Z 80.0 V 75.0 Ω = = 0.762 A. cos φ = = 0.714. P = (80.0 V)(0.762 A)(0.714) = 43.5 W. 105 Ω 105 Ω

22.16. Set Up: P = Vrms I rms cos φ. I rms = Solve: I rms

*22.17. Set Up: From Problem 22.14, R = 200 Ω. For a resistor, PR = I rms 2 R. For the source, Psource = Vrms I rms cos φ. cos φ =

R V and rms = I rms . Z Z

Solve: (a) PR = I rms2 R = (0.250 A) 2 (200 Ω) = 12.5 W ⎛ R⎞ (b) Psource = Vrms I rms ⎜ ⎟ = I rms2 R = 12.5 W ⎝Z⎠

(c) Energy is stored and released in the capacitor and inductor but not dissipated. The average rate of dissipation of energy for each is zero.

22.18. Set Up: The angular frequency and the capacitance can be used to calculate the reactance X C of the capacitor. The angular frequency and the inductance can be used to calculate the reactance X L of the inductor. Calculate the phase angle φ and then the power factor is cos φ. Calculate the impedance of the circuit and then the 45 V V = = 31.8 V. The average power is 2 2 P = Vrms I rms cos φ. On the average no power is consumed in the capacitor or the inductor, it is all consumed in the

rms current in the circuit. The source has rms voltage Vrms = resistor. Solve: X L = ω L = (360 rad/s)(15 × 10−3 H) = 5.4 Ω.

1 1 = = 794 Ω. ωC (360 rad/s)(3.5 × 10−6 F) X − X C 5.4 Ω − 794 Ω tan φ = L = and φ = −72.4°. R 250 Ω The power factor is cos φ = 0.302. XC =

(b) Z = R 2 + ( X L − X C ) 2 = (250 Ω) 2 + (5.4 Ω − 794 Ω) 2 = 827 Ω.

Vrms 31.8 V = = 0.0385 A. 827 Ω Z P = Vrms I rms cos φ = (31.8 V)(0.0385 A)(0.302) = 0.370 W. I rms =

(c) The average power delivered to the resistor is P = I rms2 R = (0.0385 A) 2 (250 Ω) = 0.370 W. The average power

delivered to the capacitor and to the inductor is zero. Reflect: On average the power delivered to the circuit equals the power consumed in the resistor. The capacitor and inductor store electrical energy during part of the current oscillation but each return the energy to the circuit during another part of the current cycle. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

22-6

Chapter 22

1 . The period, the time duration of one cycle, is LC

*22.19. Set Up: The resonance angular frequency is ω 0 = T=

2π

ω0

.

Solve: (a) ω 0 = (b) T =

2π

ω0

=

1 (5.0 × 10

−3

2π 1.0 × 104 rad/s

H)(2.0 × 10−6 C)

= 1.0 × 104 rad/s

= 6.28 × 10−4 s

22.20. Set Up: The resonance angular frequency is the angular frequency for which the reactance X L of the inductor equals the reactance X C of the capacitor. 1 and ωC 1 1 ω= = = 4.47 × 103 rad/s. 3 − LC (10.0 × 10 H)(5.00 × 10−6 F)

Solve: (a) X L = X C says ω L =

(b) The resonance frequency is the frequency calculated in part (a), ω 0 = 4.47 × 103 rad/s.

*22.21. Set Up: At resonance X L = X C , φ = 0 and Z = R. R = 150 Ω, L = 0.750 H, C = 0.0180 μ F, V = 150 V Solve: (a) At the resonance frequency X L = X C and from tan φ =

factor is cos φ = 1.00.

X L − XC we have that φ = 0° and the power R

(b) Pav = 12 VI cos φ (Equation 22.29)

At the resonance frequency Z = R, so I =

V V = Z R

V 2 1 (150 V)2 ⎛V ⎞ Pav = 12 V ⎜ ⎟ cos φ = 12 = = 75.0 W ⎝ R⎠ R 2 150 Ω (c) Evaluate: When C and f are changed but the circuit is kept on resonance, nothing changes in Pav = V 2 /(2 R), so

the average power is unchanged: Pav = 75.0 W. The resonance frequency changes but since Z = R at resonance the current doesn’t change.

22.22. Set Up: ω 0 = Solve: (a) L =

1

Cω 02 (b) Z = R = 138 Ω

=

1 . At resonance, Z = R. LC 1 (10.5 × 10−6 F)(1525 rad/s) 2

= 0.0410 H = 41.0 mH

Reflect: At the resonance frequency X L = X C and the impedance Z has its maximum value, Z = R. Since I =

V , Z

at resonance I has its maximum value.

22.23. Set Up: The resonance angular frequency is ω 0 =

1 1 . X L = ω L. X C = and Z = R 2 + ( X L − X C ) 2 . ω C LC

At the resonance frequency X L = X C and Z = R. Solve: (a) Z = R = 115 Ω 1 (b) ω 0 = = 1.33 × 104 rad/s. ω = 2ω 0 = 2.66 × 104 rad/s. −3 −6 (4.50 × 10 H)(1.26 × 10 F)


Alternating Current

22-7

X L = ω L = (2.66 × 104 rad/s)(4.50 × 10−3 H) = 120 Ω. XC =

1

ωC

=

1 (2.66 × 104 rad/s)(1.25 × 10−6 F)

= 30 Ω

Z = (115 Ω) 2 + (120 Ω − 30 Ω) 2 = 146 Ω (c) ω = ω 0 /2 = 6.65 × 103 rad/s. X L = 30 Ω. X C =

1

ωC

= 120 Ω.

Z = (115 Ω) 2 + (30 Ω − 120 Ω) 2 = 146 Ω, the same value as in part (b). 1 . VC = IX C . V = IZ . At resonance, Z = R. LC 1 1 = = 1.54 × 104 rad/s −6 LC (0.350 H)(0.0120 × 10 F)

22.24. Set Up: ω 0 = Solve: (a) ω 0 =

⎛V ⎞ ⎛V ⎞ 1 1 (b) V = IZ = ⎜ C ⎟ Z = ⎜ C ⎟ R. X C = = = 5.41 × 103 Ω. 4 ωC (1.54 × 10 rad/s)(0.0120 × 10−6 F) ⎝ XC ⎠ ⎝ XC ⎠ ⎛ 550 V ⎞ V = ⎜⎜ ⎟⎟ (400 Ω) = 40.7 V. 3 ⎝ 5.41 × 10 Ω ⎠ Reflect: The voltage amplitude for the capacitor is more than a factor of 10 times greater than the voltage amplitude of the source. 1 . V = IZ . At resonance, Z = R. LC 1 1 = = 250 rad/s LC (0.200 H)(80.0 × 10−6 F)

*22.25. Set Up: ω 0 = Solve: (a) ω 0 =

240 V V = = 400 Ω. I 0.600 A (c) X L = ω L = (250 rad/s)(0.200 H) = 50.0 Ω. X C = X L = 50.0 Ω.

(b) V = IZ = IR so R =

VL = IX L = (0.600 A)(50.0 Ω) = 30.0 V. VC = VL = 30.0 V. VR = IR = (0.600 A)(400 Ω) = 240 V.

22.26. Set Up: At the resonance frequency, Z = R. Use that V = IZ, VR = IR, VL = IX L and VC = IX C . We have Pav as given by Eq. (22.29). (a) Solve: V = IZ = IR = (0.500 A)(300 Ω) = 150 V (b) VR = IR = 150 V

X L = ω L = L(1/ LC ) = L /C = 2582 Ω; VL = IX L = 1290 V X C = 1/(ωC ) = L /C = 2582 Ω; VC = IX C = 1290 V (c) Pav = 12 VI cos φ = 12 I 2 R, since V = IR and cos φ = 1 at resonance.

Pav = 12 (0.500 A)2 (300 Ω) = 37.5 W Reflect: At resonance VL = VC . Note that VL + VC > V . However, at any instant vL + vC = 0.

*22.27. Set Up: In a parallel L-R-C circuit the impedance is a maximum at the resonance angular frequency of 1 ω0 = . At the resonance angular frequency, Z = R. LC 1 1 Solve: (a) ω 0 = = = 3.24 × 103 rad/s −3 −6 LC (11.2 × 10 H)(8.50 × 10 F) (b) Z = R = 125 Ω © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

22-8

Chapter 22

1 is the resonance angular frequency for either a series or a parallel L-R-C circuit. For the series LC circuit, at resonance the impedance is a minimum and the current is a maximum. For the parallel circuit, the impedance is a maximum and the current is a minimum.

Reflect: ω 0 =

1 V V . I rms = rms . At resonance, Z = R. I = . For a parallel circuit, VL = VC = VR = V . Z Z LC V V V , IC = , I R = . At resonance, φ = 0°. IL = R XL XC

22.28. Set Up: ω 0 =

Solve: (a) ω 0 =

1 1 ω = = 1.83 × 103 rad/s. f = = 291 Hz. −6 π 2 LC (0.500 H)(0.600 × 10 F)

(b) The phasor diagram is sketched in the figure below.

V V 120 V = = = 0.400 A Z R 300 Ω V V V 120 V (d) I R = = 0.400 A. I L = = = = 0.131 A. 3 R X L ω L (1.83 × 10 rad/s)(0.500 H)

(c) I =

IC =

V = V ωC = (120 V)(1.83 × 103 rad/s)(0.600 × 10−6 F) = 0.131 A XC

22.29. Set Up: At resonance Z = R. For a parallel circuit, V = VR = VL = VC . X L = ω L. X C = Solve: (a) ω = ω 0 =

1

ωC

. ω0 =

1 . LC

1 1 = = 500 rad/s. LC (0.800 H)(5.00 × 10−6 F)

X L = ω L = (500 rad/s)(0.800 H) = 400 Ω. V = VL = IX L = (0.400 A)(400 Ω) = 160 V. X C = X L = 400 Ω. I C = (b) I R =

V 160 V = = 0.400 A; I C = I L . VC 400 Ω

V 160 V = = 0.800 A. R 200 Ω

22.30. Set Up: At any instant of time the same rules apply to the parallel ac circuit as to parallel dc circuit: the voltages are the same and the currents add. For a resistor the current and voltage are in phase. For an inductor the voltage leads the current by 90° and for a capacitor the voltage lags the current by 90°.


Alternating Current

22-9

Solve: (a) The phase diagram is shown in Figure 22.21 in the text. From the figure we have 2

2

V ⎞ ⎛V ⎞ ⎛ . I 2 = I R2 + ( I C − I L ) 2 = ⎜ ⎟ + ⎜ V ωC − ⎝ R⎠ ⎝ ω L ⎟⎠ 1

I =V

(b) From part (a):

R2

2

1 ⎞ 1 V ⎛ + ⎜ ωC − = ⎟⎠ . But I = , so ⎝ Z ωL Z

2

1 ⎞ ⎛ + ωC − . 2 ⎜ ⎝ ω L ⎟⎠ R 1

1 . The current in the capacitor branch is much larger than the current in the ωC other branches. For small ω we have Z → ω L. The current in the inductive branch is much larger than the current in the other branches. Reflect: For large ω we have Z →

XL , since there is no capacitance. X L = ω L. R X X 62.1 Ω Solve: X L = R tan φ = (48.0 Ω) tan 52.3° = 62.1 Ω. L = L = L = = 0.124 H ω 2π f (2π )(80.0 Hz) *22.31. Set Up: tan φ =

22.32. Set Up: We have X L = ω L, Pav = Vrms I rms cos φ , f = 120 Hz, and ω = 2π f . Solve: (a) X L = ω L ⇒ L = (b) Z = R 2 + X L2 =

Vrms = Z

XL

ω

250 Ω = 0.332 H 2π (120 Hz)

=

(400 Ω) 2 + (250 Ω) 2 = 472 Ω. cos φ =

V2 R R V and I rms = rms . Pav = rms , so Z Z Z Z

Pav 800 W = (472 Ω) = 668 V. R 400 Ω

Reflect: I rms =

Vrms 668 V 2 = = 1.415 A. We can calculate Pav as I rms R = (1.415 A) 2 (400 Ω) = 800 W, which checks. Z 472 Ω

*22.33. Set Up: The capacitance of an air-filled parallel plate capacitor is C = solenoid was calculated in Problem 37 of Chapter 21 to be L = 1125 coils. The resonance frequency is f 0 = Solve: C =

⑀0 A d

=

8.00 × 10−3 m

μ0 AN 2 l

=

l

. The inductance of a long

. The coil has N = (125 coils/cm)(9.00 cm) =

= 2.24 × 10−12 F.

(4π × 10−7 T ⋅ m/A)π (0.250 × 10−2 m) 2 (1125) 2

ω0 =

d

1 . ⑀ 0 = 8.85 × 10−12 C2 /N ⋅ m 2 . μ0 = 4π × 10−7 T ⋅ m/A. 2π LC

(8.85 × 10−12 C2 /N ⋅ m 2 )(4.50 × 10−2 m) 2

L=

μ0 AN 2

⑀0 A

9.00 × 10−2 m 1 (3.47 × 10

−4

H)(2.24 × 10−12 F)

= 3.47 × 10−4 H

= 3.59 × 107 rad/s


22-10

Chapter 22

22.34. Set Up: X C = Solve: (a)

1

ω1C

1

ωC

. X L = ω L.

= ω1L and LC =

1

ω21

. At angular frequency ω 2 ,

XL ω L 1 = 2 = ω 22 LC = (2ω1 ) 2 2 = 4. X L > X C . X C 1/ω 2C ω1

2

⎛ 1 ⎞ 1 ⎜⎜ 2 ⎟⎟ = . X C > X L . ⎝ ω1 ⎠ 9 Reflect: When ω increases, X L increases and X C decreases. When ω decreases, X L decreases and X C increases. (b) At angular frequency ω 3 ,

22.35. Set Up: Vrms = VC ,rms = I rms X C . VL ,rms

XL ⎛ω ⎞ = ω23 LC = ⎜ 1 ⎟ XC ⎝ 3 ⎠

V 30.0 V V 2 = = 21.2 V. Z = R 2 + ( X L − X C ) . I rms = rms . Vrms = I rms R. Z 2 2 = I rms X L .

(a) ω = 200 rad/s, so X L = ω L = (200 rad/s)(0.400 H) = 80.0 Ω and

XC =

1

ωC

=

1 (200 rad/s)(6.00 × 10−6 F)

= 833 Ω.

Vrms 21.2 V = = 0.0272 A. 779 Ω Z = I rms R = (0.0272 A)(200 Ω) = 5.44 V. V2 reads VL,rms = I rms X L = (0.0272 A)(80.0 Ω) = 2.18 V.

Z = (200 Ω) 2 + (80.0 Ω − 833 Ω) 2 = 779 Ω. I rms = V1 reads VR ,rms

V3 reads VC ,rms = I rms X C = (0.0272 A)(833 Ω) = 22.7 V. V4 reads VL − VC = 0VL,rms − VC ,rms 0 = 02.18 V − 22.7 V 0 = 20.5 V. 2 V5 reads Vrms = 21.2 V. (b) ω = 1000 rad/s so X L = ω L = (5)(80.0 H) = 400 Ω and X C =

Z = (200 Ω) 2 + (400 Ω − 167 Ω) 2 = 307 Ω. I rms =

1

ωC

=

833 Ω = 167 Ω. 5

Vrms 21.2 V = = 0.0691 A. Z 307 Ω

V1 reads VR,rms = 13.8 V. V2 reads VL ,rms = 27.6 V. V3 reads VC ,rms = 11.5 V. V4 reads 0VL,rms − VC ,rms 0 = 027.6 V − 11.5 V 0 = 16.1 V. V5 reads Vrms = 21.2 V.

22.36. Set Up: VC = IX C . V = IZ . Z = R 2 + ( X L − X C )2 . Solve: (a) I = (b) Z =

VC 720 V = = 1.20 A X C 600 Ω

V 240 V = = 200 Ω I 1.20 A

(c) Z 2 = R 2 + ( X L − X C )2 . X L − X C = ± Z 2 − R 2 .

X L = X C ± Z 2 − R 2 = 600 Ω ± (200 Ω) 2 − (150 Ω) 2 = 600 Ω ± 132 Ω. X L = 732 Ω or 468 Ω. Reflect: The impedance of the circuit depends only on the absolute value of X L − X C , not on its sign. For

X L = 732 Ω the phase angle is positive and for X L = 468 Ω the phase angle is negative. *22.37. Set Up: The average power supplied by the source is P = Vrms I rms cos φ. The power consumed in the 2 R. ω = 2π f = 2π (1.25 × 103 Hz) = 7.854 × 103 rad/s. X L = ω L = 157 Ω. X C = resistance is P = I rms

1

ωC

= 909 Ω.


Alternating Current Solve: (a) First, let us find the phase angle between the voltage and the current: tan φ =

22-11

X L − X C 157 Ω − 909 Ω = R 350 Ω

and φ = −65.04°. The impedance of the circuit is Z = R 2 + ( X L − X C ) 2 = (350 Ω) 2 + (−752 Ω) 2 = 830 Ω. The average power provided by the generator is then P = Vrms I rms cos(φ ) =

2 Vrms (120 V)2 cos(φ ) = cos(−65.04°) = 7.32 W Z 830 Ω 2

⎛ 120 V ⎞ 2 (b) The average power dissipated by the resistor is PR = I rms R=⎜ ⎟ (350 Ω) = 7.32 W. ⎝ 830 Ω ⎠ Reflect: Conservation of energy requires that the answers to parts (a) and (b) are equal.

22.38. Set Up: The power factor is cos φ , where φ is the phase angle in Figure 22.13 in the text. We can calculate the average power by using Eq. (22.29). Use the result of part (a) to rewrite this expression. Solve: (a) The phasor diagram is sketched in the figure below.

From the diagram we see that cos φ =

VR IR R = = , which was to be shown. V IZ Z

V ⎛ R⎞ ⎛V ⎞ 2 (b) Pav = Vrms I rms cos φ = Vrms I rms ⎜ ⎟ = ⎜ rms ⎟ I rms R. But rms = I rms , so Pav = I rms R. ⎝ Z⎠ ⎝ Z ⎠ Z

Reflect: In an L-R-C circuit, electrical energy is stored and released in the inductor and capacitor but none is dissipated in either of these circuit elements. The power delivered by the source equals the power dissipated in the resistor. 2 *22.39. Set Up: We know R, X C and φ so Eq. (22.23) tells us X L . Use Pav = I rms R from Exercise 22.38 to

calculate I rms . Then calculate Z and use Eq. (22.24) to calculate Vrms for the source. Source voltage lags current so

φ = −54.0°. X C = 350 Ω, R = 180 Ω, Pav = 140 W X L − XC R X L = R tan φ + X C = (180 Ω) tan( −54.0°) + 350 Ω = −248 Ω + 350 Ω = 102 Ω

Solve: (a) tan φ =

2 (b) Pav = Vrms I rms cos φ = I rms R (Exercise 22.38). I rms =

Pav 140 W = = 0.882 A R 180 Ω

(c) Z = R 2 + ( X L − X C ) 2 = (180 Ω) 2 + (102 Ω − 350 Ω) 2 = 306 Ω

Vrms = I rms Z = (0.882 A)(306 Ω) = 270 V. Reflect: We

could

also

use

Equation

22.29:

Pav = Vrms I rms cos φ.

This

gives us

Vrms =

Pav = I rms cos φ

140 W = 270 V, which agrees with our previous result. The source voltage lags the current when (0.882 A)cos( −54.0°) X C > X L , and this agrees with what we found.


22-12

Chapter 22

22.40. Set Up: P = I rms2 R. cos φ =

R . Z = R 2 + ( X L − X C ) 2 . Vrms = I rms Z . Z

P 60.0 W = = 0.447 A. Z = (300 Ω)2 + (500 Ω − 300 Ω) 2 = 361 Ω. R 300 Ω R 300 Ω cos φ = = = 0.831. Z 361 Ω = I rms Z = (0.447 A)(361 Ω) = 161 V

Solve: (a) I rms =

(b) Vrms

X L − XC R . P = I rms2 R. Vrms = I rms Z . cos φ = . R Z Solve: (a) X L = R tan φ + X C = (200 Ω) tan 40.0° + 400 Ω = 568 Ω

22.41. Set Up: φ = +40.0°. tan φ =

P 150 W = = 0.866 A R 200 Ω R 200 Ω = = 261 Ω. Vrms = I rms Z = (0.866 A)(261 Ω) = 226 V. (c) Z = cos φ cos 40.0° (b) I rms =

1 V . At resonance, Z = R. I = . VC = IX C . VL = IX L . U C = 12 CVC2 . U L = 12 LI 2 . Z LC 1 1 = = 5.77 × 103 rad/s LC (0.300 H)(0.100 × 10−6 F)

22.42. Set Up: ω 0 = Solve: (a) ω 0 =

V V 240 V = = = 2.40 A Z R 100.0 Ω 1 1 (c) X C = = = 1.73 × 103 Ω. ωC (5.77 × 103 rad/s)(0.100 × 10−6 F) (b) I =

VC = IX C = (2.40 A)(1.73 × 103 Ω) = 4.15 × 103 V. (d) X L = ω L = (5.77 × 103 rad/s)(0.300 H) = 1.73 × 103 Ω. VL = IX L = 4.15 × 103 V. (e) U C = 12 CVC2 =

1 (0.100 2

× 10−6 F)(4.15 × 103 V)2 = 0.861 J. U L = 12 LI 2 = 12 (0.300 H)(2.40 A) 2 = 0.864 J.

Reflect: At resonance, X L = X C , VL = VC , and U L = U C

*22.43. Set Up: From Problem 22.47, R = 100.0 Ω, C = 0.100 μF, L = 0.300 H and V = 240 V. I =

V . Z

Solve: (a) X L = ω L = (400 rad/s)(0.300 H) = 120 Ω. XC =

1

ωC

=

1 (400 rad/s)(0.100 × 10−6 F)

= 2.50 × 104 Ω.

Z = R 2 + ( X L − X C ) 2 = (100.0 Ω) 2 + (120 Ω − 2.50 × 104 Ω) 2 = 2.49 × 104 Ω. I=

V 240 V = = 9.64 × 10−3 A = 9.64 mA. Z 2.49 × 104 Ω

(b) VC = IX C = (9.64 × 10−3 A)(2.50 × 104 Ω) = 241 V (c) VL = IX L = (9.64 × 10−3 A)(120 Ω) = 1.16 V (d) U C = 12 CVC 2 = 12 (0.100 × 10−6 F)(241 V) 2 = 2.90 × 10−3 J = 2.90 mJ.

U L = 12 LI 2 = 12 (0.300 H)(9.64 × 10−3 A) 2 = 1.39 × 10−5 J.


Alternating Current

22-13

Solutions to age Problems 22.44. Set Up: We are told that the platinum electrode behaves like an ideal capacitor in series with the resistance of the fluid. The impedance of an R-C circuit is Z = R 2 + X C2 , where X C = Solve: For a dc signal we have ω = 2π f = 0. Using X C =

Z → ∞. The correct answer is B.

1

ωC

.

1 we see that as ω → 0 we have X C → ∞, and so ωC

*22.45. Set Up: We are told that the platinum electrode behaves like an ideal capacitor in series with the resistance of the fluid, which is given by RA = ρ /(10a), where ρ = 100 Ω ⋅ cm = 1 Ω ⋅ m and d = 2a = 20μm. We know that XC =

1

ωC

, where we are given C = 10 nF = 10−8 F and ω = 2π f = 2π [(5000/π)Hz] = 104 rad/s. For an R-C circuit

we know that the impedance is given by Z = R 2 + X C2 . Solve: RA = ρ /(10a) = (1 Ω ⋅ m)/[10(10−5 m)] = 104 Ω.

XC =

1

ωC

=

1 (10 rad/s)(10−8 F) 4

= 104 Ω.

Thus we have

Z = R 2 + X C2 = (104 Ω) 2 + (104 Ω) 2 = 2 × (104 Ω). Thus, the correct answer is C. V , where V is the amplitude (peak value) of the voltage. 2 According to the problem, the peak-to-peak voltage, Vpp , is the difference between the two extreme values of voltage.

22.46. Set Up: From the textbook we know that Vrms =

Solve: Since the voltage oscillates between +V and −V we have Vpp = V − (−V ) = 2V = 2 2Vrms . Thus, the correct

answer is D.


23

ELECTROMAGNETIC WAVES

Answers to Multiple-Choice Problems 1. D 2. E 3. A, B, E 14. B, D 15. D

4. D

5. C

6. B

7. A

8. B

9. C

10. A, B

11. B

12. D

13. B

Solutions to Problems 23.1. Set Up: The distance of the earth from the sun is the orbital radius of the earth, r = 1.50 × 1011 m. The speed of light is c = 3.00 × 108 m/s. Solve: t =

x 1.50 × 1011 m = = 500 s = 8.33 min c 3.00 × 108 m/s

23.2. Set Up: The speed of electromagnetic waves in air is c = 3.00 × 108 m/s. Solve: A total time difference of 0.60 μs corresponds to a difference in distance of cΔt = (3.00 × 108 m/s) (0.60 × 10−6 s) = 180 m.

*23.3. Set Up: The speed of light in vacuum is c = 3.00 × 108 m/s. d = ct . 1 yr = 3.156 × 107 s. Solve: (a) t =

d 3.84 × 108 m = = 1.28 s c 3.00 × 108 m/s

(b) d = ct = (3.00 × 108 m/s)(8.61 yr)(3.156 × 107 s/yr) = 8.15 × 1016 m = 8.15 × 1013 km Reflect: The speed of light is very large, but it can still take light a long time to travel enormous astronomical distances.

23.4. Set Up: In the figure below, the signal leaves point A and travels to the satellite, S, which then retransmits the signal to point B. Note that the triangle ASB is an isosceles triangle and the length of AB is the diameter of the earth. The height, h, of the triangle is the radius of the satellite’s orbit, which is equal to the given altitude of the satellite plus the radius of the earth: h = 3.58 × 107 m + 6.38 × 106 m = 4.218 × 107 m. Solve: From

the 7

right 2

triangle 6

CSA 2

we

can

calculate

the

distance

d

between

A

and

S:

7

d = (4.218 × 10 m) + (6.38 × 10 m) = 4.266 × 10 m. For one-way communication, the total distance traveled by the signal is 2d so the time for the signal to traverse the path ASB is t = 2

4.266 × 107 m 3.00 × 108 m/s

= 0.284 s.


23-1

23-2

Chapter 23

Reflect: Assuming that communication between transmitters and receivers is by direct line-of-site, it is not possible to establish two-way communication between diametrically opposed points on the earth by relay off a low altitude satellite. However, as the figure below suggests, communication between diametrically opposed points could be nearly possible with a geostationary satellite due to its high altitude.

*23.5. Set Up: The wave speed is c = 3.00 × 108 m/s. c = f λ . Solve: (a) (i) f =

(ii) f =

λ

3.00 × 108 m/s

=

5.0 × 103 m

3.00 × 108 m/s 5.0 × 10

(iii) f =

c

λ

(b) (i) λ =

(ii) λ =

c

=

−6

m

= 6.0 × 104 Hz

= 6.0 × 1013 Hz

3.00 × 108 m/s 5.0 × 10−9 m

= 6.0 × 1016 Hz

c 3.00 × 108 m/s = = 4.62 × 10214 m = 4.62 × 10−5 nm f 6.50 × 1021 Hz

3.00 × 108 m/s 590 × 103 Hz

= 508 m = 5.08 × 1011 nm

23.6. Set Up: c = 3.00 × 108 m/s. The wavelength ranges for each color are: violet: 400 nm to 440 nm; red: 630 nm to 700 nm. c

. Largest λ corresponds to smallest f. λ (a) violet: 6.8 × 1014 Hz to 7.5 × 1014 Hz

Solve: f =

(b) red: 4.3 × 1014 Hz to 4.8 × 1014 Hz

*23.7. Set Up: For an electromagnetic wave propagating in the negative x direction, E = − Emax sin (ω t + kx).

ω = 2π f and k =

2π

λ

.T=

1 . Emax = cBmax . f

Solve: (a) Emax = 375 V/m so Bmax = (b) ω = 5.98 × 1015 rad/s so f =

Emax = 1.25 μ T. c

ω = 9.52 × 1014 Hz. k = 1.99 × 107 rad/m so 2π 2π = 3.16 × 10−7 m = 316 nm. λ= k

T=

1 = 1.05 × 10215 s. f


Electromagnetic Waves

23-3

This wavelength is too short to be visible. (c) c = f λ = (9.52 × 1014 Hz)(3.16 × 10−7 m) = 3.00 × 108 m/s. This is what the wave speed should be for an

electromagnetic wave propagating in vacuum. ⎛ ω ⎞⎛ 2π ⎞ ω Reflect: c = f λ = ⎜ is an alternative expression for the wave speed. ⎟⎜ ⎟= ⎝ 2π ⎠⎝ k ⎠ k

23.8. Set Up: Since the wave is traveling in empty space, its wave speed is c = 3.00 × 108 m/s. c = f λ . Emax = cBmax . T = Solve: (a) f =

c

λ

=

1 . f 3.00 × 108 m/s 432 × 10−9 m

= 6.94 × 1014 Hz

(b) Emax = cBmax = (3.00 × 108 m/s)(1.25 × 10−6 T) = 375 V/m (c) T =

1 = 1.44 × 10−15 s. f ⎛ ⎞ t x ⎛ t x⎞ E = Emax sin 2π ⎜ − ⎟ = (375 V/m)sin 2π ⎜⎜ − ⎟ −15 −9 T λ s 432 × 10 m ⎟⎠ ⎝ ⎠ ⎝ 1.44 × 10

E = (375 V/m)sin ([4.36 × 1015 rad/s]t − [1.45 × 107 m/s] x) ⎛ t x⎞ B = Bmax sin 2π ⎜ − ⎟ = (1.25 × 10−6 T)sin ([4.36 × 1015 rad/s]t − [1.45 × 107 m/s] x) ⎝T λ ⎠

*23.9. Set Up: c = 3.00 × 108 m/s. c = f λ . 2π f = ω. k = Solve: (a) f =

c

2π

λ

.

so f ranges from 4.29 × 1014 Hz to 7.50 × 1014 Hz.

λ (b) ω = 2π f so ω ranges from 2.70 × 1015 rad/s to 4.71 × 1015 rad/s. 2π (c) k = so k ranges from 8.98 × 106 rad/m to 1.57 × 107 rad/m. λ

23.10. Set Up: c = 3.00 × 108 m/s. c = f λ and k = Solve: (a) f = (b) k =

2π

λ

2π

λ

. Larger λ corresponds to smaller f and k.

c . UVA: 7.50 × 1014 Hz to 9.38 × 1014 Hz. UVB: 9.38 × 1014 Hz to 1.07 × 1015 Hz. λ

. UVA: 1.57 × 107 rad/m to 1.96 × 107 rad/m. UVB: 1.96 × 107 rad/m to 2.24 × 107 rad/m.

23.11. Set Up: c = 3.00 × 108 m/s. c = f λ . T = Solve: f =

c

λ

= 3.0 × 1018 Hz, T =

1 2π . k= . f λ

1 = 3.3 × 10−19 s, k = 6.3 × 1010 rad/m. f

23.12. Set Up: For an electromagnetic wave in air the wave speed is c = 3.00 × 108 m/s. c = f λ . k = 2π / λ . ω = 2π f . E = cB. Solve: (a) λ = (b) k =

2π

λ

c 3.00 × 108 m/s = = 361 m f 8.30 × 105 Hz

= 0.0174 rad/m


23-4

Chapter 23

(c) ω = 2π f = (2π )(8.30 × 105 Hz) = 5.22 × 106 rad/s (d) Emax = cBmax = (3.00 × 108 m/s)(4.82 × 10−11 T) = 1.45 × 10−2 V/m

*23.13. Set Up: Equations 23.3 apply for motion parallel to the + x-axis when the electric and magnetic field are oriented as shown in Figure 23.5 in the text. Equation 23.4 relates Emax to Bmax . We know that f = 6.10 × 1015 Hz and Bmax = 5.80 × 10−4 T. Solve: (a) We are given Bmax = 5.80 × 10−4 T so we can calculate

Emax = cBmax = (3.00 × 108 m/s)(5.80 × 10−4 T) = 1.74 × 105 V/m. We are told that the magnetic field is parallel to the y-axis. By rotating the electromagnetic wave shown in Figure 23.5 of the textbook to align its magnetic field with the +y-axis, we see that the electric field will then be parallel to the −z-axis. (b) We first calculate ω and k: ω = 2π f = 2π (6.10 × 1014 Hz) = 3.83 × 1015 rad/s, and 2π f ω 3.83 × 1015 rad/s = = = 1.28 × 107 rad/m. λ c c 3.00 × 108 m/s Note that Equations 23.3 describes motion along the +x-axis so that E has a positive value when B has a positive value. However, as explained in part (a), when B is aligned with the y-axis, E and B must have opposite signs. Thus, B = Bmax sin(ωt − kx ) = (5.80 × 10−4 T)sin[(3.83 × 1015 rad/s)t − (1.28 × 107 rad/m) x] k=

2π

=

E = − Emax sin(ωt − kx) = −(1.74 × 105 V/m ) sin[(3.83 × 1015 rad/s)t − (1.28 × 107 rad/m) x]. G G Reflect: E and B are perpendicular and oscillate in synchronization.

G

G

23.14. Set Up: The direction of propagation of an electromagnetic wave is related to the directions of E and B

G G according to the right-hand rule illustrated in Figure 23.2 in Section 23.2. The directions of E and B in each case G G are shown in Figures (a) through (d) below The direction of propagation is perpendicular to the plane of E and B.


Electromagnetic Waves Solve: (b) + z (c) − y (d) − x

23-5

(a) The wave is propagating in the + z direction. direction direction direction

*23.15. Set Up: Note that Equations 23.5 are the equations for an electromagnetic wave moving parallel to the − x-axis with a magnetic field that is parallel to the z-axis, as shown in Figure 23.6 in the textbook. The relation between Emax and Bmax is Emax = cBmax . Solve: (a) The phase of the wave is given by kx + ω t , so the wave is traveling in the − x direction. (b) We know that f =

c

λ

, where λ =

2π kc (1.38 × 104 rad/m)(3.0 × 108 m/s) = = 6.59 × 1011 Hz. . Thus we have f = k 2π 2π

(c) First we find Emax = cBmax = (3.00 × 108 m/s)(8.25 × 10−9 T) = 2.48 V/m. By rotating the electromagnetic wave shown in Figure 23.6 in the textbook so that its magnetic field is aligned with the + y -axis we see that the electric field will now be aligned with the + z -axis. Thus, E and B must have the same sign: modify Equations 23.5 to read E = cBmax sin(ωt + kx) = (2.48 V/m) sin((4.14 × 1012 rad/s)t + (1.38 × 104 rad/m) x). G G Reflect: E and B have the same phase and are in perpendicular directions.

23.16. Set Up: P = IA. A = π r 2 . Emax = cBmax . The intensity I = Sav is related to the maximum electric field by I = 12 ⑀0 cEmax 2 , where ⑀ 0 = 8.85 × 10−12 C2 /N ⋅ m 2 . The average energy density uaυ is related to the intensity I by I = uaυ c. Solve: (a) I = (b) Emax = (c) uaυ =

P 0.500 × 10−3 W = = 637 W/m 2 . A π (0.500 × 10−3 m) 2

2I 2(637 W/m 2 ) E = = 693 V/m. Bmax = max = 2.31 μ T. c ⑀0 c (8.85 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s)

I 637 W/m 2 = = 2.12 × 10−6 J/m3 . c 3.00 × 108 m/s

23.17. Set Up: The surface area of a sphere of radius r is A = 4π r 2 . I = P/A.

I = 12 ⑀0 cEmax 2 , with

⑀ 0 = 8.85 × 10−12 C2 /N ⋅ m 2 . Emax = cBmax . P (0.05)(75 W) Solve: (a) I = = = 330 W/m 2 . A 4π (3.0 × 10−2 m) 2 (b) Emax =

2I 2(330 W/m 2 ) = = 500 V/m. ⑀0 c (8.85 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s)

Emax = 1.7 × 10−6 T = 1.7 μ T. c Reflect: At the surface of the bulb the power radiated by the filament is spread over the surface of the bulb. Our calculation approximates the filament as a point source that radiates uniformly in all directions. Bmax =

23.18. Set Up: Intensity is average power per unit area and power is energy per unit time. If the energy is absorbed during one cycle of the wave, then the average power absorbed is the energy divided by the period T of the wave. The c 1 area of one end of the rod is π r 2 . f = and T = , so the wavelength determines the period. λ f Solve: T =

λ c

=

505 × 10−9 m 3.00 × 108 m/s

= 1.68 × 10−15 s.


23-6

Chapter 23

I=

3.94 × 10−19 J Paυ = = 7.46 × 107 W/m 2 . A (1.68 × 10−15 s)π (0.0010 × 10−3 m)2

23.19. Set Up: Intensity is average power per unit area and power is energy per unit time. I = 12 ⑀0 cEmax 2 and Emax = cBmax . Solve: (a) For the beam, energy = Pt = (2.0 × 1012 W)(4.0 × 10−9 s) = 8.0 × 103 J = 8.0 kJ. This is spread uniformly

over 100 cells, so the energy given to each cell is 80 J. (b) The cross sectional area of each cell is A = π r 2 , with r = 2.5 × 10−6 m. I= (c) Emax =

P 2.0 × 1012 W = = 1.0 × 1021 W/m 2 A (100)π (2.5 × 10−6 m) 2

2I 2(1.0 × 1021 W/m 2 ) = = 8.7 × 1011 V/m. −12 2 2 8 ⑀0 c (8.85 × 10 C /N ⋅ m )(3.00 × 10 m/s) Bmax =

Emax = 2.9 × 103 T c

23.20. Set Up: The radiation pressure on a totally absorbing surface is Solve:

I . c

I 8.00 W/m 2 = = 2.67 × 10−8 Pa c 3.00 × 108 m/s

*23.21. Set Up: The radiation pressure is Solve: (a)

I 2I for a totally absorbing surface and for a totally reflecting surface. c c

I 6.00 W/m 2 = = 2.00 × 1028 Pa c 3.00 × 108 m/s

2I = 2(2.00 × 10−8 Pa) = 4.00 × 10−8 Pa c Reflect: For the same intensity of light, the radiation pressure is twice as great for a totally reflecting surface versus a totally absorbing surface. For ordinary light intensities, the radiation pressure is very small. (b)

23.22. Set Up: Equation 23.11 gives the intensity of an electromagnetic wave: I = 12 ⑀0cEmax 2 = ⑀0cErms 2 , where for a sinusoidal wave Erms =

Emax . The total energy ing through a window of area A during a time t is IAt. We 2

know that ⑀ 0 = 8.85 × 10−12 F/m. Solve: Energy = ⑀0cErms 2 At = (8.85 × 10 −12 F/m)(3.00 × 108 m/s)(0.0200 V/m) 2 (0.500 m 2 )(30.0 s) = 15.9 μ J Reflect: The intensity is proportional to the square of the electric field amplitude.

*23.23. Set Up: I =

P and I = 12 ⑀ 0 cEmax 2 . A

Solve: P = 12 A⑀0 cEmax 2 .

Emax,1502 Emax,75

2

=

Emax,150 150 W = 2. and Emax,75 75 W

Reflect: The power output of the source is proportional to the square of the electric field amplitude in the emitted waves.



23.24. Set Up: For a totally reflecting surface the average pressure is

23-7

2I I and for a totally reflecting surface it is . c c

The radiation is the same so has the same intensity in the two cases. I 1 ⎛ 2I ⎞ Solve: P = = ⎜ ⎟ = P/2 c 2⎝ c ⎠ *23.25. Set Up: We have c = f λ , Emax = cBmax and I = Emax Bmax /2μ0 . Solve: (a) f = (b) Bmax

c

λ

=

3.00 × 108 m/s = 8.47 × 108 Hz. 0.354 m

E 0.0540 V/m = max = = 1.80 × 10−10 T. c 3.00 × 108 m/s

(c) I = Sav =

Emax Bmax (0.0540 V/m)(1.80 × 10−10 T) = = 3.87 × 10−6 W/m 2 . 2 μ0 2 μ0

Reflect: Alternatively, I = 12 ⑀ 0cEmax 2 .

23.26. Set Up: For reflection, θ r = θ a . The desired path of the ray is sketched in the figure below. Solve: tan φ =

14.0 cm , so φ = 50.6°. θ r = 90° − φ = 39.4° and θ r = θ a = 39.4°. 11.5 cm

*23.27. Set Up: For reflection, θ r = θ a . The angles of incidence and reflection at each reflection are shown in the figure below. For the rays to be perpendicular when they cross, α = 90°.


23-8

Chapter 23

Solve: (a) θ + φ = 90° and β + φ = 90°, so β = θ .

α 2

+ β = 90° and α = 180° − 2θ .

(b) θ = 12 (180° − α ) = 12 (180° − 90°) = 45°. Reflect: As θ → 0°, α → 180°. This corresponds to the incident and reflected rays traveling in nearly the same direction. As θ → 90°, α → 0°. This corresponds to the incident and reflected rays traveling in nearly opposite

directions.

23.28. Set Up: For reflection, θ r = θ a . Solve: The law of reflection is used to trace each ray in parts a, b, and c of the figure below. Ray 1 strikes mirror B at an angle of incidence of zero so reflects straight back and retraces its path. When it reaches point P it then continues to travel parallel to the surface of mirror B and has no further reflections. Ray 2 reflects from B and then travels parallel to the surface of A and has no further reflections. Ray 3 strikes mirror B at zero angle of incidence and reflects straight back. It reflects from mirror A and then travels parallel to the surface of mirror B and has no further reflections.



23-9

23.29. Set Up: The figure showing the incident and reflected rays is shown below. The angle of incidence, where the light strikes the horizontal mirror at point A, is θ1. Solve: The law of reflection applied to point C shows that θ5 = θ 4 . Since nABC is a right triangle, we can conclude that θ 5 = θ 4 = 90D − θ 3. Since θ 2 + θ 3 = 90D , we know that θ 5 = θ 4 = 90D − θ 3 = θ 2 . Finally, applying the law of reflection at point A we obtain θ 5 = θ 4 = 90D − θ 3 = θ 2 = θ1, which shows that θ5 = θ1 so that the reflected ray emerges parallel to the original ray. Reflect: This type of arrangement of mirrors is known as a “corner reflector” and it is useful because it returns an incident light ray back to its source with only a slight offset.

c λ and λ = 0 , where λ0 is the wavelength in vacuum and c = 3.00 × 108 m/s. v n The index of refraction for air is about 1.0003, which we take to be 1.00. c 3.00 × 108 m/s = 1.55. Solve: (a) n = = v 1.94 × 108 m/s

23.30. Set Up: We know that n =

(b) λ0 = nλ = (1.55)(3.55 × 10−7 m) = 5.50 × 10−7 m. Reflect: In quartz the speed is lower and the wavelength is smaller than in air.

c *23.31. Set Up: Use the distance and time to find the speed of light in the plastic. We know that n = , where

υ

8

c = 3.00 × 10 m/s. Solve: We have υ =

c 3.00 × 108 m/s d 2.50 m 8 = = = 1.38. thus, n = = 2 . 17 × 10 m/s; υ 2.17 × 108 m/s t 11.5 × 10−9 s

Reflect: In air light travels this same distance in

23.32. Set Up: In air, c = f λ0 . In glass, λ = Solve: (a) λ0 = (b) λ =

λ0 n

=

λ0 n

2.50 m 3.00 × 108 m/s

= 8.3 ns.

.

c 3.00 × 108 m/s = = 517 nm f 5.80 × 1014 Hz

517 nm = 340 nm 1.52


23-10

Chapter 23

23.33. Set Up: n = Solve: n =

c v

3.00 × 108 m/s 1.82 × 108 m/s

= 1.65

c

23.34. Set Up: n = . The frequency of light doesn’t change when it es from one material into another. υ λ0 υ = f λ. λ = . n

Solve: (a) λυ =

λ0,υ n

=

λ0,r 700 nm 400 nm = 299 nm. λr = = = 522 nm. Range is 299 nm to 522 nm. 1.34 n 1.34

(b) Calculate the frequency in air, where υ = c = 3.00 × 108 m/s.

fυ =

c

λυ

(c) υ =

=

3.00 × 108 m/s 400 × 10−9 m

fr =

c

λr

=

3.00 × 108 m/s 700 × 10−9 m

= 4.29 × 1014 Hz.

= 7.50 × 1014 Hz. Range is 4.29 × 1014 Hz to 7.50 × 1014 Hz.

c 3.00 × 108 m/s = = 2.24 × 108 m/s 1.34 n

*23.35. Set Up: λ =

λ0 n

. From Table 23.1, nwater = 1.333 and nbenzene = 1.501.

⎛ n ⎞ ⎛ 1.333 ⎞ Solve: (a) λwater nwater = λbenzene nbenzene = λ0 . λbenzene = λwater ⎜ water ⎟ = (438 nm) ⎜ ⎟ = 389 nm. n ⎝ 1.501 ⎠ ⎝ benzene ⎠ (b) λ0 = λwater nwater = (438 nm)(1.333) = 584 nm Reflect: λ is smallest in benzene, since n is largest for benzene.

23.36. Set Up: The figure below shows a light ray leaving the head of the fisherman at point H, striking the surface of the lake at point L where it is refracted to the eye of the trout at point T. Solve: Let d be the horizontal distance between points L and T, so that 3.50 − d is the horizontal distance between points H and L.

Using geometry and Snell’s law we can write the following three equations in three unknowns: 3.50 − d tan θi = 1.55 d tan θ t = 0.45 (1.00)sinθi = (1.33)sinθt © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


23-11

There is no obvious simple way to solve these equations; however, we can find a value for d that works by trial-anderror. After we guess a value for d we can use the first two equations to find the corresponding values for θi and θ t . If the calculated values for θi and θ t do not satisfy Snell’s law, we must modify our guess and try again. After repeated trials, we find that d = 0.408 meters gives angles that agree with Snell’s law to 3 significant figures. The corresponding angles are θi = 63.4D and θ t = 42.2D. After checking both sides of Snell’s law we obtain (1.00)sin(63.4D ) = 0.894 and (1.33)sin(42.2D ) = 0.893, which agree to the desired accuracy. Thus, the trout will see the top of the fisherman’s head at an angle of θ t = 42.2D from the vertical. Reflect: We can get a quick approximate solution to the problem by assuming that 3.50 meters is the approximate 3.50 , so θi ≈ 66D. Using Snell’s law with this value horizontal distance between points H and L. This gives tanθ i = 1.55

for θi we obtain θ t ≈ 43D , which is a reasonable approximation in this case. *23.37. Set Up: The path of the light ray as it travels through the sheet of glass is sketched below. Assume that the index of refraction of air is n1 and the index of refraction of the glass is n2 . Solve: First note that the transmitted light ray forms a transversal between the two parallel faces so that θ 2 = θ3.

Next apply Snell’s law at point A to obtain n1 sin θ1 = n2 sin θ 2 , and substitute θ 2 = θ3 to obtain n1 sin θ1 = n2 sin θ3. Finally apply Snell’s law at point B to obtain n2 sin θ3 = n1 sin θ 4 . Combining these last two equations we obtain n1 sin θ1 = n2 sin θ3 = n1 sin θ 4 , which gives n1 sin θ1 = n1 sin θ 4 . Thus we conclude that θ1 = θ 4 . The transmitted ray

emerges parallel to the incident ray, and it is only slightly displaced if the sheet is thin. Reflect: Since we did not need to specify values for n1 or n2 , this result would also be true if the glass sheet were submerged in water.

23.38. Set Up: na sin θ a = nb sin θb . The light refracts from the liquid into the glass, so na = 1.70, θ a = 62.0°. nb = 1.58.

⎛n ⎞ ⎛ 1.70 ⎞ Solve: sin θb = ⎜ a ⎟ sin θ a = ⎜ ⎟ sin 62.0° = 0.950 and θb = 71.8°. n ⎝ 1.58 ⎠ ⎝ b⎠

23.39. Set Up: The angle of incidence is θ a = 90° − 47.5° = 42.5°. The light refracts from air into the glass, so na = 1.00 and nb = 1.66. Solve: θ r = θ a = 42.5° so the reflected ray makes an angle of 90° − 42.5° = 47.5° with the surface of the glass.

⎛n ⎞ ⎛ 1.00 ⎞ (b) na sin θ a = nb sin θb . sin θb = ⎜ a ⎟ sin θ a = ⎜ ⎟ sin 42.5° = 0.407 and θb = 24.0°. The refracted ray makes an n ⎝ 1.66 ⎠ ⎝ b⎠ angle of 90° − 24.0° = 66.0° with the surface of the glass.


23-12

Chapter 23

23.40. Set Up: υ =

c n

Solve: The time for the beam traveling in air to reach the detector is t =

d 2.50 m = = 8.33 × 10−9 s. The c 3.00 × 108 m/s

light traveling in the block takes time t = 8.33 × 10−9 s + 6.25 × 10−9 s = 1.46 × 10−8 s. The speed of light in the block is υ =

d 2.50 m c 3.00 × 108 m/s 8 The refractive index of the block is = = 1 . 71 × 10 m/s . = = = 1.75. n t 1.46 × 10−8 s υ 1.71 × 108 m/s

⎛ sin 63.0° ⎞ n1 = (1.00) ⎜ ⎟ = 1.14 ⎝ sin 51.3° ⎠ *23.41. Set Up: Snell’s law is na sin θ a = nb sin θb . Light is incident in material a and refracts into material b.

⎛n ⎞ ⎛ 1.25 ⎞ Solve: (a) na = 1.25, nb = 1.66, and θ a = 51.0°. sin θb = ⎜ a ⎟ sin θ a = ⎜ ⎟ sin 51.0° = 0.585 and θb = 35.8°. ⎝ 1.66 ⎠ ⎝ nb ⎠ ⎛n ⎞ ⎛ 1.66 ⎞ (b) Now na = 1.66, nb = 1.25, and θ a = 35.8°. sin θb = ⎜ a ⎟ sin θ a = ⎜ ⎟ sin 35.8° = 0.777 and θb = 51.0°. ⎝ 1.25 ⎠ ⎝ nb ⎠ Reflect: Reflected rays are also reversible.

23.42. Set Up: The figure below shows a light ray from your cell phone, P, which grazes the corner of the pool, C, and reaches your eyes, E. We assume that the pool is full so that the light exits the water at point C. ⎛ 2.00 m ⎞ ° Solve: From the figure we can find θ t using geometry: θ t = tan −1 ⎜ ⎟ = 48.81 . Applying Snell’s law at point ⎝ 1.75 m ⎠ C we obtain

(1.000)sin 48.81D = (1.333)sin θ i ,

which gives θi = 34.37°.

Finally, we can find d from

°

d = (2.50 m) tan 34.37 = 1.71 m. Reflect: The phone would not be visible at it present location if the pool were empty.

23.43. Set Up: The light first refracts from air into the glass and then from the glass into the water. The path of the ray is sketched in the figure below. θ a = 32.0°. nglass = 1.60 and nwater = 1.333. θ ′ a = θb . Snell’s law says na sin θ a = nb sin θb .



23-13

Solve: For the refraction air → glass, nair sin θ a = nglass sin θb . For the refraction glass → water, nglass sin θ ′ a @

⎛ n ⎞ ⎛ 1.00 ⎞ nwater sin θ ′ b . But θ ′ a = θb , so nair sin θ a = nwater sin θ ′ b . sin θ ′ b = ⎜ air ⎟ sin θ a = ⎜ ⎟ sin (32.0°) = 0.3975 ⎝ 1.333 ⎠ ⎝ nwater ⎠ and θ ′ b = 23.4°. Reflect: The angle of the ray in the water is the same as if the glass plate wasn’t there.

23.44. Set Up: Snell’s law says na sin θ a = nb sin θb . Apply Snell’s law to the refraction from material X into the water and then from the water into the air. Solve: (a) material X to water: na = n X , nb = nw = 1.333. θ a = 25° and θb = 48°.

⎛ sin θb ⎞ ⎛ sin 48° ⎞ na = nb ⎜ ⎟ = (1.333) ⎜ ⎟ = 2.34 sin θ ⎝ sin 25° ⎠ a⎠ ⎝ (b) water to air: As the figure below shows, θ a = 48°. na = 1.333 and nb = 1.00. ⎛n ⎞ sin θb = ⎜ a ⎟ sin θ a = (1.333)sin 48° = 82°. ⎝ nb ⎠


23-14

Chapter 23

*23.45. Set Up: na sin θ a = nb sin θb . The light is in diamond and encounters an interface with air, so na = 2.42 and nb = 1.00. The largest θ a is when θb = 90°. Solve: (2.42)sin θ a = (1.00)sin 90°. sin θ a =

1 and θ a = 24.4°. 2.42

23.46. Set Up: The critical angle occurs when light in liquid encounters an interface with air. na = nliq , nb = nair . When θ a = 42.5°, θb = 90°. na sin θ a = nb sin θb . Solve: (a) First solve for nliq : nliq sin 42.5° = nair sin 90°.

nliq = nair

sin 90° 1 = = 1.48. sin 42.5° sin 42.5°

Then na sin θ a = nb sin θb with na = 1.48, nb = 1.00 and θ a = 35.0° gives

⎛n ⎞ (1.48)sin 35.0° sin θb = ⎜ a ⎟ sin θ a = = 0.849 and θb = 58.1°. 1.00 ⎝ nb ⎠ ⎛n ⎞ (1.00)sin 35.0° (b) Now na = 1.00, nb = 1.48 and θ a = 35.0°. sin θb = ⎜ a ⎟ sin θ a = = 0.3876 and θ b = 22.8°. n 1.48 ⎝ b⎠ *23.47. Set Up: The largest angle of incidence for which any light refracts into the air is the critical angle for water → air. The figure below shows a ray incident at the critical angle and therefore at the edge of the ring of light. The radius of this circle is r and d = 10.0 m is the distance from the ring to the surface of the water.

Solve: From the figure, r = d tan θ crit . θ crit is calculated from na sin θ a = nb sin θ b with na = 1.333, θ a = θ crit ,

nb = 1.00 and θ b = 90°. sin θ crit =

(1.00)sin 90° and θ crit = 48.6°. r = (10.0 m) tan 48.6° = 11.3 m. 1.333

A = π r 2 = π (11.3 m) 2 = 401 m 2 .

Reflect: When the incident angle in the water is larger than the critical angle, no light refracts into the air.

23.48. Set Up: 48.7° is the critical angle for glass → water. nwater = 1.333. Solve: na = nglass , nb = nwater . θ a = 48.7° when θ b = 90°. na sin θ a = nb sin θ b gives nglass =

nwater sin 90° (1.333)(1.00) = = 1.77. sin 48.7° sin 48.7°



23-15

*23.49. Set Up: The ray has an angle of incidence of 0° at the first surface of the glass, so it enters the glass without being bent, as shown in the figure below. If no light refracts out of the glass at the glass to air interface, then the incident angle at that interface is θ crit . The figure shows that α + θ crit = 90°.

Solve: (a) For the glass-air interface θ a = θ crit , na = 1.52, nb = 1.00 and θ b = 1.00. na sin θ a = nb sin θ b gives sin θ crit =

(1.00)(sin 90°) and θ crit = 41.1°. α = 90° − θ crit = 48.9°. 1.52

(b) Now the second interface is glass → water and nb = 1.333. na sin θ a = nb sin θ b gives sin θ crit =

and θ crit = 61.3°. α = 90° − θ crit = 28.7°.

(1.333)(sin 90°) 1.52

Reflect: The critical angle increases when the air is replaced by water and rays are bent as they refract out of the glass.

23.50. Set Up: The critical angle for total internal reflection is θ a that gives θ b = 90° in Snell’s law. In the figure below the angle of incidence θ a is related to angle θ by θ a + θ = 90°.

Solve: (a) Calculate θ a that gives θb = 90°. na = 1.60, nb = 1.00 so na sin θa = nb sinθb gives (1.60)sinθa = (1.00)sin90°.

sin θ a =

1.00 and θ a = 38.7°. θ = 90° − θ a = 51.3°. 1.60

(b) na = 1.60, nb = 1.333. (1.60)sin θ a = (1.333)sin 90°. sin θ a =

23.51. Set Up: The critical angle is given by sin θ crit =

1.333 and θ a = 56.4°. θ = 90° − θ a = 33.6°. 1.60

nb , where θ crit is the angle between the incident ray and na

the normal to the axis of the fiber. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

23-16

Chapter 23

Solve: If θ is the angle between the incident ray and the fiber’s longitudinal axis, we must have θ = 90D − θ crit .

Thus we have sin θ crit = cos θ =

nb 1.46 = . Solving for θ we have θ = 9.43D. na 1.48

Reflect: Light rays that do not diverge from the longitudinal axis by more than 9.43D will experience total internal reflection with the sides of the fiber and will consequently be constrained within the confines of the fiber. c n na sin θ a (1.00)sin 57.0° (1.00)sin 57.0° = = 1.36. violet: nb = = 1.40. Solve: (a) red: nb = sin θ b sin 38.1° sin 36.7°

23.52. Set Up: Snell’s law is na sin θ a = nb sin θ b . a = air, b = glass. υ = .

(b) red: υ =

c 3.00 × 108 m/s c 3.00 × 108 m/s = = 2.21 × 108 m/s; violet: υ = = = 2.14 × 108 m/s. n n 1.36 1.40 c

23.53. Set Up: f = , so when λ increases, f decreases. λ Solve: For λ = 400 nm, f =

3.00 × 108 m/s 400 × 10

−9

m

= 7.50 × 1014 Hz.

8

3.00 × 10 m/s

= 4.29 × 1014 Hz. 700 × 10−9 m The graph of n versus f for silicate flint glass is sketched in the figure below. The refractive index increases as f increases. For λ = 700 nm, f =

23.54. Set Up: na sin θ a = nb sin θ b . From Figure 23.29 in the textbook, for λ = 400 nm, n = 1.67 and for λ = 700 nm, n = 1.62. The path of a ray with a single λ is sketched in the figure below.


Electromagnetic Waves Solve: For λ = 400 nm, sin θ b =

23-17

na 1.00 1.00 sin θ a = sin 35.0° and θ b = 20.1°. For λ = 700 nm, sin θ b = sin 35.0° 1.62 nb 1.67

and θ b = 20.7°. Δθ is about 0.6°. λ0 c and υ = . na sin θ a = nb sin θ b . n n Solve: (a) red: n = 1.62; violet: n = 1.67 700 nm 400 nm = 432 nm; λ violet = = 240 nm. (b) λ red = 1.62 1.67 c c υred ⎛ c ⎞ ⎛ nviolet ⎞ nviolet 1.67 ; υ violet = so (c) υred = =⎜ = = = 1.03. The red light has a smaller n so ⎟ nred nviolet υviolet ⎝ nred ⎠ ⎝⎜ c ⎠⎟ nred 1.62

*23.55. Set Up: λ =

travels faster. (d) For red light, sin θ b =

na 1.00 sin θ a = sin 65.0° and θ b = 34.0°. For violet light, nb 1.62 sin θ b =

na 1.00 sin θα = sin 65.0° nb 1.67

and θ b = 32.9°. The angle between the two rays will be Δθ = 34.0° − 32.9° = 1.1°. Reflect: The violet light has a larger n, slows down more in glass and is bent through a greater angle when it enters the glass.

23.56. Set Up: Apply Snell’s law twice: the first time use the index of refraction for red light ( n = 2.41 ) and the second time use the index of refraction for violet light ( n = 2.46 ). Assume that the index of refraction for air is n = 1.00. Solve: For red light Snell’s law gives (1.00)sin 53.5D = (2.41)sin θ red . Solving this equation we find θ red = 19.48D.

For violet light Snell’s law gives (1.00)sin 53.5D = (2.46)sin θ violet . Solving this equation we find θ violet = 19.07D. From these two values Δθ = θ red − θ violet = 0.41D.

we

can

calculate

the

angle

between

the

two

initially

coincident

rays:

Reflect: Violet light is refracted more than red light since it has the larger index of refraction. Although the angular separation between the red and the blue rays is small, it is easily noticeable under the right circumstances.

*23.57. Set Up: For unpolarized light incident on a filter, I = 12 I 0 and the light is linearly polarized along the filter axis. For polarized light incident on a filter, I = I max (cos φ ) 2 , where I max is the intensity of the incident light, and the emerging light is linearly polarized along the filter axis. Solve: (a) After the first filter, I = 12 I 0 and the light is polarized. After the second filter I =

( 12 I0 ) (cos 41.0°)2 =

0.285I 0 . (b) The light is linearly polarized along the axis of the second filter.

23.58. Set Up: After ing through the first filter the light is linearly polarized along the filter axis. After the second filter, I = I max (cos φ ) 2 , where φ is the angle between the axes of the two filters. Solve: (a) I = I 0 (cos 22.5°) 2 = 0.854 I 0 (b) I = I 0 (cos 45.0°) 2 = 0.500 I 0 (c) I = I 0 (cos67.5°) 2 = 0.146 I 0

*23.59. Set Up: When unpolarized light es through a polarizer the intensity is reduced by a factor of

1 2

and

the transmitted light is polarized along the axis of the polarizer. When polarized light of intensity I max is incident on a


23-18

Chapter 23

polarizer, the transmitted intensity is I = I max cos 2φ , where φ is the angle between the polarization direction of the incident light and the axis of the filter. Solve: (a) At point A the intensity is I 0 /2 and the light is polarized along the vertical direction. At point B the intensity is ( I 0 /2)(cos60°) 2 = 0.125I 0 , and the light is polarized along the axis of the second polarizer. At point C the intenisty is (0.125I 0 )(cos30°) 2 = 0.0938 I 0 . (b) Now for the last filter φ = 90° and I = 0. Reflect: Adding the middle filter increases the transmitted intensity.

23.60. Set Up: For unpolarized light incident on a filter, I = 12 I 0 and the light is linearly polarized along the filter axis. For polarized light incident on a filter, I = I max (cos φ ) 2 , where I max is intensity of the incident light, and the emerging light is linearly polarized along the filter axis. Solve: With all three polarizers, if the incident

intensity

is

I0

the

transmitted

intensity

is

2

I 75.0 W/cm = = 293 W/cm 2 . ( 12 I0 ) (cos 23.0°)2 (cos[62.0° − 23.0°])2 = 0.256I0 . I0 = 0.256 0.256 and third polarizers, I = ( 12 I 0 ) (cos62.0°) 2 = 0.110 I 0 = (0.110)(293 W/cm 2 ) = 32.2 W/cm 2 .

I=

With only the first

23.61. Set Up: When unpolarized light es through a polarizer the intensity is reduced by a factor of

1 2

and the

transmitted light is polarized along the axis of the polarizer. When polarized light of intensity I max is incident on a polarizer, the transmitted intensity is I = I max cos 2φ , where φ is the angle between the polarization direction of the incident light and the axis of the filter. Solve: (a) After the first filter I = I 0 /2 and the light is polarized along the vertical direction. After the second filter we want I = I 0 /10, so I 0 /10 = ( I 0 /2)(cos φ ) 2 . cos φ = 2/10 and φ = 63.4°. (b) Now the first filter es the full intensity I 0 of the incident light. For the second filter I 0 /10 = I 0 (cos φ ) 2 . cos φ = 1/10 and φ = 71.6°. Reflect: When the incident light is polarized along the axis of the first filter, φ must be larger to achieve the same

overall reduction in intensity than when the incident light is unpolarized.

23.62. Set Up: The polarizing angle θ p is given by tan θ p =

nb . na = 1.00, nb = nglass . na

Solve: nglass = (1.00)(tan 57.6°) = 1.58

*23.63. Set Up: From Malus’s law, the intensity of the emerging light is proportional to the square of the cosine of the angle between the polarizing axes of the two filters. If the angle between the two axes is θ, the intensity of the emerging light is I = I max cos 2 θ . Solve: At the angle θ we have I = I max cos 2 θ , and at the new angle α we have

of the intensities gives Reflect:

I max cos 2 α I max cos 2 θ

=

1I 2

I

, which gives us cos α =

1 I = I max cos 2 α . Taking the ratio 2

cos θ ⎛ cos θ ⎞ . Solving for α yields α = arccos ⎜ . ⎝ 2 ⎟⎠ 2

Careful! This result is not cos 2 θ .

23.64. Set Up: The reflected beam is completely polarized when θ a = θ p , with tan θ p = Solve: (a) tan θ p =

nb . na = 1.00, nb = nglass . na

nb gives nglass = na tan θ p = (1.00) tan 54.5° = 1.40. na


Electromagnetic Waves (b) na sin θ a = nb sin θ b with θ a = 54.5° gives sin θ b =

23-19

na sin θ a (1.00)(sin 54.5°) = and θ b = 35.5°. Note that nb 1.40

θ a + θb = 90°.

23.65. Set Up: We will apply Malus’s law for the light that es through each of the two polarizing filters: I = I max cos 2 φ. Solve: Assume that the polarization of the incident light makes an angle θ with the axis of the first polarizer. The transmitted intensity through the first polarizer is thus I1 = I max cos 2 θ . The light transmitted through the first polarizer will be polarized along the axis of the first polarizer when it reaches the second polarizer, whose axis is tilted by 35.0D with respect to the first polarizer. Thus, the intensity of light transmitted through the second polarizer is I 2 = I1 cos 2 35.0D = I max cos 2 θ cos 2 35.0D. Applying the condition that only 15.0% of the original beam is transmitted we have I 2 /I max = 0.150 = cos 2 θ cos 2 35.0D. Solving for the original polarization of the light we obtain

θ = 61.8D. Reflect: The polarization of light transmitted through an ideal polarizer will always be rotated so that it is aligned with the axis of the polarizer. Although it is not possible to rotate the direction of polarization by 90D with a single polarizing filter (why?), it can be done with a pair of polarizing filters (can you see how?).

23.66. Set Up: I = 12 ⑀0 cEmax 2 . Emax = cBmax . At the earth the power radiated by the sun is spread over an area of 4π r 2 , where r = 1.50 × 1011 m is the distance from the earth to the sun. P = IA. Solve: (a) Emax =

2I 2(1.4 × 103 W/m 2 ) = = 1.03 × 103 N/C. ⑀0 c (8.854 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s)

Bmax =

Emax 1.03 × 103 N/C = = 3.43 × 10−6 T. c 3.00 × 108 m/s

(b) P = I (4π r 2 ) = (1.4 × 103 W/m 2 )(4π )(1.50 × 1011 m)2 = 4.0 × 1026 W

*23.67. Set Up: The wave speed in air is c = 3.00 × 108 m/s. c = f λ. Emax = cBmax . I = 12 ⑀ 0 cEmax 2 . For a totally absorbing surface the radiation pressure is Solve: (a) f = (b) Bmax =

I . c

c 3.00 × 108 m/s = = 7.81 × 109 Hz λ 3.84 × 10−2 m

1.35 V/m Emax = = 4.50 × 10−9 T c 3.00 × 108 m/s

(c) I = 12 ⑀ 0 cEmax 2 = 12 (8.854 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s)(1.35 V/m) 2 = 2.42 × 10−3 W/m 2 (d) F = (pressure) A =

IA (2.42 × 10−3 W/m 2 )(0.240 m 2 ) = = 1.94 × 10−12 N c 3.00 × 108 m/s

Reflect: The intensity depends only on the amplitudes of the electric and magnetic fields and is independent of the wavelength of the light.

23.68. Set Up: For a totally absorbing surface the radiation pressure is Solve: F = (pressure) A =

I . c

IA (36.0 × 103 W/m 2 )(0.500 m 2 ) = = 6.0 × 10−5 N. A 1 kg mass has a weight of 9.8 N. c 3.00 × 108 m/s

The force the light beam exerts is much too small to be felt by the man.


23-20

Chapter 23

*23.69. Set Up: Energy = Pt . For absorption the radiation pressure is

I P λ , where I = . λ = 0 . I = 12 ⑀0 cEmax 2 c A n

and Emax = cBmax . Solve: (a) Energy = Pt = (250 × 10−3 W)(1.50 × 10−3 s) = 3.75 × 10−4 J = 0.375 mJ. (b) I =

P 250 × 10−3 W = = 1.22 × 106 W/m 2 . The average pressure is A π (255 × 10−6 m) 2 I 1.22 × 106 W/m 2 = = 4.08 × 10−3 Pa. c 3.00 × 108 m/s

v c 3.00 × 108 m/s λ 0 810 nm = = 3.70 × 1014 Hz; f is the same in the air and in the = = 604 nm. f = = n λ λ 0 810 × 10−9 m 1.34 vitreus humor. (c) λ =

(d) E =

2I 2(1.22 × 106 W/m 2 ) = = 3.03 × 104 V/m. ⑀0 c (8.85 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s) Bmax =

Emax = 1.01 × 10−4 T. c

23.70. Set Up: (a) Calculate I and then use Eq. (23.11) to calculate Emax and Eq. (23.4) to calculate Bmax . The intensity is power per unit area: I =

P 3.20 × 10−3 W Emax 2 2 I = = 652 W/m . = , so Emax = 2μ0cI A π (1.25 × 10−3 m) 2 2 μ0c

Emax = 2(4π × 10−7 T ⋅ m/A)(2.998 × 108 m/s)(652 W/m 2 ) = 701 V/m Emax 701 V/m = = 2.34 × 10−6 T c 2.998 × 108 m/s Solve: The magnetic field amplitude is quite small. (b) Set Up: Equations 18.20 and 21.21 give the energy density in of the electric and magnetic field values at any time. We must average over E 2 and B 2 to get the average energy densities. For sinusoidal fields these averages are 12 Emax 2 and 12 Bmax 2 , as explained in the derivation of Eq. (23.9). Bmax =

Solve: The energy density in the electric field is the average of u E = 12 ⑀0 E 2 , which is

energy density in the electric field then is

uE ,av =

2 1⑀ E 4 0 max 2

1.09 × 10−6 J/m3. The energy density in the magnetic field is u B = (2.34 × 10−6 T) 2 4(4π × 10−7 T ⋅ m/A)

=

1 (8.854 × 10 4

2 1⑀ E . 4 0 max −12 2 2

The average

C /N ⋅ m )(701 V/m) 2 =

B B 2 and its average value is u B ,av = max = 2μ0 4 μ0

= 1.09 × 10−6 J/m3 .

Reflect: Our result agrees with the statement in Section 23.5 that the average energy density for the electric field is the same as the average energy density for the magnetic field. (c) Set Up: The total energy in this length of beam is the total energy density uav = u E ,av + u B,av = 2.18 × 10−6 J/m3

times the volume of this part of the beam. Solve: U = uav LA = (2.18 × 10−6 J/m3 )(1.00 m)π (1.25 × 10−3 m) 2 = 1.07 × 10−11 J. Reflect: This quantity can also be calculated as the power output times the time it takes the light to travel L = 1.00 m: 1.00 m ⎛ L⎞ ⎛ ⎞ = 1.07 × 10−11 J, which checks. U = P ⎜ ⎟ = (3.20 × 10−3 W) ⎜ ⎝ c⎠ ⎝ 2.998 × 108 m/s ⎟⎠

*23.71. Set Up: The minimum intensity required for detection can be determined from I = 12 ⑀0 cEmax 2 , where Emax = 10.0 mV/m = 1022 V/m. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


23-21

Solve: Assuming that the intensity of the transmission is distributed uniformly over a sphere of radius r centered on the station, the total power output of the station must be P = 4π r 2 I . Solving for r we obtain

r=

P = 4π I 4π

P

( 12 ⑀0 cEmax 2 )

=

P 2π ⑀0cEmax 2

=

50.0 × 103 W 2π ⑀ 0c(10−2 V/m) 2

= 1.73 × 105 m, which is 173 km.

Reflect: For a given power output, the intensity varies according to I =

P 4π r 2

, which is the inverse-square law for

electromagnetic radiation. 23.72. Set Up: The required average electrical power can be found from the current and voltage: Pav = Vi. Next the required beam intensity can be found from the cross-sectional area: I =

Pav . Finally the required electric field can be A

found from I = 12 ⑀0cEmax 2 . We can also find the associated magnetic field from Emax = cBmax . Pav 1 = ⑀0cEmax 2 we obtain A 2

Solve:

Using I =

Emax =

2Pav 2Vi 2(5.00 ×105 V)(1000 A) = = = 6.14 ×104 V/m. A⑀0c A⑀0c (100 m2 )⑀0 (3.00 ×108 m/s)

Bmax =

Emax 6.14 × 104 V/m = = 2.05 × 10−4 T. c 3.00 × 108 m/s

Reflect: I = Vi /A =

(5.00 × 105 V)(1000 A) 100 m

2

= 5.00 × 106 W/m 2 . This is a very intense beam spread over a large area.

2I . Find the force due to this pressure and c mM sun express the force in of the power output P of the sun. The gravitational force of the sun is Fg = G . r2 Solve: (a) The sail should be reflective, to produce the maximum radiation pressure. P ⎛ 2I ⎞ (b) Frad = ⎜ ⎟ A, where A is the area of the sail. I = , where r is the distance of the sail from the sun. ⎝ c⎠ 4π r 2

23.73. Set Up: For a totally reflective surface the radiation pressure is

PA mM sun PA ⎛ 2 A⎞ ⎛ P ⎞ =G . = . F = Fg so Frad = ⎜ ⎟ ⎜ ⎝ c ⎠ ⎝ 4π r 2 ⎟⎠ 2π r 2c rad 2π r 2c r2 A=

2π cGmM sun 2π (3.00 × 108 m/s)(6.67 × 10−11 N ⋅ m 2 /kg 2 )(10,000 kg)(1.99 × 1030 kg) = . P 3.9 × 1026 W

A = 6.42 × 106 m 2 = 6.42 km 2 .

(c) Both the gravitational force and the radiation pressure are inversely proportional to the square of the distance from the sun, so this distance divides out when we set Frad = Fg . Reflect: A very large sail is needed, just to overcome the gravitational pull of the sun. 23.74. Set Up: na sin θ a = nb sin θ b . na = noil . nb = 1.333. θ a = 30.0° and θ b = 45.0°. Solve: na =

nb sin θ b (1.333)sin (45.0°) = = 1.89 sin θ a sin (30.0°)

*23.75. Set Up: na sin θ a = nb sin θ b . na = 1.00. nb = 1.80. θ b = θ a /2. sin a = 2sin (a /2)cos(a /2). Solve: Snell’s law gives (1.00)sin θ a = (1.80)sin (θ a /2). sin θ a = 2sin (θ a /2)cos (θ a /2), so

2sin (θ a /2)cos(θ a /2) = (1.80)sin (θ a /2). cos(θ a /2) = 0.900. θ a /2 = 25.84° and θ a = 51.7°. Reflect: When the angle of incidence increases the angle of refraction increases. The angle of refraction is smaller than the angle of incidence when nb > na , which is the case here. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

23-22

Chapter 23

23.76. Set Up: The weight of the paper is mg. For a totally absorbing surface the radiation pressure is

I and for a c

2I P . Force is pressure times area. Intensity is I = . c A ⎛ I⎞ Solve: (a) The radiation force must equal the weight of the paper, so ⎜ ⎟ A = mg . ⎝ c⎠ totally reflecting surface it is

I=

mgc (1.50 × 10−3 kg)(9.80 m/s 2 )(3.00 × 108 m/s) = = 7.16 × 107 W/m 2 (0.220 m)(0.280 m) A

(b) I = 12 ⑀0 cEmax 2 . Emax =

2I 2(7.16 × 107 W/m 2 ) = = 2.32 × 105 V/m. ⑀0 c (8.85 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s) Bmax =

(c) The pressure is (d) I =

Emax 2.32 × 105 V/m = = 7.74 × 10−4 T. c 3.00 × 108 m/s

2I mgc ⎛ 2I ⎞ so ⎜ ⎟ A = mg . I = = 3.58 × 107 W/m 2 . ⎝ c⎠ c 2A

0.500 × 10−3 W P = = 637 W/m 2 . The intensity of this laser is much less than what is needed to A π (0.500 × 10−3 m) 2

a sheet of paper. And to the paper, not only must the intensity be large, it also must be over a large area. *23.77. Set Up: As the light crosses the glass-air interface along AB, it is refracted and obeys Snell’s law. Snell’s law is na sin θ a = nb sin θ b , and we know that n = 1.000 for air. At point B the angle of the prism is 30.0D. Solve: Apply Snell’s law at AB. The prism angle at A is 60.0°, so for the upper ray, the angle of incidence at AB is 60.0° + 12.0° = 72.0°. Using this value gives n1 sin 60.0° = sin 72.0° and n1 = 1.10. For the lower ray, the angle of

incidence at AB is 60.0° + 12.0° + 8.50° = 80.5°, giving n2 sin 60.0° = sin 80.5° and n2 = 1.14. Reflect: The lower ray is deflected more than the upper ray because that wavelength has a slightly greater index of refraction than the upper ray.

23.78. Set Up: The angle of incidence at A is to be the critical angle. The ray is sketched the figure below.



23-23

(1.38)sin θ crit = (1.00)sin 90° and θ crit = 46.4°. θ b = 90° − θ crit = 43.6°. Snell’s law applied to the refraction from air to glass at the top of the block gives (1.00)sin θ a = (1.38)sin (43.6°) and θ a = 72.1°.

Solve: For glass → air at point A, Snell’s law gives

23.79. Set Up: The light travels slower in the jelly than in the air and hence will take longer to travel the length of the tube when it is filled with jelly than when it contains just air. The definition of the index of refraction is n = c/v, where v is the speed of light in the jelly. Solve: First get the length L of the tube using air. In the air, we have L = ct = (3.00 × 108 m/s)(8.72 ns) = 2.616 m. L c = = (2.616 m)/(8.72 ns + 2.04 ns) = 2.431 × 108 m/s. Thus we have n = The speed in the jelly is v = v t (3.00 × 108 m/s)/(2.431 × 108 m/s) = 1.23. Reflect: A high-speed timer would be needed to measure times as short as a few nanoseconds.

23.80. Set Up: na sin θ a = nb sin θ b . If θ a is the critical angle then θ b = 90°. For air, nair = 1.00. For heart muscle, nmus =

344 m/s = 0.2324. 1480 m/s

Solve: (a) na sin θ a = nb sin θ b gives (1.00)sin (9.73°) = (0.2324)sin θ b . sin θ b = (b) (1.00)sin θ crit = (0.2324)sin 90° gives θ crit = 13.4°.

sin (9.73°) and θ b = 46.7°. 0.2324

*23.81. Set Up: The rays are incident on the prism in the normal direction so they do not change direction as they enter the prism. Apply Snell’s law to the refraction of the rays as they exit the prism. The rays are sketched in the figure below. The angle between the two rays after they emerge from the prism is 2γ .

Solve: The angle of incidence is A = 25.0°. na sin θ a = nb sin θ b gives (1.66)sin (25.0°) = (1.00)sin θ b . θ b = 44.55°.

β = 90° − θ b = 45.45°. γ + β + A = 90°, so γ = 90° − A − β = 90° − 25.0° − 45.45° = 19.55°. The angle between the emerging rays is 2γ = 39.1°. Reflect: If A → 0°, β → 90° and γ → 0°. The rays remain parallel if the prism is replaced by a rectangular slab of glass.


23-24

Chapter 23

23.82. Set Up: The path of the ray is sketched in the figure below. The ray enters the prism at normal incidence so is not bent. The incident angle at the prism → water interface is to be the critical angle. For water, nwater = 1.333.

Solve: From the figure, θ crit = 45°. na sin θ a = nb sin θ b gives nglass sin 45° = (1.333)sin 90°. nglass =

1.333 = 1.89. sin 45°

*23.83. Set Up: The path of the ray is sketched in the figure below. The problem asks us to calculate θ ′ b .

Solve: Apply Snell’s law to the air → liquid refraction. (1.00)sin (42.5°) = (1.63)sin θ b and θ b = 24.5°. θ b = φ and

φ = θ′ a , so θ ′ a = θ b = 24.5°. Snell’s law applied to the liquid → air refraction gives (1.63)sin (24.5°) = (1.00)sin θ ′ b and θ ′ b = 42.5°. Reflect: The light emerges from the liquid at the same angle from the normal as it entered the liquid.

23.84. Set Up: The angle of incidence for the glass → oil interface must be the critical angle, so θ b = 90°. Solve: na sin θ a = nb sin θ b gives (1.52)sin 57.2° = noil sin 90°. noil = (1.52)sin 57.2° = 1.28. 23.85. Set Up: The polarizing angle θ p is given by tan θ p =

nb . na = 1.00. nb = nglass . na

Solve: (a) nglass = (1.00) tan θ p . Red: nglass = tan 60.0° = 1.73. Blue: nglass = tan 70.0° = 2.75. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Electromagnetic Waves (b) sin θ b =

23-25

na sin θ a . θ b differs more from θ a when nb is larger. The blue light is refracted more upon entering the nb

glass.

23.86. Set Up: Use the change in transit time to find the speed v of light in the slab, and then apply n =

c and v

λ0 . It takes the light an additional 4.2 ns to travel 0.840 m after the glass slab is inserted into the beam. n 0.840 m 0.840 m 0.840 m Solve: − = (n − 1) = 4.2 ns. We can now solve for the index of refraction: c /n c c λ=

(4.2 × 10−9 s)(3.00 × 108 m/s) 490 nm + 1 = 2.50. The wavelength inside of the glass is λ = = 196 nm. 2.50 0.840 m Reflect: Light travels slower in the slab than in air and the wavelength is shorter. n=

*23.87. Set Up: Let the light initially be in the material with refractive index na and let the final slab have refractive index nb . In part (a) let the middle slab have refractive index n1. Apply Snell’s law to each refraction. Solve: (a) 1st interface: na sin θ a = n1 sin θ1.

2nd interface: n1 sin θ1 = nb sin θ b . Combining the two equations

gives na sin θ a = nb sin θ b . This is the equation that would apply if the middle slab were absent. (b) For N slabs, na sin θ a = n1 sin θ1 , n1 sin θ1 = n2 sin θ 2 ,…, nN − 2 sin θ N − 2 = nb sin θ b . Combining all these equations

gives na sin θ a = nb sin θ b . The final direction of travel depends on the angle of incidence in the first slab and the refractive indices of the first and last slabs.

23.88. Set Up: The reflected light is completely polarized when the angle of incidence equals the polarizing angle θ p , where tan θ p =

nb . nb = 1.66. na

Solve: (a) na = 1.00. tan θ p = (b) na = 1.333. tan θ p =

1.66 and θ p = 58.9°. 1.00

1.66 and θ p = 51.2°. 1.333

23.89. Set Up: Apply Snell’s law to the water → ice, and then to the ice → air interfaces. Consider the ray shown in the figure below. We want to find the incident angle θ a at the water-ice interface that causes the incident angle at the ice-air interface to be the critical angle.

Solve: (a) ice-air interface: we have nice sin θ crit = 1.0sin 90° = 1.0 so sin θ crit =

that θ b = θ crit , so sin θ b =

1 . But from the diagram we see nice

1 . nice

water-ice interface: nw sin θ a = nice sin θ b © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

23-26

Chapter 23

1 1 1 = = 0.7502 and θ a = 48.6°. so nw sin θ a = 1.0. Thus, we have sin θ a = nw 1.333 nice (b) Reflect: The angle calculated in part (a) is the critical angle for a water-air interface; the answer would be the same if the ice layer wasn’t there! But sin θ b =

23.90. Set Up: Both l-leucine and d-glutamic acid exhibit linear relationships between concentration and rotation angle. Solve: For l-leucine: rotation angle = (−0.11 deg ⋅ mL)C . For d-glutamic acid: rotation angle = (+0.124 deg ⋅ mL)C . *23.91. Set Up: No light enters the gas because total internal reflection must have occurred at the water-gas interface. At the minimum value of S, the light strikes the water-gas interface at the critical angle. We apply Snell’s law, na sin θ a = nb sinθb , at that surface. S = (1.09 m)/(1.10 m) = 0.991 rad = 56.77°. This is the critical angle. So, using the R refractive index for water from Table 33.1, we get n = (1.333) sin 56.77° = 1.12 (b) (i) The laser beam stays in the water all the time, so ⎛ c ⎞ Dnwater t = 2R/v = 2R/ ⎜ = (2.20 m)(1.333)/(3.00 × 108 m/s) = 9.78 ns ⎟ = c n ⎝ water ⎠ Solve: (a) In the water, θ =

(ii) The beam is in the water half the time and in the gas the other half of the time. Rngas = (1.10 m)(1.12)/(3.00 × 108 m/s) = 4.09 ns tgas = c The total time is 4.09 ns + (9.78 ns)/2 = 8.98 ns Reflect: The gas must be under considerable pressure to have a refractive index as high as 1.12.

Solutions to age Problems 23.92. Set Up: We apply Snell’s law at the point where the light enters the slab and at the point where the light exits the slab. Let the ray exit the first face at an angle of θ (measured from the normal) and let it exit the second face at an angle of φ (measured from the normal). Solve: At the point of entry the angle of transmission is given by 1.00sin 30D = 1.5sin θ . As the transmitted ray es through the slab it forms a transversal between the parallel faces of the slab. Thus, the angle of incidence at the point where the ray exits the second face is θ . Applying Snell’s law at the second face we obtain 1.5sin θ = 1.00sin φ.

Equating these two expressions we obtain 1.00sin 30D = 1.5sin θ = 1.00sin φ. Thus we have 1.00sin 30D = 1.00sin φ , which implies that φ = 30D. Note that this result does not depend on the index of refraction of either the air or the glass and so it was unnecessary to calculate the value of θ . The correct answer is B. *23.93. Set Up: The figure below shows the incident light ray as it es through the oil and then through the glass slab. Note that the light ray forms a transversal through the parallel faces of the oil layer and then through the parallel faces of the glass slab. Let α , β , and γ be the exit angles into the oil, glass, and air as shown below. Solve: At the air-oil interface we apply Snell’s law to obtain 1.00sin 30D = 1.8sin α . Next, at the oil-glass interface we apply Snell’s law to obtain 1.8sin α = 1.5sin β . Finally, at the glass-air interface we apply Snell’s law to obtain 1.5sin β = 1.00sin γ . Equating these three expressions gives 1.00sin 30D = 1.8sin α = 1.5sin β = 1.00sin γ . This implies

that 1.00sin 30D = 1.00sin γ so we now know that γ = 30D. Note that this result does not depend on the index of refraction of the air, the oil or the glass provided that the ray does not experience total internal reflection at the oilglass interface. The correct answer is C.



23-27

23.94. Set Up: We are told that the fractional intensity of light reflected from the boundary of two materials 2

⎛n −n ⎞ depends on the index of refraction for each material according to ⎜ 1 2 ⎟ . ⎝ n1 + n2 ⎠ Solve: Since the glass slab and the fluid have the same index of refraction ( n1 = n2 ) no light will be reflected from the glass-fluid interface. It will be very difficult to see the submerged glass. The correct answer is C.


24

GEOMETRIC OPTICS

Answers to Multiple-Choice Problems 1. C 2. A 3. C 14. B, D 15. B

4. A, D

5. D

6. A, D

7. A

8. B

9. B

10. A

11. E

12. A

13. B, D

Solutions to Problems 24.1. Set Up: For a plane mirror s = − s′ and m = 1.00. Solve: The image is 39.2 cm to the right of the mirror and is 4.85 cm tall.

24.2. Set Up: In the figure below, ray 1 travels from the feet of the giraffe to its eyes and ray 2 travels from the top of its head to its eyes. The total height of the giraffe is hg .


24-1

24-2

Chapter 24

Solve: The two angles labeled θ1 are equal because of the law of reflection, as are the two angles labeled θ 2 . Since

these angles are equal, the two distances labeled y1 are equal and the two distances labeled y2 are equal. The height of the giraffe is hg = 2 y1 + 2 y2 . As the drawing shows, the height of the mirror is hm = y1 + y2 . Comparing, we find that hm = hg /2. The minimum height required is 5 ft, half the height of the giraffe.

*24.3. Set Up: The two mirrors 1 and 2 and the object are shown in Figure (a) below. For each mirror the image is the same distance behind the mirror as the object is in front of the mirror.

Solve: (a) Mirror 1 forms image I1 and mirror 2 forms image I 2 . The image I1 also serves as an object for mirror 2

and image I 3 is formed and is located as shown. Image I 2 serves as an object for mirror 1 and the image is also at the location of I 3 . Three images are formed: I1 , I 2 , and I 3 . (b) Figure (b) above shows a ray from the object to the eye for each image. For images I1 and I 2 there is only one

reflection, from one mirror or the other. For I 3 the ray has two reflections, off one mirror and then off the other. Reflect: All three images are virtual, erect, and the same height as the object. Δs = 2.40 m/s. Δt Solve: Since your image is the same distance behind the mirror as you are in front of the mirror, the distance of the Δd Δs image from you is d = s + 0 s ′0 = 2s. =2 = 4.80 m/s. Your image moves away from you at a speed of 4.80 m/s. Δt Δt

24.4. Set Up: For a plane mirror, s = − s′.


Geometric Optics

*24.5. Set Up: To find the location of the image use the image use m =

y′ s′ =− . y s

24-3

1 1 2 + = . For a concave mirror R > 0. To find the height of s s′ R

1 2 1 2s − R = − = , so s′ R s Rs Rs (10.0 cm)(15.0 cm) = = 7.5 cm. s′ = 2 s − R 30.0 cm − 10.0 cm s′ 7.5 cm 1 m=− =− =− s 15.0 cm 2 1 and y′ = my = − 2 (8.00 mm) = −4.00 mm. The image is 7.5 cm in front of the mirror. The image is 4.0 mm tall and is Solve: (a) s = 15.0 cm.

inverted. 10.0 cm (10.0 cm)(10.0 cm) = 21.00. y′ = my = −8.00 mm. The image is = 10.0 cm. m = 2 10.0 cm 20.0 cm − 10.0 cm 10.0 cm in front of the mirror, at the location of the object. The image is 8.00 mm tall and is inverted. (10.0 cm)(2.50 cm) −5.00 cm (c) s = 2.50 cm. s′ = = −5.00 cm. m = 2 = +2.00. y ′ = my = 16.0 mm. The image is 5.00 cm − 10.0 cm 2.50 cm 5.00 cm behind the mirror. The image is 16.0 mm tall and is erect. (10.0 cm)(1000.0 cm) 5.00 cm (d) s = 1000.0 cm. s′ = = +5.00 cm. m = − = 25.00 × 10−3. y′ = my = −0.040 mm. 2000.0 cm − 10.0 cm 1000.0 cm The image is 5.00 cm in front of the mirror. The image is 0.040 mm tall and is inverted. Rs Reflect: From s′ = we see that the image is real if s > R/2 and virtual if s < R/2. Real images are in front of 2s − R the mirror and are inverted. Virtual images are behind the mirror and are erect.

(b)

s = 10.0 cm. s′ =

24.6. Set Up: To find the location of the image use the image use m =

y′ s′ =2 . y s

1 1 2 + = . For a convex mirror R < 0. To find the height of s s′ R

1 2 1 2s − R Rs (−10.0 cm)(15.0 cm) = = −3.75 cm. = − = , so s′ = s′ R s Rs 2s − R 30.0 cm + 10.0 cm 3.75 cm s′ m =2 =2 = +0.250. 15.0 cm s y′ = my = +2.00 mm. The image is 3.75 cm behind the mirror. It is 2.00 mm tall and is upright.

Solve: (a) s = 15.0 cm.

(−10.0 cm)(10.0 cm) s′ −3.33 cm = −3.33 cm. m = − = − = +0.333. 20.0 cm + 10.0 cm 10.0 cm s y′ = my = +2.67 mm. The image is 3.33 cm behind the mirror. It is 2.67 mm tall and is upright.

(b) s = 10.0 cm. s′ =

( −10.0 cm)(2.50 cm) s′ −1.67 cm = −1.67 cm. m = − = 2 = +0.667. 5.00 cm + 10.0 cm 2.50 cm s y′ = my = +5.33 mm. The image is 1.67 cm behind the mirror. It is 5.33 mm tall and is upright.

(c) s = 2.50 cm. s′ =

(−10.0 cm)(1000.0 cm) = −4.98 cm. 2000.0 cm + 10.0 cm s′ −4.98 cm m=− =− = +4.98 × 10−3. 1000.0 cm s y′ = my = +0.040 mm. The image is 4.98 cm behind the mirror. It is 0.040 mm tall and is upright.

(d) s = 1000.0 cm. s =

24.7. Set Up: f = +1.75 m.

1 1 1 0 y ′0 s′ + = . m = − . 0 m0 = . s s′ f y s


24-4

Chapter 24

Solve: (a) s f so s ′ = f = 1.75 m. m = −

s′ 1.75 m =− = −3.14 × 10−11. s 5.58 × 1010 m

0 y′0 = 0 m 0 y = (3.14 × 10−11 )(6.794 × 106 m) = 2.13 × 10−4 m = 0.213 mm. (b) The real image is 1.75 m in front of the mirror.

24.8. Set Up: For a concave mirror R > 0.

1 1 1 sf s′ . m=− . + = and s′ = s s′ f s− f s

R 34.0 cm = = 17.0 cm 2 2 sf (22.0 cm)(17.0 cm) 74.8 cm s′ = = 74.8 cm. m = − = − (b) s′ = = −3.40. s− f 22.0 cm − 17.0 cm 22.0 cm s

Solve: (a) f =

0 y′0 = 0 m 0 y = (3.40)(7.50 mm) = 25.5 mm. The image is 74.8 cm in front of the mirror and is 25.5 mm tall. (c) The image formation by the mirror is determined by the law of reflection and that is unaffected by the medium in which the light is traveling. The focal length remains 17.0 cm.

*24.9. Set Up: For a convex mirror, R < 0, so R = −18.0 cm and f = Solve: (a)

R = −9.00 cm. 2

1 1 1 sf (1300 cm)(−9.00 cm) = = −8.94 cm. + = . s′ = s s′ f s − f 1300 cm − ( − 9.00 cm)

−8.94 cm s′ =− = 6.88 × 10−3. 0 y′0 = 0 m 0 y = (6.88 × 10−3 )(1.5 m) = 0.0103 m = 1.03 cm. 1300 cm s (b) The height of the image is much less than the height of the car, so the car appears to be farther away than its actual distance. Reflect: Problem 24.11 shows that the image formed by a convex mirror is always virtual and smaller than the object. m =2

24.10. Set Up: Since the mirror is convex we know that f = − image by using

R = −12.5 cm. We can find the location of the 2

1 1 1 s′ + = and the magnification from m = − . s s′ f s

Solve: (a) Using s = 10.0 cm we have

1 1 1 + = , which gives s' = −5.56 cm. The magnification is 10.0 cm s′ −12.5 cm

s′ −5.56 cm = +0.556 × . thus m = − = − s 10.0 cm (b) Here we use s = 10.0 cm + 10.0 cm = 20.0 cm. We now have

1 1 1 + = , which gives s′ = −7.69 cm. 20.0 cm s′ −12.5 cm

s′ −7.69 cm =− = +0.385 × . s 20.0 cm Reflect: The image of your ears is considerably more reduced in size than the image of your nose.

The magnification is thus m = −

*24.11. Set Up: The shell behaves as a spherical mirror. The equation relating the object and image distances to the focal length of a spherical mirror is Solve:

1 1 1 s′ + = , and its magnification is given by m = − . s s′ f s

1 1 1 1 2 1 + = ⇒ = − ⇒ s = 18.0 cm s s′ f s −18.0 cm −6.00 cm

from the vertex.

m=−

s′ −6.00 cm 1 =− = ⇒ s 18.0 cm 3

1 y′ = (1.5 cm) = 0.50 cm. The image is 0.50 cm tall, erect, and virtual. 3 Reflect: Since the magnification is less than one, the image is smaller than the object.


Geometric Optics

24-5

1 1 1 sf s′ + = . m=− so s′ = so real images are always inverted and virtual images are s s′ f s− f s always upright. For a concave mirror, f > 0.

24.12. Set Up:

Solve: (a) When s > f then s − f is positive and s′ =

sf s′ is positive. The image is real and inverted. m = − = s s− f

f . As s → f , 0 m 0 → ∞ and as s becomes very large, 0 m 0 → 0. The magnification is always negative but there s− f is no limit on its magnitude. f sf (b) When s < f then s − f is negative and s′ = is negative. The image is virtual and erect. m = − = s− f s− f −

f . As s → f , 0 m 0 → ∞ and as s → 0, 0 m 0 → 1. The magnification is positive and varies between 1 and ∞. f −s R = 16.0 cm. 2 1 1 1 sf (12.0 cm)(16.0 cm) s′ −48.0 cm = = − 48.0 cm. m = − = − Solve: (a) + = . s ′ = = +4.00. 12.0 cm − 16.0 cm s− f s s′ f s 12.0 cm (b) s ′ = −48.0 cm, so the image is 48.0 cm to the right of the mirror. s ′ < 0 so the image is virtual. (c) The principal-ray diagram is sketched in the figure below. The rules for principal rays apply only to paraxial rays. Principal ray 2, that travels to the mirror along a line that es through the focus, makes a large angle with the optic axis and is not described well by the paraxial approximation. Therefore, principal ray 2 is not included in the sketch.

*24.13. Set Up: For a concave mirror, R > 0. R = 32.0 cm and f =

Reflect: A concave mirror forms a virtual image whenever s < f .

24.14. Set Up: For a concave mirror, R > 0. R = 22.0 cm and f =

R = 11.0 cm. 2

Solve: (a) The principal-ray diagram is sketched in the figure below.


24-6

(b)

Chapter 24

1 1 1 sf (16.5 cm)(11.0 cm) s′ 33.0 cm = = 33.0 cm. m = − = − + = . s′ = = −2.00. s s′ f 16.5 cm − 11.0 cm s− f s 16.5 cm

0 y′0 = 0 m 0 y = (2.00) × (0.600 cm) = 1.20 cm. The image is 33.0 cm to the left of the mirror. It is 1.20 cm tall. s′ > 0, so the image is real. m < 0, so the image is inverted.

*24.15. Set Up: For a convex mirror, R < 0. R = −22.0 cm and f =

R = −11.0 cm. 2

Solve: (a) The principal-ray diagram is sketched in the figure below.

(b)

1 1 1 sf (16.5 cm)( −11.0 cm) s′ −6.6 cm = = −6.6 cm. m = − = − = +0.400. + = . s′ = s s′ f s 16.5 cm s − f 16.5 cm − (−11.0 cm )

0 y′0 = 0 m 0 y = (0.400)(0.600 cm) = 0.240 cm. The image is 6.6 cm to the right of the mirror. It is 0.240 cm tall. s′ < 0, so the image is virtual. m > 0, so the image is erect.

24.16. Set Up: For a convex mirror, R < 0. We are given that R = 22.0 m and so f = that s = 5.0 m.

R = −1.0 m. We also know 2

1 1 1 s′ 0.83 m sf (5.0 m)(−1.0 m) = = = −0.83 m. Thus, we have m = − = − + = as s′ = 5.0 m s s′ f s s − f 5.0 m − ( −1.0 m) +0.17 × . Since s′ < 0 the image is virtual. Since 1 > m > 0 the image is erect and reduced in size compared to the object. (b) The principal-ray diagram is similar to Figure 24.19 (b) in the textbook. Solve: (a) Rewrite

24.17. Set Up: For the orientation of the shell shown in the figure in the problem, R = +12.0 cm. When the glass is reversed, so the seed faces a convex surface, R = −12.0 cm. Solve: (a) R = +12.0 cm.

1 1 2 y′ s′ + = and m = = − . s s′ R y s

1 2 1 2s − R = − = and s′ R s Rs Rs (12.0 cm)(15.0 cm) = = +10.0 cm. s′ = 2s − R 30.0 cm − 12.0 cm

s′ 10.0 cm =− = −0.667. y′ = my = −2.20 mm. The image is 10.0 cm to the left of the shell vertex and is s 15.0 cm 2.20 mm tall. (12.0 cm)(15.0 cm) −4.29 cm = −4.29 cm. m = − = +0.286. y ′ = my = 0.944 mm. The image (b) R = −12.0 cm. s′ = 15.0 cm 30.0 cm + 12.0 cm is 4.29 cm to the right of the shell vertex and is 0.944 mm tall. m=−


Geometric Optics

24-7

Reflect: In (a), s > R/2 and the mirror is concave, so the image is real. In (b) the image is virtual, because a convex mirror always forms a virtual image.

24.18. Set Up: Solve: (a) s′ =

1 1 1 y′ s′ + = and m = = − . m = +2.00 and s = 1.25 cm. An erect image must be virtual. s s′ f y s sf f and m = − . For a concave mirror, m can be larger than 1.00. For a convex mirror, s− f s− f

0 f 0 = − f so m = + (b)

0f0 and m is always less than 1.00. The mirror must be concave ( f > 0). s+0f0

1 s′ + s ss ′ s′ = . f = . m = − = +2.00 and s′ = −2.00s. f ss ′ s + s′ s

s (−2.00s ) = +2.00s = +2.50 cm. s − 2.00s R = 2 f = +5.00 cm. The principal ray diagram is drawn in the figure below. f =

*24.19. Set Up: For a convex surface, R > 0. R = +3.00 cm. na = 1.00, nb = 1.60. Solve: (a) s → ∞.

na nb nb − na + = . s s′ R

⎛ nb ⎞ nb nb − na ⎛ 1.60 ⎞ = . s′ = ⎜ ⎟R = ⎜ ⎟ (+3.00 cm) = +8.00 cm. The image is 8.00 cm to n − n 1.60 − 1.00 ⎠ s′ R ⎝ ⎝ b a⎠

the right of the vertex. 1.00 1.60 1.60 − 1.00 (b) s = 12.0 cm. + = . s′ = +13.7 cm. The image is 13.7 cm to the right of the vertex. 12.0 cm s′ 3.00 cm 1.00 1.60 1.60 − 1.00 (c) s = 2.00 cm. + = . s′ = −5.33 cm. The image is 5.33 cm to the left of the vertex. 2.00 cm s′ 3.00 cm

24.20. Set Up: R = +3.00 cm. na = n (the refractive index of the liquid). nb = 1.60. s = 90.0 cm. s′ = +160 cm. na nb nb − na + = . s s′ R 1 1 1 1 ⎞ n 1.60 1.60 − n ⎛ ⎞ ⎛ + − Solve: + = . n⎜ ⎟ = 1.60 ⎜ ⎟ . n = 1.52. 90.0 cm 160 cm 3.00 cm ⎝ 90.0 cm 3.00 cm ⎠ ⎝ 3.00 cm 160 cm ⎠

24.21. Set Up: R = +4.00 cm. na = 1.00. nb = 1.60. s = 24.0 cm.

na nb nb − na n s′ + = . m=− a . s s′ R nb s


24-8

Chapter 24

1 1.60 1.60 − 1.00 (1.00)(14.8 cm) = − 0.385. y′ 0 = 0 m 0 y = (0.385)(1.50 mm) = + = . s′ = +14.8 cm. m = 2 (1.60)(24.0 cm) 24.0 cm s′ 4.00 cm 0.578 mm. The image is 14.8 cm to the right of the vertex and is 0.578 mm tall. m < 0, so the image is inverted. Reflect: The image is real. Solve:

⎛ nb − na ⎞ na s′ ⎞ ⎟ . Note that the ⎟ and Eq. (24.12) ⎜ m = − nb s ⎠ R ⎠ ⎝ situation is similar to what is shown in Figure 24.21 in the textbook with the shark tooth located at P; however— since na > nb —the transmitted ray is refracted upward (not downward) to form a virtual image to the left of the ⎛n

n

24.22. Set Up: We will apply Eq. (24.11) ⎜ a + b = s′ ⎝ s

vertex V. We set na = 1.333, nb = 1.00, s = 0.450 m, and R = 11.75 m. 1.333 1.00 1.00 − 1.333 + = . Thus s′ = −0.317 m and we have that the image is located 0.450 m s′ 1.75 m inside the tank and 31.7 cm from the plastic. (1.333)( −0.317 m) = +0.940 × . Since m > 0 we know that the image is erect. (b) The magnification is equal to m = − (1.00)(0.450 m) (c) Since we know that m = +0.940 ×, the size of image of the shark’s tooth will be 0.940(5.00 cm) = 4.70 cm.

Solve: (a) We have

*24.23. Set Up: Light comes from the fish to the person’s eye. R = −14.0 cm. s = +14.0 cm. na = 1.333 (water). nb = 1.00 (air).

na nb nb − na n s′ + = . m = − a . The figure below shows the object and the refracting surface. s s′ R nb s

1.333 1.00 1.00 − 1.333 (1.333)( −14.0 cm) = +1.33. + = . s′ = 214.0 cm. m = − 14.0 cm s′ −14.0 cm (1.00)(14.0 cm) The fish’s image is 14.0 cm to the left of the bowl surface so it is at the center of the bowl and the magnification is 1.33. n n −n (b) The focal point is at the image location when s → ∞. b = b a . na = 1.00. nb = 1.333. R = +14.0 cm. s′ R 1.333 1.333 − 1.00 = . s′ = +56.0 cm. s′ is greater than the diameter of the bowl, so the surface facing the sunlight 14.0 cm s′ does not focus the sunlight to a point inside the bowl. The focal point is outside the bowl and there is no danger to the fish. Reflect: In part (b) the rays refract when they exit the bowl back into the air so the image we calculated is not the final image.

Solve: (a)


Geometric Optics

24.24. Set Up: The image formed by refraction at the surface of the eye is located by na = 1.00, nb = 1.35. R > 0. For a distant object, s ≈ ∞ and

24-9

na nb nb − na + = . s s′ R

1 ≈ 0. s

1.35 1.35 − 1.00 and R = 0.648 cm = 6.48 mm. = 2.5 cm R 1.00 1.35 1.35 − 1.00 1.35 = 0.500 and s′ = 2.70 cm = 27.0 mm. The (b) R = 0.648 cm and s = 25 cm: + = . s′ 25 cm 0.648 s′ image is formed behind the retina. 1.35 1.35 − 1.00 (c) Calculate s′ for s ≈ ∞ and R = 0.50 cm: = . s′ = 1.93 cm = 19.3 mm. The image is formed in 0.50 cm s′ front of the retina.

Solve: (a) s ≈ ∞ and s′ = 2.5 cm:

24.25. Set Up: The surface is flat so R → ∞ and

na nb + = 0. The light travels from the speck to the eye, so s s′

na = 1.309 and nb = 1.00. s = 3.50 cm.

⎛n ⎞ ⎛ 1.00 ⎞ Solve: s′ = − ⎜ b ⎟ s = − ⎜ ⎟ (3.50 cm) = −2.67 cm. The speck appears to be 2.67 cm below the surface. ⎝ 1.309 ⎠ ⎝ na ⎠

24.26. Set Up: The object is the bird. na = 1.00 (air). nb = 1.333 (water). s′ is the image distance, measured from the surface of the lake. s = 7.0 m. For refraction at a plane surface,

na nb + = 0. s s′

⎛n ⎞ ⎛ 1.333 ⎞ Solve: s′ = − ⎜ b ⎟ s = − ⎜ ⎟ (7.0 m) = −9.3 m. The image of the bird is 9.3 m above the surface of the lake, so it n ⎝ 1.00 ⎠ ⎝ a⎠ is 2.0 m + 9.3 m = 11.3 m above the diver.

*24.27. Set Up: Assume that the walls of the aquarium are flat and parallel so that we may use Eq. (24.13) at each surface:

⎛n ⎞ na nb + = 0, which gives s′ = − ⎜ b ⎟ s. Imagine that we are facing the tank so that we see one group of s s′ ⎝ na ⎠

spectators on our left and the other group on our right (with the tank between them). Solve: First let us calculate the location of the image of the left-hand spectators that is formed by the left face of the ⎛n ⎞ ⎛ 1.333 ⎞ tank. Here we have na = 1.00, nb = 1.333, and s = 1.20 m. We have s′ = − ⎜ b ⎟ s = − ⎜ ⎟ (1.20 m) = −1.60 m, ⎝ 1.000 ⎠ ⎝ na ⎠ so that this image appears 1.60 meters to the left of the left face of the tank. Next we treat this image as an object for the right face of the tank. In this case we use s = 5.50 m + 1.60 m = 7.10 m, na = 1.333, and nb = 1.00. We have ⎛n ⎞ ⎛ 1.00 ⎞ s′ = − ⎜ b ⎟ s = − ⎜ ⎟ (7.10 m) = −5.33 m. Thus, the second image is 5.33 meters to the left of the right face of the n ⎝ 1.333 ⎠ ⎝ a⎠ tank and this is located 1.20 m + 5.33 m = 6.53 meters from the spectators on the right side. Reflect: The apparent distance between the spectators has been reduced from 5.00 m + 2(1.20 m) = 7.40 meters to 6.53 meters due to the presence of the tank.

24.28. Set Up: The object is the diving board. na = 1.00 (air). nb = 1.333 (water). The image is 5.20 m − 0.80 m = 4.40 m above the surface of the swimming pool, so s′ = −4.40 m. Calculate s. For refraction at a plane surface, na nb + = 0. s s′ ⎛n ⎞ ⎛ 1.00 ⎞ Solve: s = − ⎜ a ⎟ s′ = − ⎜ ⎟ ( −4.40 m) = 3.30 m. The diving board is 3.30 m above the surface of the water. n ⎝ 1.333 ⎠ ⎝ b⎠


24-10

Chapter 24

*24.29. Set Up: m = Solve: m =

y′ s′ 1 1 1 = 2 . Since the image is erect, y′ > 0 and m > 0. + = . s s′ f y s

1 1 1 y′ 1.30 cm s′ gives = = +3.25. m = − = +3.25 gives s′ = −3.25s. + = s s′ f s y 0.400 cm

1 1 1 + = and s = 4.85 cm. s −3.25s 7.00 cm s′ = −(3.25)(4.85 cm) = −15.8 cm. The object is 4.85 cm to the left of the lens. The image is 15.8 cm to the left of the lens and is virtual. Reflect: The image is virtual because the object distance is less than the focal length.

24.30. Set Up: m =

y′ s′ 1 1 1 = 2 . Since the image is inverted, y′ < 0 and m < 0. + = . s s′ f y s

1 1 1 1 1 1 s′ y′ −4.50 cm + = gives + gives s′ = +1.406 s. and = = = −1.406. m = − s 1.406 s 90.0 cm s s′ f s 3.20 cm y s = 154 cm. s′ = (1.406)(154 cm) = 217 cm. The object is 154 cm to the left of the lens. The image is 217 cm to the

Solve: m =

right of the lens and is real. s′ and s both s and s′ are positive, so m < 0. The height of the image is three times the height of the object, so m = −3 and 1 1 1 s ′ = 3s. + = . s s′ f

24.31. Set Up: Since the image is projected onto a wall, the image is real and s′ > 0. s′ = 1.80 m. m = −

s ′ 1.80 m = = 0.60 m. You are 0.60 m from the lens. 3 3 (b) m < 0, so the image is inverted. 1 s′ + s (0.60 m)(1.80 m) ss′ (c) = . f = = = 0.45 m. f > 0, and the lens is converging. f ss ′ s + s′ 0.60 m + 1.80 m Reflect: Since the image is real the lens must be converging and your distance from the lens must be greater than the focal length. Solve: (a) s ′ = 3s so s =

1 1 1 y′ s′ + = . The type of lens determines the sign of f . m = = − . The sign of s′ depends on s s′ f y s whether the image is real or virtual. s′ = +15.0 cm; s′ is positive because the image is on the side of the lens opposite to the object. s = 35.0 cm − 15.0 cm = 20.0 cm. 1 s + s′ (20.0 cm)(15.0 cm) ss′ Solve: (a) = and f = = = +8.57 cm. f is positive so the lens is converging. f ss ′ s + s′ 20.0 cm + 15.0 cm 15.0 cm s′ (b) m = 2 = 2 = 20.750. y ′ = my = (−0.750)(2.50 mm) = −1.88 mm. The image is 1.88 mm tall. s′ > 0 20.0 cm s and the image is real.

24.32. Set Up:

1 1 1 y′ s′ + = . The type of lens determines the sign of f . m = = − . The sign of s′ determines s s′ f y s whether the image is real or virtual. s = +8.00 cm. s′ = −3.00 cm. s′ is negative because the image is on the same side of the lens as the object. 1 s + s′ (8.00 cm)(−3.00 cm) ss ′ Solve: (a) = and f = = = −4.80 cm. f is negative so the lens is diverging. 8.00 cm − 3.00 cm s + s′ f ss ′

*24.33. Set Up:

(b) m = −

−3.00 cm s′ =2 = +0.375. y ′ = my = (0.375)(6.50 mm) = 2.44 mm. s′ < 0 and the image is virtual. s 8.00 cm


Geometric Optics

24-11

Reflect: A converging lens can also form a virtual image, if the object distance is less than the focal length. But in that case 0 s ′0 > s and the image would be farther from the lens than the object is.

1 1 1 y′ s′ + = . The type of lens determines the sign of f . m = = − . The sign of s′ depends on s s′ f y s whether the image is real or virtual. s = 16.0 cm. s′ = −22.0 cm; s′ is negative because the image is on the same side of the lens as the object. ss ′ 1 s + s′ (16.0 cm)(−22.0 cm) = = +58.7 cm. f is positive so the lens is converging. Solve: (a) = and f = s + s′ 16.0 cm − 22.0 cm f ss ′

24.34. Set Up:

(b) m = −

s′ −22.0 cm =2 = 1.38. y ′ = my = (1.38)(3.25 mm) = 4.48 mm. s′ < 0 and the image is virtual. 16.0 cm s

24.35. Set Up: We will apply the thin-lens equation (Equation 24.20):

⎛ 1 1 1 ⎞ = (n − 1) ⎜ − ⎟ . This is a converging f ⎝ R1 R2 ⎠

lens with identical surfaces — which implies that R1 > 0, R2 < 0, and f > 0. Thus, we have R1 = +22.0 cm, R2 = −22.0 cm, and f = +20.0 cm. R1R2 − (22.0 cm) 2 +1= + 1 = 1.55. (20.0 cm)(−44.0 cm) f ( R2 − R1) Reflect: According to the sign conventions, R > 0 if the center of curvature is on the outgoing side — and negative otherwise. Since our lens has two convex faces, the front face has a positive radius of curvature and the back face has a negative radius of curvature. Solve: Solving the thin-lens equation for n we obtain n =

24.36. Set Up: We will apply the thin-lens equation (Equation 24.20):

⎛ 1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ . In this case the f R R 2⎠ ⎝ 1

center of curvature of each lens is on the outgoing side so we have R1 = +11.8 cm and R2 = +6.80 cm. Solve: We can now find the focal length:

⎛ 1 1 1 ⎞ 1 1 ⎞ ⎛ = ( n − 1) ⎜ − − ⎟ = (1.67 − 1) ⎜ ⎟ . Thus, we have f ⎝ 11.8 cm 6.80 ⎠ ⎝ R1 R2 ⎠

f = −24 cm. Reflect: According to the sign conventions, R > 0 if the center of curvature is on the outgoing side—and negative otherwise. In this case R > 0 for both faces since there is a convex face on the front and a concave face on the back.

*24.37. Set Up:

⎛ 1 1 1 ⎞ = (n − 1) ⎜ − ⎟ . 1 is the surface closest to the object. R is positive if the center of curvature f ⎝ R1 R2 ⎠

is on the side of the lens opposite the object. For a flat surface, R → ∞. 1 1 1 ⎛ ⎞ Solve: (a) L1: and f = +11.0 cm. = (0.5) ⎜ − ⎝ 11.5 cm −10.5 cm ⎟⎠ f L2 :

(b) L1:

1 1 1 ⎛ ⎞ = (0.5) ⎜ − ⎟ and ⎝ f 10.5 cm −11.5 cm ⎠ f = +11.0 cm.

1 1 ⎛1 ⎞ = (0.5) ⎜ − and f = +17.0 cm. ⎝ ∞ −8.50 cm ⎟⎠ f L2 :

1 1 1⎞ ⎛ = (0.5) ⎜ − ⎟ and ⎝ f 8.50 cm ∞ ⎠ f = +17.0 cm.


24-12

(c) L1:

Chapter 24

1 1 1 ⎞ ⎛ = (0.5) ⎜ − and f = +133 cm. ⎝ 9.80 cm 11.5 cm ⎟⎠ f 1 1 1 ⎛ ⎞ = (0.5) ⎜ − and ⎝ −9.80 cm −11.5 cm ⎟⎠ f f = −133 cm.

L2 :

(d) L1:

1 1 1⎞ ⎛ = (0.5) ⎜ − and f = −18.4 cm. ⎝ −9.20 cm ∞ ⎟⎠ f L2 :

(e) L1:

1 1 ⎛1 ⎞ = (0.5) ⎜ − ⎟ and f ⎝ ∞ 9.20 cm ⎠ f = −18.4 cm.

1 1 1 ⎞ ⎛ = (0.5) ⎜ − ⎟ and f = −11.0 cm. ⎝ f −10.4 cm 11.6 cm ⎠ L2 :

24.38. Set Up: Use Solve: (a)

1 1 1 ⎞ ⎛ = (0.5) ⎜ − ⎟ and ⎝ f −11.6 cm 10.4 cm ⎠ f = −11.0 cm.

⎛ 1 1 1 1 1 1 ⎞ to locate the image. s = 18.0 cm. = (n − 1) ⎜ − ⎟ to calculate f and then use + = s s′ f f ⎝ R1 R2 ⎠

1 1 1 s− f 1 1 1 ⎛ ⎞ = − = . = (0.5) ⎜ − and f = +12.0 cm. ⎝ 10.0 cm −15.0 cm ⎟⎠ s′ f s sf f s′ =

(18.0 cm)(12.0 cm) f = = +36.0 cm. 18.0 cm − 12.0 cm s− f

The image is 36.0 cm to the right of the lens. 1 1 1⎞ ⎛ (b) = (0.5) ⎜ − ⎟ and f = +20.0 cm. f ⎝ 10.0 cm ∞ ⎠ s′ =

(18.0 cm)(20.0 cm) sf = = −180 cm. 18.0 cm − 20.0 cm s− f

The image is 180 cm to the left of the lens. 1 1 1 ⎞ ⎛ (c) = (0.5) ⎜ − and f = −12.0 cm. ⎝ −15.0 cm 10.0 cm ⎟⎠ f s′ =

(18.0 cm)(−12.0 cm) sf = = −7.2 cm. 18.0 cm + 12.0 cm s− f

The image is 7.2 cm to the left of the lens. 1 1 1 ⎛ ⎞ (d) = (0.5) ⎜ − and f = −60.0 cm. ⎝ −10.0 cm −15.0 cm ⎟⎠ f (18.0 cm)( −60.0 cm) sf = = −13.8 cm. 18.0 cm + 60.0 cm s− f The image is 13.8 cm to the left of the lens. s′ =

*24.39. Set Up: m=

⎛ 1 1 1 1 1 1 ⎞ + = . = ( n − 1) ⎜ − ⎟ . If R is the radius of the lens, then R1 = R and R2 = − R. s s′ f f R R 2⎠ ⎝ 1

y′ s′ =− . y s


Geometric Optics

Solve: (a)

24-13

⎛ 1 1 1 ⎞ 1 ⎞ 2(n − 1) ⎛1 = (n − 1) ⎜ − . ⎟ = (n − 1) ⎜ − ⎟= f R ⎝ R −R ⎠ ⎝ R1 R2 ⎠ R = 2(n − 1) f = 2(0.44)(8.0 mm) = 7.0 mm.

(b)

1 1 1 s− f (30.0 cm)(0.80 cm) sf = − = . s′ = = = 0.82 cm = 8.2 mm. The image is 8.2 mm from the lens, s′ f s sf 30.0 cm − 0.80 cm s− f

s′ 0.82 cm =− = −0.0273. y ′ = my = ( −0.0273)(16 cm) = 0.44 cm = 4.4 mm. 30.0 cm s s′ > 0 so the image is real. m < 0 so the image is inverted.

on the side opposite the object. m = −

Reflect: The lens is converging and has a very short focal length. As long as the object is farther than 7.0 mm from the eye, the lens forms a real image.

24.40. Set Up: Solve: (a) (b)

⎛ 1 1 1 1 y′ s′ 1 1 ⎞ = ( n − 1) ⎜ − ⎟ . R1 = +5.0 mm. + = . m = = − . s s ′ f y s f ⎝ R1 R2 ⎠

1 1 1 1 1 1 1 1 = − = − and R2 = 18.6 mm. = − . f (n − 1) R1 R2 R2 R1 f (n − 1) +5.0 mm (18.0 mm)(0.38)

1 1 1 s− f (25 cm)(1.8 cm) sf = − = . s′ = = = 1.9 cm = 19 mm. s′ f s sf 25 cm − 1.8 cm s− f

1.9 cm s′ =− = −0.076. 25 cm s so the image is inverted. (c) m = −

24.41. Set Up: R2 → ∞.

y ′ = my = (−0.076)(8.0 mm) = −0.61 mm. s′ > 0 so the image is real. m < 0

⎛ 1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ . R1 → ∞. R2 = −13.0 cm. If the lens is reversed, R1 = +13.0 cm and f ⎝ R1 R2 ⎠

1 1 1 y′ s′ + = . m = =2 . s s′ f y s

Solve: (a)

1 1 1 s− f 1 1 0.70 ⎛1 ⎞ = − = . = (0.70) ⎜ − = and f = 18.6 cm. ⎝ ∞ −13.0 cm ⎟⎠ 13.0 cm s′ f s sf f s′ =

(22.5 cm)(18.6 cm) sf = = 107 cm. 22.5 cm − 18.6 cm s− f

107 cm s′ =− = −4.76. 22.5 cm s y ′ = my = (−4.76)(3.75 mm) = −17.8 mm. The image is 107 cm to the right of the lens and is 17.8 mm tall. The m=−

image is real and inverted. 1 1 1⎞ ⎛ = (n − 1) ⎜ − and f = 18.6 cm. The image is the same as in part (a). ⎝ 13.0 cm ∞ ⎟⎠ f Reflect: Reversing a lens does not change the focal length of the lens (Problem 24.44).

(b)

24.42. Set Up: For a distant object the image is at the focal point of the lens. Therefore, f = 1.87 cm. For the double-convex lens, R1 = + R and R2 = − R, where R = 2.50 cm. Solve:

⎛ 1 1 1 ⎞ 1 ⎞ 2(n − 1) ⎛1 = (n − 1) ⎜ − . ⎟ = (n − 1) ⎜ − ⎟= f R ⎝ R −R ⎠ ⎝ R1 R2 ⎠ n=

2.50 cm R +1= + 1 = 1.67. 2f 2(1.87 cm)


24-14

Chapter 24

*24.43. Set Up: Use the thin-lens equation to calculate the focal length of the lens:

⎛ 1 1 1 1 ⎞ + = ( n − 1) ⎜ − ⎟ . The s s′ ⎝ R1 R2 ⎠

s′ . s ⎛ 1 ⎛ ⎞ 1 1 1 ⎞ 1 1 1 1 Solve: (a) + = (n − 1) ⎜ − + = (1.52 − 1) ⎜ − ⎟⇒ ⎟ s s′ ⎝ −7.00 cm −4.00 cm ⎠ ⎝ R1 R2 ⎠ 24.0 cm s′ ⇒ s ′ = 71.2 cm, to the right of the lens.

magnification of the lens is m = −

71.2 cm s′ =− = −2.97 24.0 cm s Reflect: Since the magnification is negative, the image is inverted. (b) m = −

24.44. Set Up: Apply

y′ s′ 1 1 1 + = and m = = − . We know that f = +12.0 cm and s′ = −17.0 cm. s s′ f y s

1 1 1 1 1 1 ( −17.0) s′ y ′ 0.800 cm = + = ⇒ = − ⇒ s = 7.0 cm. m = − = − = +2.4 ⇒ y = = +2.4 7.0 s m s s′ f s 12.0 cm −17.0 cm +0.34 cm, so the object is 0.34 cm tall and on the same side as the image. The image is erect since m > 0.

Solve:

Reflect: When the object is inside the focal point, a converging lens forms a virtual, enlarged image.

*24.45. Set Up: Apply Solve: (a) Lens 1:

1 1 1 y= y= + = to each lens. m1 = 1 and m2 = 2 . s s′ f y1 y2

1 1 1 + = gives s s′ f s1= =

(50.0 cm)(40.0 cm) s1 f1 = = +200 cm. s1 − f1 50.0 cm − 40.0 cm m1 = −

200 cm s1= =− = −4.00. 50 cm s1

y1= = m1 y1 = (−4.00)(1.20 cm) = −4.80 cm. The image I1 is 200 cm to the right of lens 1, is 4.80 cm tall and is

inverted. (b) Lens 2: y2 = −4.80 cm. The image I1 is 300 cm − 200 cm = 100 cm to the left of lens 2, so s2 = +100 cm. s2= =

(100 cm)(60.0 cm) s2 f 2 = = +150 cm. s2 − f 2 100 cm − 60.0 cm m2 = −

150 cm s2= =− = −1.50. 100 cm s2

y2= = m2 y2 = (−1.50)( −4.80 cm) = +7.20 cm. The image is 150 cm to the right of the second lens, is 7.20 cm tall, and is inverted with respect to the original object. Reflect: The overall magnification of the lens combination is mtot = m1 m2 . 1 1 1 s′ + = with f = ±35.0 cm. We know that the magnification is m = − . s s s′ f Solve: (a) We want the size of the image to be twice that of the object and so we must have m = ±2. Since the image s′ is real we know that s′ > 0, which implies that m = −2 = − . Thus we conclude that s ′ = 2s. Now we can s 1 1 1 1 3 1 3 = = . Solving for s we obtain s = f . Since we know that determine the location of the object: + = + 2 s s′ s 2s 2s f


s > 0 we must have that f = +35.0 cm, and thus s =

3 3 f = (35.0 cm) = 52.5 cm. 2 2


Geometric Optics

24-15

(b) We again want the image to be twice the size as the object; however, in this case we have a virtual image so s′ 1 1 1 s′ < 0 and m = +2 = − . Thus, we have s ′ = −2s. Now we can determine the location of the object: + = + s s s′ s 1 1 1 1 = = . Solving for s we obtain s = f . Since we know that s > 0 we must have that f = +35.0 cm and −2s 2s f 2

1 thus s = (35.0 cm) = 17.5 cm. 2 Reflect: For a diverging lens we have

than

1 1 1 1 + = < 0. This can only occur if is negative and larger in magnitude s′ s s′ f

1 s′ . Thus we have qmq = − < 1. It follows that the image is always smaller than the object for a diverging lens. s s

In this exercise qmq = 2 > 1, so only a converging lens will work.

24.47. Set Up: The image formed by the first lens serves as the object for the second lens. mtot = m1 m2 . gives s ′ =

1 1 1 + = s s′ f

sf . s− f

Solve: (a) Lens 1: f1 = −12.0 cm, s1 = 20.0 cm.

s′ =

(20.0 cm)(−12.0 cm) = −7.5 cm. 20.0 cm + 12.0 cm

m1 = −

s1= −7.5 cm =− = +0.375. 20.0 cm s1

Lens 2: The image of lens 1 is 7.5 cm to the left of lens 1 so is 7.5 cm + 9.00 cm = 16.5 cm to the left of lens 2. s2 = +16.5 cm. f 2 = +12.0 cm. s2= =

(16.5 cm)(12.0 cm) = 44.0 cm. 16.5 cm − 12.0 cm

m2 = −

44.0 cm s2= =− = −2.67. 16.5 cm s2

The final image is 44.0 cm to the right of lens 2 so is 53.0 cm to the right of the first lens. (b) s2= > 0 so the final image is real. (c) mtot = m1 m2 = ( +0.375)(−2.67) = −1.00. The image is 2.50 mm tall and is inverted. Reflect: The light travels through the lenses in the direction from left to right. A real image for the second lens is to the right of that lens and a virtual image is to the left of the second lens.

1 1 1 s′ + = with the conditions that s + s ′ = ±214 cm and m = − . s s s′ f Solve: Since the size of the image is greater than the size of the object we know that the image must be farther from the lens than the object. This implies that the focal length of the lens is positive and the lens is converging. s9 We know that the image is real so that s′ > 0 . In this case we have s + s ′ = 1214 cm and m = − = −1 2 3 × . Thus, s

24.48. Set Up: We will apply

we may write s ′ = 53 s and s + s ′ = 83 s = +214 cm. Solving for s and s′ we obtain s = 80.25 cm and s′ = 133.75 cm. This gives f =

ss ′ (80.25 cm)(133.75 cm) = = 50.2 cm. s + s′ 214 cm


24-16

Chapter 24

Reflect: For a diverging lens we have

1 1 1 1 + = < 0. This can only occur if is negative and larger in magnitude s s′ f s′

1 s′ . Thus we have m = − < 1. It follows that the image is always smaller than the object for a diverging lens. s s In this exercise m = 5 3 > 1, so only a converging lens will work.

than

*24.49. Set Up: f = +14.0 cm.

1 1 1 sf y′ s′ + = gives s ′ = . m = =2 . y s s s′ f s− f

Solve: s = 18.0 cm sf (18.0 cm)(14.0 cm) = = 63.0 cm. The image is 63.0 cm to the right of the lens. (a) s ′ = s− f 18.0 cm − 14.0 cm s′ 63.0 cm =− = −3.50 s 18.0 cm (c) s′ > 0 so the image is real. (d) m < 0 so the image is inverted. The principal-ray diagram is sketched in Figure (a) below.

(b) m = −

s = 7.00 cm


Geometric Optics

(e) s ′ =

24-17

sf (7.00 cm)(14.0 cm) = = −14.0 cm. The image is 14.0 cm to the left of the lens. s− f 7.00 cm − 14.0 cm

−14.0 cm s′ =− = +2.00 s 7.00 cm (g) s′ < 0 so the image is virtual. (h) m > 0 so the image is erect. The principal-ray diagram is sketched in Figure (b) above. Reflect: For a converging lens, when s > f the image is real and when s < f the image is virtual.

(f) m = −

24.50. Set Up: The thin-lens equation applies in this case. The thin-lens equation is magnification is m = − Solve: (a) m =

1 1 1 + = , and the s s′ f

s′ y′ = . s y

−12.0 cm y ′ 34.0 mm s′ 1 1 = = 4.25 = − = − ⇒ s = 2.82 cm. The thin-lens equation gives + = y 8.00 mm s s s s′

1 ⇒ f = 3.69 cm. f (b) The object is located at s = 2.82 cm, which is 2.82 cm to the left of the lens and on the same side as the image. (c) The principal-ray diagram is similar to Figure 24.37 (e) in the text. Reflect: Since the focal length is positive, this is a converging lens. The image distance is negative because the object is inside the focal point of the lens.

24.51. Set Up: Use Eq. (24.21) to calculate the object distance s. Calculate m from Eq. (24.18) and use that to determine the size and orientation of the image. We are given f = 248.0 cm and we have a virtual image 17.0 cm from lens so s′ = −17.0 cm. 1 1 1 1 1 1 s− f Solve: (a) + = , so = − = s s′ f s′ f s sf s=

s′ f (−17.0 cm)( −48.0 cm) = = +26.3 cm s ′ − f −17.0 cm − (−48.0 cm)

−17.0 cm s′ =− = +0.646 +26.3 cm s y′ q y q 8.00 mm = = 12.4 mm m= so q y q = y 0.646 qmq (b) The principal-ray diagram is sketched in the figure below. Reflect: We have a virtual image and a real object (s > 0), so image and object are on same side of the lens. Since m > 0 we know that the image is erect with respect to the object. The height of the object is 12.4 mm. m=−


24-18

Chapter 24

24.52. Set Up:

1 1 1 + = . The sign of f determines whether the lens is converging or diverging. s = 16.0 cm. s s′ f

s ′ = −12.0 cm. Solve: (a) f = (b) m = −

ss ′ (16.0 cm)(−12.0 cm) = = −48.0 cm. f < 0 and the lens is diverging. s + s ′ 16.0 cm + (−12.0 cm)

s′ −12.0 cm =− = +0.750. 0 y ′0 = 0 m 0 y = (0.750)(8.50 mm) = 6.38 mm. m > 0 and the image is erect. s 16.0 cm

(c) The principal-ray diagram is sketched in the figure below.

*24.53. Set Up: s = 12.0 cm. The ray intersects the optic axis 8.0 cm to the right of the lens. Solve: (a) The ray is bent toward the optic axis by the lens so the lens is converging. (b) The ray is parallel to the optic axis and is bent so that it es through the focal point to the right of the lens; f = 8.0 cm. (c) The principal ray diagram is drawn in the figure below. The diagram shows that the image is 24.0 cm to the right of the lens.

sf 1 1 1 (12.0 cm)(8.0 cm) + = gives s ′ = = = +24.0 cm. The calculated image position agrees with the s− f 12.0 cm − 8.0 cm s s′ f principal ray diagram. Reflect: The lens is converging and s > f , so the image is real. (d)

24.54. Set Up: s = 6.0 cm. The ray is traveling toward a point on the optic axis that is 6.0 cm to the right of the lens. Solve: (a) The ray is bent away from the axis by the lens so the lens is diverging. (b) The ray is parallel to the optic axis after it es through the lens, so it was initially traveling toward the focus. f = 26.0 cm.


Geometric Optics

24-19

(c) The principal ray diagram is drawn in the figure below. The diagram shows that the image is 3.0 cm to the left of the lens.

sf 1 1 1 (6.0 cm)(−6.0 cm) + = gives s ′ = = = −3.0 cm. The calculated image position agrees with the s − f 6.0 cm − ( −6.0 cm) s s′ f principal ray diagram.

(d)

*24.55. Set Up: s = 10.0 cm. If extended backwards the ray comes from a point on the optic axis 18.0 cm from the lens and is parallel to the optic axis after it es through the lens. Solve: (a) The ray is bent toward the optic axis by the lens so the lens is converging. (b) The ray is parallel to the optic axis after it es through the lens so it comes from the focal point; f = 18.0 cm. (c) The principal ray diagram is drawn in the figure below. The diagram shows that the image is 22.5 cm to the left of the lens.

sf 1 1 1 (10.0 cm)(18.0 cm) + = gives s ′ = = = −22.5 cm. The calculated image position agrees with the s− f 10.0 cm − 18.0 cm s s′ f principal ray diagram.

(d)

24.56. Set Up: s = 16.0 cm. The ray is initially parallel to the optic axis and is bent by the lens so it appears to come from a point 10.0 cm to the left of the lens. Solve: (a) The ray is bent away from the optic axis by the lens so the lens is diverging. (b) The ray is initially parallel to the optic axis so it is bent such that if it is extended backwards it appears to come from the focus. f = 210.0 cm.


24-20

Chapter 24

(c) The principal ray diagram is drawn in the figure below. The diagram shows that the image is 6.2 cm to the left of the lens.

sf 1 1 1 (16.0 cm)(−10.0 cm) + = so s ′ = = = −6.2 cm. The calculated image position agrees with the s − f 16.0 cm − ( −10.0 cm) s s′ f principal ray diagram.

(d)

na nb nb − na with R → ∞, since we assume that the surfaces are flat. The image + = s s′ R formed by the first interface serves as the object for the second interface. Solve: For the water-benzene interface to get the apparent water depth: na nb 1.33 1.50 + = 0 ⇒ s ′ = −7.33 cm. + =0⇒ s s′ 6.50 cm s′ For the benzene-air interface, to get the total apparent distance to the bottom: na nb 1.50 1 + = 0 ⇒ s ′ = −6.62 cm. + =0⇒ s s′ (7.33 cm + 2.60 cm) s ′ Reflect: At the water-benzene interface the light refracts into material of greater refractive index and the overall effect is that the apparent depth is greater than the actual depth.

*24.57. Set Up: Apply

1 1 2 + = . R > 0. s s′ R s′ 1 1 2 0.600 2 Solve: m = − = +2.50. s ′ = −2.50 s. + = and s = 0.300 R. s ′ = −2.50 s = (−2.50)(0.300 R ) = = . s s −2.50 s R s R −0.750R. The object is a distance of 0.300R in front of the mirror and the image is a distance of 0.750R behind the mirror.

24.58. Set Up: m = +2.50.

s′ so m is negative and s m = −2.25. The object, mirror, and wall are sketched in the figure below. The sketch shows that s ′ − s = 400 cm.

24.59. Set Up: Since the image is projected onto the wall it is real and s ′ > 0. m = −


Geometric Optics

24-21

s′ and s ′ = 2.25s. s ′ − s = 2.25s − s = 400 cm and s = 320 cm. s ′ = 400 cm + 320 cm = 720 cm. s 1 1 2 1 1 2 + = . R = 4.43 m. The mirror should be 7.20 m from the wall. + = . s s ′ R 320 cm 720 cm R Reflect: The focal length of the mirror is f = R /2 = 222 cm. s > f , as it must if the image is to be real. Solve: m = −2.25 = −

24.60. Set Up: Since the image is projected onto a screen the image is real and s ′ > 0. s ′ = 800 cm. m = −

s′ so s

36.0 cm = −60.0. 0.600 cm s′ s′ 800 cm Solve: (a) m = −60.0 = − . s ′ = 60.0 s. s = = = 13.3 cm. 60.0 60.0 s 1 1 2 1 1 2 (b) + = . + = . R = 26.2 cm. s s ′ R 13.3 cm 800 cm R

m is negative; m = −

*24.61. Set Up: We will apply the thin-lens equation Eq. (24.21):

⎛ 1 1 1 1 1 ⎞ + = = (n − 1) ⎜ − ⎟ . If the object is s s′ f ⎝ R1 R2 ⎠

on the convex side of the lens the center of curvature of each lens is on the outgoing side so the radius of curvatures are both positive: R1 = +12.2 cm and R2 = +15.4 cm. Solve: (a) Using the given value we have

1 1 1 1 ⎞ ⎛ . Solving for s ′ we obtain + = (1.67 − 1) ⎜ − ⎝ 12.2 cm 15.4 cm ⎟⎠ 35.0 cm s ′

s ′ = −58 cm. −58 cm s′ =− = −1.7 × . s 35.0 cm Reflect: If the object were on the concave side of the lens the center of curvature of each lens would be on the incoming side, and so the radius of curvatures would then both be negative and switched in order: 1 1 ⎞ ⎛ 1 1 ⎛ ⎞ R1 = −15.4 cm and R2 = −12.2 cm. However, since ⎜ − = − the answer ⎝ 12.2 cm 15.4 cm ⎟⎠ ⎜⎝ −15.4 cm −12.2 cm ⎟⎠ would be the unchanged.

(b) The magnification is m = −

24.62. Set Up: The mirror surface is convex so R = −3.00 cm. s = 24.0 cm − 3.00 cm = 21.0 cm. Solve: f =

R 1 1 1 = −1.50 cm. + = . 2 s s′ f

sf (21.0 cm)(−1.50 cm) = = −1.40 cm. s − f 21.0 cm − (−1.50 cm) The image is 1.40 cm behind the surface so it is 3.00 cm − 1.40 cm = 1.60 cm from the center of the ornament, on the same side as the object. s′ −1.40 cm m=− =− = +0.0667. s 21.0 cm 0 y ′0 = 0 m 0 y = (0.0667)(3.80 mm) = 0.253 mm. s′ =

*24.63. Set Up: We will apply the thin-lens equation (Equation 24.21):

⎛ 1 1 1 ⎞ = (n − 1) ⎜ − ⎟ , when n = 1.55 and f ⎝ R1 R2 ⎠

f = 20.0 cm. Solve: Since f > 0 we choose R1 = R and R2 = − R, where R is the magnitude of the radius of curvature. Thus we

have

1 1 ⎞ 2(n − 1) ⎛1 . Solving for R we obtain R = 2(n − 1) f = 2(1.55 − 1)(20.0 cm) = 22 cm. = ( n − 1) ⎜ − = ⎝ R − R ⎟⎠ f R


24-22

Chapter 24

Reflect: For identical convex surfaces, the relation between f and R is f =

relation for spherical mirrors, which is f =

1 R ⋅ . This is reminiscent of the n −1 2

R . 2

24.64. Set Up: f = +3.00 cm. The positions of the object, lens, and screen are sketched in the figure below. Since the image is projected onto a screen, the image is real and s ′ > 0. The figure shows that s ′ + s = 18.0 cm and s ′ = 18.0 cm − s.

Solve: (a)

1 1 1 1 1 1 18.0 cm 1 + = . + = . = . s (18.0 cm − s ) = 54.0 cm 2 . s2 − (18.0 cm)s + s s ′ f s 18.0 cm − s f s (18.0 cm − s ) 3.00 cm

54.0 cm 2 . s =

1 2

(18.0 ±

)

(18.0) 2 − 4(54.0) cm. s = (9.00 ± 5.20) cm. s = 14.2 cm or s = 3.8 cm. The lens can be

14.2 cm or 3.8 cm from the object. (b) s = 14.2 cm: s′ = 18.0 cm − 14.2 cm = 3.8 cm. m = −

14.2 cm. m = −

s′ 3.8 cm =− = −0.268. s = 3.8 cm: s′ = 18.0 cm − 3.8 cm = s 14.2 cm

s′ 14.2 cm =− = −3.74. s 3.8 cm

*24.65. Set Up: For the second image, the image formed by the mirror serves as the object for the lens. For the mirror, f m = +10.0 cm. For the lens, f = 32.0 cm. The center of curvature of the mirror is R = 2 f m = 20.0 cm to 1 1 1 sf + = . gives s ′ = s s′ f s− f Solve: (a) The principal-ray diagrams from the two images are sketched in Figures (a) and (b) below. In Figure (b), only the image formed by the mirror is shown. This image is at the location of the candle so the principal ray diagram that shows the image formation when the image of the mirror serves as the object for the lens is analogous to that in Figure (a) and is not drawn.

the right of the mirror vertex.


Geometric Optics

24-23

(b) Image formed by the light that es directly through the lens: The candle is 85.0 cm to the left of the lens. sf (85.0 cm)(32.0 cm) s′ = = = +51.3 cm. s− f 85.0 cm − 32.0 cm s′ 51.3 cm =− = −0.604. s 85.0 cm This image is 51.3 cm to the right of the lens. s′ > 0 so the image is real. m < 0 so the image is inverted. Image formed by the light that first reflects off the mirror: First consider the image formed by the mirror. The candle is 20.0 cm to the right of the mirror, so s = +20.0 cm. sf (20.0 cm)(10.0 cm) s′ = = = 20.0 cm. s− f 20.0 cm − 10.0 cm m=−

m1 = −

s1= 20.0 cm =− = −1.00. s1 20.0 cm

The image formed by the mirror is at the location of the candle, so s2 = +85.0 cm and s2= = 51.3 cm. m2 = −0.604. mtot = m1 m2 = (−1.00)(−0.604) = 0.604. The second image is 51.3 cm to the right of the lens. s2= > 0, so the final image is real. mtot > 0, so the final image is erect. Reflect: The two images are at the same place. They are the same size. One is erect and one is inverted.

24.66. Set Up: Refer to Figure 24.15a in Section 24.2. R and s′ are negative, so R = 20 R 0 and s ′ = 20 s ′.0 h h h h h h tan α = . tan β = . sin φ = . For small angles, α = , β = and φ = . β = θ + φ and θ = α + φ. s 0 s ′0 0 R0 s 0 s ′0 0 R0 Eliminating θ gives β = α + 2φ. Substituting in the expressions for the angles gives replacing 0 s′0 by − s ′ and 0 R 0 by − R gives

h h 2h = + . Dividing by h and 0 s ′0 s 0 R 0

1 1 2 1 1 2 = + . Rearranging gives + = , which is Equation (24.4). − s′ s − R s s′ R


24-24

Chapter 24

Now refer to Figure 24.15b in Section 24.2. Triangles PVQ and P ′VQ ′ are similar, so

0 s ′0 = − s ′ so m = −

y y′ y ′ 0 s ′0 = . m= = . s 0 s ′0 y s

s′ , which is Eq. (24.7). s

24.67. Set Up: The image formed by the first surface serves for the object for the second surface. Since the lens is thin, s2 = − s1= . nliq n n − nliq nb nc na nb nb − na , where na = nliq and nb = n (refractive index of the lens). + = + = + = . s s′ R s2 s2= s1 s1= R1 n nliq nliq − n n n − nliq nliq n nliq n − nliq n nc − nb n − . 2 = = + , where nb = n and nc = nliq . so + = . == . = =− R R s s s2 s2= R2 s R s1 s1 s2 1 1 2 2 2

Solve:

n − nliq R1

−

nliq s1

=+

nliq s2=

+

n − nliq R2

.

⎛ 1 1 1 1 1 1 1 1 ⎞ + = (n/nliq − 1) ⎜ − gives = (n /nliq − 1) + = = ⎟ . Comparing to s f ′ f ′ s1 s2= R R s 1 2⎠ ⎝ 1 2

⎛ 1 1 ⎞ ⎜ − ⎟ , as was to be shown. ⎝ R1 R2 ⎠ (b)

⎛ n − nliq ⎞ ⎛ nliq (n − 1) ⎞ ⎛ 1 1 1 ⎞ = ( n − 1) ⎜ − ⎟. f ′ = ⎜ ⎟ f , as was to be shown. ⎟ . f ( n − 1) = f ′(n/nliq − 1) = f ′ ⎜⎜ ⎟ ⎜ n − nliq ⎟ f ⎝ R1 R2 ⎠ ⎝ nliq ⎠ ⎝ ⎠

Solutions to age Problems ⎞⎛ 1 1 ⎛ n 1 ⎞ =⎜ − 1⎟ ⎜ − ⎟. ⎜ ⎟ f ⎝ nliq ⎠ ⎝ R1 R2 ⎠ Solve: Note that the case of a glass lens immersed in air can be obtained as a special case of the given equation by 1 choosing nliq = 1. As the value of nliq increases to approach n, we see that approaches zero—which mean that f f

24.68. Set Up: We will apply the given equation:

increases without bound. Thus, if we move a glass lens (n = 1.5) from air (nliq = 1.00) into water (nliq = 1.33) the focal length of the lens will increase. The correct answer is C.

*24.69. Set Up: We will apply the given equation:

⎞⎛ 1 1 ⎛ n 1 ⎞ =⎜ − 1⎟ ⎜ − . ⎟ ⎝ R1 R2 ⎟⎠ f ⎜⎝ nliq ⎠

⎛ 1 1 ⎞ Solve: Since the lens is concave we know that ⎜ − ⎟ < 0 and so f < 0 when the lens is immersed in air R R 2⎠ ⎝ 1 ⎛ n ⎞ (nliq = 1.00). If the lens is now placed in a liquid where nliq > n, we have ⎜ − 1⎟ < 0 so that ⎜ nliq ⎟ ⎝ ⎠ ⎞⎛ 1 1 ⎛ n 1 ⎞ =⎜ − 1⎟ ⎜ − ⎟ > 0. Thus, the concave lens is a converging lens when placed in a liquid with a high enough ⎜ ⎟ f ⎝ nliq ⎠ ⎝ R1 R2 ⎠ index of refraction. The correct answer is D. R . 2 Solve: The law of reflection does not depend on the index of refraction and the focal length of the mirror is given by R f = independent of the surrounding material. The correct answer is D. 2

24.70. Set Up: The focal length of a mirror is given by f =


25

OPTICAL INSTRUMENTS

Answers to Multiple-Choice Problems 1. C 2. B 3. C 14. A 15. D

4. C

5. C

6. B, D

7. D

8. C

9. A, D

10. B, C, E

11. D

12. B

13. D

Solutions to Problems *25.1. Set Up: The f-number N is N =

f and is expressed as f/N. The exposure time is inversely proportional to D

the aperture diameter. Solve: (a) f /4 means

f 300 mm f = 75 mm. = 4 and D = = 4 4 D 2

2

1πD ⎛D ⎞ f t A 2 (b) If t is the exposure time, then 1 − 2 = 4 = ⎜ 2 ⎟ . For f /8, D = = 37.5 mm. 8 t2 A1 1 π D12 ⎝ D1 ⎠ 4

2

⎛ D ⎞ ⎛ 1 ⎞⎛ 75 mm ⎞2 4 1 t2 = t1 ⎜ 1 ⎟ = ⎜ s ⎟⎜ s= s ⎟ = D 250 37.5 mm 250 62.5 ⎠⎝ ⎠ ⎝ 2⎠ ⎝ Reflect: Larger f-number means smaller aperture diameter and larger exposure time.

25.2. Set Up: The exposure time is inversely proportional to the aperture area. 2

2

Solve: If t is the exposure time, then

⎛D ⎞ t1 A2 14 π D2 − = =⎜ 2 ⎟ . 2 1 t2 A1 π D1 ⎝ D1 ⎠ 4 2

⎛D ⎞ ⎛ 1 t2 = t1 ⎜ 1 ⎟ = ⎜ ⎝ D2 ⎠ ⎝ 30

25.3. Set Up: We have the relation: f-number =

2

1 ⎞⎛ 8.00 mm ⎞ s ⎟⎜ s. ⎟ = 250 ⎠⎝ 23.1 mm ⎠

f . The notation f/N means that the lens or mirror has an D

f-number of N. f 1.325 m = = f /22.1. D 0.0600 m (b) f = (f-number) (diameter) = 3.3(200 inches) = 660 inches

Solve: (a) f-number =

(c) The minimum f-number is

2.50 cm 2.50 cm = f/3.1 and the maximum f-number is = f /13. 0 .20 cm 0.80 cm


25-1

25-2

Chapter 25

Reflect: The wide range of f-numbers available to the human eye (which is comparable to that of many camera lenses) allows the eye to function well under a wide range of light intensities.

25.4. Set Up: f-number =

f . f/N means an f-number of N. f = 135 mm for the telephoto lens and f = 50 mm D

for the second lens. The light intensity is proportional to D 2 . 135 mm 135 mm 135 mm Solve: (a) Dmin = = 6.1 mm; Dmax = = 48 mm. For f/11, D = = 12 mm. 2.8 11 22 50 mm 50 mm (b) Dmin = = 2.3 mm; Dmax = = 18 mm 2.8 22 (c)

Dmax 2 Dmin 2

2

⎛ 22 ⎞ =⎜ = 62 ⎝ 2.8 ⎟⎠

(d) Keep D 2t constant, where t is the time the shutter is open. Dmin 2 tmin = Dmax 2 tmax so 2

⎛ D ⎞ ⎛ 1 ⎞⎛ 2.8 ⎞2 1 s. tmax = tmin ⎜ min ⎟ = ⎜ ⎟⎜ ⎟ = 30 s 22 1900 D ⎠⎝ ⎠ ⎝ max ⎠ ⎝ 1 The closest shutter speed setting is s. 2000

25.5. Set Up: The image is real since it is projected onto the film. s′ = 20.4 cm. Solve:

1 1 1 s′ f (20.4 cm)(20.0 cm) = = 10.2 m + = . s= s s′ f s′ − f 20.4 cm − 20.0 cm

Reflect: The object distance s is much greater than f, so the image is just beyond the focal point of the lens.

25.6. Set Up: Apply

1 1 1 + = and calculate s′ for each s. f = 90 mm s s′ f

1 1 1 1 1 1 + = ⇒ s ′ = 96.7 mm. + = ⇒ s s′ f 1300 mm s ′ 90 mm 1 1 1 1 1 1 + = ⇒ s′ = 91.3 mm. + = ⇒ s s′ f 6500 mm s′ 90 mm ⇒ Δs′ = 96.7 mm − 91.3 mm = 5.4 mm toward the film

Solve:

Reflect: s′ =

sf . For f > 0 and s > f , s′ decreases as s increases. s− f

*25.7. Set Up: Solve: (a) s′ =

s′ y′ 1 1 1 + = . m = 2 = . The photocells must be at the image location. s y s s′ f

sf (4.25 m)(0.0750 m) = = 0.0763 m = 7.63 cm s− f 4.25 m − 0.0750 m

s′ 0.0763 m =− = −0.0180. y′ = my = ( −0.0180)(1.85 m) = −0.0333 m = −3.33 cm. The image is inverted s 4.25 m and 3.33 cm tall. s′ > 0 so the image is real.

(b) m = −

(c)

0.0333 m 8.0 × 10−6 m/pixel

= 4.2 × 103 pixels

Reflect: Since the image is projected onto the photocell, it must be real. Any real image formed by a single lens is inverted. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Optical Instruments

25-3

25.8. Set Up: From the known height of the giraffe and the space available for its image we can determine the 1 1 1 s′ magnification, m = − . From the relation + = we can then determine s. s s s′ f Solve: We want the inverted image of the 4.50-meter tall giraffe to fit within 4.82 mm. Thus we have s ′ y ′ −0.00482 1 1 1 to obtain m=− = = = −1.071 × 10−3. This gives, s ′ = 1.071 × 10−3 s. Next we apply + = s y 4.50 m s s′ f 1 1 934.6 1 + = = , which gives us s = 934.6 f = 934.6(50 mm) = 46.7 m. − 3 s 1.071 × 10 s s f 1 1 1 s′ y ′ + = . m = − = . You want 0 y′0 = 36 mm. s s′ f s y s′ > 0, m < 0 and y′ = −36 mm.

*25.9. Set Up:

Solve: (a) m =

y ′ −36 × 10−3 m 1 1 1 = = −0.0180. s ′ = − ms = −( −0.0180) s = 0.0180 s. Use this in + = : s s′ f y 2.00 m 1 1 1 + = s 0.0180 s 50.0 × 10−3 m

and s = 2.83 m. (b) s ′ = 0.0180 s = 5.09 cm

1 1 1 sf s′ y ′ + = so s ′ = . m=− = . s s′ f s− f s y

25.10. Set Up: Solve: (a) s ′ =

f = 50.0 mm. The image will be real, so

(1.50 m)(6.50 × 10−3 m) 1.50 m − 6.50 × 10−3 m

= 6.53 mm. m = −

s′ 6.53 × 10−3 m =− = −4.35 × 10−3. s 1.50 m

0 y ′0 = 0 m 0 y = (4.35 × 10−3 )(2.00 m) = 8.70 mm (b) s ′ =

(1.50 m)(19.5 × 10−3 m) 1.50 m − 19.5 × 10

−3

m

= 19.8 mm. m = −

s′ 19.8 × 10−3 m =− = −1.32 × 10−2. s 1.50 m

0 y ′0 = 0 m 0 y = (1.32 × 10−2 )(2.00 m) = 26.4 mm (c) The 19.5 mm distance is the telephoto focal length, since it produces greater magnification of the image.

*25.11. Set Up: Solve: (a) s ′ =

1 1 1 s′ y′ + = . m =2 = . s s′ f s y

sf (12.0 cm)(11.5 cm) = = 276 cm = 2.76 m s− f 12.0 cm − 11.5 cm

s′ 276 cm =− = −23.0. The dimensions of the image are (23.0)(24 mm) = 552 mm = 0.552 m and s 12.0 cm (23.0)(36 mm) = 828 mm = 0.828 m

(b) m = −

25.12. Set Up: We know that on the array the width of each pixel is

3.30 cm and the height of each pixel is 800 pixels

3.30 cm s′ . We can then use the magnification of the lens, m = − , to calculate the width and height of each pixel 600 pixels s on the screen. We are given s′ = 8.00 m, so we will need to find s from the equation need to find f from the relation f-number =

1 1 1 + = . To do this we will s s′ f

f . D


25-4

Chapter 25

Solve: We have f = (f-number) (diameter) = 1.8(46 mm) = 82.8 mm. We can now solve for s: s =

s′ f = s′ − f

(8,000 mm)(82.8 mm) s′ 8000 mm = −95.6 × . This gives the width of each = 83.7 mm. Thus we have m = − = − 8,000 mm − 82.8 mm s 83.7 mm pixel on the screen to be (95.6) (95.6)

3.30 cm = 0.39 cm = 3.9 mm. Similarly, the height of each pixel on the screen is 800 pixels

3.30 cm = 0.53 cm = 5.3 mm. 600 pixels

Reflect: The fact that the magnification is negative simply means that the image is inverted relative to the object—to produce an upright image on the screen the object must therefore be inverted.

25.13. Set Up: The image is real since it is projected onto the screen; s′ = +900 cm. Solve: (a)

1 1 1 s′ f (900 cm)(15.0 cm) + = . s= = = 15.3 cm s s′ f s′ − f 900 cm − 15.0 cm

s′ 900 cm =− = −58.8. 0 y ′0 = 0 m 0 y, so the image has dimensions (24 mm)(58.8) = 1.41 m by (36 mm) s 15.3 cm (58.8) = 2.12 m.

(b) m = −

Reflect: The object is just outside the focal point of the lens. s ′ s and the image is greatly enlarged.

25.14. Set Up: The magnification should be 0 m 0 = m=−

4.00 m 36 × 1023 m

= 111. The image is real so s′ > 0 and m = −111.

s′ 1 1 1 . + = . s′ = 15.0 m. s s s′ f

Solve: (a) s = −

s′ 15.0 m ss ′ (0.135 m)(15.0 m) =− = 0.135 m. f = = = 134 mm. m −111 s + s ′ 0.135 m + 15.0 m

(b) s = 13.5 cm (c) One dimension of the image is (111)(36 mm) = 4.00 m. The other is (111)(24 mm) = 2.66 m.

*25.15. Set Up: The magnification has magnitude m < 0. s′ = 6.0 m. Solve: (a) s = −

48.0 in. 60.0 in. = = 12. The image is real, so s′ > 0 and 4.0 in. 5.0 in.

1 1 1 s′ + = . m=− . s s′ f s

s′ 600 cm ss ′ (0.50 m)(6.0 m) =− = +50 cm. f = = = 462 mm. m −12 s + s ′ 0.50 m + 6.0 m

(b) s = 50 cm. Reflect: Only this particular focal length gives both an image distance of 6.0 m and the desired image size.

25.16. Set Up:

⎛ 1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ . From the shape of the cornea shown in the figure, R1 = +7.80 mm and f ⎝ R1 R2 ⎠

R2 = +7.30 mm. 1 1 1 1 ⎛ ⎞ = (1.38 − 1) ⎜ − = −3.33 diopters. f < 0 ⎟ and f = −300 mm. The power is ⎝ ⎠ f 7.80 mm 7.30 mm −0.300 m and the lens is diverging.

Solve:

25.17. Set Up: s′ = 2.50 cm.

1 1 1 + = . From Table 25.1, the near point is 10 cm. s s′ f


Optical Instruments

25-5

ss ′ (10 cm)(2.50 cm) . For s = 10 cm, f = = 2.00 cm. For s → ∞, f = s ′ = 2.50 cm. 10 cm + 2.50 cm s + s′ Reflect: The eye changes the focal length of its lens by using muscles to change the shape of the lens. Solve: f =

25.18. Set Up: From Table 25.1, the near point is 22 cm. Solve: (a) s = 22 cm, s′ = 2.50 cm. f =

1 1 1 s′ y ′ + = . m=− = . s s′ f s y

ss ′ (22 cm)(2.50 cm) = = 2.24 cm. The power is s + s ′ 22 cm + 2.50 cm 1 = 44.6 diopters. 0.0224 m

s′ 2.50 cm =− = −0.114. 0 y ′0 = 0 m 0 y = (0.114)(6.50 cm) = 7.41 mm. m < 0 so the image is inverted. s′ > 0 s 22 cm so the image is real. s′ 2.50 cm = 20.0111. 0 y ′0 = 0 m 0 y = (0.0111)(6.50 cm) = 0.722 mm (c) m = 2 = 2 s 225 cm (b) m = −

*25.19. Set Up:

⎛ 1 1 1 1 ⎞ = (n − 1) ⎜ − in diopters, ⎟ . For a double convex lens, R1 > 0 and R2 < 0. To compute f f ⎝ R1 R2 ⎠

R1 and R2 must be expressed in meters. Maximum power corresponds to minimum focal length. (a) maximum power:

1 1 1 ⎛ ⎞ = (1.43 − 1) ⎜ − = 150 diopters ⎝ 6.0 × 10−3 m −5.5 × 10−3 m ⎟⎠ f

1 1 1 ⎛ ⎞ = (1.43 − 1) ⎜ − = 115 diopters ⎝ 10.0 × 10−3 m −6.0 × 10−3 m ⎟⎠ f 1 1 m = 6.67 mm. f max = m = 8.70 mm. (b) f min = 150 115 (c) For a distant object s → ∞ and s ′ = f . At maximum power, f = f max so s′ = 8.70 mm.

minimum power:

(d) At minimum power, f = f min = 6.67 mm.

1 1 1 + = so s s′ f

sf (25 cm)(0.667 cm) = = 0.685 cm = 6.85 mm. s− f 25 cm − 0.667 cm Reflect: Since the distance from the lens to the retina is fixed, this eye won’t focus clearly on both objects. s′ =

25.20. Set Up: For an object 25.0 cm from the eye, the corrective lens forms a virtual image at the near point of the eye. Solve: (a) The person is farsighted. (b) A converging lens is needed. 1 1 1 ss ′ (25.0 cm)(−45.0 cm) 1 = = +56.2 cm. The power is (c) + = . f = = +1.78 diopters. s + s′ s s′ f 25.0 cm − 45.0 cm 0.562 m *25.21. Set Up: For an object 25.0 cm from the eye, the corrective lens forms a virtual image at the near point of the eye. The distances from the corrective lens are s = 23.0 cm and s′ = −43.0 cm. 1 1 1 ss ′ (23.0 cm)(−43.0 cm) 1 = = +49.4 cm. The power is Solve: + = . f = = 2.02 diopters. s + s′ 23.0 cm − 43.0 cm 0.494 m s s′ f Reflect: In Problem 25.20 the lenses have power 1.78 diopters. The power of the lenses is different for ordinary glasses versus lenses.


25-6

Chapter 25

25.22. Set Up: For an object very far from the eye, the corrective lens forms a virtual image at the far point of the eye. Solve: (a) The person is nearsighted. (b) A diverging lens is needed. 1 1 1 1 = −1.33 diopters. (c) + = . s → ∞ so f = s ′ = −75.0 cm. The power is −0.750 m s s′ f

*25.23. Set Up: Apply

1 1 1 + = . The near point is at infinity, so that is where the image must be formed for any s s′ f

objects that are close. The power in diopters is equal to Solve:

1 , with f in meters. f

1 1 1 1 1 1 = + = + = = 4.17 diopters. f s s ′ 24 cm −∞ 0.24 m

Reflect: To focus on closer objects, the power must be increased.

25.24. Set Up: To correct the close vision, for an object 25.0 cm from the eye the lens forms a virtual image 35.0 cm from the eye. For a distant object the corrective lens forms a virtual image 50.0 cm from the eye. 1 1 1 Solve: (a) s = 25.0 cm, s′ = −35.0 cm. + = . s s′ f f =

ss ′ (25.0 cm)(−35.0 cm) = = 87.5 cm. s + s′ 25.0 cm − 35.0 cm

1 = 1.14 diopters. 0.875 m 1 1 1 1 (b) s → ∞, s′ = −50.0 cm. + = gives f = s ′ = −50.0 cm. The power is = −2.00 diopters. −0.500 m s s′ f

The power is

1 1 1 1 + = to determine . s s′ f f Solve: We want the image of the computer screen that is formed by the student’s eyeglasses to be located at the far point of her vision. Assuming that the distance between her glasses and eye is negligible this means that s′ = −17 cm 1 1 1 = + = −3.66 diopters. when s = 45 cm. Thus we have f 0.45 m −0.17 m

*25.25. Set Up: Apply the equation

Reflect: The eyeglass lens forms a virtual image at your far point so that your relaxed eye can comfortably view the computer screen, which is beyond your far point.

25.26. Set Up: To correct near vision, the lens forms a virtual image at the near point when the object is 25 cm from the eye. To correct far vision, the lens forms a virtual image at the far point of the eye. The power in diopters equals 1/f , when f is in meters. Solve: (a)

1 1 1 1 1 m = 36.4 cm. Find s′ for s = 25 cm. + = . = +2.75 diopters, so f = 2.75 f s s′ f

s′ =

sf (25 cm)(36.4 cm) = = −79.8 cm. s− f 25 cm − 36.4 cm

The near point is 79.8 cm from the eye. 1 1 1 1 = −1.30 diopters so f = −0.769 m = −76.9 cm. Find s′ for s → ∞. + = gives s ′ = f = −76.9 cm. f s s′ f The far point is 76.9 cm from the eye. (b)


Optical Instruments

25-7

25.27. Set Up: To correct the near vision of the hyperopic eye, the corrective lens forms a virtual image at the near point for an object that is 25 cm from the eye. To correct the far vision of the myopic eye, the corrective lens forms a virtual image at the far point for an object that is very far from the eye. Power in diopters equals 1/f , when f is in meters. (a) s′ = −60.0 cm, s = +25.0 cm. (b) s′ = −60.0 cm, s → ∞.

1 1 1 1 1 = + = + = +2.33 diopters. f s s ′ 0.250 m −0.600 m

1 1 1 1 = + = = −1.67 diopters. f s s ′ −0.600 m

25.28. Set Up: For a thin lens, −

s′ y′ y′ y = , so = , and the angular size of the image equals the angular size of s y s′ s

the object. The object has angular size θ = Solve: θ =

y , with θ in radians. f

y y 2.00 mm ⇒ f = = = 80.0 mm = 8.00 cm. θ 0.025 rad f

Reflect: If the insect is at the near point of a normal eye, its angular size is

2.00 mm = 0.0080 rad. 250 mm

*25.29. Set Up: When the object is at the focal point of the lens and the image therefore is at infinity, then M=

25 cm . f

25 cm = 2.17 11.5 cm (b) The object is at the focal point of the lens, so the lens is 11.5 cm from the stamp.

Solve: (a) M =

25.30. Set Up: When the object is at the focal point the image is at infinity and M =

25 cm . f

25 cm = 4.17 6.00 cm (b) The object can be brought as close as the focal length of the lens, 6.00 cm. Solve: (a) M =

*25.31. Set Up: The image is not at infinity, so the equation M =

25 cm doesn’t apply. The image is virtual, so f

s′ = −25.0 cm. Solve: (a)

(b) m = −

1 1 1 s′ f (−25.0 cm)(8.00 cm) = = +6.06 cm + = . s= s′ − f −25.0 cm − 8.00 cm s s′ f

s′ −25.0 cm =− = +4.13. 0 y ′0 = 0 m 0 y = (4.13)(1.00 mm) = 4.13 mm s 6.06 cm

Reflect: The lens allows the object to be brought much closer to the eye (6.06 cm) than the near point.

25.32. Set Up: We can use

1 1 1 + = to determine the location of the image formed by the objective lens and s s′ f

s′ to determine the magnification of the objective. The magnification of the eyepiece can then be determined s 25 cm . by using M 2 = f m1 = −


25-8

Chapter 25

Solve: (a) Solving for the location of the image we have s ′ =

sf (12.8 mm)(12.0 mm) = = 192 mm. s− f 12.8 mm − 12.0 mm

s′ 192 mm =− = −15×, which may be written s 12.8 mm as 15 × if we are not interested in the orientation of the image relative to the object. 25 cm (c) The overall angular magnification is M = m1M1 = 15 = 150 × . 2.5 cm Reflect: It is customary to ignore the sign of the magnification in cases where the relative orientation of the image relative to that of the object is not of interest.

(b) We can find the linear magnification of the objective using m1 = −

25.33. Set Up: For the objective, f1 = 10.0 mm and s1 = f1 + 2.0 mm = 12.0 mm. M = m1 M 2 . 1 1 1 s f (12.0 mm)(10.0 mm) = 60.0 mm. + = . s1= = 1 1 = s1 − f1 12.0 mm − 10.0 mm s1 s1= f1

Solve: (a)

s1= 60.0 mm 25 cm 25 cm =− = −5.00. M 2 = = = 16.7. M = (−5.00)(16.7) = −83.3. The minus sign is s1 12.0 mm 1.50 cm f2 customarily ignored. (25 cm) s1= (25 cm)(6.00 cm) = = 100. Eq. (25.4) assumes that s1 ≈ f1 that this is Reflect: Eq. (25.4) gives M = f1 f 2 (1.00 cm)(1.50 cm) only approximately the case here. (b) m1 = −

25.34. Set Up: For each lens,

1 1 1 s′ y′ + = . m = 2 = . The image formed by the objective serves as the object s s′ f s y

for the eyepiece. Solve: (a) s ′ =

sf (13 mm)(12 mm) s′ 156 mm = = 156 mm. m = − = − = −12. s− f s 13 mm − 12 mm 13 mm

0 y′0 = 0 m 0 y = 12(1.2 mm) = 14.4 mm. (b) s′ = ∞ and f = 2.5 cm. s = 2.5 cm. (c) m1 = −

s1= −156 mm ⎛ 25 cm ⎞ =− = −12. M = m1 M 2 = (−12) ⎜ = −120. Eq. (25.4) gives ⎝ 2.50 cm ⎟⎠ s1 12 mm

(25 cm) s1= (25 cm)(15.6 cm) = = 130. f1 f 2 (1.2 cm)(2.5 cm) Eq. (25.4) omits the minus sign and is only approximately correct.

M=

*25.35. Set Up: We will assume that the given lateral magnification of the objective was calculated using the equation qm1q =

s1′ s′ (and not the approximate equation qm1q ≈ 1 ) and the angular magnification of the eyepiece is s1 f1

given by M 2 =

(25 cm) . The overall magnification of the microscope is then M = m1M 2 . To solve this problem we f2

must assume that the image of the objective is located very close to the focal point of the eyepiece: thus, we have 16.5 cm = s1 ′ + f 2 . Note that we will not use the approximation s1 ≈ f1 since it is not necessary given the information we have. Solve: (a) The overall magnification of the microscope is M = m1M 2 = (62)(10) = 620 × . (b) We can immediately determine the focal length of the eyepiece from f 2 =

(25 cm) (25 cm) = = 2.5 cm. M2 10


Optical Instruments

25-9

Since we assume that the image of the objective is located very close to the focal point of the eyepiece we have s ′ 14.0 cm = 0.226 cm. Finally we obtain the focal length s1 ′ = 16.5 cm − 2.5 cm = 14.0 cm. Thus, we obtain s1 = 1 = qm1q 62 of the objective: f1 =

s1s1 ′ (0.226 cm)(14.0 cm) = = 0.222 cm. s1 + s1′ 0.226 cm + 14.0 cm

(c) If the object is located at s = 0.226 cm, the image of the objective will be located at the focal point of the eyepiece—this produces a final image at infinity, which allows for comfortable viewing of a normal eye. Thus, the object should be placed approximately 0.226 cm − 0.222 cm = 0.004 cm outside the focal point of the objective lens for comfortable viewing. Reflect: If we assume that the magnification of the objective was calculated by using the approximate equation s′ qm1q = 62 ≈ 1 (with s1 ′ = 14.0 cm ) we obtain slightly different answers for f1 and s1 : f1 = 0.226 cm and f1

s1 = 0.230 cm. However, note that the object must still be placed roughly 0.004 cm outside the focal point of the objective for comfortable viewing.

25.36. Set Up: Since the image formed by the eyepiece is at infinity, the angular magnification for the eyepiece is M2 =

25 cm . The overall angular magnification of the microscope is M = m1 M 2 . The image formed by the f

objective is the object for the eyepiece and is at the focal point of the eyepiece. Solve: (a) The image formed by the objective is 19.7 cm − 1.80 cm = 17.9 cm from the lens.

s=

1 1 1 + = gives s s′ f

s′ f (17.9 cm)(0.800 cm) = = 0.837 cm = 8.37 mm. s′ − f 17.9 cm − 0.800 cm

s′ 17.9 cm =− = −21.4 0.837 cm s ⎛ 25 cm ⎞ ⎛ 25 cm ⎞ (c) M = m1 M 2 = m1 ⎜ = (−21.4) ⎜ = −297. Eq. (25.4) gives ⎟ ⎝ 1.80 cm ⎟⎠ f ⎝ ⎠ (b) m1 = −

2

(25 cm) s1= (25 cm)(17.9 cm) = = 311. f1 f 2 (0.800 cm)(1.80 cm) Eq. (25.4) omits the minus sign and is only approximately correct.

M=

⎛ s1= ⎞ = ⎟ M 2 . s1 = 120 mm + f1 . f ⎝ 1⎠

*25.37. Set Up: M = m1 M 2 = ⎜

Solve: (a) The largest M is for the smallest f1 and largest M 2 . M 2 = 10.

f = 1.9 mm and s1= = 120 mm + 1.9 mm = 121.9 mm. ⎛ 121.9 mm ⎞ M =⎜ (10) = 640. ⎝ 1.9 mm ⎟⎠

(b) The smallest M is for the largest f1 and the smallest M 2 . M 2 = 5,

f1 = 16 mm and s1= = 120 mm + 16 mm = 136 mm.

⎛ 136 mm ⎞ (5) = 43. M =⎜ ⎝ 16 mm ⎟⎠

25.38. Set Up: M = Solve: (a) M =

(25 cm) s1= = . s1 = 16.0 cm + 0.500 cm = 16.50 cm. f1 f 2

(25 cm)(16.50 cm) = 317 (0.500 cm)(2.60 cm)


25-10

Chapter 25

(b) The minimum separation is

*25.39. Set Up: M = − Solve: M max = −

0.10 mm 0.10 mm = = 3.2 × 10−4 mm. M 317

f1 . The largest M is for the smallest f 2 , where f 2 is the focal length of the eyepiece. f2

(16.0 in.)(2.54 cm/in.) = −27; 1.5 cm

(16.0 in.)(2.54 cm/in.) = −4.8 8.5 cm Reflect: The focal length of the objective is f1 = 16.0 in. = 40.6 cm and f1 > f 2 for each eyepiece.

M min = −

25.40. Set Up: Since the final image is at infinity, M = −

f1 . The length of the telescope is f1 + f 2 . f2

Solve: f1 + f 2 = 1.80 m, so f1 = 1.80 m − f1 = 180 cm − 9.00 cm = 171 cm.

M =−

25.41. Set Up: M = −

f1 171 cm =− = −19.0. f2 9.00 cm

f1 . The length of the telescope is L = f1 + f 2 . f2

Solve: (a) M is largest when f 2 < f1 . Use the 980 mm lens as the eyepiece and the 1710 mm lens as the objective.

L = f1 + f 2 = 1710 mm + 980 mm = 2690 mm = 2.69 m. 1710 mm = −1.74. 980 mm (c) θ ′ = M θ = (−1.74)(0.5°) = 0.87°

(b) M = −

25.42. Set Up: We will use

1 1 1 y′ s′ + = to find the location of the image and m = = − to determine the size y s s s′ f

of the image. From the appendix we find that the diameter of Mars is 2(3.40 × 106 m) = 6.80 × 106 m and that meters ⎞ ⎛ 10 1 mile = 1.609 × 103 m. At its closest approach Mars is (35 × 106 miles) ⎜1.609 × 103 ⎟ = 5.63 × 10 meters ⎝ mile ⎠ from the earth. Solve: Note that s is much larger than f so that

(57.6 m)(6.80 × 106 m) 5.63 × 1010 m

1 1 1 1 s′ = − ≈ gives s ′ = f = 57.6 meters. This gives | y ′ | = q yq = s s′ f s f

= 7.0 × 10−3 m = 7.0 mm.

Reflect: You may easily the validity of the approximation used since

1 5.63 × 1010 m

= 1.8 × 10−11 (m)−1.

*25.43. Set Up: The unaided angular size of Jupiter is θ = M=

1 = 0.0174 (m) −1 and 57.6 m

θ′ f = 2 1 . f1 = 19.4 m. θ f2

Solve: Need M =

138,000 km 6.28 × 108 km

= 2.20 × 10−4 rad = 0.0126°.

f 19.4 m 0.500° = 0.49 m. = 39.7. f 2 = 1 = 39.7 0M 0 0.0126°


Optical Instruments

25-11

Reflect: The angular magnification is the ratio of the angular size of the image when the optical device is used divided by the angular size of the object.

25.44. Set Up: Use

⎛ 1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ to find f for each wavelength. Double concave means R1 = −32.0 cm f R R 2⎠ ⎝ 1

and R2 = +24.0 cm. Then use

1 1 1 + = to find the image location. s s′ f

1 1 1 ⎞ ⎛ and f = −31.2 cm. = (1.44 − 1) ⎜ − ⎝ −0.320 m 0.240 m ⎟⎠ f

Solve: red light:

s′ =

sf (50.0 cm)(−31.2 cm) = = −19.2 cm. s − f 50.0 cm − (−31.2 cm)

The red light image will be 19.2 cm in front of the lens. 1 1 1 ⎞ ⎛ violet light: = (1.57 − 1) ⎜ − and f = −24.1 cm. ⎝ −0.320 m 0.240 m ⎟⎠ f

s′ =

sf (50.0 cm)(−24.1 cm) = = −16.3 cm. s − f 50.0 cm − (−24.1 cm)

The violet light image will be 16.3 cm in front of the lens.

*25.45. Set Up: From Figure 23.29 in Section 23.9, for 700 nm, n = 1.61 and for 400 nm, n = 1.66. Use ⎛ 1 1 1⎞ 1 1 1 to find s′. Double convex means = (n − 1) ⎜ − ⎟ to find f for each wavelength. Then use + = s s′ f f ⎝ R1 R2 ⎠ R1 = 25.0 cm and R2 = −35.0 cm. Solve: (a) (i) red light:

(ii) violet light:

1 1 1 ⎛ ⎞ = (1.66 − 1) ⎜ − = 22.1 cm ⎝ 0.250 m −0.350 m ⎟⎠ f

(b) red light: s ′ =

violet light: s ′ =

1 1 1 ⎛ ⎞ = (1.61 − 1) ⎜ − = 23.9 cm ⎝ 0.250 m −0.350 m ⎟⎠ f

(30.0 cm)(23.9 cm) sf = = 118 cm. 30.0 cm − 23.9 cm s− f

(30.0 cm)(22.1 cm) sf = = 83.9 cm. 30.0 cm − 22.1 cm s− f

25.46. Set Up: Use

1 1 1 + = to calculate f . Since the light is focused each image is real and s′ > 0. Then use s s′ f

⎛ 1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ to calculate n. For a plano-convex lens, R1 → ∞ and R2 < 0. f ⎝ R1 R2 ⎠ Solve: green: f =

⎛ 1 ⎞ (48.0 cm)(277 cm) ss ′ R −22.5 cm 1 = = 40.9 cm. = (n − 1) ⎜ − ⎟ so (n − 1) = − 2 = − = 0.550. f 40.9 cm s + s ′ 48.0 cm + 277 cm f ⎝ R2 ⎠

n = 1.55. blue: f =

(48.0 cm)(171 cm ) R −22.5 cm = 0.600. n = 1.60. = 37.5 cm. (n − 1) = − 2 = − 48.0 cm + 171 cm 37.5 cm f

25.47. Set Up: Figure 23.29 in Section 23.9 gives n as a function of λ. For λ = 550 nm, n = 1.63, for λ = 700 nm, n = 1.61 and for λ = 400 nm, n = 1.66.

⎛ 1 1 1 1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ , with R1 → ∞ and R2 = −28.0 cm. + = . s s′ f f ⎝ R1 R2 ⎠


25-12

Chapter 25

Solve: (a)

1 1 ⎛ ⎞ = (1.63 − 1) ⎜ 0 − ⎟ and f = 44.4 cm. ⎝ f −0.280 cm ⎠ s′ =

(b) red: n = 1.61.

(65.0 cm)(44.4 cm) sf = = 140 cm. 65.0 cm − 44.4 cm s− f

1 1 ⎛ ⎞ and f = 45.9 cm. = (1.61 − 1) ⎜ − ⎝ −0.280 cm ⎟⎠ f s′ =

(65.0 cm)(45.9 cm) sf = = 156 cm, 65.0 cm − 45.9 cm s− f

and therefore 16 cm beyond the screen. 1 1 ⎛ ⎞ = (1.66 − 1) ⎜ − violet: n = 1.66. ⎟ and f = 42.4 cm. ⎝ −0.280 cm ⎠ f (65.0 cm)(42.4 cm) sf = = 122 cm, 65.0 cm − 42.4 cm s− f and therefore 18 cm in front of the screen. Reflect: The chromatic aberration of this lens is very large. Silicate flint glass is not good lens material, because of its large dispersion. s′ =

25.48. Set Up: The object is very far from the lens, so the image distance equals the focal length. Solve: m = −

5.00 m s′ =− = −5.26 × 10−4. s 9.50 × 103 m

0 y ′0 = 0 m 0 y = (5.26 × 10−4 )(70.7 m) = 3.72 × 10−2 m = 3.72 cm. na nb nb − na , with na = 1.00, nb = 1.40. s = 40.0 cm, s′ = 2.60 cm. + = s s′ R 1 1.40 0.40 Solve: + = and R = 0.710 cm. 40.0 cm 2.60 cm R Reflect: The cornea presents a convex surface to the object, so R > 0.

*25.49. Set Up: We have

na nb nb − na . na = 1.00, nb = 1.40. + = s s′ R 1 1.40 0.40 Solve: + = ⇒ s ′ = 2.77 cm. 36.0 cm s′ 0.75 cm Reflect: This distance is greater than the normal eye, which has a cornea vertex to retina distance of about 2.6 cm.


*25.51. Set Up: m = −

s′ y′ = . s′ = 2.50 cm. 0 y ′0 = 5.0 μm. The angle subtended (in radians) is height divided by s y

distance from the eye. s′ 2.50 cm y ′ 5. 0 μ m Solve: (a) m = − = − = −0.10. y = 0 0 = = 50 μ m. s 25 cm m 0.10 y 50 μm 50 × 10−6 m = = = 2.0 × 10−4 rad = 0.0115° = 0.69 min. s 25 cm 25 × 10−2 m This is only a bit smaller than the typical experimental value of 1.0 min. 0 y′ 0 5.0 × 10−6 m = = 2.0 × 10−4 rad of the image. Reflect: The angle subtended by the object equals the angular size s′ 2.50 × 10−2 m (b) θ =

25.52. Set Up: When a single lens is used as a magnifier, a virtual image is formed and m > 0. m = 16.50. Solve: m = −

s′ 1 1 1 1 1 1 0.8462 1 gives + = . = and and s ′ = − ms = − (6.50) s = −6.50s. Then + = s s s′ f s −6.50s f s f


Optical Instruments

25-13

s = (0.8462) f = (0.8462)(4.00 cm) = 3.38 cm. s′ = −6.50(3.38 cm) = −22.0 cm. The flea is 3.38 cm from the lens and the image is 22.0 cm from the lens, on the same side of the lens as the flea.

25.53. Set Up: Your near point is the closest distance at which you can clearly focus on a object and your far point is the greatest distance at which you can clearly focus on an object. To illustrate the solution, assume a near point of 75 cm and a far point of 4.0 m. Solve: (a) Measure the near point and far point distance. The answers will vary from person to person. (b) Assume lenses. Close-up vision: For an object at s = 25 cm from the eye, the lens is to form a virtual image at s′ = −75 cm. 1 1 1 ss ′ (25 cm)(−75 cm) 1 + = gives f = = = +37.5 cm. The power is = +2.67 diopters. 0.375 m s s′ f s + s ′ 25 cm + (−75 cm)

Distant vision: For an object at s → ∞, the lens forms a virtual image at s′ = −400 cm. f = s ′ = −400 cm. The power is

1 1 1 + = gives s s′ f

1 = −0.25 diopters. −4.00 m

25.54. Set Up: In the simplified model of the eye all of the refraction occurs at the outside surface of the eye and we can apply Eq. (24.11):

na nb nb − na with na = 1 and nb = 1.40. In this case we take s′ = 2.75 cm when s = ∞. + = s s′ R

Solve: We can make the given substitutions to obtain R = 0.40

1.00 1.40 1.40 − 1.00 + = . Solving for R we obtain ∞ 2.75 cm R

2.75 cm = 0.786 cm ≈ 0.79 cm. 1.40

Reflect: In the simplified model of the eye we take advantage of the fact that the index of refraction of the crystalline lens, the aqueous humor, and the vitreous humor are all nearly equal to 1.40. This allows us to solve the problem by using Eq. (24.11).

*25.55. Set Up: The image is the same distance behind the mirror as the object is in front of the mirror. 1 1 1 s′ y′ + = . m=− = . s s′ f s y Solve: (a) s = 1.50 m. s ′ = (b) m = −

sf (1.50 m)(19.5 × 10−3 m) = = 19.8 mm. s− f 1.50 m − 19.5 × 10−3 m

s′ 19.8 × 10−3 m =− = −0.0132. 0 y ′0 = 0 m 0 y = (0.0132)(8.0 cm) = 0.106 cm. s 1.50 m

Reflect: The camera focuses on the image, behind the mirror, not on the mirror surface.

25.56. Set Up: s f so s ′ ≈ f = 0.300 m. Solve: (a) m = −

s′ 0.300 m =− = −7.77 × 10−10. s 3.86 × 108 m 0 y ′0 = 0 m 0 y = (7.77 × 10−10 )(3.48 × 106 m) = 2.70 × 10−3 m = 2.70 mm.

(b)

2.70 mm = 7.5, 36 mm

*25.57. Set Up: The object for the camera lens is the image formed by the double convex lens. m=−

1 1 1 + = . s s′ f

s′ y′ = . s y


25-14

Chapter 25

Solve: (a) image formed by the convex lens: sf (10.0 cm)(17.0 cm) s′ = = = −24.3 cm. s− f 10.0 cm − 17.0 cm The image is 24.3 cm to the left of the convex lens, so it is 24.3 cm + 5.00 cm = 29.3 cm to the left of the camera lens. s′ −24.3 cm m=− =− = +2.43. s 10.0 cm The image is (2.43)(1.40 cm) = 3.40 cm tall. image formed by camera lens: s = 29.3 cm. sf (29.3 cm)(1.50 cm) s′ = = = 1.58 cm. s− f 29.3 cm − 1.50 cm

s′ 1.58 cm =− = −0.0539. y = 3.40 cm. 0 y ′0 = 0 m 0 y = (0.0539)(3.40 cm) = 0.183 cm. s 29.3 cm (c) No. For light from the seedling to reach the camera lens it must through the convex lens. The camera cannot view the seedling directly. Reflect: The overall magnification is the product m1 m2 of the magnifications of the two lenses. m1 m2 = (2.43) (−0.0539) = −0.131 and the final image is 0.131 times the height of the seedling. (b) m = −

25.58. Set Up: The overall lateral magnification of the final image is m = m1 m2 . The image formed by the objective is the object for the eyepiece and the distance between the two lenses is s1= + s2 . Solve: (a) For the objective, f1 = 0.800 cm and s1= = 18.0 cm.

s1 =

s1= f1 s1= − f1

=

1 1 1 + = gives s s′ f

(18.0 cm)(0.800 cm) = 0.837 cm. 18.0 cm − 0.800 cm

m1 = −

s1= 18.0 cm =− = −21.5. s1 0.837 cm

For the eyepiece, f 2 = 7.50 cm and s2= = 200 cm. s2 = m2 = −

s2= f 2 s2= − f 2

=

(200 cm)(7.50 cm) = 7.79 cm. 200 cm − 7.50 cm

s2= 200 cm =− = −25.7. s2 7.79 cm

m = m1 m2 = (−21.5)( −25.7) = 553 (b) s1= + s2 = 18.0 cm + 7.79 cm = 25.8 cm

25.59. Set Up: The person’s eye cannot focus on anything closer than 85.0 cm. The problem asks us to find the location of an object such that his old lenses produce a virtual image 85.0 cm from his eye. 1 = 2.25 diopters so f = 44.4 cm. The image is 85.0 cm from his eye so is 83.0 cm from the eyeglass lens. Solve: (a) f s′ f (−83.0 cm)(44.4 cm) = = +28.9 cm. s′ − f −83.0 cm − 44.4 cm The object is 28.9 cm from the eyeglasses so is 30.9 cm from his eyes. (c) Now s′ = −85.0 cm. s′ f (−85.0 cm)(44.4 cm) s= = = +29.2 cm. s′ − f −85.0 cm − 44.4 cm s=

Reflect: The old glasses allow him to focus on objects as close as about 30 cm from his eyes. This is much better than a closest distance of 85 cm with no glasses, but his current glasses probably allow him to focus as close as 25 cm.


Optical Instruments

25-15

θ′ f = 2 1 . f1 = 95.0 cm, f 2 = 15.0 cm. f2 θ f 95.0 cm Solve: (a) M = − 1 = − = −6.33 f2 15.0 cm (b) y = 60.0 m. s f so s ′ ≈ f = 95.0 cm. s′ 0.950 m m=− =− = −3.17 × 10−4. s 3.00 × 103 m

25.60. Set Up: M =

0 y ′0 = 0 m 0 y = (3.17 × 10−4 )(60.0 m) = 1.90 × 10−2 m = 1.90 cm

(c) θ ′ = 0 M 0θ . θ is the angular size of the object;

θ= θ = (6.33)(0.0200 rad) = 0.127 rad = 7.28°.

*25.61. Set Up:

y 60.0 m = = 0.0200 rad. s 3.00 × 103 m

⎛ 1 f1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ . For a double-convex lens, R1 > 0 and R2 < 0. M = 2 . The length of f f R R 2 2⎠ ⎝ 1

the telescope is f1 + f 2 . Solve: (a) Find the focal length of each lens. L1:

1 1 1 ⎛ ⎞ and f = 133 cm. L2 : = (1.528 − 1) ⎜ − ⎝ 0.950 m −2.700 m ⎟⎠ f

1 1 1 ⎛ ⎞ and f = 96.3 cm. = (1.550 − 1) ⎜ − ⎝ 0.535 m −50.5 m ⎟⎠ f The largest magnification would be with L1 as the objective and L2 as the eyepiece. f1 133 cm =− = −1.38. f2 96.3 cm (b) The length of the telescope would be 133 cm + 96.3 cm = 229 cm. M =−

25.62. Set Up: We will apply the thin lens equation

⎛ 1 1 1 ⎞ 1 1 = ( n − 1) ⎜ − ⎟ = + , with n = 1.333 (for water), f ⎝ R1 R2 ⎠ s s′

R1 = 1.6 cm, and R2 = ∞. The angular magnification of a lens is given by M = focal point of the lens) and its lateral magnification is given by m = − Solve: (a)

25 cm (for an object placed near the f

s′ . s

⎛ 1 1 ⎞ 0.40 R 1.60 cm 1 = (1.333 − 1.000) ⎜ − ⎟ = . Thus we have f = 1 = = 4.80 cm. 0 . 333 0.333 f R ∞ R 1 ⎝ 1 ⎠

(b) For an object placed near the focal point of the lens we have M = (c) We may use the equation

25 cm 25 cm = = 5.2 × . f 4.80 cm

1 1 1 fs (48.0 mm)(1.0 mm) + = to obtain s ′ = = = −1.02 mm. The resulting s s′ f s− f 1.0 mm − 48.0 mm

s′ −1.02 mm =− = + 1. 0 × . s 1.0 mm Reflect: The large angular magnification that occurs when you hold an object near the focal point of the magnifying lens results from the fact that the magnifier allows you to focus on objects that are closer to you than your eye’s near point.

lateral magnification is m = −


26

INTERFERENCE AND DIFFRACTION

Answers to Multiple-Choice Problems 1. C 2. A, C, D 14. C, D 15. B

3. B

4. C

5. A

6. A

7. B

8. A

9. C

10. B

11. A

12. B

13. D

Solutions to Problems *26.1. Set Up: Constructive interference occurs for r2 − r1 = mλ , m = 0, ±1, ±2, … . Destructive interference

(

)

occurs for r2 − r1 = m + 12 λ , m = 0, ±1, ±2, … . For this problem, r2 = 150 cm and r1 = x. The path taken by the

person ensures that x is in the range 0 ≤ x ≤ 150 cm. Solve: (a) 150 cm − x = m(34 cm). x = 150 cm − m(34 cm). For m = 0, 1, 2, 3, 4 the values of x are 150 cm, 116 cm, 82 cm, 48 cm, 14 cm. (b) 150 cm − x = m + 12 (34 cm). x = 150 cm − m + 12 (34 cm). For m = 0, 1, 2, 3 the values of x are 133 cm, 99 cm,

(

(

)

)

65 cm, 31 cm. Reflect: When x = 116 cm the path difference is 150 cm − 116 cm = 34 cm, which is one wavelength. When x = 133 cm the path difference is 17 cm, which is one-half wavelength.

26.2. Set Up: For radio waves, υ = 3.00 × 108 m/s. λ=

υ f

=

3.00 × 108 m/s 92.0 × 106 Hz

Constructive interference occurs for r2 − r1 = mλ ,

(

r2 − r1 = m +

1 2

) λ,

m = 0,

= 3.26 m.

±1,

±2, … . Destructive interference occurs for

m = 0, ±1, ±2, … . For this problem, r2 = x and r1 = 50.0 m. The path taken by the person

ensures that x is greater than 50.0 m. Solve: (a) x − 50.0 m = m(3.26 m). x = 50.0 m + m(3.26 m). For m = 0, 1, 2, 3, 4, x = 50.0 m, 53.3 m, 56.5 m, 59.8 m, 63.0 cm. (b) x − 50.0 m = m + 12 (3.26 m). x = 50.0 m + m + 12 (3.26 m). For m = 0, 1, 2, 3, 4, x = 51.6 m, 54.9 m, 58.2 m,

(

)

(

)

61.4 m, 64.7 m.

26.3. Set Up: The path difference is 120 m. For destructive interference the path difference is ( m + 12 ) λ (where m = 0, ± 1, ± 2, . . .), and longest wavelength occurs when m = 0. For constructive interference the path difference is mλ (where m = 0, ± 1, ± 2, . . .), and the longest wavelength occurs when m = 1.


26-1

26-2

Chapter 26

λ

= 120 m ⇒ λ = 240 m. 2 (b) The longest wavelength for constructive interference is λ = 120 m. Solve: (a) For destructive interference

Reflect: The path difference doesn't depend on the distance of point Q from B.

26.4. Set Up: When she is at the midpoint between the two speakers the path difference r2 − r1 is zero. When she walks a distance d toward one speaker, r2 increases by d and r1 decreases by d, so the path difference changes by 2d.

(

)

path difference = mλ , m = 0, ±1, ±2, … . gives constructive interference and path difference = m + 12 λ , m = 0, ±1, ±2, … . gives destructive interference.

v 340.0 m/s = = 1.36 m f 250.0 Hz (a) Path difference is zero, so the interference is constructive. Solve: λ =

λ

λ

1.36 m = = 34.0 cm. 4 4 λ 1.36 m (c) Constructive interference, so path difference equals λ . 2d = λ and d = = = 68.0 cm. 2 2 (b) Destructive interference, so the path difference equals λ /2. 2d =

2

and d =

*26.5. Set Up: When she is at the midpoint between the two speakers the path difference r2 − r1 is zero. When she walks a distance d toward one speaker, r2 increases by d and r1 decreases by d, so the path difference changes by 2d. Since the speakers are 180° out of phase, path difference = mλ , m = 0, ±1, ±2, … . gives destructive interference

(

)

and path difference = m + 12 λ , m = 0, ±1, ±2, … . gives constructive interference.

υ

340.0 m/s = = 1.36 m f 250.0 Hz (a) Path difference is zero, so the interference is destructive. Solve: λ =

λ

1.36 m = 68.0 cm. 2 λ λ 1.36 m = 34.0 cm. (c) Constructive interference, so the path difference equals λ /2. 2d = and d = = 4 4 2 Reflect: When she moves from a point of destructive interference to a point of constructive interference the path difference changes by λ /2.

(b) Destructive interference, so the path difference equals λ . 2d = λ and d =

2

=

26.6. Set Up: Since the values of m are rather large, don’t use Eq. (26.6). ym = R tan θ m . For bright fringes,

(

)

d sin θ = mλ . For dark fringes, d sin θ = m + 12 λ . For the fifth bright fringe (not counting the central bright spot),

m = 5 and for the eighth dark fringe, m = 7. ⎛ 525 × 10−9 m ⎞ ⎛λ⎞ Solve: (a) sin θ = 5 ⎜ ⎟ = 5 ⎜ ⎟⎟ = 0.0633 and θ = 3.63°. −3 ⎜ ⎝d ⎠ ⎝ 0.0415 × 10 m ⎠

y = R tan θ = (75.0 cm) tan 3.63° = 4.75 cm

⎛λ⎞ (b) sin θ = 7 + 12 ⎜ ⎟ = 0.0949 and θ = 5.45°. y = R tan θ = 7.16 cm. ⎝d ⎠

(

)

26.7. Set Up: The values of ym are much smaller than the distance R to the screen, so the approximate expression mλ ym = R

d

is accurate.

d Δy (0.460 × 10−3 m)(2.82 × 10−3 m) Rλ . λ= = = 590 nm d R 2.20 m Reflect: This is a visible wavelength. Solve: Δy = ym11 − ym =


Interference and Diffraction

26.8. Set Up: The value of y20 is much smaller than R and the approximate expression ym = R

26-3

mλ is accurate. d

y20 = 10.6 × 10−3 m. Solve: d =

20 Rλ (20)(1.20 m)(502 × 10−9 m) = = 1.14 × 10−3 m = 1.14 mm y20 10.6 × 10−3 m

*26.9. Set Up: λ =

c 3.00 × 108 m/s mλ , when ym R. = = 4.75 × 10−7 m. Bright fringes are located at ym = R d f 6.32 × 1014 Hz

(

)

Dark fringes are at d sin θ = m + 12 λ and y = R tan θ . For the third bright fringe (not counting the central bright spot), m = 3. For the third dark fringe, m = 2. Solve: (a) d =

mλ R 3(4.75 × 1027 m)(0.850 m) = = 3.89 × 10−5 m = 0.0389 mm 0.0311 m ym

(

(b) sin θ = 2 + 12

−7

⎛

⎞

75 × 10 m ⎟ = 0.0305 and θ = 1.75°. ) λd = (2.5) ⎜⎜ 34..89 × 10−5 m ⎟

y = R tan θ = (85.0 cm) tan1.75° = 2.60 cm. ⎝ ⎠ Reflect: The third dark fringe is closer to the center of the screen than the third bright fringe on one side of the central bright fringe.

26.10. Set Up: Use the information given about the bright fringe to find the distance d between the two slits. Then use Eq. (26.5) and y = R tan θ to calculate λ for which there is a first-order dark fringe at this same place on the screen. Solve:

y1 =

Rλ1 Rλ (3.00 m)(600 × 10−9 m) = 3.72 × 10−4 m. (R is much greater than d, so Eq. (26.6) is , so d = 1 = d y1 4.84 × 10−3 m

1⎞ ⎛ valid.) The dark fringes are located by d sin θ = ⎜ m + ⎟ λ (where m = 0 , ± 1, ± 2, . . .). The first-order dark fringe is 2⎠ ⎝ located by sin θ = λ2 /2d , where λ2 is the wavelength we are seeking. y = R tan θ ≈ R sin θ =

λ2 R

2d Rλ1 Rλ2 We want λ2 such that y = y1 . This gives and λ2 = 2λ1 = 1200 nm. = 2d d Reflect: For λ = 600 nm the path difference from the two slits to this point on the screen is 600 nm. For this same path difference (point on the screen) the path difference is λ /2 when λ = 1200 nm.

λ

26.11. Set Up:

d

expression ym = R Solve: Δy =

=

500 × 10−9 m 0.450 × 10−3 m

( m + 12 ) λ d

= 1.11 × 10−3. θ is very small for these dark lines and the approximate

is accurate. Adjacent dark lines are separated by Δy = ym11 − ym =

(0.750 m)(500 × 10−9 m) (0.450 × 10

−3

m)

Rλ . d

= 8.33 × 10−4 m = 0.833 mm

mλ . First-order means m = 1. d Solve: The distance between corresponding bright fringes is (5.00 m)(1) Rm Δy = Δλ = (660 − 470) × (10−9 m) = 3.17 mm. d (0.300 × 10−3 m) Reflect: The separation between these fringes for different wavelengths increases when the slit separation decreases.

26.12. Set Up: Use Eq. (26.6) for small angles: ym = R


26-4

Chapter 26

*26.13. Set Up: The largest value sin θ can have is 1.00. Bright fringes are located at angles θ given by d sin θ = mλ . Solve: (a) m =

d sin θ

. For sin θ = 1, m =

d

0.0116 × 10−3 m

= 19.8. Therefore, the largest m for fringes on the 5.85 × 10−7 m screen is m = 19. There are 2(19) + 1 = 39 bright fringes, the central one and 19 above and 19 below it. (b) The most distant fringe has m = ±19. ⎛ 5.85 × 10−7 m ⎞ λ sin θ = m = ±19 ⎜ = ±0.958 ⎜ 0.0116 × 10−3 m ⎟⎟ d ⎝ ⎠ and θ = ±73.3°. Reflect: For small θ the spacing Δy between adjacent fringes is constant but this is no longer the case for larger

λ

λ

=

angles.

υ

340 m/s = 1.23 m. The interference of these two coherent sources obeys the same equation f 277 Hz as in the Young two-slit experiment. Solve: (a) Constructive interference occurs when d sin θ = mλ , m = 0, ±1, … .

26.14. Set Up: λ =

=

λ

⎛ 1.23 m ⎞ = m⎜ ⎟ = 0.224m. d ⎝ 5.50 m ⎠ m = 0 : θ = 0. m = ±1: θ = ±12.9°. m = ±2 : θ = ±26.6°. m = ±3 : θ = ±42.2°. m = ±4 : θ = ±63.6°. Since 0 sin θ 0 can’t be larger than 1.00, the equation isn’t satisfied for m > 4. sin θ = m

(

)

(b) Destructive interference occurs when d sin θ = m + 12 λ , m = 0, ±1, … .

(

sin θ = m + 12

) λd = 0.224 ( m + 12 ) .

m = 0, −1: θ = ±6.4°. m = 1, −2 : θ = ±19.6°. m = 2, −3: θ = ±34.1°. m = 3, −4 : θ = ±51.6°.

*26.15. Set Up: Strongly reflected means constructive interference so we will consider the interference between rays reflected from the two surfaces of the soap film. Take into the phase difference due to the path difference of 2t and any phase difference due to phase changes upon reflection. As shown in the figure below there is a 180° phase change when the light is reflected from the outside surface of the bubble and no phase change when the light is reflected from the inside surface.

Solve: (a) The reflections produce a net 180° phase difference and for there to be constructive interference the path difference 2t must correspond to a half-integer number of wavelengths to compensate for the λ /2 shift due to the 1⎞ ⎛ reflections. Hence the condition for constructive interference is 2t = ⎜ m + ⎟ (λ0 /n) (where m = 0, 1, 2, . . .). Here 2⎠ ⎝

λ0 is the wavelength in air and (λ0 /n) is the wavelength in the bubble, where the path difference occurs. Thus we 2tn 2(290 nm)(1.33) 771.4 nm = = . For m = 0, λ = 1543 nm; for m = 1, λ = 514 nm; for m = 2, 1 1 1 m+ m+ m+ 2 2 2 λ = 308 nm; . . . Only 514 nm is in the visible region; the color for this wavelength is green.

have λ0 =



26-5

2tn 2(340 nm)(1.33) 904.4 nm = = 1 1 1 m+ m+ m+ 2 2 2 For m = 0, λ = 1809 nm; for m = 1, λ = 603 nm; for m = 2, λ = 362 nm; . . . Only 603 nm is in the visible region; the color for this wavelength is orange. Reflect: The dominant color of the reflected light depends on the thickness of the film. If the bubble has varying thickness at different points, these points will appear to be different colors when the light reflected from the bubble is viewed. (b) λ0 =

26.16. Set Up: The interference is destructive since the film appears black. The waves reflected from the front surface of the film (waves in air reflecting from water) undergo a half-cycle reflection phase shift. The waves reflected from the back surface of the film (waves in water reflecting from air) do not undergo a reflection phase shift. Therefore, the condition for destructive interference is 2t = mλ . The thinnest film is for m = 1. λ = Solve: t =

λ0 n

.

mλ λ0 480 nm = = = 180 nm 2 2n 2(1.33)

26.17. Set Up: For both reflections, from the top and bottom surfaces of the film, the waves reflect from a material of greater n and undergo a half-cycle reflection phase shift. A nonreflective coating requires destructive interference

(

)

and the condition for destructive interference is 2t = m + 12 λ . λ = Solve: t =

λ 4

=

λ0 4n

=

λ0 n

. The minimum thickness is for m = 0.

480 nm = 80.5 nm. 4(1.49)

26.18. Set Up: Consider light reflected at the inner and outer surfaces of the film. At the inner surface of the film, light in air (n = 1.00) reflects from the film (n = 1.38) and there is a 180° phase shift due to the reflection. At the outer surface of the film, light in the film (n = 1.38) reflects from glass (n = 1.51) and there is also a 180° phase shift. Both rays have a 180° phase shift due to reflection, so there is no net phase difference produced. The path 650 nm difference for these two rays is 2t, where t is the thickness of the film. The wavelength in the film is λ = . n Solve: (a) Since the reflections produce no net phase difference, for constructive interference of the reflected light, ⎛ 650 nm ⎞ 2t = mλ . t = m ⎜ = (236 nm)m. The minimum thickness is 236 nm. ⎜ 2 [1.38] ⎟⎟ ⎝ ⎠ (b) The next three thicknesses that give constructive interference are for m = 2, 3, 4: 472 nm, 708 nm and 944 nm. *26.19. Set Up: Consider light reflected at the front and rear surfaces of the film. At the front surface of the film, light in air (n = 1.00) reflects from the film (n = 2.62) and there is a 180° phase shift due to the reflection. At the back surface of the film, light in the film (n = 2.62) reflects from glass (n = 1.62) and there is no phase shift due to reflection. Therefore, there is a net 180° phase difference produced by the reflections. The path difference for these 505 nm two rays is 2t, where t is the thickness of the film. The wavelength in the film is λ = . 2.62 Solve: (a) Since the reflection produces a net 180° phase difference, destructive interference of the reflected light ⎛ 505 nm ⎞ occurs when 2t = mλ . t = m ⎜ ⎟ = (96.4 nm)m. The minimum thickness is 96.4 nm. ⎝ 2[2.62] ⎠ (b) The next three thicknesses are for m = 2, 3, and 4: 192 nm, 289 nm, and 386 nm. Reflect: The minimum thickness is for t = λ /2n. Compare this to Problem 26.15, where the minimum thickness for destructive interference is t = λ /4n.


26-6

Chapter 26

26.20. Set Up: Since the lenses reflect 525 nm strongly, we must have constructive interference at that wavelength. The light reflected from the air-coating interface experiences a 180D phase inversion (since nair < ncoating ) and the light reflected from the coating-glass interface also experiences a 180D phase inversion (since ncoating < nglass ). Thus,

λ

λ

is the wavelength of the light the condition for constructive interference is 2t = m , where m = 0, 1, 2, . . . and n n in the coating (which is where the path-difference occurs). λ 525 nm Solve: The minimum (non-zero) required thickness occurs when m = 1 . This gives, t = = = 190 nm . 2n 2(1.38) Reflect: Since the path difference occurs within the coating, and not within the glass, the exact value of the index of refraction of the glass is not needed (provided we know that ncoating < nglass ).

*26.21. Set Up: The fringes are produced by interference between light reflected from the top and from the bottom surfaces of the air wedge. The refractive index of glass is greater than that of air, so the waves reflected from the top surface of the air wedge have no reflection phase shift and the waves reflected from the bottom surface of the air wedge do have a half-cycle reflection phase shift. The condition for constructive interference (bright fringes) therefore is 2t = m + 12 λ . The geometry of the air wedge is sketched in the figure below.

(

Solve: tan θ =

)

t 0.0800 mm λ = 8.89 × 10−4. tan θ = so t = (8.89 × 10−4 ) x. tm = m + 12 . x 2 90.0 mm

(

(

) 2(8.89λ× 10−4 ) and

(

) 2(8.89λ× 10−4 ) .

xm = m + 12

xm11 = m + 32

)

The distance along the plate between adjacent fringes is λ 656 × 10−9 m Δx = xm11 − xm = = = 3.69 × 10−4 m = 0.369 mm. −4 2(8.89 × 10 ) 2(8.89 × 10−4 ) 1.00 1.00 = = 27.1 fringes/cm. Δx 0.0369 cm Reflect: As t → 0 the interference is destructive and there is a dark fringe at the line of between the two

The number of fringes per cm is plates.

26.22. Set Up: The fringes are produced by interference between light reflected from the top and bottom surfaces of the air wedge. The refractive index of glass is greater than that of air, so the waves reflected from the top surface of the air wedge have no reflection phase shift and the waves reflected from the bottom surface of the air wedge do have a half-cycle reflection phase shift. The condition for constructive interference (bright fringes) is therefore



(

26-7

)

2t = m + 12 λ . The geometry of the air wedge is sketched in the figure below. At a distance x from the point of of the two plates, the thickness of the air wedge is t.

Solve: tan θ =

λ t λ so t = x tan θ . tm = m + 12 . xm = m + 12 x 2 2 tan θ

(

(

)

)

along the plate between adjacent fringes is Δx = xm11 − xm = Δx = tan θ =

λ 2Δx

=

546 × 10−9 m 2(0.0667 × 10

−2

m)

λ 2 tan θ

(

and xm11 = m + 32

. 15.0 fringes/cm =

λ . ) 2 tan θ

The distance

1.00 and Δx

1.00 = 0.0667 cm. 15.0 fringes/cm

= 4.09 × 10−4. The angle of the wedge is 4.09 × 10−4 rad = 0.0234°

26.23. Set Up: Since the film reflects 575 nm strongly, we must have constructive interference at that wavelength. The light reflected from the air-benzene interface experiences a 180D phase inversion (since nair < nbenzene ), but the light reflected from the benzene-water interface does not experience a phase inversion (since nbenzene > nwater ). Thus, the condition for constructive interference is 2t = m

λ

, where m = 1, 3, 5, . . . and

2n in the benzene (which is where the path-difference occurs).

λ n

is the wavelength of the light

λ 575 nm = = 95.8 nm. 4n 4(1.50) Reflect: Since the path difference occurs within the benzene, and not within the water, the exact value of the index of refraction of water is not needed (provided we know that nbenzene > nwater ). Solve: The minimum required thickness occurs when m = 1 . This gives, t =

26.24. Set Up: Both reflections occur for waves in the plastic substrate reflecting from the reflective coating, so they both have the same phase shift upon reflection and the condition for destructive interference (cancellation) is λ 2t = m + 12 λ , where t is the depth of the pit. λ = 0 . The minimum pit depth is for m = 0. n λ λ 790 nm λ Solve: 2t = . t = = 0 = = 110 nm = 0.11 μ m. 2 4 4n 4(1.8)

(

)

*26.25. Set Up: The condition for a dark fringe is sin θ =

mλ , m = ±1, ±2, … . a

⎛ 632.8 × 10−9 m ⎞ Solve: sin θ = m ⎜ = m(0.1687). m = ±1: θ = ±9.71°. m = ±2 : θ = ±19.7°. m = ±3 : θ = ±30.4°. ⎜ 0.00375 × 10−3 m ⎟⎟ ⎝ ⎠ m = ±4 : θ = ±42.4°. m = ±5 : θ = ±57.5°. mλ Reflect: There are a finite number of dark fringes because = sin θ can’t be larger than 1.00. This establishes a a maximum value for m. mλ 26.26. Set Up: λ a so the approximate expression ym = R is accurate. a


26-8

Chapter 26

λR

Solve: y1 =

a

=

(546 × 10−9 m)(1.75 m) 0.437 × 10−3 m

= 2.19 × 10−3 m = 2.19 mm

ym mλ is very small, so the approximate expression ym = R is accurate. y1 is the distance from the R a center of the screen to the first dark fringe on either side of the central maximum. 2 y1 = 4.50 mm and y1 = 2.25 mm.

26.27. Set Up: λR

λR

(600 × 10−9 m)(3.00 m)

= 8.00 × 10−4 m = 0.800 mm 2.25 × 10−3 m Reflect: The central bright fringe is twice as wide as the other bright fringes. y mλ 26.28. Set Up: m is very small, so the approximate expression ym = R is accurate. y1 = 1.35 mm. R a Solve: y1 =

Solve: λ =

a

. a=

y1

=

y1a (1.35 × 10−3 m)(0.750 × 10−3 m) = = 506 nm R 2.00 m

*26.29. Set Up: Calculate the angular positions of the minima and use y = x tan θ to calculate the distance on the screen between them. The central bright fringe is shown in Figure (a) below and the first bright fringe on one side of the central bright fringe is shown in Figure (b) below.

Solve: (a) The first minimum is located by sin θ1 =

λ a

=

633 × 10−9 m 0.350 × 10−3 m

= 1.809 × 10−3. Solving we obtain

θ1 = 1.809 × 10−3 rad. We can now find the distance between the center of the central fringe and the first minimum: y1 = x tan θ1 = (3.00 m) tan(1.809 × 10−3 rad) = 5.427 × 10−3 m.

2(5.427 × 10

−3

m) = 1.09 × 10

−2

The

width

of

the

central

bright

fringe

is:

w = 2 y1 =

m = 10.9 mm.

(b) The width of the first bright fringe adjacent to the central fringe is w = y2 − y1, where y1 = 5.427 × 10−3 m

from part (a). To find

y2

we note that sin θ 2 =

2λ = 3.618 × 10−3 so that θ 2 = 3.618 × 10−3 rad. Thus, a



26-9

y2 = x tan θ 2 = 1.085 × 10−2 m and the width of the bright fringe adjacent to the central one is w = y2 − y1 = 1.085 × 10−2 m − 5.427 × 10−3 m = 5.4 mm. Reflect: The central bright fringe is twice as wide as the other bright fringes.

26.30. Set Up: λ /a is very small, so the approximate expression ym = R

mλ is accurate. The distance between the a

two dark fringes on either side of the central maximum is 2 y1. Solve:

y1 =

λR a

=

(633 × 10−9 m)(3.50 m) 0.750 × 10−3 m

= 2.95 × 10−3 m = 2.95 mm. 2 y1 = 5.90 mm.

*26.31. Set Up: υ = f λ gives λ . The person hears no sound at angles corresponding to diffraction minima. The diffraction minima are located by sin θ = mλ /a (where m = ±1, ± 2, . . . ). Solve: λ = υ /f = (344 m/s)/(1250 Hz) = 0.2752 m; a = 1.00 m. m = ±1, θ = ±16.0°; m = ±2, θ = ±33.4°; m = ±3, θ = ±55.6°; no solution for larger m Reflect: λ /a = 0.28 so for the large wavelength sound waves diffraction by the doorway is a large effect.

Diffraction would not be easily observable for visible light because its wavelength is much smaller and λ /a 1.

26.32. Set Up: The dark fringes are located at angles θ that satisfy sin θ =

mλ , m = ±1, ±2, … . The largest value a

of 0 sin θ 0 is 1.00. Solve: (a) Solve for m that corresponds to sin θ = 1: m =

a

λ

=

0.0666 × 10−3 m 585 × 10−9 m

= 113.8. The largest value m can

have is 113. m = ±1, ±2, … , ±113 gives 226 dark fringes. ⎛ 585 × 10−9 m ⎞ (b) For m = ±113, sin θ = ±113 ⎜ = ±0.9926 and θ = ±83.0°. ⎜ 0.0666 × 10−3 m ⎟⎟ ⎝ ⎠

mλ . The dark fringes are located at sin θ = , n a m = ±1, ±2, … . The first dark fringes on either side of the central bright fringe correspond to m = +1 and m = −1. If a feature is located by angle θ , its distance from the center of the central maximum is y = R tan θ . In this problem θ is not small and Eq. (26.10) cannot be used. y 11.2 cm Solve: Each dark fringe is a distance y = 11.2 cm from the center of the screen. tan θ = = and R 30.0 cm

*26.33. Set Up: The wavelength of light in the liquid is λ =

λair

a sin θ (0.00125 × 10−3 m)sin 20.5° = = 438 nm. λair = nλ = (1.45)(438 nm) = 635 nm. 1 1 Reflect: Since the path difference occurs in the liquid it must be compared to the wavelength in the liquid.

θ = 20.5°. λ =

26.34. Set Up: For a grating the bright fringes are located by d sin θ = mλ , m = 0, ±1, ±2, … . The spots at ±15.4° correspond to m = ±1. 5510 lines/cm means d =

1 cm = 1.81 × 10−4 cm = 1.86 × 10−6 m is the spacing 5510

of adjacent lines of the grating. d sin θ (1.81 × 10−6 m)sin15.4° = = 481 nm. Solve: (a) λ = m 1 ⎛ 481 × 10−9 m ⎞ λ (b) The next angle is for m = ±2. sin θ = m = ±2 ⎜ = ±0.531 and θ = ±32.1°. ⎜ 1.81 × 10−6 m ⎟⎟ d ⎝ ⎠


26-10

Chapter 26

26.35. Set Up: The bright fringes are located at angles θ given by sin θ =

mλ . The separation d between adjacent d

1.00 × 10−3 m = 2.50 × 10−6 m. 400.0 ⎛ 600.0 × 10−9 m ⎞ Solve: sin θ = m ⎜ = m(0.240). First-order (m = 1) : θ = 13.9°. Second-order (m = 2) : θ = 28.7°. ⎜ 2.50 × 10−6 m ⎟⎟ ⎝ ⎠ Third-order (m = 3) : θ = 46.1°. slits in the grating is d =

26.36. Set Up: The line density in lines/cm is 1/d , with d in cm. The bright spots are located by d sin θ = mλ , m = 0, ±1, ±2, … . Solve: (a) d = (b) sin θ =

mλ (1)(632.8 × 10−9 m) 1 = = 2.07 × 10−6 m = 2.07 × 10−4 cm. = 4830 lines/cm. d sin θ sin17.8°

⎛ 632.8 × 10−9 m ⎞ mλ = m⎜ = m(0.3057). For m = ±2, θ = ±37.7°. For m = ±3, θ = ±66.5°. ⎜ 2.07 × 10−6 m ⎟⎟ d ⎝ ⎠

*26.37. Set Up: The bright spots are located by d sin θ = mλ , m = 0, ±1, ±2, … . The second pair correspond to 1 = 1.79 × 1024 cm = 1.79 × 1026 m. 5580 lines/cm y 26.3 cm = 0.6188 and θ = 31.75°. Solve: (a) y = R tan θ . tan θ = = R 42.5 cm m = ±2. d =

λ=

d sin θ (1.79 × 10−6 m)(sin 31.75°) = = 471 nm m 2

mλ 3(471 × 10−9 m) = and θ = 52.13°. d 1.79 × 10−6 m y = R tan θ = (42.5 cm)(tan 52.13°) = 54.7 cm.

(b) The next bright spots are for m = ±3. sin θ =

Reflect:

λ d

=

471 × 10−9 m 1.79 × 10−6 m

= 0.263 and is not small. The approximation sin θ ≈ tan θ is not accurate.

26.38. Set Up: The maxima are located by d sin θ = mλ where the order corresponds to the values of m. d sinθ4 4λ Solve: First-order: d sin θ1 = λ . Fourth-order: d sin θ 4 = 4λ. = , sinθ4 = 4sinθ1 = 4sin8.94° and θ4 = 38.4°. d sinθ1 λ Reflect: We did not have to solve for d.

*26.39. Set Up: The maxima are located by d sin θ = mλ, where d = 1.60 × 1026 m for a CD and d = 0.740 × 1026 m for a DVD. ⎛ m[6.328 × 10−7 m] ⎞ ⎛ mλ ⎞ Solve: (a) For a CD we have: θ = arcsin ⎜ ⎟⎟ = arcsin([0.396]m). For m = 1 we have ⎟ = arcsin ⎜⎜ −6 ⎝ d ⎠ ⎝ 1.60 × 10 m ⎠

θ1 = 23.3D. For m = 2 we have θ 2 = 52.3D. There are no other maxima. ⎛ m[6.328 × 10−7 m] ⎞ ⎛ mλ ⎞ (b) For a DVD we have: θ = arcsin ⎜ = arcsin ⎜⎜ ⎟⎟ = arcsin([0.855]m). For m = 1 we have ⎟ −6 ⎝ d ⎠ ⎝ 0.740 × 10 m ⎠

θ1 = 58.8D. There are no other maxima. Reflect: The reflective surface produces the same interference pattern as a grating with slit separation d.



26-11

26.40. Set Up: Maxima are located by d sin θ = mλ , m = 0, ±1, ±2, … . The largest value 0 sin θ 0 can have is 1.00. d=

1 = 2.06 × 10−3 mm = 2.06 × 10−6 m. 485 lines/mm

Solve: (a) Set sin θ = 1.00 and solve for m: m =

d sin θ

λ

=

(2.06 × 10−6 m)(1.00) 631 × 10−9 m

= 3.3. The largest 0 m 0 is 3, so there

are 7 bright spots (m = 0, ±1, ±2, ±3). (b) m = 3. sin θ =

mλ 3(631 × 10−9 m) = and θ = 66.8°. d 2.06 × 10−6 m

26.41. Set Up: The bright spots are located by d sin θ = mλ . Third-order means m = 3 and second-order means m = 2. Solve:

mλ m λ mλ = d = constant, so r r = v v . sin θ r sin θ v sin θ

⎛ m ⎞⎛ λ ⎞ ⎛ 2 ⎞⎛ 400 nm ⎞ sin θ v = sin θ r ⎜ v ⎟⎜ v ⎟ = (sin 65.0°) ⎜ ⎟⎜ ⎟ = 0.345 and θ v = 20.2°. λ m ⎝ 3 ⎠⎝ 700 nm ⎠ ⎝ r ⎠⎝ r ⎠ Reflect: The third-order line for a particular λ occurs at a larger angle than the second-order line. In a given order, the line for violet light (400 nm) occurs at a smaller angle than the line for red light (700 nm).

26.42. Set Up: The maxima are at angles θ given by 2d sin θ = mλ , where d is the spacing between adjacent planes in the crystal.

2λ 0.0850 nm = = 0.232 nm 2sin θ sin 21.5° *26.43. Set Up: The maxima are at angles θ given by 2d sin θ = mλ , where d = 0.440 nm. Solve: m = 2. d =

Solve: m = 1. λ =

2d sin θ = 2(0.440 nm)sin 39.4° = 0.559 nm. 1

26.44. Set Up: Maxima are located by 2d sin θ = mλ , where θ is the angle of incidence and d is the separation of crystal planes. m = 1, 2, 3, … Solve: (a) First maximum, so m = 1. d = (b) sin θ =

mλ (1)(0.173 × 10−9 m) = = 0.227 nm. 2sin θ 2(sin 22.4°)

⎛ 0.173 × 10−9 m ⎞ mλ = m⎜ = (0.381)m. For m = 2, θ = 49.6°. For m > 2 the equation has no solution. ⎜ 2[0.227 × 10−9 m] ⎟⎟ 2d ⎝ ⎠

0.713 nm λ mλ = 0.762. Since is not small, we find that = 1 for m = 2.6 and there are only two d d 0.227 nm 2d angles for which constructive interference occurs. Reflect:

λ

=

λ y *26.45. Set Up: Rayleigh’s criterion says θ res = 1.22 . D = 7.20 cm. θ res = , where s is the distance of the D

s

object from the lens and y = 4.00 mm. Solve:

λ yD (4.00 × 10−3 m)(7.20 × 10−2 m) y = 1.22 . s = = = 429 m. s D 1.22λ 1.22(550 × 10−9 m)

Reflect: The focal length of the lens doesn’t enter into the calculation. In practice, it is difficult to achieve resolution that is at the diffraction limit.


26-12

Chapter 26

26.46. Set Up: The diffraction is by a single slit of width a, not by a circular aperture. The diffraction minima are mλ y λ and Rayleigh’s criterion is θ res = . θ res = , where s is the distance of the object from the located by sin θ = a

a lens and y = 2.50 m.

Solve:

s

y λ ya (2.50 m)(0.350 × 10−3 m) = = 1.46 km = . s= s a λ 600 × 10−9 m

λ W 26.47. Set Up: Apply sin θ = 1.22 , where θ = , W = 28 km, and h = 1200 km. Since θ is small we have D

sin θ ≈ θ . Solve: D =

h

1.22λ h 1.2 × 106 m = 1.22λ = 1.22(0.036 m) = 1.88 m sin θ W 2.8 × 104 m

Reflect: D must be significantly larger than the wavelength, so a much larger diameter is needed for microwaves than for visible wavelengths.

λ 26.48. Set Up: Rayleigh’s criterion says sin θ = 1.22 . The best resolution is 0.3 arcseconds, which is about D

−5

(8.33 × 10 )°. 1.22λ 1.22(5.5 × 10−7 m) = = 0.46 m sin θ sin(8.33 × 10−5°) Reflect: (b) The Keck telescopes are able to gather more light than the Hale telescope; hence they can detect fainter objects. However, their larger size does not allow them to have greater resolution⎯atmospheric conditions limit the resolution. Solve: (a) D =

λ *26.49. Set Up: θ res = 1.22 . The angular size of the object is its height divided by its distance from the eye. D

−9

⎛ 550 × 10 m ⎞ y Solve: θ res = 1.22 ⎜ = 3.36 × 10−4. θ res = and ⎜ 2.00 × 10−3 m ⎟⎟ s ⎝ ⎠ y = s θ res = (25.0 cm)(3.36 × 10−4 ) = 8.4 × 10−3 cm = 0.084 mm.

Diffraction doesn’t play a significant role.

λ 26.50. Set Up: θ res = 1.22 . The angular size of the object is its height divided by its distance from the eye. D Solve: (a) The angular size of the object is

θ=

50 × 10−6 m 25 × 10

θ res = 1.22

−2

m

= 2.0 × 10−4 rad.

⎛ 550 × 10−9 m ⎞ = 1.22 ⎜ = 3.4 × 10−4 rad. ⎜ 2.0 × 10−3 m ⎟⎟ D ⎝ ⎠

λ

θ < θ res and the object cannot be resolved. (b) θ res =

y and y = sθ res = (25 cm)(3.4 × 10−4 rad) = 8.5 × 10−3 cm = 85 μ m. s

(c) θ = θ res = 3.4 × 10−4 rad = 0.019° = 1.1 min. This is very close to the experimental value of 1 min. (d) Diffraction is more important.

λ 26.51. Set Up: Take the width of a typical license plate to be about 30 cm. θ res = 1.22 . D



Solve: The angular size of the object is θ =

D=

0.30 m 400 × 103 m

1.22λ

=

= 7.5 × 10−7 rad. Setting θ res = θ gives

(1.22)(500 × 10−9 m)

7.5 × 10−7 rad Reflect: A mirror of this diameter (less than one meter) is feasible.

θ

26-13

= 81 cm.

26.52. Set Up: υ = f λ . Constructive interference occurs when the path difference r2 − r1 from the two sources is r2 − r1 = mλ , m = 0, ±1, D ±2, … Destructive interference occurs when the path difference r2 − r1 is r2 − r1 = mλ , m = 0, ±1, ±2, … Solve: (a) The path difference from the two speakers is a half-integer number of wavelengths and the interference is destructive. λ λ υ 340 m/s (b) The path difference changes by , so = 0.398 m and λ = 0.796 m. f = = = 427 Hz, λ 0.796 m 2 2 (c) The speaker must be moved a distance λ = 0.796 m, so the path difference will change by λ.

(

)

*26.53. Set Up: The interference minima are located by d sin θ = m + 12 λ . For a liquid with refractive index n, λliq = Solve:

λair n

.

sin θ

λ

=

( m + 12 ) = constant, so sin θair = sin θliq . λair

d

λliq

sin θ air

λair

(

=

Reflect: In the liquid the wavelength is shorter and sin θ = m + 12

sin θ liq

λair /n

) λd

and n =

sin θ air sin 35.20° = = 1.730. sin θ liq sin19.46°

gives a smaller θ than in air, for the same m.

26.54. Set Up: Consider the interference of the rays reflected from each side of the film. At the front of the film light in air reflects off the film (n = 1.432) and there is a 180° phase shift. At the back of the film light in the film (n = 1.432) reflects off the glass (n = 1.62) and there is a 180° phase shift. Therefore, the reflections introduce no net phase shift. The path difference 2t, where t is the thickness of the film. The wavelength in the film is λ =

λair

. n Solve: (a) Since there is no net phase difference produced by the reflections, the condition for destructive λ λ λ 550 nm and the minimum thickness is t = = air = = 96.0 nm. interference is 2t = m + 12 λ . t = m + 12 2 4 4 4[1.432]

(

)

(

)

(

(b) For destructive interference, 2t = m + 12

) λnair

and λair =

2tn 275 nm = . m = 0 : λair = 550 nm. m = 1: m + 12 m + 12

λair = 183 nm. All other λair values are shorter. The only visible wavelength in air for which there is destructive λ 2tn 275 nm interference is 550 nm. For constructive interference, 2t = m air and λair = = . For m = 1, m

n

m

λair = 275 nm and all other λair values are shorter. There are no visible wavelengths in air for which there is constructive interference.

*26.55. Set Up: Consider reflection from either side of the film. (a) At the front of the film, light in air (n = 1.00) reflects off the film (n = 1.45) and there is a 180° phase shift. At the back of the film, light in the film (n = 1.45) reflects off the cornea (n = 1.38) and there is no phase shift. The reflections produce a net 180° phase difference and

(

)

the condition for constructive interference is 2t = m + 12 λ , where λ = for m = 0, and is given by t =

λair 4n

=

λair n

(

. t = m + 12

) λ2airn . Minimum thickness is

600 nm = 103 nm. 4(1.45)


26-14

(b) λair =

Chapter 26

2nt 2(1.45)(103 nm) 299 nm = = . For m = 0, λair = 598 nm. For m = 1, λair = 199 nm and all other m + 12 m + 12 m + 12

values are smaller. No other visible wavelengths are reinforced. The condition for destructive interference is λ 2tn 299 nm 2t = m air . λ = = . For m = 1, λair = 299 nm and all other values are shorter. There are no visible m m n wavelengths for which there is destructive interference. (c) Now both rays have a 180° phase change on reflection and the reflections don’t introduce any net phase shift. The expression for constructive interference in parts (a) and (b) now gives destructive interference and the expression in (a) and (b) for destructive interference now gives constructive interference. The only visible wavelength for which there will be destructive interference is 600 nm and there are no visible wavelengths for which there will be constructive interference. Reflect: Changing the net phase shift due to the reflections can convert the interference for a particular thickness from constructive to destructive, and vice versa.

26.56. Set Up: Rayleigh’s criterion limits the angular resolution according to sin θ ≈ θ = 1.22 λ/D. Solve: (a) Using Rayleigh’s criterion we have sin θ ≈ θ = 1.22 λ/D = (1.22)(550 nm)/(135/4 mm) = 1.99 × 10–5 rad. On the bear this angle subtends a distance x. θ = x/R and x = Rθ = (11.5 m)(1.99 × 10–5 rad) = 2.29 × 10–4 m = 0.23 mm. (b) At f /22, D is 4/22 times as large as at f /4. Since θ is proportional to 1/D, and x is proportional to θ, x is 1/(4/22) = 22/4 times as large as it was at f /4. Thus, x = (0.229 mm)(22/4) = 1.3 mm. Reflect: A wide-angle lens, such as one having a focal length of 28 mm, would have a much smaller opening at f /22 and hence would have an even less resolving ability.

λ *26.57. Set Up: The dark fringes are located by sin θ = m . The first dark fringes are for m = ±1. y = R tan θ is a the distance from the center of the screen. From the center to one minimum is 2.61 cm.

Solve: tan θ =

y 2.61 cm λ 632.8 × 10−9 m = = 0.02088 and θ = 1.20°. a = = = 30.2 μm. R 125 cm sin θ sin1.20°

26.58. Set Up: For the waves reflected at the top surface of the oil film there is a half-cycle reflection phase shift. For the waves reflected at the bottom surface of the oil film there is no reflection phase shift. The condition for constructive interference is 2t = m + 12 λ . The condition for destructive interference is 2t = mλ . Wavelengths that

(

)

are predominant in the transmitted light are those for which there is destructive interference in the reflected light. The range of visible wavelengths is approximately 400 nm to 700 nm. In the oil film, λ =

(

) (

Solve: (a) 2t = m + 12 λ = m + 12

) λn0 .

λ0 =

λ0 n

.

2tn 2(380 nm)(1.45) 1102 nm = = . m = 0 : λ0 = 2200 nm. m = 1: 1 m+ 2 m + 12 m + 12

λ0 = 735 nm. m = 2 : λ0 = 441 nm. m = 3 : λ0 = 315 nm. The visible wavelength for which there is constructive interference in the reflected light is 441 nm.

λ0

2tn 1102 nm . λ0 = = . m = 1 : λ0 = 1102 nm. m = 2 : λ0 = 551 nm. m = 3 : λ0 = 367 nm. The m m n visible wavelength for which there is destructive interference in the reflected light is 551 nm. This is the visible wavelength predominant in the transmitted light.

(b) 2t = mλ = m

Reflect: At a particular wavelength the sum of the intensities of the reflected and transmitted light equals the intensity of the incident light.



26-15

26.59. Set Up: If the glass plate is illuminated from above, the waves reflected at the top surface of the plate have a half-cycle reflection phase shift and the waves reflected at the bottom surface of the plate have no reflection phase shift. The condition for constructive interference is 2t = m + 12 λ and the condition for destructive interference is

(

2t = mλ . λ =

λ0 n

)

.

(

) (

Solve: (a) 2t = m + 12 λ = m + 12

) λn0 .

λ0 =

2tn 2(0.485 × 10−6 m)(1.53) 1484 nm = = . m = 0 : λ0 = 2968 nm. m + 12 m + 12 m + 12

m = 1: λ0 = 989 nm. m = 2 : λ0 = 594 nm. m = 3 : λ0 = 424 nm. m = 4 : λ0 = 330 nm. The visible wavelengths intensified in the reflected light are 594 nm and 424 nm. λ 2tn 1484 nm (b) 2t = mλ = m 0 . λ0 = = . m = 1: λ0 = 1484 nm. m = 2 : λ0 = 742 nm. m = 3 : λ0 = 495 nm. n m m m = 4 : λ0 = 371 nm. The visible wavelength cancelled in the reflected light is 495 nm.

26.60. Set Up: Consider the interference between light reflected from the top and bottom surfaces of the air film between the lens and the glass plate. For maximum intensity, with a net half-cycle phase shift due to reflections, we 1⎞ ⎛ have 2t = ⎜ m + ⎟ λ . We have t = R − R 2 − r 2 , where R is the radius of curvature of the lens and r is measured ⎝ 2⎠ horizontally from the point of of the lens and glass plate. (2m + 1)λ (2m + 1)λ Solve: = R − R2 − r 2 ⇒ R2 − r 2 = R − 4 4 2

(2m + 1)λ R (2m + 1)λ R ⎡ (2m + 1)λ ⎤ ⎡ (2m + 1)λ ⎤ − ⇒r= −⎢ ⇒ R2 − r 2 = R2 + ⎢ ⎥ ⎥⎦ 4 2 2 4 ⎣ ⎦ ⎣

2

(2m + 1)λ R , for R λ . 2 The second bright ring is when m = 1: ⇒r≈

(2(1) + 1) (5.80 × 10−7 m) (0.952 m) = 9.10 × 10−4 m = 0.910 mm. 2 So the diameter of the second bright ring is 1.82 mm. Reflect: The diameter of the m th ring is proportional to 2m + 1, so the rings get closer together as m increases. This agrees with Figure 26.12b in the textbook. r≈

*26.61. Set Up: Maxima are given by 2d sin θ = mλ , where d is the separation between crystal planes. ⎛ 0.125 nm ⎞ ⎛ mλ ⎞ Solve: (a) θ = arcsin ⎜ ⎟ = arcsin(0.2216m). ⎟ = arcsin ⎜ m ⎝ 2d ⎠ ⎝ 2(0.282 nm) ⎠ For m = 1: θ = 12.8°, m = 2 : θ = 26.3°, m = 3 : θ = 41.7°, and m = 4 : θ = 62.4°. No larger m values yield answers. ⎛ 2mλ ⎞ a , then θ = arcsin ⎜⎜ ⎟⎟ = arcsin(0.3134m). 2 ⎝ 2a ⎠ So for m = 1: θ = 18.3°, m = 2: θ = 38.8°, and m = 3: θ = 70.1°. No larger m values yield answers. Reflect: In part (b), where d is smaller, the maxima for each m are at larger θ . (b) If the separation d =

λ y 26.62. Set Up: The diameter of Jupiter is 1.38 × 108 m. θ res = 1.22 . The angular size of an object is θ = . D

s

1 light { year = 9.461 × 1015 m.


26-16

Chapter 26

Solve: θ =

y 1.38 × 108 m = = 3.39 × 10−9 rad. Setting θ res = θ gives s 4.3(9.461 × 1015 m) D=

1.22λ

θ

⎛ 550 × 10−9 m ⎞ = 1.22 ⎜ = 198 m. ⎜ 3.39 × 10−9 rad ⎟⎟ ⎝ ⎠

*26.63. Set Up: The bright fringes are located by d sin θ = mλ . First-order means m = 1. The line density is 1/d. The largest λ appears at the largest θ . If the λ1 = 400 nm line is at θ , then the λ 2 = 700 nm line is to be at θ + 30.0°. Solve: sin θ =

(0.500) =

λ2 d

λ1 d

. sin (θ + 30.0°) =

λ2 d

. cos θ = 1 − sin 2 θ = 1 −

. sin (θ + 30.0°) = sin θ cos30.0° + cos θ sin 30.0°, so (sin θ )(0.866) + (cos θ )

λ1 2 d

2

, so 0.866

λ1 d

+ 0.500 1 −

λ1 2 d

2

=

λ2 d

. This gives 700 nm − (0.866)

(400 nm) = 0.500 d 2 − (400 nm)2 and d = 812 nm = 8.12 × 10−4 mm. The line density is

1 = 1230 lines/mm. d

λ2

> 1 and the second-order pattern doesn’t include all of the visible spectrum. The first order is the only d order that displays the entire visible spectrum. Reflect: 2

26.64. Set Up: Interference occurs due to the path difference of light in the thin film. Originally the path difference was an odd number of half-wavelengths for cancellation to occur. If the path difference decreases by ½ wavelength, it will be a multiple of the wavelength, so constructive interference will occur. Solve: Calling ΔT the thickness that must be removed, we have the path difference = 2ΔT = ½ λ/n and ΔT = λ/4n = (525 nm)/[4(1.40)] = 93.75 nm. At 4.20 nm/yr, we have (4.20 nm/yr)t = 93.75 nm and t = 22.3 yr. Reflect: If you were giving a warranty on this film, you certainly could not give it a “lifetime guarantee”!

26.65. Set Up: The maxima are given by d sin θ = mλ. We need sin θ = wavelengths to be seen. For 650 slits/mm ⇒ d =

6.50 × 105 m −1

= 1.53 × 10−6 m.

2λ1 3λ = 0.52; m = 3 : 1 = 0.78. d d d λ2 2λ 2 3λ −7 = 0.92; m = 3 : 2 = 1.37. λ 2 = 7.00 × 10 m: m = 1: = 0.46; m = 2 : d d d So, the third order does not contain the violet end of the spectrum, and therefore only the first- and second-order diffraction patterns contain all colors of the spectrum. Reflect: θ for each maximum is larger for longer wavelengths. Solve: λ1 = 4.00 × 10−7 m: m = 1:

λ1

1

mλ ≤ 1 in order for all the visible d

= 0.26; m = 2 :

Solutions to age Problems 26.66. Set Up: If two speakers are connected to a single oscillator it is likely that some students will hear constructive interference and others will hear destructive interference. Solve: Changing either the volume or the phase relationship of the speakers does not alter the fact that some students will hear constructive interference while others will hear destructive interference. However, if one speaker is unplugged the interference effects will disappear. Any student who was originally located at a point of destructive interference will now be able to hear a tone, and any student who was originally at a point of constructive interference will hear the intensity of the sound diminish. The correct answer is D.



26-17

*26.67. Set Up: If two speakers are connected to a single oscillator it is likely that some students will hear constructive interference and others will hear destructive interference. Solve: Both speakers must still be connected since some students are still unable to hear a tone. Furthermore, if the volume were simply increased or decreased the points of destructive interference would not change. However, if the phase relationship between the speakers is changed (e.g., the polarity of the wires on one speaker is switched so that that speaker vibrates 180D out of phase with the other speaker) then points of destructive and constructive interference will be switched. The correct answer is C.

26.68. Set Up: For constructive interference the path difference must be a multiple of the wavelength and for destructive interference the path difference must be an odd multiple of a half-wavelength (this is reversed if the speakers are driven out of phase). The relation between frequency and wavelength is given by f λ = υ , where we are given the speed of sound to be υ = 340 m/s. Solve: To switch from constructive to destructive interference the speaker must be moved so that its sound travels an additional half-wavelength to reach you. Thus f =

λ

2

= 0.34 m and we have λ = 0.68 m. The frequency must then be

340 m/s = 500 Hz. The correct answer is C. 0.68 m


27

RELATIVITY

Answers to Multiple-Choice Problems 1. A, B 13. A, C

2. B, C 14. A

3. A, D 15. A

4. D

5. C

6. A

7. D

8. B

9. D

10. A

11. C

12. B

Solutions to Problems *27.1. Set Up: We will work in the rest frame of the spaceship. In this rest frame the enger sees her distance from the asteroid increase as the asteroid moves away from her at 0.9c. Simultaneously, she sees her distance from the earth decrease as the earth moves toward her at 0.9c. Solve: The enger receives the radio messages simultaneously when her distance from the asteroid is equal to her distance from the earth. Therefore, she deduces that the signal from the earth was sent at a time when she was farther from the earth than she was from the asteroid. Since the signal from the earth must travel a greater distance than the signal from the asteroid, the enger deduces that the signal from the earth must have been sent before the signal from the asteroid. Reflect: In the rest frame of an observer on Mars the two signals travel identical distances. Thus, in order to arrive simultaneously at the spaceship, the signals were sent simultaneously from Asteroid 1040A and from the earth.

27.2. Set Up: Consider the distance traveled by the light to reach each wall. Solve: (a) With respect to the astronaut riding with the rocket the light travels the same distance to each wall and the events are simultaneous. (b) Relative to the person at rest on earth, the walls of the rocket move to the right as the light travels to them so the light travels a shorter distance to reach the rear wall and this person observes B occurring first.

27.3. Set Up: The observer on Pluto measures the proper time Δt0 since the light blinks on and off at the same point in her frame. Δt0 = 80.0 μ s and u = 0.964c. Solve: Δt =

Δt0 2

1 − u /c

2

=

80.0 × 10−6 s 1 − (0.964)

2

= 3.01 × 10−4 s = 0.301 ms

Reflect: The pilot measures a much longer time for the duration of the pulse than is measured by an observer at rest with respect to the signal light.

27.4. Set Up: The astronaut in the spaceship measures the proper time, since the end of a swing occurs at the same location in his frame. Δt0 1.50 s Solve: (a) Δt0 = 1.50 s. Δt = = = 2.27 s 2 2 1 − u /c 1 − (0.75c/c) 2 © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

27-1

27-2

Chapter 27

(b) Δt = 1.50 s. Δt0 = Δt 1 − u 2 /c 2 = (1.50 s) 1 − (0.75c/c) 2 = 0.992 s

*27.5. Set Up: As seen by your friend in the earth’s frame of reference the round-trip distance is d0 = 11.5 × 109 km. As seen from your frame of reference the round-trip distance is contracted to d = d 0 1 −

u2 c2

. Your speed

m/s ⎞ ⎛ 4 8 relative to the earth is u = (45,000 km/h) ⋅ ⎜ 0.2778 ⎟ = 1.25 × 10 m/s and we know that c = 3.00 × 10 m/s. ⎝ km/h ⎠ Solve: (a) As seen by your friend on the earth the trip takes tearth = (b) As seen by you the round-trip distance is d = d 0 1 −

d 0 11.5 × 109 km = = 2.56 × 105 h = 29.2 y. u 45,000 km/h

u2

, and the round-trip time is t you = 2

d0 u2 1− 2 u c

c Thus, the difference in time between your friend’s watch and yours is 2⎞ ⎛ ⎛ 1.25 × 104 m/s ⎞ ⎟ d ⎛ u 2 ⎞ 11.5 × 1012 m ⎜ tearth − t you = 0 ⎜1 − 1 − 2 ⎟ = − − = 0.80 s. 1 1 ⎜ ⎟ ⎜ 3.00 × 108 ms ⎟ ⎟ u ⎜ c ⎟⎠ 1.25 × 104 m/s ⎜⎜ ⎝ ⎠ ⎟ ⎝ ⎝ ⎠

Reflect: This problem can also be solved using the time dilation formula, as explained in Example 27.2 and the subsequent discussion of the Twin Paradox.

27.6. Set Up: The average lifetime measured in the pion frame is the proper time; Δt0 = 2.60 × 10−8 s. Δt = 4.20 × 10−7 s. Solve: (a) Δt =

Δt0 1 − u 2 /c 2

gives u = c 1 − ( Δt0 /Δt ) 2 = c (1 − (2.60 × 10−8 /4.20 × 10−7 )2 = 0.998c.

(b) d = vt = (0.998)(3.00 × 108 m / s)(4.20 × 10−7 s) = 126 m

27.7. Set Up: The proper time is measured in the frame where the two events occur at the same point. Solve: (a) The time of 12.0 ms measured by the first officer on the craft is the proper time. (b) Δt =

Δt0

gives u = c 1 − ( Δt0 /Δt ) 2 = c 1 − (12.0 × 10−3 /0.190)2 = 0.998c.

1 − u 2 /c 2

Reflect: The observer at rest with respect to the searchlight measures a much shorter duration for the event.

27.8. Set Up: The problem asks for u such that Δt0 /Δt = 12 . Solve: Δt =

Δt0 2

2

gives u = c 1 − (Δt0 /Δt ) 2 = (3.00 × 108 m/s) 1 −

( 12 )

2

= 2.60 × 108 m/s;

u = 0.867 c

1 − u /c Jet planes fly at less than ten times the speed of sound, less than about 3000 m/s. Jet planes fly at much lower speeds than we calculated for u.

*27.9. Set Up: A clock moving with respect to an observer appears to run more slowly than a clock at rest in the observer’s frame. The clock in the spacecraft measurers the proper time Δt0 . Δt = 365 days = 8760 hours. Solve: The clock on the moving spacecraft runs slow and shows the smaller elapsed time. Δt0 = Δt 1 − u 2 /c 2 =

(8760 h) 1 − (4.80 × 106 /3.00 × 108 )2 = 8758.88 h. The difference in elapsed times is 8760 h − 8758.88 h = 1.12 h.


Relativity

27-3

27.10. Set Up: The proper length is measured in the frame in which the object is at rest, so l0 = 3.60 m. Solve: l = l0 1 − u 2 /c 2 = (3.60 m) 1 − (0.900)2 = 1.57 m

*27.11. Set Up: When the meterstick is at rest with respect to you, you measure its length to be 1.000 m, and that is its proper length, l0 . l = 0.3048 m. Solve: l = l0 1 − u 2 /c 2 gives u = c 1 − (l/l0 )2 = c 1 − (0.3048/1.00) 2 = 0.9524c = 2.86 × 108 m/s.

27.12. Set Up: The length measured when the spacecraft is stationary is its proper length, l0 . l = 140 m. Solve: l0 =

l 2

1 − u /c

2

=

140 m 1 − (0.800)2

= 233 m

27.13. Set Up: The doctor in the rocket measures his proper length l0 . Solve: (a) l0 = 2.00 m. l = l0 1 − u 2 /c 2 = (2.00 m) 1 − (0.850) 2 = 1.05 m

The person on earth would measure his height to be 1.05 m. (b) l = 2.00 m. l0 =

l 2

1 − u /c

2

=

2.00 m 1 − (0.850)2

= 3.80 m

This is not a reasonable height for a human. (c) There is no length contraction in a direction perpendicular to the motion and both observers measure the same height, 2.00 m. Reflect: The length of an object moving with respect to the observer is shortened in the direction of the motion, so in (a) and (b) the observer on earth measures a shorter height.

27.14. Set Up: The proper distance between Alpha Centauri and the earth ( l0 = 4.37 light years ) is measured in the earth’s frame of reference. As explained in Example 27.2, the proper time for the trip, Δt0 , is measured in the plane’s frame of reference. The speed of light is c = (1 light year)/(year). Solve: (a) As seen from the earth the trip takes t =

l0 4.37 ly = = 4.58 y u 0.955 ly/y

(b) Using the time dilation formula we obtain: Δt0 = Δt 1 − u 2 /c 2 = (4.58 y) 1 − (0.955) 2 = 1.36 y. (c) As seen from the spaceship the distance to Alpha Centauri is

l = l0 1 − u 2 /c 2 = (4.37 ly) 1 − (0.955) 2 = 1.30 ly. (d) As seen by the engers in the spaceship, Alpha Centauri approaches at 0.955c and must cover a distance of l 1.30 ly 1.30 ly. Thus, from the enger’s point of view the required time for the trip is t = = = 1.36 y. u 0.955 ly/y Reflect: The answers from (b) and (d) agree, as they must.

*27.15. Set Up: The height measured in the earth’s frame is a proper length. The lifetime measured in the muon’s frame is the proper time. Solve: (a) l0 = 55.0 km. l = l0 1 − u 2 /c 2 = (55.0 km) 1 − (0.9860) 2 = 9.17 km. (b) In the muon’s frame its lifetime is 2.20 μs, so the distance it travels during its lifetime is

(2.20 × 10−6 s)(0.9860)(3.00 × 108 m/s) = 651 m. This is

651 m 9.17 × 103 m

= 7.1, of its initial height.


27-4

(c) Δt =

Chapter 27

Δt0 1 − u 2 /c 2

=

2.20 × 10−6 s 1 − (0.9860) 2

= 1.32 × 10−5 s. The distance it travels in the earth’s frame during this time is

(1.32 × 10−5 s)(0.9860)(3.00 × 108 m/s) = 3.90 km. This is

3.90 km = 7.1, of its initial height measured in the 55.0 km

earth’s frame. Reflect: There are two equivalent views. In the muon’s frame, its distance above the surface of the earth is contracted because the earth is moving relative to the muon. In the earth’s frame the lifetime of the muon is dilated due to the motion of the muon relative to the earth.

27.16. Set Up: Let the starfighter’s frame be S and let the enemy spaceship’s frame be S′. Let the positive x direction for both frames be from the enemy spaceship toward the starfighter. Then u = 10.400c. υ ′ = +0.700c. υ is the velocity of the missile relative to you. υ′ + u 0.700c + 0.400c Solve: (a) υ = = = 0.859c 2 1 + (0.400)(0.700) 1 + uυ ′/c (b) Use the distance it moves as measured in your frame and the speed it has in your frame to calculate the time it 8.00 × 109 m = 31.0 s. takes in your frame. t = (0.859)(3.00 × 108 m/s) *27.17. Set Up: Let the frame attached to Arrakis be S and let the frame attached to the imperial spaceship be S′. Let the positive x direction for both frames be from the spaceship toward Arrakis. Then υ ′ = +0.920c, υ = +0.360c and we want to solve for u. υ − υ′ 0.360c − 0.920c υ −u = = −0.837c. The speed of the starship relative to gives u = Solve: υ ′ = 2 2 1 − (0.360)(0.920) 1 − υυ ′ /c 1 − uυ /c Arrakis is 0.837c. Since u < 0 and the direction from the spaceship toward Arrakis is positive, the spaceship is moving away from Arrakis. Reflect: The Galilean transformation gives that the spaceship is moving away from Arrakis with a speed of 0.920c − 0.360c = 0.560c. The Lorentz transformation gives a larger speed. 27.18. Set Up: Let S be the laboratory frame and let S′ be the frame of one of the particles, as shown in the figure below. Let the positive x direction for both frames be from particle 1 to particle 2. In the lab frame particle 1 is moving in the +x direction and particle 2 is moving in the −x direction. Then u = 0.9520c and υ = −0.9520c. υ ′ is the velocity of particle 2 relative to particle 1.

Solve: υ ′ =

υ −u −0.9520c − 0.9520c = = −0.9988c. The speed of particle 2 relative to particle 1 is 2 1 − uυ /c 1 − (0.9520c)(−0.9520c)/c 2

0.9988c. υ ′ < 0 shows particle 2 is moving toward particle 1.


Relativity

27-5

27.19. Set Up: Let S be a frame attached to planet Tatooine and let S′ be a frame attached to the pursuit ship. Let the positive x direction for both frames be from the planet toward the pursuit ship. Then u = +0.800c and υ = +0.600c. υ ′ is the velocity of the cruiser relative to the pursuit ship. Solve: υ ′ =

υ −u 1 − uυ /c 2

=

0.600c − 0.800c 1 − (0.800c)(0.600c)/c 2

= −0.385c. υ ′ < 0 means that the cruiser is moving toward the

pursuit ship. Reflect: The Galilean transformation gives that the cruiser is moving toward the pursuit ship with a speed of 0.800c − 0.600c = 0.200c. The Lorentz transformation gives a larger speed.

27.20. Set Up: Let S be the laboratory frame and let S′ be the frame of one of the particles, as shown in the figure below. Let the positive x direction for both frames be the direction from particle 1 to particle 2. Then u = −0.650c and υ ′ = +0.950c. υ is the velocity of particle 2 as measured in the lab frame.

υ′ + u +0.950c + ( −0.650c) = = +0.784c. The speed measured in the laboratory for the second 1 + uυ ′/c 2 1 + (−0.650c)(+0.950c )/c 2 particle is 0.784c. υ > 0 and u < 0 since in the laboratory the two particles move in opposite directions.

Solve: υ =

*27.21. Set Up: The rotation speed can be calculated from the radius of the star and its period: υ = Lorentz transformation for velocity is given by υ ′ = Solve: (a) υ /c =

υ −u 1 − uυ /c 2

2π r . The T

.

2π r 2π (10.0 × 103 m) = = 0.116; thus we have υ = 0.116c. cT (3.00 × 108 m/s)(1.80 × 10−3 s)

(b) Let the frame S be located at the center of the star, let S′ be a frame located at any point on the star’s equator, and let P be a point on the equator that is diametrically opposite S′ . If the velocity of P relative to S is υ = +0.116c then the velocity of S′ as seen by S must be u = −υ = −0.116c. Thus, the velocity of P relative to S′ is υ −u 2υ 2(0.116c) = = = 0.229c. υ′ = 1 − uυ /c 2 1 + (υ /c) 2 1 + (0.116) 2 Reflect: If we ignore relativity, the relative velocity between these two points would be υ − u = 2υ = 2(0.116c ) = 0.232c, which is larger than the correct relativistic result as expected.

27.22. Set Up: The momentum of a particle has magnitude p = Solve: p = 3mv means

1 2

1 − υ /c

2

mυ 1 − υ 2 /c 2

.

= 3. υ = c 1 − (1/3) 2 = 0.943c.


27-6

Chapter 27

*27.23. Set Up: The momentum of a particle has magnitude p = Solve: (a) The problem specifies that p = 1.010mυ . 1.010 =

mυ 2

1 − υ /c

1 1 − υ 2 /c 2

2

. 1 − υ 2 /c 2 is less than 1, so p > mυ .

.

υ = c 1 − (1/1.010)2 = 0.140c = 4.20 × 107 m/s. (b) p > mυ ; the relativistic value is greater than the nonrelativistic value.

27.24. Set Up: The force can be determined by using Eq. (27.17): F = γ 3ma. Solve: (a) Indistinguishable from F = ma = 0.145 N. (b) γ 3ma = 1.75 N. (c) γ 3ma = 51.7 N.

27.25. Set Up: The nonrelativistic momentum is pnr = mυ and the relativistic momentum is pr =

mυ

1 − υ 2 /c 2 Solve: (a) and (b) The graphs of pnr and pr are shown in Figure (a) and (b) below. In both cases p = 0 when υ = 0. The nonrelativistic graph extends beyond υ = c.

Reflect: The graph of pnr is a straight line with slope m. As υ → c, pnr → mc. pr is the same as pnr at small v

but becomes larger than pnr as υ increases. pr increases without limit as υ → c.

27.26. Set Up: When the force and velocity are parallel, as in part (b), F =

ma

(1 − υ 2 /c 2 )3/ 2

. In part (a), υ c and

F = ma. Solve: (a) a = (b) γ =

F 5.00 × 10−15 N = = 5.49 × 1015 m/s 2 m 9.11 × 10−31 kg

1 2

2 1/2

(1 − υ /c )

=

1 8

(1 − [2.50 × 10 /3.00 × 108 ]2 )1/ 2

a=

F mγ

3

=

= 1.81.

5.49 × 1015 m/s 2 (1.81)3

= 9.26 × 1014 m/s 2 .

⎛ ⎞ 1 − 1⎟ mc 2 . For an electron, ⎜ ⎟ 2 2 ⎝ 1 − υ /c ⎠

*27.27. Set Up: The relativistic kinetic energy is K = ⎜

mc 2 = (9.11 × 10−31 kg)(3.00 × 108 m/s)2 = 8.20 × 10−14 J. The nonrelativistic expression for the kinetic energy is K nr = 12 mυ 2 .


Relativity

27-7

Solve: (a) K nr = 12 (9.11 × 10−31 kg)(5.00 × 107 m/s) 2 = 1.14 × 10−15 J. 1 2

1 − υ /c

2

=

1 1 − (5.00 × 107 /3.00 × 108 ) 2

and K = (1.014 − 1.00)(8.20 × 10−14 J) = 1.15 × 10−15 J.

= 1.014

K = 1.01. K nr

(b) K nr = 12 (9.11 × 10−31 kg)(2.60 × 108 m/s) 2 = 3.08 × 10−14 J.

1 2

1 − υ /c

2

=

1 1 − (2.60 × 108 /3.00 × 108 ) 2

and K = (2.00 − 1.00)(8.20 × 10−14 J) = 8.20 × 10−14 J.

= 2.00

K = 2.66. K nr

Reflect: The relativistic value of the kinetic energy, K, approaches the nonrelativistic value, K nr , for small speeds.

For large speeds, K > K nr and the ratio K/K nr increases as the speed increases. ⎛ ⎞ 1 − 1⎟ Erest . ⎜ ⎟ 2 2 ⎝ 1 − υ /c ⎠

27.28. Set Up: Erest = mc 2 . K = ⎜ 1

Solve: (a) K = Erest says (b) K = 5Erest says

1 − υ 2 /c 2

1 2

1 − υ /c

2

= 2.00. υ = c 1 − (1/2) 2 = 0.866c.

= 6.00. υ = c 1 − (1/6)2 = 0.986c.

27.29. Set Up: The total energy of the radiation equals the total rest mass energy of the particles. Solve: Erest = 2mc 2 = 2(1.67 × 10−27 kg)(3.00 × 108 m/s) 2 = 3.01 × 10−10 J = 1.88 × 109 eV = 1.88 × 103 MeV

27.30. Set Up: We have E = mc 2 , so m = E/c 2 . According to the appendix, the mass of the sun is 1.99 × 1030 kg. Solve: (a) m = E /c 2 = (3.8 × 1026 J)/(2.998 × 108 m/s)2 = 4.2 × 109 kg.

1 kg is equivalent to 2.2 lbs, so m = 4.6 × 106 tons (b) The current mass of the sun is 1.99 × 1030 kg, so it would take it (1.99 × 1030 kg)/(4.2 × 109 kg/s) = 4.7 × 1020 s = 1.5 × 1013 years to use up all its mass.

Reflect: The nuclear processes that occur inside our sun are only able to covert a fraction of its mass into energy; therefore, the estimate of part (b) is much greater than the projected lifespan of our sun.

*27.31. Set Up: Use Equations 27.22 and 27.24. Solve: (a) E = mc 2 + K , so E = 4.00mc 2 means K = 3.00mc 2 = 4.50 × 10−10 J (b) E 2 = (mc 2 )2 + ( pc) 2 ; E = 4.00mc 2 , so 15.0(mc 2 ) 2 = ( pc) 2 p = 15mc = 1.94 × 10−18 kg ⋅ m/s

(c) E = mc 2 / 1 − v 2 /c 2

E = 4.00mc 2 gives 1 − v 2 /c 2 = 1/16 and v = 15/16c = 0.968c Reflect: The speed is close to c since the kinetic energy is greater than the rest energy. Nonrelativistic expressions relating E, K, p, and v will be very inaccurate.

27.32. Set Up: If Δm is the mass change in the reaction, the energy released is E = (Δm)c 2 . Solve: The number of deuterium nuclei in 1.0 kg of deuterium is

1.0 kg (2.0126 u)(1.661 × 10−27 kg/u)

= 2.99 × 1026.


27-8

Chapter 27

The fusion forms 12 (2.99 × 1026) =1.495 × 1026 helium nuclei. The mass of the helium nuclei that are produced is (1.495 × 1026 )(4.0015 u)(1.661 × 10−27 kg/u) = 0.99365 kg. Δm = 1.0 kg − 0.99365 kg = 6.35 × 10−3 kg. E = (Δm)c2 = 5.7 × 1014 J.

(b) 1 kg of deuterium yields 5.7 × 1014 J. To produce 1.0 × 1019 J requires

1.0 × 1019 J 5.7 × 1014 J/kg

= 1.8 × 104 kg.

*27.33. Set Up: The energy equivalent of mass is E = mc 2 . ρ = 7.86 g/cm3 = 7.86 × 103 kg/m3 . For a cube, V = L3 . Solve: (a) m =

E c

2

=

1.0 × 1020 J (3.00 × 108 m/s) 2

= 1.11 × 103 kg

m 1.11 × 103 kg m = 0.141 m3 . L = V 1/3 = 0.521 m = 52.1 cm so V = = ρ 7.86 × 103 kg/m3 V Reflect: Particle/antiparticle annihilation has been observed in the laboratory, but only with small quantities of antimatter. (b) ρ =

27.34. Set Up: E 2 = (mc 2 )2 + ( pc)2 . Erest = mc 2 . K = E − Erest . Solve: (a) Erest = mc 2 = (6.64 × 10−27 kg)(3.00 × 108 m/s) 2 = 5.98 × 10−10 J.

E = (mc 2 ) 2 + ( pc) 2 = (5.98 × 10−10 J) 2 + ([2.10 × 10−18 kg][3.00 × 108 m/s])2 = 8.69 × 10−10 J (b) K = E − Erest = 8.69 × 10−10 J − 5.98 × 10−10 J = 2.71 × 10−10 J (c)

K 2.71 × 10−10 J = = 0.453 Erest 5.98 × 10−10 J

27.35. Set Up: Use Eq. (18.10) and conservation of energy to relate the potential difference to the kinetic energy gained by the electron. Use Eq. 27.21, to calculate the kinetic energy from the speed. Solve: (a) K = qΔV = eΔV ⎛ ⎞ 1 K = mc 2 ⎜ − 1⎟ = 4.025mc 2 = 3.295 × 10213 J = 2.06 MeV ⎜ ⎟ 2 2 ⎝ 1 − v /c ⎠ ΔV = K/e = 2.06 × 106 V (b) From part (a), K = 3.30 × 10213 J = 2.06 MeV Reflect: The speed is close to c and the kinetic energy is four times the rest mass.

27.36. Set Up: The nonrelativistic kinetic energy is K nr = 12 mυ 2 . The relativistic kinetic energy is Kr =

mc 2 2

2

− mc 2 .

1 − υ /c Solve: (a) and (b) The graphs of K nr and K r are shown in Figures (a) and (b) below. In both cases K is zero when

υ = 0. The nonrelativistic graph extends beyond υ = c. K r is the same as K nr at small υ but becomes larger than K nr as υ increases. As υ → c, K nr → 12 mc 2 . As υ → c, K r increases without limit.


Relativity

27-9

*27.37. Set Up: If an object has length l0 when it is at rest relative to the observer making the measurement, then when it is moving at speed u its length along the direction of motion is measured by the observer to be l, where

l = l0 1 − u 2 /c 2 . For the markings on an empire ship to appear circular, l = b when l0 = 1.40b. Solve: b = 1.40b 1 − u 2 /c 2 . u = c 1 − (1/1.40) 2 = 0.700c = 2.10 × 108 m/s. Reflect: The length in the direction of motion is contracted but lengths perpendicular to this direction are unaffected by the motion.

27.38. Set Up: The proper time Δt0 is the time that elapses in the frame of the space probe. Δt is the time that elapses in the frame of the earth. The distance traveled is 42.2 light years, as measured in the earth frame. c ⎞ ⎛ Solve: (a) Light travels 42.2 light years in 42.2 yr, so Δt = ⎜ (42.2 yr) = 42.6 yr. ⎝ 0.9910c ⎟⎠ Δt0 = Δt 1 − u 2 /c 2 = (42.6 yr) 1 − (0.9910)2 = 5.7 yr. She measures her biological age to be 19 yr + 5.7 yr = 24.7 yr. (b) Her age measured by someone on earth is 19 yr + 42.6 yr = 61.6 yr.

*27.39. Set Up: We will use the equation for time dilation to determine the relative speed of the two frames: Δt =

Δt0

1 − u 2 /c 2

. Since S is the rest frame of the event we know that Δt0 = 1.80 s, while Δt ′ = 2.35 s as seen in S ′. 2

2

u ⎛ Δt ⎞ ⎛ 1.80 s ⎞ = 1− ⎜ 0 ⎟ = 1− ⎜ = 0.643. So ⎝ Δt ′ ⎠ ⎝ 2.35 s ⎟⎠ c u = 0.643c. For convenience, imagine that both events occur at the origin of the S frame. As seen in the S ′ frame the

Solve: Solving the time dilation equation for u we obtain:

origin moves a distance x′ = uΔt ′ = (0.643c)(2.35 s) = 4.53 × 108 m during the time between the events—and this is the separation of the events as seen in the S ′ frame. Reflect: It is also possible to solve this problem using the Lorentz transformation equations.

27.40. Set Up: The 2.2 μs lifetime is Δt0 and the observer on earth measures Δt. The atmosphere is moving relative to the muon so in its frame the height of the atmosphere is l and l0 is 10 km. Solve: (a) The greatest speed the muon can have is c, so the greatest distance it can travel in 2.2 × 1026 s is d = υ t = (3.00 × 108 m/s)(2.2 × 10−6 s) = 660 m = 0.66 km.

(b) Δt =

Δt0 2

1 − u /c

2

=

2.2 × 10−6 s 1 − (0.999)

2

= 4.9 × 10−5 s

d = υt = (0.999)(3.00 × 108 m/s)(4.9 × 10−5 s) = 15 km

In the frame of the earth the muon can travel 15 km in the atmosphere during its lifetime. (c) l = l0 1 − u 2 /c 2 = (10 km) 1 − (0.999) 2 = 0.45 km

In the frame of the muon the height of the atmosphere is less than the distance it moves during its lifetime.

27.41. Set Up: We know that the lifetime of the muon in its rest frame is Δt0 = 2.2 μs. If the muon moves at a speed of u relative to an observer, the observer will measure the muon’s lifetime to be Δt = time the observer will see the muon move a distance equal to x = uΔt =

u Δt 0 1 − u 2 /c 2

Δt0

1 − u 2 /c 2

. During this

.


27-10

Chapter 27

Solve: Solving for u we obtain x 2 (1 − u 2 /c 2 ) = u 2 (Δt0 ) 2 , which, after some rearrangement, gives u=

cx x2 + (cΔt0 )2

=

c(1.0 × 103 m) (1.0 × 103 m) 2 + (3.00 × 108 m/s)2 (2.2 × 10−6 s)2

= 0.83c.

Reflect: In the muon’s rest frame, the observer moves a distance equal to x′ = uΔt0 = (0.83c)(2.2 × 10−6 s) = 550 m = 0.55 km during the lifetime of the muon.

27.42. Set Up: The proper length of a side is l0 = a. The side along the direction of motion is shortened to l = l0 1 − υ 2 /c 2 . The sides in the two directions perpendicular to the motion are unaffected by the motion and still have a length a. Solve: V = a 2l = a3 1 − υ 2 /c 2

*27.43. Set Up: Let S be the lab frame and S ′ be the frame of the proton that is moving in the + x direction, so u = +c/2. The reference frames and moving particles are shown in the figure below. The other proton moves in the

− x direction in the lab frame, so υ = −c/2. A proton has rest mass mp = 1.67 × 10−27 kg and rest energy mp c 2 = 938 MeV.

Solve: (a) υ ′ =

υ −u −c/2 − c/2 4c = =− 2 2 5 1 − uυ /c 1 − (c/2)( −c/2)/c

The speed of each proton relative to the other is

4 c. 5

(b) In nonrelativistic mechanics the speeds just add and the speed of each relative to the other is c. mc 2 (c) K = − mc 2 2 2 1 − υ /c (i) Relative to the lab frame each proton has speed υ = c /2. The total kinetic energy of each proton is

K=

938 MeV 1−

( )

1 2 2

− (938 MeV) = 145 MeV.

(ii) In its rest frame one proton has zero speed and zero kinetic energy and the other has speed

4 c. 5

In this frame the

kinetic energy of the moving proton is K=

938 MeV 1−

( ) 4 5

2

− (938 MeV) = 625 MeV


Relativity

( 12 m ) (c/2)2 = mc8

2

27-11

938 MeV = 117 MeV 8 (ii) One proton has speed υ = 0 and the other has speed c. The kinetic energy of the moving proton is

(d) (i) Each proton has speed υ = c/2 and kinetic energy K = 12 mυ 2 =

K = 12 mc 2 =

=

938 MeV = 469 MeV 2

Reflect: The relativistic expression for K gives a larger value than the nonrelativistic expression. The kinetic energy of the system is different in different frames.

27.44. Set Up: Since the final speed is close to the speed of light, there will be a considerable difference between 1 1 the relativistic and nonrelativistic results. The nonrelativistic work-energy theorem is F Δx = mv 2 − mv02 , and the 2 2 relativistic formula for a constant force is F Δx = (γ − 1)mc 2 . Solve: (a) Using the classical work-energy theorem and solving for Δx, we obtain Δx =

m(v 2 − v02 ) (0.100 × 1029 kg)[(0.900)(3.00 × 108 m/s)]2 = = 3.65 m. 2F 2(1.00 × 106 N)

(b) Using the relativistic work-energy theorem for a constant force, we obtain (γ − 1)mc 2 Δx = . F 1 For the given speed, γ = = 2.29, thus 1 − 0.9002

Δx =

(2.29 − 1)(0.100 × 10−9 kg)(3.00 × 108 m/s) 2 (1.00 × 106 N)

= 11.6 m.

Reflect: (c) The distance obtained from the relativistic treatment is greater. As we have seen, more energy is required to accelerate an object to speeds close to c, so that force must act over a greater distance.

*27.45. Set Up: The energy that enters the ice is E = Q = mLf . The mass increase is the mass equivalent of this amount of energy, m = Solve: m =

E c2

=

mLf c2

E c2

=

.

(4.00 kg)(3.34 × 105 J/kg) (3.00 × 108 m/s)2

= 1.48 × 10−11 kg

27.46. Set Up: Let S be the lab frame and let S′ be the frame of the nucleus. Let the 1x direction be the direction the nucleus is moving. u = 0.7500c. Solve: (a) υ ′ = +0.9995c.

υ′ + u 0.9995c + 0.7500c = = 0.999929c 2 1 + (0.7500)(0.9995) 1 + uυ ′/c −0.9995c + 0.7500c = −0.9965c (b) υ ′ = −0.9995c. υ = 1 + (0.7500)( −0.9995) (c) emitted in same direction: υ=

⎛ ⎞ ⎛ ⎞ 1 1 − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 42.4 MeV (i) K = ⎜ ⎜ ⎟ 2 2 ⎜ 1 − (0.999929) 2 ⎟ ⎝ 1 − υ /c ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ 1 1 (ii) K ′ = ⎜ − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 15.7 MeV ⎜ ⎟ 2 2 ⎜ 1 − (0.9995)2 ⎟ ⎝ 1 − υ /c ⎠ ⎝ ⎠


27-12

Chapter 27

(d) emitted in opposite direction: ⎛ ⎞ ⎛ ⎞ 1 1 − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 5.60 MeV (i) K = ⎜ ⎜ ⎟ 2 2 ⎜ 1 − (0.9965)2 ⎟ ⎝ 1 − υ /c ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ 1 1 (ii) K ′ = ⎜ − 1⎟ mc 2 = (0.511 MeV) ⎜ − 1⎟ = 15.7 MeV ⎜ ⎟ 2 2 ⎜ 1 − (0.9995)2 ⎟ ⎝ 1 − υ /c ⎠ ⎝ ⎠

27.47. Set Up: Equation 27.24 says E = (mc 2 ) 2 + ( pc) 2 . Erest = mc 2 and the classical (nonrelativistic) kinetic energy is K nr = 12 mυ 2 = Solve: E = mc 2 1 +

p2 . 2m

2 ⎞ 2 2 p2 2⎛ 2 1 p c 2 1 ( pc ) 1 ≈ mc + = mc + = mc + = Erest + K nr , as was to be shown. ⎜ ⎟ 2 ⎜ 2 (mc 2 ) 2 ⎟ 2m (mc 2 ) 2 mc 2 ⎝ ⎠

( pc) 2

27.48. Set Up: 500 light years = 4.73 × 1018 m. The proper distance l0 to the star is 500 light years. The energy needed is the kinetic energy of the rocket at its final speed. d 4.73 × 1018 m = 3.2 × 1010 s = 1000 yr Solve: (a) u = 0.50c. Δt = = u (0.50)(3.00 × 108 m/s) The proper time is measured by the astronauts. Δt0 = Δt 1 − u 2 /c 2 = 866 yr

⎛ ⎞ 1 − mc 2 = (1000 kg)(3.00 × 108 m/s) 2 ⎜ − 1⎟ = 1.4 × 1019 J ⎜ 1 − (0.500) 2 ⎟ 1 − υ 2 /c 2 ⎝ ⎠ This is 14% of the U.S. yearly use of energy. 4.73 × 1018 m d (b) u = 0.99c. Δt = = = 1.6 × 1010 s = 505 yr, Δt0 = 71 yr u (0.99)(3.00 × 108 m/s) K=

mc 2

⎛ ⎞ 1 K = (9.00 × 1019 J) ⎜ − 1⎟ = 5.5 × 1020 J ⎜ 1 − (0.99) 2 ⎟ ⎝ ⎠ This is 5.5 times the U.S. yearly use. 4.73 × 1018 m d (c) u = 0.9999c. Δt = = = 1.58 × 1010 s = 501 yr, Δt0 = 7.1 yr u (0.9999)(3.00 × 108 m/s) ⎛ ⎞ 1 K = (9.00 × 1019 J) ⎜ − 1⎟ = 6.3 × 1021 J ⎜ 1 − (0.9999) 2 ⎟ ⎝ ⎠ This is 63 times the U.S. yearly use. The energy cost of accelerating a rocket to these speeds is immense. E ⎛ 1 ⎞ (8.00 kg). Pav = . The change in gravitational 4⎟ t ⎝ 10 ⎠

*27.49. Set Up: The energy released is E = ( Δm)c 2 . Δm = ⎜ potential energy is mg Δy.

⎛ 1 ⎞ Solve: (a) E = ( Δm)c 2 = ⎜ 4 ⎟ (8.00 kg)(3.00 × 108 m/s) 2 = 7.20 × 1013 J ⎝ 10 ⎠ (b) Pav =

E 7.20 × 1013 J = = 1.80 × 1019 W t 4.00 × 1026 s

(c) E = ΔU = mg Δy. m =

E 7.20 × 1013 J = = 7.35 × 109 kg g Δy (9.80 m/s 2 )(1.00 × 103 m)


Relativity

27-13

⎛ ⎞ 1 − 1⎟ mc2. For an electron mc 2 = (9.11 × 10−31 kg)(3.00 × 108 m/s) 2 = 8.20 × 10−14 J. ⎜ ⎟ 2 2 ⎝ 1 −υ /c ⎠

27.50. Set Up: K = ⎜

K = 7.50 × 105 eV = 1.20 × 10−13 J. The classical expression for kinetic energy is K = 12 mυ 2 .

Solve: (a)

K mc 2

1

+1 =

1 − υ 2 /c 2

.

1 1 − υ 2 /c 2

=

1.20 × 10−13 J 8.20 × 10−14 J

+ 1 = 2.46.

υ = c 1 − (1/2.46) 2 = 0.914c = 2.74 × 108 m/s. (b) K = 12 mυ 2 gives υ =

2K 2(1.20 × 10−13 J) = = 5.13 × 108 m/s. m 9.11 × 10−31 kg

Reflect: At a given speed the relativistic value of the kinetic energy is larger than the nonrelativistic value. Therefore, for a given kinetic energy the relativistic expression for kinetic energy gives a smaller speed than the nonrelativistic expression.

*27.51. Set Up: In the rest frame of the spaceship the trip takes Δt0 = 3.35 years. As seen from the earth the trip takes Δt =

Δt0

. The distance to the star is 7.11 ly and the speed of light is c = 1 ly/y. Let u be the speed of 1 − u 2 /c 2 the spaceship (in ly/y ) as seen from the earth. Solve: (a) The time for the trip as seen from the earth will be Δt =

7.11 ly 3.35 y = . Solving for u we obtain u 1 − u 2 /c 2

2

⎛ 7.11 ly ⎞ ⎛ 7.11 ly ⎞ 2 2 2 ⎟⋅ ⎜ ⎟ (1 − u /c ) = u , which reduces to u = ⎜ ⎝ 3.35 y ⎠ ⎝ 3.35 y ⎠

seen from the earth, the trip takes

1 ⎛ ⎞ 7.11 ly 1+ ⎜ ⎟ (3 . 35 y)(1 ly/y) ⎝ ⎠

2

= 0.905 ly/y = 0.905c. Thus, as

7.11 ly 7.11 ly = = 7.86 years. u 0.905 ly/y

(b) According to the engers the distance is given by x′ = u Δt0 = (0.905 ly/y)(3.35 y) = 3.03 ly. Reflect: The distance to the star as seen by the engers could also be calculated by using the length contraction

formula: l = l0 1 − u 2 /c 2 . c

⎛

⎞

1

− 1⎟ mc 2 . For an electron, 27.52. Set Up: n = . K = ⎜ ⎜ ⎟ 2 2 υ 1 / − c υ ⎝ ⎠ mc 2 = (9.11 × 10−31 kg)(3.00 × 108 m/s)2 = 8.20 × 10−14 J = 0.512 MeV. ⎛ ⎞ 1 c = 0.658c. K = (0.512 MeV) ⎜ − 1⎟ = 0.168 MeV ⎜ 1 − (0.658)2 ⎟ n ⎝ ⎠ Reflect: A particle’s speed cannot exceed the speed of light in vacuum, but in a material the speed of a particle can exceed the speed of light in that material. Solve: υ =

27.53. Set Up: We will use the equation p = Solve: Solving for m we obtain m =

p

υ

2

mυ 1 − υ 2 /c 2 2

⋅ 1 − υ /c =

(2.52 × 10−19 kg ⋅ m/s) 1.35 × 108 m/s

2

⎛ 1.35 × 108 m/s ⎞ ⋅ 1− ⎜ ⎟ = ⎝ 3.00 × 108 m/s ⎠

1.67 × 10−27 kg. The only known positively charged particle with this mass is a proton. Reflection: We cannot use p = mυ to find the rest mass of the particle due to its high speed.


27-14

Chapter 27

Solutions to age Problems 27.54. Set Up: The speed of light in our universe is c = 3.00 × 108 m/s. Solve: The speed of light in the alternate universe is 106 times smaller than the speed of light in our universe. Thus, calt = c/106 = 300 m/s, which is roughly the speed of sound in our universe. The correct answer is D. 2 *27.55. Set Up: We will use the formula for length contraction: l = l0 1 − u 2 /calt . 2

⎛ 180 m/s ⎞ 2 = (60 m) 1 − ⎜ Solve: l = l0 1 − u 2 /calt ⎟ = 48 m. The correct answer is C. ⎝ 300 m/s ⎠

27.56. Set Up: The relativistic mass is given by: m = Solve: m =

m0 1− u

2

2 /calt

=

20,000 kg ⎛ 180 m/s ⎞ 1− ⎜ ⎝ 300 m/s ⎟⎠

2

m0 2 1 − u 2 /calt

.

= 25,000 kg. The correct answer is D.

*27.57. Set Up: The rest energy in our universe is given by E = mc 2 and the rest energy in the alternate universe is 2 . given by Ealt = mcalt

Solve: Comparing the rest energy of the electron in our universe to the alternate universe we have:

2 Ealt mcalt = , E mc 2

2

⎛c ⎞ which gives Ealt = E ⋅ ⎜ alt ⎟ = (8.2 × 10−14 J)(10−6 ) 2 = 8.2 × 10−26 J. The correct answer is B. ⎝ c ⎠


28

PHOTONS , ELECTRONS, AND ATOMS

Answers to Multiple-Choice Problems 1. A, C 2. B 15. D 16. D

3. B

4. A

5. C

6. C

7. D

8. D

9. A

10. B

11. D

12. C

13. B

14. B

Solutions to Problems *28.1. Set Up: c = f λ relates frequency and wavelength and E = hf relates energy and frequency for a photon. c = 3.00 × 108 m/s. 1 eV = 1.60 × 10−16 J. Solve: (a) f =

c

λ

=

3.00 × 108 m/s 505 × 10−9 m

= 5.94 × 1014 Hz

(b) E = hf = (6.626 × 10−34 J ⋅ s)(5.94 × 1014 Hz) = 3.94 × 10−19 J = 2.46 eV (c) K = 12 mυ 2 so υ =

2K 2(3.94 × 10−19 J) = = 9.1 mm/s m 9.5 × 10−15 kg

28.2. Set Up: For a photon, E = hf and c = f λ . h = 4.136 × 10−15 eV ⋅ s. Solve: (a)

f = (b) λ =

2.45 × 106 eV E = = 5.92 × 1020 Hz h 4.136 × 10−15 eV ⋅ s

c 3.00 × 108 m/s = = 5.07 × 10−13 m f 5.92 × 1020 Hz

(c) This wavelength is comparable to a typical nuclear diameter.

*28.3. Set Up: Pav =

energy hc . 1 eV = 1.60 × 10−19 J. For a photon, E = hf = . h = 6.63 × 10−34 J ⋅ s. t λ

Solve: (a) energy = Pav t = (0.600 W)(20.0 × 10−3 s) = 1.20 × 10−2 J = 7.5 × 1016 eV (b) E =

hc

λ

=

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) 652 × 10−9 m

= 3.05 × 10−19 J = 1.91 eV


28-1

28-2

Chapter 28

(c) The number of photons is the total energy in a pulse divided by the energy of one photon: 1.20 × 10−2 J = 3.93 × 1016 photons. 3.05 × 10−19 J/photon Reflect: The number of photons in each pulse is very large.

28.4. Set Up: In 1.00 s the energy output is 50.0 kJ. The energy of a photon is E = hf . h = 6.63 × 10−34 J ⋅ s. Solve: (a) E = hf = (6.63 × 10234 J ⋅ s)(92.0 × 106 Hz) = 6.10 × 10−26 J = 3.81 × 10−7 eV (b) The number of photons emitted per second is

50.0 × 103 J 226

6.10 × 10

J/photon

= 8.20 × 1029 photons.

28.5. Set Up: In 1.00 s the energy output is 12.0 J. The energy of a photon is E = hf = Solve: E =

hc

λ

=

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) 248 × 10

12.0 J 8.02 × 10−19 J/photon

−9

m

hc

λ

. h = 6.63 × 10−34 J ⋅ s.

= 8.02 × 10−19 J. The number of photons emitted per second is

= 1.50 × 1019 photons.

Reflect: A large number of photons are emitted each second.

28.6. Set Up: A photon has zero rest mass, so Eq. (27.25) relates its energy to its momentum. Equation 28.6 then relates its momentum and wavelength. Solve: (a) E = pc = (8.24 ×10−28 kg ⋅ m/s)(2.998 ×108 m/s) = 2.47 ×10−19 J = (2.47 × 10219 J)(1 eV/1.602 × 10−19 J) = 1.54 eV

h 6.626 × 10−34 J ⋅ s = = 8.04 × 10−7 m = 804 nm λ p 8.24 × 10−28 kg ⋅ m/s Reflect: This wavelength is longer than visible wavelengths; it is in the infrared region of the electromagnetic spectrum. To check our result we could that the same E is given by Eq. (28.5), using the λ we have calculated. (b) p =

h

so λ =

2 1 mυ max 2

*28.7. Set Up:

= hf − φ

Solve: f → f 0 , the threshold frequency, when

28.8. Set Up: Solve: hf =

hc

λ

2 1 mυ max 2

=

2 1 mυ max 2

→ 0. f 0 = φ /h.

= hf − φ . For nickel, φ = 5.1 eV. h = 4.136 × 10−15 eV ⋅ s. c = f λ .

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 235 × 10−9 m 2 1 mυ max 2

= 5.28 eV.

= 5.28 eV − 5.1 eV = 0.18 eV = 2.9 × 10−20 J.

υmax =

2(2.9 × 10−20 J) 9.11 × 10−31 kg

28.9. Set Up: λth = 272 nm. c = f λ . 12 mυmax 2 = hf − φ .

= 2.5 × 105 m/s.

At

the

threshold

frequency,

f th ,

υmax → 0.

h = 4.136 × 10−15 eV ⋅ s. Solve: (a) f th =

c

λth

=

3.00 × 108 m/s 272 × 10−9 m

= 1.10 × 1015 Hz.


Photons, Electrons, and Atoms

28-3

(b) φ = hf th = (4.136 × 10−15 eV ⋅ s)(1.10 × 1015 Hz) = 4.55 eV. (c)

2 1 mυ max 2

= hf − φ = (4.136 × 10215 eV ⋅ s)(1.45 × 1015 Hz) − 4.55 eV = 6.00 eV − 4.55 eV = 1.45 eV

Reflect: The threshold wavelength depends on the work function for the surface. hc

28.10. Set Up: φ = hf th = hc

Solve: φ =

λth

=

λth

. The minimum φ corresponds to the minimum λ .

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)

= 1.77 eV

700 × 10−9 m 2 1 mυ max 2

*28.11. Set Up:

= hf − φ

Solve: Use the data for 400.0 nm to calculate φ :

φ=

hc

λ

− 12 mυmax 2 =

(4.136 × 10215 eV ⋅ s)(3.00 × 108 m/s) 400.0 × 1029 m

− 1.10 eV = 3.10 eV − 1.10 eV = 2.00 eV.

Then for 300.0 nm, hc

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)

2 1 mυ max 2

= hf − φ =

2 1 mυ max 2

= 4.14 eV − 2.00 eV = 2.14 eV.

λ

−φ =

300.0 × 10−9 m

− 2.00 eV.

Reflect: When the wavelength decreases the energy of the photons increases and the photoelectrons have a larger minimum kinetic energy.

28.12. Set Up:

2 1 mυ max 2

= eV0 , where V0 is the stopping potential. eV0 = 0.181 eV. eV0 =

hc

λ

− φ . c = f λ.

hc

λth

= φ,

where λth is the threshold wavelength. Solve: (a) eV0 =

hc

λth

=

hc

λ

− eV0 =

Then λth = (b) φ =

hc

λth

hc

λ

−

hc

λth

.

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 254 × 10−9 m

− 0.181 eV = 4.89 eV − 0.181 eV = 4.71 eV.

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 263 nm. 4.71 eV 4.71 eV = 4.71 eV

28.13. Set Up: eV0 = 12 mυmax 2 , where V0 is the stopping potential. The stopping potential in volts equals eV0 in electron volts.

2 1 mυ max 2

= hf − φ .

Solve: (a) eV0 = 12 mυmax 2 so

eV0 = hf − φ =

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 250 × 10−9 m

− 2.3 eV = 4.96 eV − 2.3 eV = 2.7 eV.

The stopping potential is 2.7 volts. (b) 12 mυmax 2 = 2.7 eV (c) υmax =

2(2.7 eV)(1.60 × 10−19 J/eV) 9.11 × 10−31 kg

= 9.7 × 105 m/s


28-4

Chapter 28

28.14. Set Up: We will apply the equation eV0 = hf − φ . For electrons to be ejected we must have hf − φ $ 0, which gives a threshold frequency of f 0 = φ /h. We are told that there are no ejected electrons for wavelengths longer than λ0 = 289 nm, which implies a threshold frequency of f 0 =

c

λ0

.

From the text we know that h = 4.136 × 10−15 eV ⋅ s. Solve: (a) The work function of the metal is φ = hf0 =

hc

λ0

=

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 289 × 10−9 m

= 4.29 eV.

(b) We can calculate the energy of the ejected electron using (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) eV0 = hf − φ = − 4.29 eV = 1.22 eV. Thus we know that 1.22 volts will be 225 × 10−9 m required to stop the ejected electrons under these circumstances. Reflect: The threshold frequency is the lowest frequency that will eject electrons, which corresponds to the longest wavelength that will eject electrons.

*28.15. Set Up: We will apply the equation K max = hf − φ . For electrons to be ejected we must have hf − φ ≥ 0, which gives a threshold frequency of f 0 = φ /h. From the textbook we find the work function of silicon to be φ = 4.8 eV and h = 4.14 × 10−15 eV ⋅ s. Solve: (a) The threshold frequency is given by f 0 = φ /h, which corresponds to a maximum wavelength of

λ0 =

c ch (3.00 × 108 m/s)(4.14 × 10−15 eV ⋅ s) = = = 259 × 10−9 m = 259 nm. 4.8 eV f0 φ

(b) The maximum kinetic energy occurs at the shortest wavelength, which is 145 nm. The maximum kinetic energy of the electrons just after they are ejected is K max = hf − φ = (4.14 × 10−15 eV ⋅ s)(3.00 × 108 m/s)/(145 × 10−9 m) −

4.8 eV = 3.77 eV. As the electrons move to the anode they experience a reversing potential of 3.50 volts, which implies that they lose 3.50 eV of kinetic energy. Thus, the electrons reach the anode with a maximum kinetic energy of 3.77 eV − 3.50 eV = 0.27 eV. Reflect: It would take a potential of 3.77 volts to stop the electrons ejected from light with a wavelength of 145 nm.

28.16. Set Up: We will apply the formula En =

−13.6 eV n2

to find the energy of a transition and E = hf = h

c

λ

to

find the wavelength of the emitted photon. Solve: (a) For the n = 4 to n = ∞ (ionization) transition we require ΔE = E∞ − E4 = 0 −

−13.6 eV 42

= 0.850 eV.

(b) We can find the energy of the emitted photon for the n = 4 to n = 2 transition: Ephoton = −ΔE = E4 − E2 =

E c ch 1⎞ ⎛ 1 and its wavelength is λ = = = (13.6 eV) ⎜ 2 − 2 ⎟ = 2.55 eV. The frequency of this photon is f = ⎝2 h f E 4 ⎠ (3.00 × 108 m/s)(4.14 × 10−15 eV ⋅ s) = 487 × 10−9 m = 489 nm. 2.55 eV 1 1⎞ ⎛ 1 Reflect: We could have used the Balmer formula, = R ⎜ 2 − 2 ⎟ , to find the wavelength of a transition into the ⎝2 λ n ⎠

n = 2 level.

*28.17. Set Up: Balmer’s formula is

1

λ

1 ⎞ ⎛ 1 = R ⎜ 2 − 2 ⎟ , where R = 1.097 × 107 m −1. The H γ line corresponds to n ⎠ ⎝2

n = 5. c = f λ. E = hf .



1

Solve: (a) (b) f =

c

λ

λ =

28-5

100 100 ⎛ 1 1 ⎞ 21 = R⎜ − ⎟ = R. λ = = = 434 nm 4 25 100 R 21 21(1 097 . × 107 m −1) ⎝ ⎠

3.00 × 108 m/s 434 × 10−9 m

= 6.91 × 1014 Hz

(c) E = hf = (4.136 × 10−15 eV ⋅ s)(6.91 × 1014 Hz) = 2.86 eV

28.18. Set Up: For the Lyman series the final state is n = 1 and the wavelengths are given by n = 2, 3, … . For the Paschen series the final state is n = 3 and the wavelengths are given by

⎛1 1 ⎞ = R⎜ 2 − 2 ⎟, λ ⎝1 n ⎠ 1

1

λ

1 ⎞ ⎛ 1 = R⎜ 2 − 2 ⎟, n ⎠ ⎝3

n = 4, 5, … . R = 1.097 × 107 m −1. The longest wavelength is for the smallest n and the shortest wavelength is for n → ∞.

Solve: Lyman Longest:

Shortest:

1

λ

1

λ

4 ⎛ 1 1 ⎞ 3R = R⎜ 2 − 2 ⎟ = . λ= = 121.5 nm. 4 3(1.097 × 107 m −1 ) ⎝1 2 ⎠

1 1 ⎞ ⎛1 = R ⎜ 2 − 2 ⎟ = R. λ = = 91.16 nm. 1.097 × 107 m21 ⎝1 ∞ ⎠

Paschen Longest: Shortest:

1

λ

1

λ

1 ⎞ 7R 144 ⎛ 1 = R⎜ 2 − 2 ⎟ = . λ= = 1875 nm. 144 4 ⎠ 7(1.097 × 107 m −1 ) ⎝3

1 ⎞ R 9 ⎛ 1 = R⎜ 2 − 2 ⎟ = . λ = = 820.4 nm. ∞ ⎠ 9 1.097 × 107 m21 ⎝3

⎛ 1 1 ⎞ = R ⎜ 2 − 2 ⎟ , where ⎜ λ n ⎟⎠ ⎝ n0 n0 = 2 for the Balmer series, n0 = 1 for the Lyman series, and n0 = 4 for the Brackett series. The visible spectrum

28.19. Set Up: The wavelengths of the light for each line in one of the series is given by

1

extends from about 400 nm to 700 nm. Solve: (a) Balmer series The longest wavelength is for n = 3: 1 1 ⎞ 5R 36 ⎛ 1 = R⎜ 2 − 2 ⎟ = so λ = = 656 nm 36 λ R 5 2 3 ⎝ ⎠ The shortest wavelength is for n → ∞ : 1 ⎞ R 4 ⎛ 1 so λ = = 365 nm = R⎜ 2 − 2 ⎟ = λ R ∞ ⎠ 4 ⎝2 Lyman series The longest wavelength is for n = 2: 1 4 ⎛ 1 1 ⎞ 3R = R⎜ 2 − 2 ⎟ = so λ = = 122 nm λ 3R ⎝1 2 ⎠ 4 The shortest wavelength is for n → ∞ : 1

1 ⎞ 1 ⎛1 = R ⎜ 2 − 2 ⎟ = R so λ = = 91.2 nm λ R ⎝1 ∞ ⎠ Brackett series The longest wavelength is for n = 5: 1 1 ⎞ 9R 400 ⎛ 1 = R⎜ 2 − 2 ⎟ = so λ = = 4051 nm λ 9R 5 ⎠ 400 ⎝4 The shortest wavelength is for n → ∞ : 1

1 ⎞ R 16 ⎛ 1 so λ = = 1459 nm = R⎜ 2 − 2 ⎟ = λ R ∞ ⎠ 16 ⎝4 1


28-6

Chapter 28

(b) The Balmer series is mostly in the visible. The Lyman series is in the ultraviolet and the Brackett series is in the infrared. Reflect: From Problem 28.18, the wavelength range for the Paschen series (n0 = 3) is 820 nm to 1875 nm. The

Lyman, Balmer, and Paschen spectral series do not overlap in wavelength; the longest wavelength for one is shorter than the shortest wavelength of the next. But, the Paschen series partially overlaps the Brackett series.

28.20. Set Up: The ionization threshold is at E = 0. The energy of an absorbed photon equals the energy gained by the atom and the energy of an emitted photon equals the energy lost by the atom. Solve: (a) ΔE = 0 − (−20 eV) = 20 eV (b) When the atom in the n = 1 level absorbs an 18 eV photon, the final level of the atom is n = 4. The possible transitions from n = 4 and corresponding photon energies are n = 4 → n = 3, 3 eV; n = 4 → n = 2, 8 eV; n = 4 → n = 1, 18 eV. Once the atom has gone to the n = 3 level, the following transitions can occur: n = 3 → n = 2, 5 eV; n = 3 → n = 1, 15 eV. Once the atom has gone to the n = 2 level, the following transition can occur: n = 2 → n = 1, 10 eV. The possible energies of emitted photons are: 3 eV, 5 eV, 8 eV, 10 eV, 15 eV, and 18 eV. (c) There is no energy level 8 eV higher in energy than the ground state, so the photon cannot be absorbed. (d) The photon energies for n = 3 → n = 2 and for n = 3 → n = 1 are 5 eV and 15 eV. The photon energy for n = 4 → n = 3 is 3 eV. The work function must have a value between 3 eV and 5 eV.

*28.21. Set Up: The wavelength of the photon is related to the transition energy Ei − Ef of the atom by Ei − Ef =

hc

λ

where hc = 1.240 × 10−6 eV ⋅ m.

Solve: (a) The minimum energy to ionize an atom is when the upper state in the transition has E = 0, so

E1 = −17.50 eV. For n = 5 → n = 1, λ = 73.86 nm and E5 − E1 =

1.240 × 10−6 eV ⋅ m 73.86 × 10−9 m

= 16.79 eV. E5 = −17.50 eV +

16.79 eV = −0.71 eV. For n = 4 → n = 1, λ = 75.63 nm and E4 = 21.10 eV. For n = 3 → n = 1, λ = 79.76 nm and

E3 = −1.95 eV. For n = 2 → n = 1, λ = 94.54 nm and E2 = −4.38 eV. (b) Ei − Ef = E4 − E2 = −1.10 eV − (−4.38 eV) = 3.28 eV and

λ=

hc 1.240 × 10−6 eV ⋅ m = = 378 nm 3.28 eV Ei − Ef

Reflect: The n = 4 → n = 2 transition energy is smaller than the n = 4 → n = 1 transition energy so the wavelength is longer. In fact, this wavelength is longer than for any transition that ends in the n = 1 state.

28.22. Set Up: K n =

1 2

me4 2 2

⑀0 8n h

state has n = 1. Solve: (a) K1 = 13.6 eV

=

hcR n

2

=

13.6 eV

n

2

. Un = −

1 2

me4

⑀ 0 4n 2 h 2

=−

2(13.6 eV)

n2

13.6 eV . En = 2 . The ground n2

(b) U1 = −27.2 eV (c) E1 = −13.6 eV (d) The minimum energy of the atom when the electron is no longer bound is E = 0. ΔE = 0 − E1 = 13.6 eV. (e) ΔE =

hc

λ

.

λ=

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 91.2 nm. ΔE 13.6 eV

This photon is in the ultraviolet.



28-7

13.6 eV

*28.23. Set Up: Section 28.3 calculates that r1 = 0.5293 × 10−10 m. rn = n 2r1 and En = 2 Solve: (a) v1 = vn =

e2 2⑀0 h

=

(1.602 × 10−19 C)2 2(8.854 × 10

−12

C2 /N ⋅ m 2 )(6.626 × 10−34 J ⋅ s)

n2

= 2.19 × 106 m/s

v1 so v2 = 1.09 × 106 m/s and v3 = 7.29 × 105 m/s n

(b) rn = n 2r1 , so r1 = 0.5293 × 10−10 m, r2 = 2.117 × 10−10 m and r3 = 4.764 × 10−10 m (c) E1 = −13.6 eV, E2 = −3.40 eV, E3 = −1.51 eV Reflect: As n increases, r increases, υ decreases and E becomes less negative.

28.24. Set Up: If an atom gains energy in a transition, conservation of energy requires that a photon is absorbed. If the atom loses energy in a transition, a photon is emitted. The transition energy of the atom equals the energy of the photon that is absorbed or emitted. 13.6 eV Solve: (a) En = 2 so E5 = −0.544 eV and E2 = −3.40 eV. In the n = 5 to n = 2 transition the atom loses n2 energy and a photon is emitted. (b) Ei − Ef = E5 − E2 = 2.86 eV Ei − Ef =

hc

λ

so λ =

28.25. Set Up: For a hydrogen atom En = −

hc 1.240 × 10−6 eV ⋅ m = = 434 nm 2.86 eV E5 − E2

13.6 eV

. ΔE =

hc

, where ΔE is the magnitude of the energy change λ n for the atom and λ is the wavelength of the photon that is absorbed or emitted. ⎛ 1 1⎞ Solve: ΔE = E4 − E1 = −(13.6 eV) ⎜ 2 − 2 ⎟ = +12.75 eV. ⎝4 1 ⎠ 2

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 97.3 nm. ΔE 12.75 eV c f = = 3.08 × 1015 Hz.

λ=

λ

28.26. Set Up: We will apply E = hf = h

c

λ

to find the energy of the photon and En =

−13.6 eV

n2

to find the

ionization energy of hydrogen. Solve: To ionize hydrogen in its ground state we must provide enough energy to move its electron from the n = 1 −13.6 eV = 13.6 eV. The energy of level to the n = ∞ level. This requires a minimum energy of ΔE = E∞ − E1 = 0 − 12 the given incident photon is E = h

c

λ

=

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 59 × 10−9 m

= 21 eV. Thus, the energy of the ejected

electron must be 21 eV − 13.6 eV = 7.4 eV. Reflect: The ionization energy of hydrogen −15 8 hc (4.136 × 10 eV ⋅ s)(3.00 × 10 m/s) = 91 nm. λ= = 13.6 eV E

corresponds

to

a

photon

*28.27. Set Up: In the expression for hydrogen replace e2 by Ze2 . For hydrogen, En = − so for a one-electron ion with atomic number Z, En = −(13.6 eV)

Z2 n2

1 2

wavelength

me4 2 2

⑀ 0 8n h

=−

of

13.6 eV

n2

,

. For Be3+ , Z = 4.


28-8

Chapter 28

Solve: (a) E1 = −(13.6 eV)

42 12

= −218 eV. This is a factor of 16 times the hydrogen atom value.

(b) The ionization energy is E∞ − E1 = +218 eV. This is a factor of 16 times the hydrogen atom value. (c) 0 ΔE 0 =

⎛ 1 1⎞ . 0 ΔE 0 = E2 − E1 = −(13.6 eV)(42 ) ⎜ 2 − 2 ⎟ = 163.2 eV. λ ⎝2 1 ⎠

hc

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 7.60 nm. 0 ΔE 0 163.2 eV This is the hydrogen value divided by 16. n2h2 n2h2 r (d) For the hydrogen atom, rhyd = ⑀ 0 . For a one-electron ion with atomic number Z , = ⑀ = rhyd /Z . Z 0 π me2 π mZe2

λ=

The radius for Be3+ is

1 4

times the hydrogen atom value.

Reflect: The level energies are proportional to Z 2 , the wavelength for a specific transition is proportional to 1/Z 2 , and the radius of a particular orbit is proportional to 1/Z .

28.28. Set Up: E =

hc

λ

. From the figure, E5s = 20.66 eV, E3 p = 18.70 eV and E3s = 16.70 eV.

Solve: (a) ΔE = E5s − E3 p = 20.66 eV − 18.70 eV = 1.96 eV = 3.14 × 10−19 J.

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 633 nm. This agrees with the observed wavelength of 632.8 nm ΔE 1.96 eV for the laser light. (b) λ =

(c) ΔE = E3 p − E3s = 18.70 eV − 16.70 eV = 2.00 eV

λ=

hc (4.136 × 10215 eV ⋅ s)(3.00 × 108 m/s) = = 620 nm. ΔE 2.00 eV

*28.29. Set Up: The energy of each photon is E = hf = energy Etot Solve: P =

hc

. The power is the total energy per second and the total λ is the number of photons N times the energy E of each photon.

Etotal Nhf Nhc (4.50 × 1017 )(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = = = 4.63 × 10−3 W = 4.63 mW. t t tλ (30 s)(645 × 10−9 m)

Reflect: Despite the very large number of photons over this short time, the average power output is relatively small.

28.30. Set Up: The energy of each photon is E = hf =

hc

λ

=

1.99 × 10−25 J ⋅ m

λ

. The power is the total energy per

second and the total energy Etot is the number of photons N times the energy E of each photon. Solve: λ = 10.6 × 10−6 m so E = 1.88 × 10−20 J. P =

Etot NE = so t t

N P 0.100 × 103 W = = = 5.32 × 1021 photons/s t E 1.88 × 10−20 J

28.31. Set Up: Visible light has wavelengths from about 400 nm to about 700 nm. The energy of each photon is E = hf =

hc

λ

=

1.99 × 10−25 J ⋅ m

λ

. The power is the total energy per second and the total energy Etot is the number of

photons N times the energy E of each photon. Solve: (a) 193 nm is shorter than visible light so is in the ultraviolet.



(b) E = (c) P =

hc

λ

28-9

= 1.03 × 10−18 J = 6.44 eV

Etot NE = so t t N=

Pt (1.50 × 10−3 W)(12.0 × 10−9 s) = = 1.75 × 107 photons E 1.03 × 10−18 J

Reflect: A very small amount of energy is delivered to the lens in each pulse, but this still corresponds to a large number of photons.

28.32. Set Up: The energy of each photon is E = hf =

hc

λ

=

1.99 × 10−25 J ⋅ m

λ

. The power is the total energy per

second and the total energy Etot is the number of photons N times the energy E of each photon. The photon beam is spread over an area A = π r 2 , with r = 2.5 mm. hc Solve: (a) λ = 585 nm and E = = 3.40 × 10−19 J = 2.12 eV

λ

(b) The number of photons in a pulse is P =

N=

Etot NE = so t t

Pt (20.0 W)(0.45 × 10−3 s) = = 2.65 × 1016 photons. E 3.40 × 10−19 J

These photons are spread over an area π r 2 so the number of photons per mm 2 is 2.65 × 1016 photons

π (2.5 mm) 2

*28.33. Set Up: Solve: (a) V = (b) λ =

hc

λ

= 1.35 × 1015 photons/mm 2

= eV , where λ is the wavelength of the x-ray and V is the accelerating voltage.

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 8.29 kV eλ (1.60 × 10−19 C)(0.150 × 10−9 m)

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 4.14 × 10−11 m = 0.0414 nm eV (1.60 × 10219 C)(30.0 × 103 V)

(c) No. A proton has the same magnitude of charge as an electron and therefore gains the same amount of kinetic energy when accelerated by the same magnitude of potential difference.

28.34. Set Up: Solve: λ =

hc

λ

= eV , where λ is the wavelength of the x-ray and V is the accelerating voltage.

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 8.29 × 10−11 m = 0.0829 nm eV (1.60 × 10−19 C)(15.0 × 103 V)

28.35. Set Up: Energy conservation gives

hc

λ

=

hc + Ke . λ′

⎛ ⎞ 1 1 ⎛1 1 ⎞ Solve: K e = hc ⎜ − ⎟ = (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) ⎜⎜ − ⎟⎟ . −9 −9 λ λ ′ . × . × 0 100 10 m 0 11 10 m ⎝ ⎠ ⎝ ⎠ K e = 1.81 × 10−16 J = 1.13 keV.


28-10

Chapter 28

28.36. Set Up: The shift in wavelength of the photon is λ ′ − λ = the scattering and

h (1 − cos φ ) where λ ′ is the wavelength after mc

h = λc = 2.426 × 10−12 m. The energy of a photon of wavelength λ is mc E=

hc

λ

=

1.24 × 10−6 eV ⋅ m

λ

.

Conservation of energy applies to the collision, so the energy lost by the photon equals the energy gained by the electron. Solve: (a) λ ′ − λ = λc (1 − cos φ ) = (2.426 × 10−12 m)(1 − cos35.0°) = 4.39 × 10−13 m = 4.39 × 10−4 nm (b) λ ′ = λ + 4.39 × 10−4 nm = 0.04250 nm + 4.39 × 10−4 nm = 0.04294 nm (c) Eλ =

hc

λ

= 2.918 × 104 eV and Eλ ′ =

hc = 2.888 × 104 eV so the photon loses 300 eV of energy. λ′

(d) Energy conservation says the electron gains 300 eV of energy.

h (1 − cos φ ) = λc (1 − cos φ ). Solve for λ ′ : λ ′ = λ + λc (1 − cos φ ). mc The largest λ ′ corresponds to φ = 180°, so cos φ = −1.

*28.37. Set Up: Apply Equation 28.19: λ ′ − λ =

Solve: λ ′ = λ + 2λC = 0.0665 × 10−9 m + 2(2.426 × 10−12 m) = 7.135 × 10−11 m = 0.0714 nm. This wavelength occurs at a scattering angle of φ = 180°. Reflect: The incident photon transfers some of its energy and momentum to the electron from which it scatters. Since the photon loses energy its wavelength increases, λ ′ > λ .

28.38. Set Up: Equation 28.19 gives the change in wavelength of the scattered photon: λ ′ − λ = momentum of a photon is given by p =

h

λ

.

Solve: (a) Solving for λ we obtain λ = λ ′ −

λ = (0.0830 × 10−9 m) −

h (1 − cos φ ). The mc

h (1 − cos φ ). Thus, we have mc

(6.63 × 10−34 J ⋅ s)

(1 + 1) = 0.0781 nm. (9.11 × 10−31 kg)(3.00 × 108 m/s) (b) Since the collision is one-dimensional, the magnitude of the electron’s momentum must be equal to the magnitude of the change in the photon’s momentum. Thus, 1 ⎞ 9 −1 ⎛ 1 −1 ⎞ ⎛ 1 + pe = h ⎜ − ⎟ = (6.63 × 10−34 J ⋅ s) ⎜ ⎟ (10 m ) λ λ ′ . . 0 0781 0 0830 ⎝ ⎠ ⎝ ⎠ @1.65 × 10−23 kg ⋅ m/s ≈ 2 × 10−23 kg ⋅ m/s. (c) Since the electron is non-relativistic (υ /c = 0.06), K e =

pe 2 = 1.49 × 10−16 J ≈ 10−16 J. 2m

Reflect: Note that the wavelength of the incident photon is less than the wavelength of the scattered photon. To conserve momentum and energy, the wavelength of the photon cannot decrease after collision with a stationary electron.

*28.39. Set Up: An electron or a proton accelerated through a potential VAC = 4.00 kV has a maximum kinetic energy of K = 4.00 keV . We will assume that this is nearly the maximum energy of the ejected photon, which is also hc given by Ephoton = hf max = .

λmin



28-11

Solve: Solving for the minimum wavelength of the ejected photons we obtain hc (4.14 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = 3.11 × 10−10 m λmin = = Ephoton 4.00 × 103 eV Reflect: In this case the energy of the ejected photons is the same for both the electron and proton beam. However, electron beams are much more easily produced and accelerated than proton beams.

28.40. Set Up: λ = Solve: (a) λ =

h h = . For an electron, m = 9.11 × 10−31 kg. For a proton, m = 1.67 × 10−27 kg. p mυ 6.63 × 10−34 J ⋅ s

(9.11 × 10−31 kg)(4.70 × 106 m/s)

(b) λ is proportional to

28.41. Set Up: λ = Solve: υ =

= 1.55 × 10−10 m = 0.155 nm

⎛m ⎞ ⎛ 9.11 × 10−31 kg ⎞ 1 , so λp = λe ⎜ e ⎟ = (1.55 × 10−10 m) ⎜ = 8.46 × 10−14 m. ⎜ 1.67 × 10−27 kg ⎟⎟ ⎜ ⎟ m m ⎝ ⎠ ⎝ p⎠

h h = . m = 9.11 × 10−31 kg. p mυ

h 6.63 × 10−34 J ⋅ s = = 0.728 m/s mλ (9.11 × 10−31 kg)(1.00 × 10−3 m)

28.42. Set Up: The visible spectrum is from around λ = 400 nm to around λ = 700 nm. For a photon, E =

hc

λ

.

h p2 and E = , with m = 9.11 × 10−31 kg. p 2m Solve: (a) For λ = 400 nm,

For an electron, λ =

λ=

h 6.63 × 10−34 J ⋅ s = = 0.245 nm p 2.70 × 10−24 kg ⋅ m/s

⎛ 400 nm ⎞ = 1.8 eV. The visible spectrum corresponds to photon energies between For λ = 700 nm, E = (3.1 eV) ⎜ ⎝ 700 nm ⎟⎠ 1.8 eV and 3.1 eV. (b) p = 2mE . E = 1.8 eV = 2.88 × 10−19 J. For E = 1.8 eV, Δυ x =

h 6.63 × 10−34 J ⋅ s = = 3.2 × 104 m/s 2π m Δx 2π (1.67 × 10−27 kg)(2.0 × 10−12 m)

1.8 eV = 0.70 nm. The electron would have energies between 1.8 eV and 3.1 eV 3.1 eV and wavelengths between 0.70 nm and 0.92 nm. For E = 3.1 eV, λ = (0.92 nm)

*28.43. Set Up: The de Broglie wavelength is λ =

h h = . p mv

In the Bohr model,

so mv = nh /(2π rn ). Combine these two expressions and obtain an equation for λ

mvrn = n(h /2π ),

in of n. Then

⎛ 2π rn ⎞ 2π rn = . ⎝ nh ⎟⎠ n

λ = h⎜

Solve: (a) For n = 1, λ = 2π r1 with r1 = a0 = 0.529 × 10−10 m, so λ = 2π (0.529 × 10−10 m) = 3.32 × 10−10 m

λ = 2π r1; the de Broglie wavelength equals the circumference of the orbit. (b) For n = 4, λ = 2π r4 /4.

rn = n 2 a0 so r4 = 16a0 .


28-12

Chapter 28

λ = 2π (16a0 )/4 = 4(2π a0 ) = 4(3.32 × 10−10 m) = 1.33 × 10−9 m 1 1 = times the circumference of the orbit. n 4 Reflect: As n increases the momentum of the electron increases and its de Broglie wavelength decreases. For any n, the circumference of the orbits equals an integer number of de Broglie wavelengths.

λ = 2π r4 /4; the de Broglie wavelength is

28.44. Set Up: When a particle with magnitude of charge e is accelerated through 800 V, the final kinetic energy of the particle is 800 eV. λ =

p2 h . E= so p = 2mE and λ = p 2m

h . mE

Solve: (a) E = 800 eV = 1.28 × 10−16 J.

λ= (b) λ p = λe

6.63 × 10−34 J ⋅ s 2(9.11 × 10−31 kg)(1.28 × 10−16 J)

= 4.34 × 10−11 m = 0.0434 nm

me 9.11 × 10−31 kg = (4.34 × 10−11 m) = 1.01 × 10−12 m = 0.00101 nm mp 1.67 × 10−27 kg

*28.45. Set Up: Both for particles with mass (electrons) and for massless particles (photons) the wavelength is related to the momentum p by λ =

h . But for each type of particle there is a different expression that relates the p

energy E and momentum p. For an electron E = 12 mυ 2 = Solve: photon p =

p2 but for a photon E = pc. 2m

hc 1.24 × 10−6 eV ⋅ m E h h E = = 49.6 nm and p = so = and λ = λ λ c E 25 eV c

electron p = 2mE = 2(9.11 × 10−31 kg)(25 eV)(1.60 × 10−19 J/eV) = 2.70 × 10−24 kg ⋅ m/s

λ=

h 6.63 × 10−34 J ⋅ s = = 0.245 nm p 2.70 × 10−24 kg ⋅ m/s

h hc and for the photon λ = so for an E 2mE E and for a photon λ is proportional to E −1. It is incorrect to say p = for a c

Reflect: The wavelengths are quite different. For the electron λ =

electron λ is proportional to E −1/2

particle such as an electron that has mass; the correct relation is p =

E 2 − ( mc 2 ) 2 . c

28.46. Set Up: The Heisenberg Uncertainty Principle says Δx Δpx $

h . The minimum allowed Δx Δp x is h /2π . 2π

Δpx = m Δvx . Solve: (a) m Δx Δυ x = (b) Δx =

h h 6.63 × 10−34 J ⋅ s = = 3.2 × 104 m/s . Δυ x = 2π 2π m Δx 2π (1.67 × 10−27 kg)(2.0 × 10−12 m)

h 6.63 × 10−34 J ⋅ s = = 4.6 × 10−4 m 2π m Δvx 2π (9.11 × 10−31 kg)(0.250 m/s)

*28.47. Set Up: The wavelength λ of the photon is related to the transition energy E of the atom by E = ΔE Δt $

hc

λ

.

h h . The minimum uncertainty in energy is ΔE = . 2π 2π Δt



28-13

Solve: (a) The photon energy equals the transition energy of the atom, 3.50 eV. hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 355 nm. λ= E 3.50 eV (b) ΔE =

6.63 × 10−34 J ⋅ s 2π (4.0 × 10−6 s)

= 2.69 × 10−29 J = 1.6 × 10−10 eV

28.48. Set Up: Since we know only that the mosquito is somewhere in the room, there is an uncertainty in its position. The Heisenberg uncertainty principle tells us that there is an uncertainty in its momentum. The uncertainty principle is Δ xΔp x ≥ U. Solve: (a) You know the mosquito is somewhere in the room, so the maximum uncertainty in its horizontal position is Δ x = 5.0 m. (b) The uncertainty principle gives Δ xΔp x ≥ U, and Δp x = mΔvx since we know the mosquito’s mass. This gives

(Δ x)(mΔvx ) ≥ U , which we can solve for Δvx to get the minimum uncertainty in vx : Δv x =

U 1.055 × 10−34 J ⋅ s = 1.4 × 10−29 m/s, which is hardly a serious impediment! = mΔx (1.5 × 10−6 kg)(5.0 m)

Reflect: For something as “large” as a mosquito, the uncertainty principle places a negligible limitation on our ability to measure its speed. h . Δx = r1 , where r1 is the radius of the n = 1 2π h h . Bohr orbit. In the n = 1 Bohr orbit, mυ1 r1 = and p1 = mυ1 = 2π 2π r1

28.49. Set Up: The minimum uncertainty product is Δx Δp x =

h h 6.63 × 10−34 J ⋅ s = = = 2.0 × 10−24 kg ⋅ m/s. This is the same as the magnitude of the 2π Δx 2π r1 2π (0.529 × 10−10 m) momentum of the electron in the n = 1 Bohr orbit. Reflect: Since the momentum is the same order of magnitude as the uncertainty in the momentum, the uncertainty principle plays a large role in the structure of atoms. Solve: Δp x =

28.50. Set Up: The kinetic energy K of the electron is related to the accelerating voltage V by K = eV . For an electron E = 12 mυ 2 =

h p2 hc and λ = . For a photon E = . p 2m λ

Solve: (a) For an electron p =

h

λ

=

6.63 × 10−34 J ⋅ s 5.00 × 10

−9

m

= 1.33 × 10−25 kg ⋅ m/s and

E=

p (1.33 × 10−25 kg ⋅ m/s) 2 = = 9.71 × 10−21 J. −31 2m 2(9.11 × 10 kg)

V=

K 9.71 × 10−21 J = = 0.0607 V e 1.60 × 10−19 C

2

The electrons would have kinetic energy 0.0607 eV. hc 1.24 × 10−6 eV ⋅ m = = 248 eV (b) E = λ 5.00 × 10−9 m (c) E = 9.71 × 10−21 J so

λ=

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 20.5 μ m E 9.71 × 10−21 J


28-14

Chapter 28

*28.51. Set Up: We can use the de Broglie relation to find the momentum corresponding to the given wavelength: p=

h

λ

. The required accelerating potential can be found from eV = K =

Solve: (a) eV = K =

p2 . 2m

p 2 ( h /λ ) 2 . Solving for V we obtain = 2m 2m

( h /λ ) 2 (6.63 × 10−34 J ⋅ s) 2 = = 419 V. 2me 2(6.00 × 10−11 m) 2 (9.11 × 10−31 kg)(1.60 × 10−19 C) (b) We repeat our calculation for a proton—noting that only the mass changes: ( h /λ ) 2 (6.63 × 10−34 J ⋅ s) 2 V= = = 0.228 V. − 11 2me 2(6.00 × 10 m) 2 (1.67 × 10−27 kg)(1.60 × 10−19 C) V=

Reflect: We could also calculate the answer to part (b) by recognizing that the voltage is reduced by the ratio of the 9.11 × 10−31 kg particle masses, (419 V) = 0.229 V. 1.67 × 10−27 kg

28.52. Set Up: For a photon Eph = Solve: (a) E =

h2

hc

λ

=

1.24 × 10

−6

hc

λ

eV ⋅ m

500 × 10−9 m

=

1.24 × 10−6 eV ⋅ m

λ

. For an electron Ee =

p2 h2 = . 2m 2mλ 2

= 248 eV

(6.63 × 10−34 J ⋅ s) 2

= 9.65 × 10−21 J = 0.0603 eV 2mλ 2 2(9.11 × 10−31 kg)(5.00 × 1029 m) 2 The electron has much less energy than a photon of the same wavelength.

(b) Ee =

=

hc

. 1 mole = 6.02 × 1023 molecules. λ Solve: (a) The dissociation energy of one AgBr molecule is 1.00 × 105 J/mol E= = 1.66 × 10219 J = 1.04 eV. 6.02 × 1023 molecules/mol The photon energy is 1.04 eV.

*28.53. Set Up: The energy of a photon is E = hf =

(b) λ = (c) f =

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 1.19 × 10−6 m = 1.19 μ m E 1.04 eV c

λ

=

3.00 × 108 m/s 1.19 × 10

−6

m

= 2.52 × 1014 Hz

(d) For f = 100 MHz = 1.00 × 108 Hz. E = hf = (4.136 × 10−15 eV ⋅ s)(1.00 × 108 Hz) = 4.1 × 10−7 eV. The radio station

produces a large number of photons but they individually have far too little energy to dissociate a AgBr molecule. Reflect: The frequency or wavelength of the radiation determines the energy of each photon. For fixed frequency, increasing the power output of the source increases the number of photons emitted per second but doesn’t change the energy of each photon.

28.54. Set Up: The photoelectric effect occurs, so the energy of the photon is used to eject an electron, with any excess energy going into kinetic energy of the electron. Conservation of energy gives hf = hc/λ = K max + φ. Solve: (a) Using hc/λ = K max + φ, we solve for the work function:

φ = hc/λ – K max = (4.136 × 10−15 eV ⋅ s )(3.00 × 108 m/s)/(124 nm) – 4.16 eV = 5.85 eV (b) The number N of photoelectrons per second is equal to the number of photons per second that strike the metal per second. N × (energy of a photon) = 2.50 W. N(hc/λ) = 2.50 W. N = (2.50 W)(124 nm)/[(6.626 × 10−34 J ⋅ s )(3.00 × 108 m/s)] = 1.56 × 1018 electrons/s (c) N is proportional to the power, so if the power is cut in half, so is N, which gives N = (1.56 × 1018 el/s)/2 = 7.80 × 1017 el/s © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


28-15

(d) If we cut the wavelength by half, the energy of each photon is doubled since E = hc/λ. To maintain the same power, the number of photons must be half of what they were in part (b), so N is cut in half to 7.80 × 1017 el/s. We could also see this from part (b), where N is proportional to λ. So if the wavelength is cut in half, so is N. Reflect: In part (c), reducing the power does not reduce the maximum kinetic energy of the photons; it only reduces the number of ejected electrons. In part (d), reducing the wavelength does change the maximum kinetic energy of the photoelectrons because we have increased the energy of each photon.

28.55. Set Up: For the hydrogen atom the ionization energy, the energy to just free an electron that is initially in the ground state, is 13.6 eV. The kinetic energy of the released electrons is K max = Solve: K max =

hc

λ

= 13.6 eV.

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)

− 13.6 eV = 14.5 eV − 13.6 eV = 0.9 eV 85.5 × 10−9 m Reflect: The electron must receive enough energy (the ionization energy) from one photon to free it from the atom. Any excess energy then appears as kinetic energy of the free electron.

28.56. Set Up: The energy of the photon absorbed or emitted equals the transition energy of the atom.

Solve: (a) The ground level has energy E = −10.0 eV, if the ionization threshold is where E = 0. The other levels are 2.0 eV, 5.0 eV, and 9.0 eV above the ground level, so are at −8.0 eV, −5.0 eV, and −1.0 eV. The energy level diagram is sketched in Figure (a) below.

(b) After the 9.0 eV photon is absorbed the atom is in the state with E = 21.0 eV. The subsequent transitions and the transition energies are shown in Figure (b) above. The possible photon energies are 2.0 eV, 3.0 eV, 4.0 eV, 5.0 eV, 7.0 eV, and 9.0 eV.

13.6 eV

*28.57. Set Up: The Hα line in the Balmer series corresponds to the n = 3 to n = 2 transition. En = 2 hc

λ

n2

.

= ΔE .

Solve: (a) The atom must be given an amount of energy ⎛ 1 1⎞ E3 − E1 = −(13.6 eV) ⎜ 2 − 2 ⎟ = 12.1 eV. ⎝3 1 ⎠

hc = 103 nm; n = 3 → n = 2 : ΔE 1⎞ ⎛ 1 ⎛ 1 1⎞ ΔE = − (13.6 eV) ⎜ 2 − 2 ⎟ = 1.89 eV and λ = 657 nm; n = 2 → n = 1: ΔE = − (13.6 eV) ⎜ 2 − 2 ⎟ = 10.2 eV and ⎝3 ⎝2 1 ⎠ 2 ⎠ λ = 122 nm.

(b) There are three possible transitions. n = 3 → n = 1: ΔE = 12.1 eV and λ =

28.58. Set Up: The de Broglie wavelength is given by λ =

1 p2 h . . Kinetic energy is given by K = mυ 2 = 2 2m mυ

Solve: (a) The speed of the microorganism is υ = 2.9(0.0020 cm/s) = 5.8 × 10−5 m/s. Thus we have λ =

(6.63 × 10234 J ⋅ s) (8.0 × 10−12 kg)(5.8 × 10−5 m/s)

h = mυ

= 1.4 × 10−18 m = 1.4 × 10−9 nm.


28-16

Chapter 28

(b) When swimming at top speed the kinetic energy of the organism is 1 1 K = mυ 2 = (8.0 ×10−12 kg)(5.8 ×10−5 m/s)2 (1 eV/1.60 ×10−19 J) = 0.084 eV. 2 2 Reflect: The de Broglie wavelength of the microorganism is very small compared to the wavelength of visible light. h h (1 − cos φ ), with = 2.426 × 10−12 m. Conservation of energy and conservation of mc mc momentum apply to the collision. Momentum is a vector and each of its components is conserved. A photon has hc h energy and momentum . φ = 180°.

*28.59. Set Up: λ ′ = λ +

λ

λ

Solve: (a) λ ′ = 0.1800 × 10

−9

m + (2.426 × 10−12 m)(1 − cos180°) = 0.1849 nm

(b) Energy conservation gives

hc

λ

=

hc + K e , where K e is the final kinetic energy of the electron. K e = λ′

⎛1 1⎞ ⎛ λ′ − λ ⎞ hc ⎜ − ⎟ = hc ⎜ . ⎝ λ λ′⎠ ⎝ λ ′ λ ⎟⎠ K e = (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) (c) K e = 12 mυ 2 , so υ =

2Ke = m

0.0049 × 10−9 m (0.1800 × 10−9 m)(0.1849 × 10−9 m)

2(2.93 × 10−17 J) 9.11 × 10

−31

kg

= 2.93 × 10−17 J = 183 eV

= 8.0 × 106 m/s. υ c, so it is not necessary to use the

relativistic kinetic energy relationship. Reflect: We can also find υ from conservation of momentum. Let 1x be the direction the photon is traveling h h initially. Then conservation of the x component of momentum gives = − + mυ. λ λ′ ⎛ ⎞ h ⎛ 1 1 ⎞ h ⎛ λ + λ ′ ⎞ ⎛ 6.63 × 10−34 J ⋅ s ⎞ ⎜ 0.1849 × 10−9 m + 0.1800 × 10−9 m ⎟ . υ = ⎜ + ⎟= ⎜ ⎟ ⎟ = ⎜⎜ m ⎝ λ λ ′ ⎠ m ⎝ λλ ′ ⎠ ⎝ 9.11 × 10−31 kg ⎟⎠ ⎜⎜ ⎡0.1849 × 10−9 m ⎤ ⎡ 0.1800 × 10−9 m ⎤ ⎟⎟ ⎦⎣ ⎦⎠ ⎝⎣

υ = 8.0 × 106 m/s, which is the same result as computed from energy conservation. h h (1 − cos φ ). = 2.426 × 10−12 m = 2.426 pm. mc mc Solve: (a) The largest Δλ is for φ = 180° and Δλ max = 4.85 pm.

28.60. Set Up: Δλ =

(b) If λ ′ = 2λ , then λ = Δλ max and λ = 4.85 × 10−12 m.

E=

hc

λ

=

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) 4.85 × 10−12 m

= 2.56 × 105 eV = 0.256 MeV.

h 2h h to calculate the (1 − cos φ ), where φ = 180° so λ ′ = λ + = 0.09485 m. Use p = λ mc mc momentum of the scattered photon. Apply conservation of energy to the collision to calculate the kinetic energy of hc the electron after the scattering. The energy of the photon is given by E = hf = .

28.61. Set Up: λ ′ = λ +

λ

Solve: (a) p ′ = h /λ ′ = 6.99 × 10−24 kg ⋅ m/s. (b) E = E ′ + Ee ; hc /λ = hc /λ ′ + Ee

λ′ − λ ⎛1 1⎞ = 1.129 × 10−16 J = 705 eV Ee = hc ⎜ − ⎟ = (hc) ⎝ λ λ′ ⎠ λλ ′ Reflect: The energy of the incident photon is 13.8 keV, so only about 5% of its energy is transferred to the electron. This corresponds to a fractional shift in the photon’s wavelength that is also 5%.



28.62. Set Up: Conservation of energy applied to the collision gives

hc

λ

=

28-17

hc hc + K e . For a photon, E = . λ′ λ

⎛1 1⎞ ⎛ λ′ − λ ⎞ . Solve: (a) K e = hc ⎜ − ⎟ = hc ⎜ ⎝ λ λ′ ⎠ ⎝ λ ′ λ ⎟⎠ ⎛ ⎞ 0.0032 × 10−9 m ⎟ K e = (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) ⎜ ⎜⎜ ⎡0.1100 × 10−9 m ⎤ ⎡ 0.1132 × 10−9 m ⎤ ⎟⎟ ⎦⎣ ⎦⎠ ⎝⎣ = 5.11 × 10217 J = 319 eV

υe =

2Ke 2(5.11 × 10−17 J) = = 1.06 × 107 m/s. m 9.11 × 10−31 kg

(b) The photon has energy 319 eV, so hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) λ= = = 3.89 × 10−9 m = 3.89 nm. E 319 eV

*28.63. Set Up: For a photon, E =

hc

λ

. T = 27°C = 300 K. k = 1.38 × 10−23 J/K.

Solve: E = 32 (1.38 × 10−23 J/K)(300 K) = 6.21 × 10−21 J.

λ=

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 3.20 × 10−5 m = 32.0 μm. −21 E 6.21 × 10 J

Reflect: This photon is in the infrared region of the electromagnetic spectrum.

28.64. Set Up: Use Equation 28.20 to relate your wavelength and speed. (a) λ =

Solve:

h h 6.626 × 10−34 J ⋅ s , so v = = = 1.1 × 10−35 m/s mv mλ (60.0 kg)(1.0 m)

distance 0.80 m = = 7.3 × 1034 s(1 y/3.156 × 107 s) = 2.3 × 1027 y velocity 1.1 × 10−35 m/s Since you walk through doorways much more quickly than this, you will not experience diffraction effects. Reflect: A 1 kg object moving at 1 m/s has a de Broglie wavelength λ = 6.6 × 10−34 m, which is exceedingly small. An object like you has a very, very small λ at ordinary speeds and does not exhibit wavelike properties. (b) t =

28.65. Set Up: λ =

h h = . m = 1.00 × 10−14 kg. v = 0.400 × 10−2 m/s. p mv 6.63 × 10−34 J ⋅ s

Solve: λ =

−14

(1.00 × 10 kg)(0.400 × 10 concerned with the wave nature.

−2

m/s)

= 1.66 × 10−17 m. λ is very small, so no, we do not need to be

28.66. Set Up: The energy added to mass m of the blood to heat it to Tf = 100 °C and to vaporize it is Q = mc(Tf − Ti ) + mLv , E=

hc

λ

=

with

1.99 × 10−25 J ⋅ m

λ

c = 4190 J/kg ⋅ K

and

Lv = 2.256 × 106 J/kg.

The energy of one photon is

.

Solve: (a) Q = (2.0 × 10−9 kg)(4190 J/kg ⋅ K)(100°C − 33°C) + (2.0 × 10−9 kg)(2.256 × 106 J/kg)

= 5.07 × 10−3 J The pulse must deliver 5.07 mJ of energy.


28-18

Chapter 28

energy 5.07 × 10−3 J = = 11.3 W t 450 × 10−6 s (c) One photon has energy (b) P =

hc

1.99 × 10−25 J ⋅ m

= 3.40 × 10−19 J. 585 × 10−9 m The number N of photons per pulse is the energy per pulse divided by the energy of one photon: 5.07 × 10−3 J N= = 1.49 × 1016 photons 3.40 × 10−19 J/photon E=

λ

=

*28.67. Set Up: The energy of a photon is given by E = hf = through a potential difference ΔV is K = Solve: (a) E = hc /λ = 12 eV

hc . The kinetic energy of an electron accelerated λ

p2 = − qΔV = eΔV . 2m

(b) Find E for an electron with λ = 0.10 × 10−6 m. λ = h /p so p = h /λ = 6.626 × 10−27 kg ⋅ m/s.

E = p 2 /(2m) = 1.5 × 10−4 eV. E = eΔV so ΔV = 1.5 × 10−4 V v = p /m = (6.626 × 10−27 kg ⋅ m/s)/(9.109 × 10−31 kg) = 7.3 × 103 m/s

(c) The required wavelength is the same so we need the same momentum. E = p 2 /(2m) but now m = 1.673 × 10227 kg

so E = 8.2 × 1028 eV and qΔV q = 8.2 × 1028 V. v = p /m = (6.626 × 10−27 kg ⋅ m/s)/(1.673 × 10−27 kg) = 4.0 m/s. Reflect: The voltage required to accelerate the proton could also be found by noting that, for a given wavelength (momentum), the ratio of the required accelerating voltages is the inverse ratio of the masses: V p /Ve = me /m p , thus

V p = (1.5 × 10−4 V)(9.109 × 10−31 kg)/(1.673 × 10−27 kg) = 8.2 × 10−8 V.

28.68. Set Up: Let the xy plane be horizontal and let the ball be thrown in the x direction. Δy Δp y ≥ minimum uncertainty product, so Δp y =

h . Take the 2π

h . Let Δy be the length of one side of the cubical volume, so 2π Δy

Δy = 10 cm = 0.10 m.

Solve: (a) Δp y =

0.0663 J ⋅ s = 0.1 kg ⋅ m/s 2π (0.10 m)

(b) It takes the ball

Δp y 0.1 kg ⋅ m/s 50 m = = 1.0 m/s. = 10 s to travel from one student to the other. Δv y = m 0.10 kg 5.0 m/s

During 10 s the uncertainty in v y introduces an uncertainty of (1.0 m/s)(10 s) = 10 m in the y component of the position of the ball.

*28.69. Set Up: ΔE Δt $ m = 264me . Δm = Solve: ΔE =

ΔE c2

h h , with Δt = 8.4 × 10217 s. . Take the minimum uncertainty product, so ΔE = 2π Δt 2π

.

6.63 × 10−34 J ⋅ s 2π (8.4 × 10−17 s)

= 1.26 × 10−18 J. Δm =

1.26 × 10−18 J (3.00 × 108 m/s)2

= 1.4 × 10−35 kg.



28-19

Δm 1.4 × 10−35 kg = = 5.8 × 10−8 m (264)(9.11 × 10−31 kg)

28.70. Set Up: The electrons behave like waves and produce a double-slit interference pattern after ing through

the slits. The first angle at which destructive interference occurs is given by d sin θ = λ /2. The de Broglie wavelength of each of the electrons is λ = h /mv. Solve: (a) First find the wavelength of the electrons. For the first dark fringe, we have d sin θ = λ /2, which gives (1.25 nm)sin18.0D = λ /2, and λ = 0.7725 nm. Now solve the de Broglie wavelength equation for the speed of the

electron: h 6.626 × 10−34 J ⋅ s v= = = 9.42 × 105 m/s. mλ (9.11 × 10−31 kg)(0.7725 × 10−9 m) which is about 0.3% the speed of light, so they are nonrelativistic. (b) Energy conservation gives eV = 12 mv 2 and V = mv 2 /2e = (9.11 × 10−31 kg)(9.42 × 105 m/s) 2 /[2(1.60 × 10−19 C)] = 2.52 V. Reflect: The hole must be much smaller than the wavelength of visible light for the electrons to show diffraction.

Solutions to age Problems *28.71. Set Up: The treatment dosage is

70 Gy 2 Gy 2 J/kg = = of absorbed radiation. We know that there are 35 days day day

108 cells/cm3 and the density of the tumor is 1 g/cm3 = 10−3 kg/cm3. We are told that 1 J = 6 × 1018 eV. 2 J 10−3 kg 1 cm3 6 × 1018 eV ⋅ ⋅ ⋅ = 1.20 × 108 eV/cell. Thus, in kg cm3 108 cells 1J one day the average dosage is 120 MeV per cell. The correct answer is B. Solve: The absorbed energy per cell in one day is

28.72. Set Up: We are told that each Compton interaction imparts an energy of 1 MeV to the electrons and each cell absorbs an average of 12 MeV each day (note that we are told to use this value so we do not use the 120 MeV that we calculated in the previous problem). Solve: Using the given values, the number of Compton interactions in a cell each day is (12 MeV) ⋅ 1 Compton interaction = 12 Compton interactions. The correct answer is D. 1 MeV

*28.73. Set Up: We are told that each ionization in the tissue requires about 40 eV and we know (from Problem 71) that each cell receives an average of 120 MeV each day. Solve: The number of ionizations in a day is roughly (120 × 106 eV)

(1 ionization) = 3 × 106 ionizations. The correct (40 eV)

answer is D.


29

ATOMS, MOLECULES, AND SOLIDS

Answers to Multiple-Choice Problems 1. A, C

2. A, D

3. B, D

4. A

5. A, D

6. A, B, E

7. C

8. C

9. C

10. D

Solutions to Problems *29.1. Set Up: Equation 29.1 relates the magnitude of the orbital angular momentum L to the quantum number l: L = l (l + 1)

h , l = 0, 1, 2, . . . 2π 2

2 −34 kg ⋅ m 2 /s ⎞ ⎛ L ⎞ ⎛ 4.716 × 10 Solve: l (l + 1) = ⎜ 2π ⎟ = ⎜ 2π ⎟ = 20. − 34 h ⎠ ⎜⎝ 6.626 × 10 J ⋅ s ⎟⎠ ⎝ And then l (l + 1) = 20 gives that l = 4.

Reflect: l must be integer.

29.2. Set Up: Lz = ml U. ml = 0, ±1, ±2, … , ±l. L = l (l + 1) U. cosθ = Lz /L. Solve: (a) The largest Lz is for ml = 3 and is Lz = 3U. (b) L = 3(3 + 1) U = 2 3 U = 3.46 U. L is larger than the maximum possible Lz . (c) For Lz = 3U , cosθ =

3U 2U U and θ = 30.0D. For Lz = 2 U , cosθ = and θ = 54.7D. For Lz = U , cosθ = 2 3U 2 3U 2 3U

and θ = 73.2D. For Lz = 0, cosθ = 0 and θ = 90.0D. For Lz = −U , cosθ = cosθ =

-U and θ = 106.8D. For Lz = −2U, 2 3U

−2 U −3U and θ = 125.3D. For Lz = −3U , cosθ = and θ = 150.0D. 2 3U 2 3U

*29.3. Set Up: L = l (l + 1) U. Lz = ml U. l = 0, 1, 2, … , n − 1. ml = 0, ±1, ±2, … , ±l. cosθ = Lz /L. Solve: (a) l = 0 : L = 0, Lz = 0. l = 1: L = 2 U , Lz = U , 0, -U. l = 2 : L = 6 U , Lz = 2 U , U , 0, -U , −2 U. (b) In each case cosθ = Lz /L. L = 0 : θ not defined. L = 2 U : 45.0D , 90.0D , 135.0D. L = 6 U : 35.3D , 65.9D ,

90.0D , 114.1D , 144.7D.

G Reflect: There is no state where L is totally aligned along the z axis.

29.4. Set Up: L = l (l + 1) U. Lz = ml U. l = 0, 1, 2, … , n − 1. ml = 0, ±1, ±2, … , ±l. cosθ = Lz /L. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

29-1

29-2

Chapter 29

Solve: (a) l = 0 : L = 0, Lz = 0. l = 1: L = 2 U , Lz = U , 0, -U. l = 2 : L = 6 U , Lz = 2U , U , 0, -U, 22U.

l = 3 : L = 2 3 U , Lz = 3U , 2 U , U , 0, -U , −2U , −3U. l = 4 : L = 2 5 U , Lz = 4U , 3U , 2U , U , 0, -U , −2U , −3U , −4U. (b) L = 0 : θ not defined. L = 2 U : 45.0D , 90.0D , 135.0D. L = 6 U : 35.3D , 65.9D , 90.0D , 114.1D , 144.7D. L = 2 3 U : 54.7D , 73.2D , 90.0D , 106.8D , 125.3D , 150.0D. L = 2 5 U : 26.6D , 47.9D , 63.4D , 77.1D , 90.0D ,

102.9D , 116.6D , 132.1D , 153.4D. (c) The minimum angle is 26.6D and occurs for l = 4, ml = +4. The maximum angle is 153.4D and occurs for l = 4,

ml = −4

29.5. Set Up: For the N shell n = 4. 0 ≤ l ≤ n − 1. 0 m 0 ≤ l. s = ±1. 1 J ⋅ s = 1 kg ⋅ m 2 /s Solve: (a) The smallest l is l = 0. L = l (l + 1) U so Lmin = 0. (b) The largest l is n − 1 = 3 so Lmax = 3(4) U = 2 3 U = 3.65 × 10−34 kg ⋅ m 2 /s (c) Let the chosen direction be the z axis. The largest m is m = l = 3. Lz,max = mU = 3U = 3.16 × 10−34 kg ⋅ m 2 /s (d) S z = ± 12 U. The maximum value is S z = U/2 = 5.27 × 10−35 kg ⋅ m 2 /s. 1 Sz 2 U 1 = = Lz 3U 6 Reflect: The orbital and spin angular momenta are of comparable sizes.

(e)

29.6. Set Up: 0 ≤ l ≤ n − 1. ml = 0, ±1, ±2, … , ±l. s = ± 12 . d means l = 2 and f means l = 3. Solve: (a) There are ten 3d states: ml = 0, ±1, ±2, with s = ± 12 for each. (b) There are no 3 f states. These states would have l = 3, n = 3. But the maximum l is n − 1 and l = 3 doesn’t occur for n = 3.

*29.7. Set Up: ml = 0, ±1, ±2, … , ±l. s = ± 12 . g means l = 4. cosθ = Lz /L, with L = l (l + 1) U and Lz = ml U , Solve: (a) There are eighteen 5g states: ml = 0, ±1, ±2, ±3, ±4, with s = ± 12 for each. (b) The largest θ is for the most negative ml . L = 2 5 U. The most negative Lz is Lz = −4U. cosθ =

−4 U and 2 5U

θ = 153.4°. (c) The smallest θ is for the largest positive ml , which is ml = +4. cosθ =

4U and θ = 26.6D. 2 5U

G Reflect: The minimum angle between L and the z axis is for ml = +l and for that ml , cosθ =

29.8. Set Up: We have cosθ =

l . l (l + 1)

Lz ml = . L l (l + 1)

Solve: The smallest nonzero angle for a given l occurs for ml = +l. Thus we have cos 26.6D =

l . This can be l (l + 1)

written l (l + 1)cos 2 (26.6D ) = l 2 . After cancelling a common factor of l we can solve this linear equation to obtain

l=

cos 2 (26.6D ) 1 − cos 2 (26.6D )

= 4.


Atoms, Molecules, and Solids

29-3

Reflect: For l = 4 we see that the angle between the angular momentum vector and the positive z-axis ranges from

26.6J ( ml = +l ) to 180D − 26.6D = 153.4D (ml = −l ).

*29.9. Set Up: 0 ≤ l ≤ n − 1. 0 ml 0 ≤ l . s = ± 12 . Solve: n = 1, l = 0, ml = 0, s = 12 . n = 1, l = 0, ml = 0, s = 2 12 . n = 2, l = 0, ml = 0, s = 12 . n = 2, l = 0,

ml = 0, s = − 12 . n = 2, l = 1, ml = 1, s = 12 . n = 2, l = 1, ml = 1, s = − 12 . n = 2, l = 1, ml = 0, s = 12 . n = 2, l = 1, ml = 0, s = − 12 . n = 2, l = 1, ml = 21, s = 12 . n = 2, l = 1, ml = −1, s = − 12 . n = 3, l = 0, ml = 0, s = 12 . n = 3, l = 0, ml = 0, s = 2 12 .

29.10. Set Up: 0 ≤ l ≤ n − 1. 0 ml 0 ≤ l . s = ± 12 . Solve: (a) l = 0 : ml = 0. l = 1: ml = 0, ±1. l = 2 : ml = 0, ±1, ±2. l = 3 : ml = 0, ±1, ±2, ±3. l = 4 : ml = 0, ±1, ±2, ±3, ±4. (b) There are 25 combinations of l, ml values. For each, s can be + 12 or − 12 . Therefore, 2(25) = 50 electrons can

be placed into the n = 5 shell.

29.11. Set Up: Fill the shells in order. Each l = 0 subshell can hold 2 electrons, each l = 1 subshell holds 6 and each l = 2 subshell holds 10. 0 ≤ l ≤ n − 1.

Solve: 1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4s 2 4 p5 .

29.12. Set Up: Li has 3 electrons, Na has 11 electrons. Solve: (a) The electron configurations are

Li: 1s 2 2s Na: 1s 2 2s 2 2 p 6 3s (b) Each has one electron in its outer shell.

*29.13. Set Up: Carbon has 6 electrons. Solve: (a) 1s 2 2 s 2 2 p 2 (b) The element of next larger Z with a similar electron configuration has configuration 1s 2 2s 2 2 p 6 3s 2 3 p 2 . Z = 14

and the element is silicon. Reflect: Carbon and silicon are in the same column of the periodic table.

29.14. Set Up: Beryllium has 4 electrons. Solve: (a) 1s 2 2 s 2 (b) 1s 2 2s 2 2 p 6 3s 2 . Z = 12 and the element is magnesium. (c) 1s 2 2s 2 2 p 6 3s 2 3 p 6 4 s 2 . Z = 20 and the element is calcium.

*29.15. Set Up: Ne has 10 electrons; Ar has 18 electrons, Kr has 36 electrons. Solve: The electron configurations are

Ne: 1s 2 2s 2 2 p 6 Ar: 1s 2 2s 2 2 p 6 3s 2 3 p 6 Kr: 1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 These atoms have filled outer shells and are chemically inert. Reflect: These three elements are in the same column of the periodic table.


29-4

Chapter 29

29.16. Set Up: The angular momentum L is related to the quantum number l by Eq. (29.1), L = l (l + 1) U, where U=

h . The maximum l, lmax , for a given n is lmax = n − 1. 2π

Solve: For n = 2, lmax = 1 and L = 2 U = 1.414 U.

For n = 20, lmax = 19 and L = (19)(20) U = 19.49U. For n = 200, lmax = 199 and L = (199)(200) U = 199.5U. Reflect: As n increases, the maximum L gets closer to the value nU postulated in the Bohr model.

29.17. Set Up: L = l (l + 1) U. For a classical particle in a circular orbit of radius r, L = I ω = mυ r. Solve: (a) s subshell means l = 0 so L = 0 (b) In a classical orbit, L = mυ r so L = 0 means υ = 0. υ = 0 gives arad = 0. But there is a net force on the

electron, the Coulomb attraction exerted by the positively charged nucleus. The electron would be accelerated into the nucleus. Reflect: The Bohr model gives the correct energy levels but gives an incorrect description of the angular momentum of the electron.

29.18. Set Up: ΔE = M → K transition.

hc

λ

. A Kα x-ray is produced in a L → K transition and a K β x-ray is produced in a

Solve: Kα : ΔE = EL − EK = −12 ,000 eV − ( −69 ,500 eV) = +57 ,500 eV. hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 0.0216 nm. ΔE 57 ,500 eV − EK = −2200 eV − ( −69,500 eV) = +67 ,300 eV.

λ=

K β : ΔE = E M

λ=

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 0.0184 nm. ΔE 67 ,300 eV

29.19. Set Up: The energy of the photon is related to its wavelength by E=

hc

λ

=

1.24 × 10−6 eV ⋅ m

λ

.

Visible light has wavelengths between 400 nm and 700 nm. Solve: The photon must have energy 4.48 eV. hc 1.24 × 10−6 eV ⋅ m = 277 nm λ= = E 4.48 eV This wavelength is shorter than the wavelengths of visible light so lies in the ultraviolet.

29.20. Set Up: ΔE =

hc

λ

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 4.96 μ m. This radiation is in the infrared. Solve: (a) λ = ΔE 0.250 eV (b) λ =

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 146 nm. This radiation is in the ultraviolet. ΔE 8.50 eV

(c) λ =

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 388 μ m. This radiation is in the microwave. ΔE 3.20 × 10−3 eV

*29.21. Set Up: The electric potential energy for two point charges is given by U = k

q1q2 e2 = −k . r r



29-5

(1.60 × 10−19 C) 2 e2 = − (8.99 × 109 N ⋅ m 2 /C2 ) = −7.9 × 10−19 J = −5.0 eV. r 0.29 × 10−9 m (b) To create a potassium bromide molecule we remove an electron from the potassium atom (add energy) and add it to the bromine atom (release of energy). In addition, the resulting ions release potential energy as they combine: Binding Energy ≈ +(4.3 eV − 3.5 eV) − 5.0 eV = −4.2 eV. (a) U = − k

Reflect: Our calculations assume we can treat each ion as a point charge; however, the electron charge distribution around each atom will distort as the ions approach one another—this should result in a further release of energy.

29.22. Set Up: The density ρ is the mass per unit volume. Assume each atom occupies a volume of a3 , where a = 0.282 nm. Solve: There are equal numbers of Na and Cl atoms and the average mass is 1 (3.82 × 10−26 kg + 5.89 × 10 −26 kg) = 4.855 × 10−26 kg. 2

ρ=

m 4.855 × 10−26 kg = = 2.16 × 103 kg/m3 . V (0.282 × 10−9 m)3

*29.23. Set Up: The density ρ is the mass per unit volume. Assume each atom occupies a volume of a3 , where a is the spacing between adjacent atoms. Solve: There are equal numbers of K and Br atoms and the average mass is 1 (6.49 × 10−26 kg + 1.33 × 10−25 kg) = 9.90 × 10−26 kg. 2 1/3

1/3 ⎛ 9.90 × 10−26 kg ⎞ ⎛m⎞ = = 3.30 × 10−10 m = 0.330 nm. and a ⎜ ⎟ = ⎜⎜ 3 3 3⎟ ⎟ ρ 2 75 10 kg/m a . × ⎝ ⎠ ⎝ ⎠ Problem 29.22 states that the spacing of adjacent atoms for NaCl is 0.282 nm. The spacing is larger for KBr, because the K and Br atoms contain more electrons and have larger atomic radii than Na and Cl.

ρ=

m

Reflect: In a solid the atoms are closely packed together and the average spacing between atoms is approximately equal to the average diameter of the atoms.

29.24. Set Up: R = V/I . If the circuit element obeys Ohm’s law, R is independent of I. Solve: This behavior of the graph shows that I is in the reverse direction if the applied voltage is in the reverse direction, but for reverse bias the current is small. For an ordinary resistor that obeys Ohm’s law, a reverse voltage gives the same magnitude of I as for a forward voltage of the same magnitude. The I versus V graph for an ohmic resistor is sketched in the figure below.

*29.25. Set Up: The energy of a photon is given by Ephoton = hf =

hc

λ

. The gap in the valence band and the

conduction band is ΔE = 5.47 eV. Solve: (a) We will assume that the energy of the photon is equal to the band gap energy: Ephoton = ΔE.


29-6

Chapter 29

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s) = = 2.27 × 10−7 m = 227 nm, which is in the ultraviolet. ΔE 5.47 eV (b) Visible light lacks enough energy to excite the electrons into the conduction band, so visible light es through the diamond unabsorbed. Reflect: (c) Impurities can lower the gap energy, making it easier for the material to absorb shorter wavelength visible light. This allows longer wavelength visible light to through, giving the diamond color. Thus we have λ =

29.26. Set Up: Let the polarity of the power supply voltage as shown in the figure in the problem be positive. Let clockwise currents be positive. Solve: (a) The I versus V graph is shown in Figure (a) below. For this orientation of the diode a positive voltage for the power supply corresponds to a positive bias for the diode.

(b) The I versus V graph is shown in Figure (b) above. Now a positive applied voltage corresponds to a negative bias for the diode. (c) For a resistor a certain magnitude of applied voltage produces the same magnitude of current, regardless of the polarity of the voltage. The graph of I versus V would be as sketched in Figure (c) above. (d) (i) In one case a positive voltage is forward bias for the diode and produces a large positive current and in the other case a negative voltage is forward bias for the diode and produces a large negative current. In each case the current is in the same direction as the polarity of the voltage. (ii) For a resistor there is no forward or reverse bias direction.

29.27. Set Up: The energy of a photon is given by Ephoton = hf = conduction band is ΔE = 1.12 eV. Solve: Ephoton =

hc

λ

=

(6.63 × 10−34 J ⋅ s) (3.00 × 108 m/s) 9.31 × 10−13 m

hc

λ

. The gap in the valence band and the

= 2.14 × 10−13 J = 1.34 × 106 eV. So the number of electrons

1.34 × 106 eV = 1.20 × 106 electrons. ΔE 1.12 eV Reflect: The gamma ray can excite a large number of electrons into the conduction band due to its high energy. that can be excited to the conduction band is n =

Ephoton

=

29.28. Set Up: For an ordinary metallic conductor the resistance increases as T increases. The resistance of an ordinary conductor is not zero at T = 0 K. For a superconductor the resistance is zero below the critical temperature, has a nonzero value just above Tc , and increases as T increases. Solve: The graphs are sketched in Figures (a) and (b) below.



29-7

*29.29. Set Up: The ionization potential is the energy required to remove an electron from the atom and is the negative of the binding energy of the atom. Solve: The least bound and outermost electron in the atom moves under the action of the total electric field due to the nucleus and the averaged-out electron cloud of all the electrons. The other electrons approximately screen the outermost electron from the charge of the nucleus. As electrons are removed, this screening is reduced and the outermost electron in each successive charge state is more tightly bound. Reflect: We could describe the binding energy for a particular electron as (13.6 eV) Z eff 2 /n 2 , where Z eff is the effective nuclear charge of the screened nucleus. Z eff increases as electrons are removed.

29.30. Set Up: For a sphere with an axis through its center I = 25 mR 2 and L = Iω . The tangential speed υ of a point on the surface is given by υ = Rω . Solve: (a) S z = ± U /2 = ±5.27 × 10−35 kg ⋅ m 2 /s (b) I = 52 mR 2 = 52 (9.11 × 10−31 kg)(5.0 × 10−16 m)2 = 9.11 × 10−62 kg ⋅ m 2

ω=

L 5.27 × 10−35 kg ⋅ m 2 /s = = 5.78 × 1026 rad/s −62 2 I 9.11 × 10 kg ⋅ m

(c) υ = Rω = (5.0 × 10−16 m)(5.78 × 1026 rad/s) = 2.9 × 1011 m/s. This is much larger than the speed of light. (d) L =

(

2 mR 2 5

) ⎛⎜⎝ υR ⎞⎟⎠ = 2m5υR and R=

5L 5(5.27 × 10−35 kg ⋅ m 2 /s) = = 4.82 × 10−13 m 2mυ 2(9.11 × 10−31 kg)(3.00 × 108 m/s)

The diameter would be 2 R = 9.6 × 10−13 m. This would mean the diameter of the electron would be about 1000 times greater than the diameter of the nucleus; not plausible. (e) No, the hollow shell has I = 23 mR 2 , only 53 times larger than for a solid sphere. The results in (c) and (d) are bizarre by orders of magnitude.

29.31. Set Up: An electron has spin angular momentum along the z direction equal to S z = ± U /2 and total orbital angular momentum L = l (l + 1) U. For a d electron, l = 2. Solve:

S z ± U /2 1 = =± = ±20.4, L 6U 2 6

29.32. Set Up: The average translational kinetic energy is related to the Kelvin temperature T by K av = 32 kT . k = 1.38 × 10−23 J/K. 4.48 eV = 7.17 × 10−19 J.


29-8

Solve:

Chapter 29

3 kT 2

= 7.17 × 10−19 J. T =

2(7.17 × 10−19 J) 3(1.38 × 10−23 J/K)

= 3.46 × 104 K.

*29.33. Set Up: The energy gap is the energy of the maximum-wavelength photon. The energy difference is equal to the energy of the photon, so ΔE = hc /λ . Solve:

(a) Using the photon wavelength to find the energy difference gives

ΔE = hc /λ = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)/(1.11 × 10−6 m) = 1.12 eV (b) A wavelength of 1.11 µm = 1110 nm is in the infrared, shorter than that of visible light. Reflect: Since visible photons have more than enough energy to excite electrons from the valence to the conduction band, visible light will be absorbed, which makes silicon opaque.

29.34. Set Up: The electron configurations for He, Ne, and Ar are given in Table 29.3. Solve: Each of these atoms has a filled outer shell so they do not readily accept or donate electrons to other atoms in order to form a chemical bond.

29.35. Set Up: The electron configurations for H and O are given in Table 29.3. Solve: H has one electron in an unfilled shell and O needs two more electrons to have a filled outer shell, so the compound is H 2 O. Reflect: The most energetically favored electron configurations are those for which each atom has a filled outer shell.

29.36. Set Up: The electron configurations for Si and O are given in Table 29.3. Solve: Si has two electrons outside a filled shell (two electrons in the 3s subshell and two in the 3p subshell) and O needs two more electrons to have a filled outer shell, so each atom can achieve filled shells by combining in a one-toone ratio to form SiO.

*29.37. Set Up: The total energy determines what shell the electron is in, which limits its angular momentum. The electron’s orbital angular momentum is given by L = l (l + 1) U (where U =

h ), and its total energy in the nth shell 2π

is En = −(13.6 eV)/n 2 . Solve:

(a) First find n: En = −(13.6 eV)/n 2 = −0.5440 eV which gives n = 5, so l = 4, 3, 2, 1, 0. Therefore the

possible values of L are given by L = l (l + 1) U , giving L = 0,

2U,

6U,

12 U ,

20 U.

2

(b) E6 = – (13.6 eV)/ 6 = –0.3778 eV. ΔE = E6 − E5 = –0.3778 eV – (–0.5440 eV) = +0.1662 eV

This must be the energy of the photon, so ΔE = hc/λ, which gives λ = hc/ΔE = (4.136 × 10−15 eV ⋅ s) (3.00 × 108 m/s)/(0.1662 eV) = 7.47 × 10−6 m = 7470 nm, which is in the infrared and hence not visible. Reflect: The electron can have any of the five possible values for its angular momentum, but it cannot have any others.

29.38. Set Up: The average translational kinetic energy is K = 32 kT , with k = 1.38 × 10223 J/K = 8.62 × 1025 eV/K Solve: (a) K = 0.50 eV so

T=

2K 2(0.50 eV) = = 3.87 × 103 K 3k 3(8.62 × 10−5 eV/K)

(b) At T = 3.87 × 103 K and 1 atm of pressure the water would be gas; this temperature is well above the normal

boiling point of water. At ordinary temperatures then the molecules don’t have enough kinetic energy to break this bond. Van der Waals forces play an important role in the behavior of water. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


29-9

*29.39. Set Up: The shell that the electron is in limits the possibilities for its orbital angular momentum. For an electron in the n shell, its orbital angular momentum quantum number l is limited by 0 ≤ l < n, and its orbital angular h momentum is given by L = l (l + 1) U , where U = . The z-component of its angular momentum is Lz = ml U , 2π where m1 = 0, ±1, … , ±l, and its spin angular momentum is S = 3/4U for all electrons. Its energy in the nth shell is En = − (13.6 eV)/n 2 . Solve:

(a) L = l (l + 1) U = 12 U ⇒ l = 3. Therefore the smallest that n can be is 4, so En = – (13.6 eV)/ n 2 =

– (13.6 eV)/ 42 = –0.8500 eV. (b) For l = 3, m1 = ±3, ±2, ±1, 0. Since Lz = ml U , the largest Lz can be is 3 U and the smallest it can be is –3 U. (c) S = 3/4 U for all electrons. (d) In this case, n = 3, so l = 2, 1, 0. Therefore the maximum that L can be is Lmax = 2(2 + 1) U = 6 U. The minimum

L can be is zero when l = 0. Reflect: At the quantum level, electrons in atoms can have only certain allowed values of their angular momentum.

29.40. Set Up: For the 5 f state, l = 3. Lz = ml U. ml = 0, ±1, … ± l . L = l (l + 1) U. Solve: (a) The largest possible ml is ml = 3. Lz = 3U. (b) Lx 2 + Ly 2 + Lz 2 = L2 . L2 = 3(4) U 2 = 12U 2 .

Lx 2 + Lx 2 = L2 − Lz 2 = 12U 2 − 9 U 2 = 3 U.


30

NUCLEAR AND HIGH -ENERGY PHYSICS

Answers to Multiple-Choice Problems 1. A 2. C 3. A, C, D 14. B 15. A

4. C, D

5. C

6. C

7. A, D

8. B

9. D

10. E

11. B

12. A 13. D

Solutions to Problems 30.1. Set Up: In the symbol

A Z El,

Z is the number of protons and A is the number of nucleons. The number of

neutrons is N = A − Z . Solve: (a) Z = 10. N = 21 − 10 = 11. There are 10 protons and 11 neutrons. (b) Z = 30. N = 65 − 30 = 35. There are 30 protons and 35 neutrons. (c) Z = 47. N = 108 − 47 = 61. There are 47 protons and 61 neutrons.

30.2. Set Up: We know that R = R0 A1/3 , with R0 = 1.2 × 10−15 m; V = 43 π R3 ; and ρ = M/V . Ignoring the binding energy and the slight difference between the mass of the neutron and proton we have M ≈ Amp , where

mp = 1.67 × 10−27 kg. Solve: (a) For

197 79 Au

we have A = 197, so that R = (1.2 × 10−15 m)(197)1/3 = 7.0 × 10−15 m

(b) The corresponding volume for this radius is V = 43 π (7.0 × 10−15 m)3 = 1.4 × 10−42 m3. (c) The corresponding nuclear density is ρ =

M 197(1.67 × 10−27 kg) = = 2.3 × 1017 kg/m3. V 1.4 × 10−42 m3

Reflect: An alternate calculation shows that the density of the nucleus is roughly the same for all elements: 3mp Amp Amp 3(1.67 × 10−27 kg) ρ= = = = = 2.3 × 1017 kg/m3. 3 −15 3 4 π R3 4πR 3 A π π . × 4 R 4 (1 2 10 m) 0 0 3 3

*30.3. Set Up: The volume of a sphere is V = 43 πR3 . The empirical formula for the radius is R = R0 A1/3 where R0 = 1.2 × 10−15 m. Density is ρ =

m . mp = 1.67 × 10−27 kg V

Solve: (a) V = 43 π R3 = 43 π R03 A, so V is proportional to A. (b) Neutrons and protons have very similar masses and A is the total number of neutrons and protons in the nucleus. mp A 3mp 3(1.67 × 10−27 kg) , (c) ρ = = which is independent of A. ρ = = 2.3 × 1017 kg/m3 3 −15 3 4 π R 3A 4 π R π 4 (1 . 2 × 10 m) 0 0 3 © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30-1

30-2

Chapter 30

This is 2.0 × 1013 times the density of lead and is similar to the density of a neutron star. Reflect: The mass of an atom is close to the mass of the nucleus but the volume of the nucleus is much less than the volume of the atom, so the average density of the atom is much less than the density of the nucleus. Neutron stars are thought to be composed of nuclear matter; they are essentially one very large nucleus.

30.4. Set Up:

14

N has 7 protons, 7 neutrons, and 7 electrons in the neutral atom. Table 30.2 gives the atomic mass to be M = 14.003074 u. 1 u is equivalent to 931.5 MeV. The mass defect is ΔM = ZmH + Nmn − M .

Solve: (a) ΔM = 7(1.007825 u) + 7(1.008665 u) − 14.003074 u = 0.1124 u (b) EB = (0.1124 u)(931.5 MeV/u) = 105 MeV (c) EB /A =

105 MeV = 7.5 MeV/nucleon 14 nucleons

*30.5. Set Up: The constituents of the 12 H neutral atom are one 11 H atom and one neutron. Table 30.2 gives the atomic mass to be M neutrons. M

( H) = 2.014101 u. The constituents of the 2 1

4 2 He

neutral atom are two 11 H atoms and two

( He) = 4.002603 u. 1 u is equivalent to 931.5 MeV. The mass defect is ΔM = Zm 4 2

H

+ Nmn − M .

Solve: (a) ΔM = mH + mn − M = 1.007825 u + 1.008665 u − 2.014101 u = 0.002389 u.

EB = (0.002389 u)(931.5 MeV/u) = 2.23 MeV. E B /A =

2.23 MeV = 1.11 MeV/nucleon. 2 nucleons

(b) ΔM = 2mH + 2mn − M = 2(1.007825 u + 1.008665 u) − 4.002603 u = 0.03038 u.

EB = (0.03038 u)(931.5 MeV/u) = 28.3 MeV. E B /A =

28.3 MeV = 7.08 MeV/nucleon. 4 nucleons

(c) The binding energy per nucleon is much larger for 42 He than for 12 H. Reflect: The masses in Table 30.2 are atomic masses and include the masses of the electrons that are present in the neutral atom. It is important to correctly for the electron masses in the calculation.

Fe has 28 electrons and 56 nucleons, 28 protons and 28 neutrons. 207 Pb has 82 electrons and 207 nucleons, 82 protons and 125 neutrons. m p = 1.0072765 u, mn = 1.0086649 u and me = 0.0005486 u. 1 u is

30.6. Set Up:

56

equivalent to 931.5 MeV. Solve: (a) ΔM = Zmp + Nmn − M 56

Fe: M = 55.934937 u − 26me = 55.920673 u

ΔM = (26)mp + 30(mn ) − M = 0.528463 u and the binding energy is 492.3 MeV 207

Pb: M = 206.975897 u − 82me = 206.930912 u

ΔM = (82)mp + 125( mn ) − M = 1.748874 u and the binding energy is 1629 MeV

492.3 MeV = 8.791 MeV/nucleon 56 nucleons 1629 MeV 207 Pb: binding energy per nucleon is = 7.870 MeV/nucleon 207 nucleons (c) The binding energy is the energy needed to take the nucleus completely apart. 56 Fe : 492.3 MeV; 207 Pb: 1629 MeV (b)

56

Fe : binding energy per nucleon is

(d) The binding energy per nucleon is greater for

56

Fe.


Nuclear and High-Energy Physics

30-3

30.7. Set Up: The text calculates the binding energy of the deuteron to be 2.23 MeV. A photon that breaks the deuteron up into a proton and a neutron must have at least this much energy. The energy of a photon is given hc hc by E = so we have λ = . E λ (4.14 × 10−15 eV ⋅ s)(3.00 × 108 m/s)

= 5.57 × 10−13 m = 0.557 pm. 2.23 × 106 eV Reflect: This photon has gamma-ray wavelength.

Solve: λ =

30.8. Set Up: Conservation of energy tells us that the initial energy (photon plus deuteron) is equal to the energy after the split (kinetic energy plus energy of the proton and neutron). Therefore the kinetic energy released is equal to the energy of the photon minus the binding energy of the deuteron. The binding energy of a deuteron is 2.23 MeV and the energy of the photon is E = hc /λ . Kinetic energy is K = 12 mv 2 . Solve: (a) The energy of the photon is hc

Eph =

λ

=

(6.626 × 10−34 J ⋅ s)(3.00 × 108 m/s) 3.50 × 10213 m

= 5.68 × 10−13 J.

The binding of the deuteron is EB = 2.23 MeV = 3.57 × 10−13 J. Therefore the kinetic energy is K = (5.68 − 3.57) × 10−13 J = 2.11 × 10−13 J = 1.32 MeV. (b) The particles share the energy equally, so each gets half. Solving the kinetic energy for v gives

v=

2K 2(2.11 × 10−13 J)/2 = = 1.13 × 107 m/s. m 1.6605 × 10−27 kg

Reflect: Considerable energy has been released, because the particle speeds are in the vicinity of the speed of light.

*30.9. Set Up: In α -decay the nucleon number decreases by 4 and the atomic number decreases by 2. The atomic number of Po is 84. An α -particle is the nucleus of

4 2 He,

and the atomic mass of

4 2 He

is 4.00260 u. 1 u is

equivalent to 931.5 MeV. 4 214 Solve: (a) 218 84 Po → 2 He + 82 Pb. The daughter nucleus is lead with A = 214, Z = 82, and N = 132. (b) The mass change is

m

(

218 84 Po

) − ⎡⎣m (

214 82Pb

) + m ( He)⎤⎦ = 218.008973 u − 213.999805 u − 4.00260 u = 0.00657 u 4 2

84 electrons are included on each side of the decay equation so their masses cancel. The energy released is 6.12 MeV. This is the kinetic energy of the α { particle, if recoil of the daughter nucleus is ignored. Reflect: For a radioactive decay to be energetically possible the mass of the products must be less than the mass of the parent.

30.10. Set Up: In beta decay an electron, e− , is emitted by the nucleus. The beta decay reaction is 3 − 1H → e

+ 32He. If neutral atom masses are used, 13 H includes one electron and

electron mass cancels and the other electron mass in

3 2 He

3 2 He

includes two electrons. One

represents the emitted electron. Or, we can subtract the

electron masses and use the nuclear masses. The atomic mass of 32 He is 3.016029 u. Solve: (a) The mass of the

3 1H

nucleus is 3.016049 u − 0.000549 u = 3.015500 u. The mass of the

is 3.016029 u − 2(0.000549 u) = 3.014931 u. The nuclear mass of

3 2 He

3 2 He

nucleus

plus the mass of the emitted electron

is 3.014931 u + 0.000549 u = 3.015480 u. This is slightly less than the nuclear mass for

3 1 H,

so the decay is

energetically allowed.


30-4

Chapter 30

(b) The mass decrease in the decay is 3.015500 u − 3.015480 u = 2.0 × 1025 u. Note that this can also be calculated

( ) ( He) , where atomic masses are used. The energy released is

as m 13 H − m

4 2

(2.0 × 10−5 u)(931.5 MeV/u) = 0.019 MeV. The total kinetic energy of the decay products is 0.019 MeV, or 19 keV.

*30.11. Set Up: In the α { decay the nucleon number decreases by 4 and the atomic number decreases by 2. In β 2 decay the nucleon number doesn’t change and the atomic number increases by 1. The atomic number for each nucleus is identified by the chemical symbol and can be found using the periodic table in Appendix D. 4 228 228 Solve: (a) 232 90 Th → 2 He + 88 Ra; X 1 is 88Ra; A = 228, Z = 88, N = 140 228 0 228 228 89 Ac → 21e + 90 Th; X 2 is 89 Ac; A = 228, Z = 89, N = 139 228 4 224 90 Th → 2 He + 88 Ra; X 3 is α { particle; A = 4, Z = 2, N = 2 220 4 216 216 86 Rn → 2 He + 84 Po; X 4 is 84 Po; A = 216, Z = 84, N = 132 216 4 212 216 84 Po → 2 He + 82 Pb; X 5 is 84 Po; A = 216, Z = 84, N = 132 212 0 212 − 82 Pb → 21e + 83 Bi; X 6 is β ; A = 0, Z = −1, N = 0

(b)

α 4 212 → 2 He 83 Bi ⎯⎯

+ 208 81Tl and then

0 β− 208 →21 e 81 Tl ⎯⎯⎯

0 β− 212 → −1 e 83 Bi ⎯⎯⎯

The end product in each case is

+ 208 82 Pb or

+ 212 84 Po and then

4 α 212 → 2 He 84 Po ⎯⎯

+ 208 82Pb

208 82 Pb.

(c) The Segre chart is given in the figure below.

Reflect: Each decay conserves charge and nucleon number.



30.12. Set Up: The mass of a single nucleus is 105mp = 1.75 × 10225 kg. 0 ΔN /Δt 0 = λ N and T1/2 = Solve: N =

8.50 × 1023 kg 225

1.75 × 10

kg

= 4.86 × 1022 ; λ =

T1/2 =

*30.13. Set Up: T1/2 = Solve: (a) λ =

ln 2

λ

ln 2

λ

=

ln 2

λ

30-5

.

0 ΔN /Δt 0 6.24 × 1011 Bq = = 1.28 × 10−11 s −1 N 4.86 × 1022 ln 2

1.28 × 10−11 s −1

= 5.42 × 1010 s = 1720 yr

After one half-life the activity decreases by a factor of

1 . 2

ln 2 ln 2 = = 2.31 × 10−3 s −1 T1/ 2 300.0 s

(b) (i) After 5.00 min the decay rate is (6.00 × 103 Bq)/2 = 3.00 × 103 Bq

(ii) After an additional 5.00 min the decay rate is (3.00 × 103 Bq)/2 = 1.50 × 103 Bq (iii) 25.0 min is three half-lives after 10.0 min so the decay rate is (1.50 × 103 Bq)/(23 ) = 188 Bq

30.14. Set Up: The activity is T1/ 2 = 76.0 min = 4560 s. λ = Solve: (a) λ =

ΔN = λN. Δt

In one

T1/2

the number of radioactive nuclei halves.

ln 2 . T1/2

ΔN /Δt 0 16.0 × 1010 Bq ln 2 = = 1.05 × 1015 nuclei. = 1.52 × 10−4 s −1. N 0 = λ 4560 s 1.52 × 10−4 s −1

(b) 76.0 min is one T1/2 , so N = N 0 /2 = 5.25 × 1014.

ΔN is proportional to N, so the activity is Δt

(16.0 × 1010 Bq)/2 = 8.0 × 1010 Bq.

(c) 152 min is one T1/ 2 after the time in part (b), so N = (5.25 × 1014 )/2 = 2.62 × 1014 and the activity is

(8.0 × 1010 )/2 = 4.0 × 1010 Bq. ln 2

The mass of a single nucleus is 45mp = 7.52 × 10−26 kg. ΔN /Δt = − λ N λ 2.24 × 10−3 kg ln 2 ln 2 Solve: N = = = 1.78 × 10−6 s −1 = 2.98 × 1022 ; λ = −26 T1/2 (4.5 days)(24 h/day)(3600 s/h) 7.52 × 10 kg

30.15. Set Up: T1/2 =

ΔN = − λ N = − (1.78 × 10−6 s −1)(2.98 × 1022 ) = −5.30 × 1016 Bq Δt Reflect: The shorter the half-life the greater the decay rate for the same number of radioactive nuclei.

30.16. Set Up: According to Example 30.4 the half-life of λ=

14

C is 5730 years, which gives a decay constant of

ln 2 = 3.833 × 10−12 s −1 (not λ = 3.38 × 10−12 s −1 as reported in the example). As explained in the hint, the T1/2

decay rate, A =

ΔN , obeys an equation similar to Equation 30.6: A = A0e − λt . Note that A0 = 180.0 decays/min. Δt

From the appendix we find that 1 y = 3.156 × 107 s. Solve:

(a) We can now find the decay rate at t = 1000 y = 3.156 × 1010 s:

A = A0e − λt = (180.0 decays/min)e2[(3.833×10

−12 s −1 )(3.156 ×1010 s)]

= 159 decays/min


30-6

Chapter 30

(b) The decay rate at t = 50,000 y = 1.578 × 1012 s is

A = A0e− λt = (180.0 decays/min)e2[(3.833×10

−12 s −1 )(1.578 ×1012 s)]

= 0.43 decays/min.

Reflect: The time in part (b) is 8.73 half-lives, so the decay rate has decreased by a factor or

( 12 )

8.73

from the initial

value of A0 = 180.0 decays/min. Also, the mass of the sample is not relevant in the solution of this problem (since it is primarily

12

14

C and not

C ).

*30.17. Set Up: The decay rate decreases by a factor of 2 in a time of one half-life.

( )

Solve: (a) 24 d is 3T1/ 2 so the activity is (375 Bq)/ 23 = 46.9 Bq (b) The activity is proportional to the number of radioactive nuclei, so the percent is (c)

131 53I

→ 201e + 131 54 Xe. The nucleus

131 54 Xe

17.0 Bq = 36.2, 46.9 Bq

is produced.

Reflect: Both the activity and the number of radioactive nuclei present decrease by a factor of 2 in one half-life.

30.18. Set Up: λ =

ln 2 ΔN . T1/2 = 4.47 × 109 yr = 1.41 × 1017 s. The activity is = λ N . The mass of one T1/2 Δt

238

U is

approximately 238mp . 1 Ci = 3.70 × 1010 decays/s. Solve: (a) λ = (b) N =

ln 2 1.41 × 1017 s

0 ΔN /Δt 0

=

λ

= 4.92 × 10−18 s −1

3.70 × 1010 Bq 4.92 × 10−18 s −1

= 7.52 × 1027 nuclei. The mass m of uranium is the number of nuclei times the

mass of each one. m = (7.52 × 1027 )(238)(1.67 × 10−27 kg) = 2.99 × 103 kg. (c) N =

10.0 × 10−3 kg 10.0 × 10−3 kg = = 2.52 × 1022 nuclei −27 238mp 238(1.67 × 10 kg)

0

ΔN 0 = λ N = (4.92 × 10−18 s −1 )(2.52 × 1022 ) = 1.24 × 105 decays/s Δt

*30.19. Set Up: Let A denote the activity of the sample as a function of time: According to Equation 30.5 we have A=

ΔN = λ N , which says that the activity is proportional to N. Since N decreases according to equation 30.6, so Δt

does the activity: A = A0e2λt . Note that A0 = 7.56 × 1011 Bq. Solve: (a) After 92.4 minutes the activity has decreased to A = 9.45 × 1010 Bq. The decay constant can be found

from

⎛ 7.56 × 1011 Bq ⎞ 1 ⎛ A0 ⎞ 1 −1 ln ⎜ ⎟ = 0.0225 (min) . ⎟= t ⎝ A ⎠ 92.4 min ⎝⎜ 9.45 × 1010 Bq ⎠⎟

λ = ln ⎜ ln 2

0.0225 (min) −1

Thus,

the

half-life

is

T1/2 =

ln 2

λ

=

= 30.8 min.

(b) We can determine the number of radioactive nuclei initially present from the initial activity: A0 = λ N 0 , so we

have N 0 =

A0

λ

=

7.56 × 1011 Bq −1

(0.0225 (min) )(1 min/60 s)

= 2.02 × 1015 radioactive nuclei.

Reflect: Note that since 1 Bq = 1 decay/s we must convert λ from (min) −1 to s21 in our answer to part (b).



30.20. Set Up: The activity of carbon in the atmosphere is 0.255 Bq per gram of carbon. The half-life of

30-7 14

C is

5730 yr. 0.877 Bq 1 = 14.0 Bq 16

Solve: The activity of a present-day sample with 55.0 g of carbon is (55.0 g)(0.255 Bq/g) = 14.0 Bq. 1 16

and

=

( 12 )

4

so the age of the sample is 4T1/ 2 = 2.29 × 104 yr.

30.21. Set Up: One-half of the sample decays in a time of T1/ 2 . Solve: (a) (b)

( 12 )

10 × 109 yr = 5.0 × 104 200,000 yr

5.0 ×104

This exponent is too large for most hand-held calculators. But

( 12 )

5.0 ×104

( 12 ) = 10−0.301 so

4

= (10−0.301 )5.0×10 = 10−15,000

30.22. Set Up: Let A denote the activity of the sample as a function of time: According to Equation 30.5 we have A=

ΔN = λ N , which says that the activity is proportional to N. Since N decreases according to Equation 30.6, so Δt

does the activity: A = A0e− λt . Note that A0 = 5000 Ci. Solve: First we calculate the decay constant from the half-life: λ =

ln 2 ln 2 = = 0.1315 y−1. Note that the time T1/2 5.271 y

period is roughly 2.5 years, so the expected activity is A = A0e− λt = (5000 Ci)e− (0.1315 y

−1 )(2.5 y)

= 3600 Ci. The

source is barely usable. Reflect: We can use an alternate form of Equation 30.6 to solve this problem: setting λ =

gives the alternate form N = N 0 2

−t / T1/ 2

ln 2 in Equation 30.6 T1/2

.

*30.23. Set Up: 1 Gy = 1 J/kg and is the SI unit of absorbed dose. 1 rad = 0.010 Gy. Sv is the SI unit for equivalent dose. Equivalent dose = RBE × absorbed dose. Rem is the equivalent dose when the absorbed dose is in rad. For x-rays, RBE = 1.0. For protons, RBE = 10. Solve: (a) 5.0 Gy, 500 rad. RBE = 1.0 so equivalent dose = absorbed dose. 5.0 Sv and 500 rem. (b) (70.0 kg)(5.0 J/kg) = 350 J (c) The absorbed dose and total absorbed energy are the same but the equivalent dose is 10 times larger. So the answers are: 5.0 Gy, 500 rad, 50 Sv, 5000 rem, 350 J. Reflect: The same energy deposited by protons as x-rays is ten times greater in its biological effect.

30.24. Set Up: rem is equivalent dose corresponding to an absorbed dose in rad. For α { particles, RBE = 20. 125 rem = 6.25 rad 20 ⎛ 0.01 J/kg ⎞ (b) (6.25 rad) ⎜ = 0.0625 J/kg and (5.0 × 10−3 kg)(0.0625 J/kg) = 3.1 × 10−4 J ⎝ 1 rad ⎟⎠ Solve: (a)

(c)

125 rem = 100 rad 1.25

*30.25. Set Up: We have the following conversions: 1 Gy = 1 J/kg and 1 rad = 10−2 Gy. Also, we have dose in rem = (RBE)(absorbed dose in rad). The RBE for slow neutrons is given to be 4.0 in this case.


30-8

Chapter 30

Solve: (5.0 Gy)(1 rad/10−2 Gy) = 500 rad, dose in rem = (RBE)(absorbed dose in rad) = 4.0(500 rad) = 2000 rem, and (5.0 Gy)(1 J/kg)/Gy = 5.0 J/kg. Reflect: The RBE is an important factor in determining the effect of a given type of radiation.

30.26. Set Up: equivalent dose (Sv, rem) = RBE × absorbed dose (Gy, rad); 100 rad = 1 Gy Solve: (a) RBE = 1, so 0.25 mSv corresponds to 0.25 mGy. Energy = (0.25 × 10 −3 J/kg)/(5.0 kg) = 1.2 × 10−3 J (b) RBE = 1 so 0.10 mGy = 10 mrad and 10 mrem

(0.10 × 10−3 J/kg)(75 kg) = 7.5 × 10−3 J (c)

7.5 × 1023 J 1.2 × 1023 J

= 6.2

*30.27. Set Up: For x rays RBE = 1 so the equivalent dose in Sv is the same as the absorbed dose in J/kg. Solve: One whole-body scan delivers (75 kg)(12 × 10−3 J/kg) = 0.90 J. One chest x ray delivers

(5.0 kg)(0.20 × 10−3 J/kg) = 1.0 × 10−3 J. It takes

0.90 J 1.0 × 1023 J

= 900 chest x rays to deliver the same total energy.

30.28. Set Up: We have that 1 rem = 0.01 Sv and dose in rem = (RBE)(absorbed dose in rad). The RBE for x rays is equal to 1 (Table 30.3). From Chapter 14 we have Equation 14.10: Q = mcΔT , with c = 4190 J/kg ⋅ K for water. Solve: (a) 5.4 Sv (100 rem/Sv) = 540 rem. (b) The RBE of 1 gives an absorbed dose of 540 rad. (c) The absorbed dose is 5.4 Gy, so the total energy absorbed is (5.4 Gy) (65 kg) = 351 J. The energy required to raise the temperature of 65 kg by 0.010° C is (65 kg) (4190 J/kg ⋅ K) (0.01C°) = 3 kJ. Reflect: The overall heating caused by this lethal dose of radiation is negligible.

30.29. Set Up: For x rays RBE = 1 and the equivalent dose equals the absorbed dose. Solve: (a) 175 krad = 175 krem = 1.75 kGy = 1.75 kSv. (1.75 × 103 J/kg)(0.150 kg) = 2.62 × 102 J. (b) 175 krad = 1.75 kGy; (1.50)(175 krad) = 262 krem = 2.62 kSv

The energy deposited would be 2.62 × 102 J, the same as in (a). Reflect: The energy required to raise the temperature of 0.150 kg of water 1 C° is 628 J, and 2.62 × 102 J is less

than this. The energy deposited corresponds to a very small amount of heating.

30.30. Set Up: equivalent dose (rem) = RBE × absorbed does (rad). 1 rad = 0.010 J/kg. For x rays, RBE = 1. Each photon has energy E = Solve: (a) E =

hc

λ

=

hc

λ

.

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) 0.0200 × 10−9 m

= 9.94 × 10−15 J. The absorbed energy is

(5.00 × 1010 photons)(9.94 × 10−15 J/photon) = 4.97 × 10−4 J = 0.497 mJ. (b) The absorbed dose is

4.97 × 10−4 J = 8.28 × 10−4 J/kg = 0.0828 rad. Since RBE = 1, the equivalent dose is 0.600 kg

0.0828 rem.



30-9

*30.31. Set Up: Since this source has a long lifetime its activity will be nearly constant: number of decays = (activity)(time). The absorbed dose is measured in grays, where 1 Gy = 1 J/kg = 100 rad. Also we have: dose in rem = (RBE) (absorbed does in rad), where RBE = 20 for alpha particles. Finally, we know that 1 Ci = 3.70 × 1010 Bq, 1 y = 3.156 × 107 s, and 1 eV = 1.602 × 10−19 J. Solve: The number of alpha particles generated in 1 year is (0.72 × 10−6 Ci) (3.7 × 1010 Bq/Ci)(3.156 × 107 s) =

8.41 × 1011. The absorbed dose is

(8.41 × 1011) (4.0 × 106 eV) (1.602 × 10−19 J/eV) = 1.08 Gy = 108 rad. The equivalent (0.50 kg)

dose is (20)(108 rad) = 2160 rem. Reflect: The equivalent does is relatively high due to the large RBE for alpha particles.

30.32. Set Up: The amount of kinetic energy released is the energy equivalent of the mass change in the decay. me = 0.0005486 u and the atomic mass of

14 7N

is 14.003074 u. The energy equivalent of 1 u is 931.5 MeV.

14

C has

a half-life of T1/2 = 5730 yr = 1.81 × 1011 s. The RBE for an electron is 1.0. Solve: (a)

14 6C

→ e2 + 147 N + υe

(b) The mass decrease is ΔM = m

( C) − ⎡⎣m + m ( N)⎤⎦ . 14 6

e

14 7

Use nuclear masses, to avoid difficulty in ing for atomic electrons. The nuclear mass of 14.003242 u − 6me = 13.999950 u. The nuclear mass of

14 7N

14 6C

is

is 14.003074 u − 7 me = 13.999234 u.

ΔM = 13.999950 u − 13.999234 u − 0.000549 u = 1.67 × 1024 u. The energy equivalent of ΔM is 0.156 MeV. (c) The mass of carbon is (0.18)(75 kg) = 13.5 kg. From Example 30.4, the activity due to 1 g of carbon in a living organism is 0.255 Bq. The number of decay/s due to 13.5 kg of carbon is (13.5 × 103 g)(0.255 Bq/g) = 3.4 × 103 decays/s. (d) Each decay releases 0.156 MeV so 3.4 × 103 decays/s releases 530 MeV/s = 8.5 × 10211 J/s. (e) The total energy absorbed in 1 yr is (8.5 × 10−11 J/s)(3.156 × 107 s) = 2.7 × 10−3 J. The absorbed dose is

2.7 × 10−3 J = 3.6 × 10−5 J/kg = 36 μGy = 3.6 mrad. With RBE = 1.0, the equivalent dose is 36 μ Sv = 3.6 mrem. 75 kg

*30.33. Set Up: The reaction energy is Q = ( M A + M B − M C − M D )c 2 , where A = 42 He, B = 37 Li, C = 105B and D = 10n. If neutral atom masses from Table 30.2 are used, the electron masses cancel. 1 u is equivalent to 931.5 MeV. If Q > 0, energy is liberated in the reaction. If Q < 0, energy is absorbed in the reaction.

Solve: M A + M B − M C − M D = 4.002603 u + 7.016005 u − 10.012937 u − 1.008665 u = 22.994 × 1023 u

Q = (−2.994 × 10−3 u)(931.5 MeV/u) = −2.79 MeV. The reaction absorbs 2.79 MeV.

30.34. Set Up: (a) Determine X by balancing the charge and nucleon number on the two sides of the reaction equation. Solve: X must have A = 2 + 14 − 10 = 6 and Z = 1 + 7 − 5 = 3. Thus X is 36 Li and the reaction is 12 H + 147N → 36 Li + 105 B (b) Set Up:

Calculate the mass decrease and find its energy equivalent.


30-10

Chapter 30

Solve: The neutral atoms on each side of the reaction equation have a total of 8 electrons, so the electron masses cancel when neutral atom masses are used. The neutral atom masses are found in Table 30.2.

mass of 12 H + 147 N is 2.014102 u + 14.003074 u = 16.017176 u mass of 36 Li + 105B is 6.015121 u + 10.012937 u = 16.028058 u The mass increases, so energy is absorbed by the reaction. The Q value is (16.017176 u − 16.028058 u) (931.5 MeV/u) = −10.14 MeV Reflect: Since Q < 0 the reaction is endoergic and it cannot occur unless the reactants have sufficient kinetic energy.

30.35. Set Up: Assume the separation of the proton and proton. The

17

O is the radius of the

17

O nucleus has charge +8e. Treat the proton and

17

O nucleus plus the radius of the

17

O nucleus as point charges when calculating the

electrical potential energy of the two particles. To reach the surface of the 17O nucleus the proton must have initial kinetic energy equal to the electrical potential energy when its center is a distance R from the center of the nucleus, where R is the sum of radius of the nucleus and the radius of the proton. Solve: R = (1.2 × 10−15 m)(17)1/3 + 1.2 × 10−15 m = 4.3 × 10−15 m

U=

kq1 q2 (8.99 × 109 N ⋅ m 2 /C 2 )(1)(8)(1.60 × 10−19 C) 2 = = 4.3 × 10−13 J = 2.7 MeV −15 R 4.3 × 10 m

Reflect: Our analysis assumes that the charge distribution within the 17O nucleus is spherically symmetric so outside the nucleus its potential is the same as if all its charge were concentrated at its center. We have also assumed that the 17O remains at rest; it must in fact recoil slightly in order to conserve linear momentum.

30.36. Set Up: The number of fissions per second can be found from the power output of the reactor (100 MW) and the energy released by the fission of one atom (200 MeV). We know that 1 eV = 1.60 × 10−19 J. 1 eV ⎛ 1 fission ⎞ ⎛ ⎞ Solve: (1.00 × 108 J/s) ⎜ = 3.13 × 1018 fissions/s. ⎟ ⎜ 8 ⎝ 2.00 × 10 eV ⎠ ⎝ 1.60 × 10−19 J ⎟⎠ Reflect: Note that the number of fissions per second is relatively small compared to Avogadro’s number. This means that the reactor does not consume a large amount of material to generate power at this level.

*30.37. Set Up: 0.7% of naturally occurring uranium is the isotope

235

U. The mass of one

235

U nucleus is about

235mp . 1.0 × 1019 J

Solve: (a) The number of fissions needed is

(200 × 106 eV)(1.60 × 10−19 J/eV)

= 3.13 × 1029. The mass of

235

U

required is (3.13 × 1029 )(235mp ) = 1.23 × 105 kg. 1.23 × 105 kg

= 1.76 × 107 kg 0.7 × 1022 Reflect: The calculation assumes 100% conversion of fission energy to electrical energy. (b)

30.38. Set Up: The energy released is the energy equivalent of the mass decrease. 1 u is equivalent to 931.5 MeV. The mass of one Solve:

(a)

235 92 U

235

U nucleus is 235mp .

+ 10n

89 1 → 144 56 Ba + 36 Kr + 30n

We can use atomic masses since the same number of electrons are included on each side of the reaction equation and the electron masses cancel. The mass decrease is 1 89 1 ⎤ ⎡ 144 ΔM = m 235 92 U + m 0n − ⎣ m 56Ba + m 36 Kr + 3m 0 n ⎦ ΔM = 235.043930 u + 1.0086649 u − 143.922953 u − 88.917630 u − 3(1.0086649 u )

(

) ( )

(

) (

)

( )

ΔM = 0.1860 u. The energy released is (0.1860 u)(931.5 MeV/u) = 173.3 MeV. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


(b) The number of

235

U nuclei in 1.00 g is

30-11

1.00 × 10−3 kg = 2.55 × 1021. The energy released per gram is 235mp

(173.3 MeV/nucleus)(2.55 × 1021 nuclei/g) = 4.42 × 1023 MeV/g.

30.39. Set Up: Q = ( M A + M B − M C − M D )c 2 , where A = 12H, B = 13H, C = 42He and D = 10n. If neutral atom masses from Table 30.2 are used, the electron masses cancel. Solve: M A + M B − M C − M D = 2.014101 u + 3.016049 u − 4.002603 u − 1.008665 u = 1.888 × 10−2 u

The energy

released is (1.888 × 10−2 u)(931.5 Mev/u) = 17.6 MeV.

30.40. Set Up: Q = ( M A + M B − M C − M D )c 2 , where A = 12H, B = 12H, C = 32He and D = 01n. If neutral atom masses from Table 30.2 are used, the electron masses cancel. 1 mole of deuterium gas contains 6.02 × 1023 molecules and hence 2(6.02 × 1023 ) nuclei. Each reaction requires two deuterium nuclei, so 1 mole of gas produces 6.02 × 1023 reactions. Solve: M A + M B − M C − M D = 2(2.014101 u) − 3.016029 u − 1.008665 u = 3.508 × 10−3 u. The energy liberated is (3.508 × 10−3 u)(931.5 Mev/u) = 3.27 MeV = 5.23 × 10−13 J.

(b) The energy liberated by 1 mole of deuterium is (6.02 × 1023 )(5.23 × 10−13 J) = 3.15 × 1011 J.

*30.41. Set Up: m = ρV . 1 gal = 3.788 L = 3.788 × 10−3 m3 . The mass of a 1 MeV = 1.60 × 10

−13

235

U nucleus is 235mp .

J

Solve: (a) For 1 gallon, m = ρV = (737 kg/m3 )(3.788 × 10−3 m3 ) = 2.79 kg = 2.79 × 103 g

1.3 × 108 J/gal 2.79 × 103 g/gal (b) 1 g contains

= 4.7 × 104 J/g

1.00 × 1023 kg = 2.55 × 1021 nuclei 235mp

(200 MeV/nucleus)(1.60 × 10−13 J/MeV)(2.55 × 1021 nuclei) = 8.2 × 1010 J/g (c) A mass of 6mp produces 26.7 MeV.

(26.7 MeV)(1.60 × 10−13 J/MeV) = 4.26 × 1014 J/kg = 4.26 × 1011 J/g 6mp (d) The total energy available would be (1.99 × 1030 kg)(4.7 × 107 J/kg) = 9.4 × 1037 J

energy 9.4 × 1037 J energy = = 2.4 × 1011 s = 7600 yr so t = power 3.92 × 1026 W t Reflect: If the mass of the sun were all proton fuel, it would contain enough fuel to last ⎛ 4.3 × 1011 J/g ⎞ (7600 yr) ⎜ = 7.0 × 1010 yr. ⎜ 4.7 × 104 J/g ⎟⎟ ⎝ ⎠ power =

30.42. Set Up: The reactions in the proton-proton fusion chain are 1 1 2 + 1H + 1H → 1 H + β + υe 2 1 3 1 H + 1H → 2 He + γ 3 3 4 1 2 He + 2He → 2He + 1H

(1) (2)

+ 11H

(3)


30-12

Chapter 30

Solve: Add the reaction equations as 2(1) + 2(2) + (3). This gives

211H + 211H + 212H + 211H + 32He + 32He → 212H + 2β + + 2υe + 232He + 2γ + 42He + 11H + 11H The two

3 2 He +

and the two

2 1H

an each side of the reaction equation cancel and we are left with 611 H → 42He +

211 H + 2β + 2υe + 2γ . Note that both nucleon number and charge are conserved in the overall reaction.

*30.43. Set Up: Apply conservation of linear momentum to the collision. A photon has momentum p = h /λ , in the direction it is traveling. The energy of a photon is E = pc =

hc

λ

. All the mass of the electron and positron is

converted to the total energy of the two photons, according to E = mc 2 . The mass of an electron and of a positron is me = 9.11 × 10−31 kg.

Solve: (a) In the lab frame the initial momentum of the system is zero, since the electron and positron have equal speeds in opposite directions. According to momentum conservation, the final momentum of the system must also be zero. A photon has momentum, so the momentum of a single photon is not zero. (b) For the two photons to have zero total momentum they must have the same magnitude of momentum and move in opposite directions. Since E = pc, equal p means equal E. (c) 2 Eph = 2me c 2 so Eph = me c 2

Eph =

hc

λ

so

hc

λ

= me c 2 and λ =

h 6.63 × 10−34 J ⋅ s = = 2.43 pm me c (9.11 × 10−31 kg)(3.00 × 108 m/s)

These are gamma ray photons. Reflect: The total charge of the electron/positron system is zero and the photons have no charge, so charge is conserved in the particle-antiparticle annihilation.

30.44. Set Up: Section 30.7 says that a charged pion decays into a muon plus a neutrino. The muon in turn decays into an electron or positron plus two neutrinos. Solve: (a) π − → μ − + neutrino → e − + three neutrinos (b) If we neglect the mass of the neutrinos, the mass decrease is m(π − ) − m(e − ) = 273me − me = 272me = 2.480 × 10−28 kg. E = mc 2 = 2.23 × 10−11 J = 139 MeV

(c) The total energy delivered to the tissue is (50.0 J/kg)(10.0 × 10−3 kg) = 0.500 J. The number of π − mesons

0.500 J = 2.24 × 1010. 2.23 × 10−11 J (d) The RBE for the electrons that are produced is 1.0, so the equivalent dose is 1.0(50.0 Gy) = 50.0 Sv = 5.0 × 103 rem.

required is

30.45. Set Up: Find the energy equivalent of the mass decrease. Solve:

The mass decrease is m(∑ + ) − m(p) − m(π 0 ) and the energy released is mc 2 (∑ + ) − mc 2 (p) − mc 2 (π 0 ) =

1189 MeV − 938.3 MeV − 135.0 MeV = 116 MeV. (The mc 2 values for each particle were taken from Table 30.5.) Reflect: The mass of the decay products is less than the mass of the original particle, so the decay is energetically allowed and energy is released.

30.46. Set Up: (a) In section 30.8 we find the mass of the π + particle is 273me and the mass of the μ + particle is 207 me . The rest mass of the electron is me = 0.511 MeV/c 2 and we assume that the neutrino has negligible rest mass.



30-13

Solve: The loss of mass during the reaction is Δm = (273 − 207)me = 66me , which corresponds to a released energy of E = 66(0.511 MeV) = 34 MeV. Reflect: (b) A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you could always find a frame where energy was not conserved. This cannot occur.

*30.47. Set Up: The mass of the proton and antiproton are both equal to 938.3 MeV according to Table 30.5. The energy of a photon is given by E = hf =

hc

λ

, where h = 4.136 × 10−15 eV ⋅ s.

Solve: (a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of f =

938.3 × 106 eV 4.136 × 10

−15

eV ⋅ s

= 2.27 × 1023 Hz and a wavelength of λ =

c 3.00 × 108 m/s = = 1.32 × 10−15 m. f 2.27 × 1023 Hz

(b) The energy of each photon will be 938.3 MeV + 830 MeV = 1768 MeV. Following the method our previous work

this gives a frequency 4.28 × 1023 Hz and wavelength 7.02 × 10216 m. Reflect: Note that the frequency is directly proportional to the released energy. Thus, we have ⎛ 1768 MeV ⎞ = 4.28 × 1023 Hz. (2.27 × 1023 Hz) ⎜ ⎝ 938.3 MeV ⎠⎟

30.48. Set Up: p and n have baryon number +1 and p has baryon number −1. e+ , e− , υe and γ all have baryon number zero. Baryon number is conserved if the total baryon number of the products equals the total baryon number of the reactants. Solve: (a) reactants: B = 1 + 1 = 2. Products: B = 1 + 0 = 1. Not conserved. (b) reactants: B = 1 + 1 = 2. Products: B = 0 + 0 = 0. Not conserved. (c) reactants: B = +1. Products: B = 1 + 0 + 0 = +1. Conserved. (d) reactants: B = 1 − 1 = 0. Products: B = 0. Conserved.

*30.49. Set Up: Section 30.9 says the strong interaction is 100 times as strong as the electromagnetic interaction and that the weak interaction is 10−9 times as strong as the strong interaction. The Coulomb force is Fe = the gravitational force is Fg = G Solve: (a) Fe =

m1 m2 r2

r2

and

.

(9.0 × 109 N ⋅ m 2 /C 2 )(1.60 × 10−19 C) 2 (1 × 10−15 m) 2

Fg =

kq1 q2

= 200 N

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.67 × 10−27 kg)2 (1 × 10−15 m)2

= 2 × 10−34 N

(b) Fstr ≈ 100 Fe ≈ 2 × 104 N. Fweak ≈ 10−9 Fstr ≈ 2 × 10−5 N (c) Fstr > Fe > Fweak > Fg (d) Fe ≈ 1 × 1036 Fg . Fstr ≈ 100 Fe ≈ 1 × 1038 Fg . Fweak ≈ 1029 Fstr ≈ 1 × 1029 Fg Reflect: The gravity force is much weaker than any of the other three forces. Gravity is important only when one very massive object is involved.

30.50. Set Up: The properties of the quarks are given in Table 30.7. An antiquark has charge and quantum numbers of opposite sign from the corresponding quark. Solve: (a) Q /e = 23 + 23 + − 13 = +1. B = 13 + 13 + 13 = 1. S = 0 + 0 + ( −1) = −1. C = 0 + 0 + 0 = 0.

( ) (b) Q /e = 23 + 13 = +1. B = 13 + (2 13 ) = 0. S = 0 + 1 = 1. C = 1 + 0 = 1.


30-14

Chapter 30

( ) ( ) ( ) 1 1 1 + ( − 3 ) = −1. B = − 3 + 3 = 0. S = 0 + 0 = 0. C = −1 + 0 = −1.

(c) Q /e = 13 + 13 + − 23 = 0. B = − 13 + − 13 + − 13 = −1. S = 0 + 0 + 0 = 0. C = 0 + 0 + 0 = 0. (d) Q /e =

− 23

*30.51. Set Up: mH = 1.67 × 10227 kg. The ideal gas law says pV = nRT . Normal pressure is 1.013 × 105 Pa and normal temperature is about 27 °C = 300 K. 1 mole is 6.02 × 1023 atoms. Solve: (a)

5.8 × 10−27 kg/m3 1.67 × 10−27 kg/atom

= 3.5 atoms/m3

(b) V = (4 m)(7 m)(3 m) = 84 m3 and (3.5 atoms/m3 )(84 m3 ) = 294 atoms (c) With p = 1.013 × 105 Pa, V = 84 m3 , T = 300 K and the ideal gas law gives the number of moles to be

n=

pV (1.013 × 105 Pa)(84 m3 ) = = 3.4 × 103 moles RT (8.3145 J/mol ⋅ K)(300 K) (3.4 × 103 moles)(6.02 × 1023 atoms/mol) = 2.0 × 1027 atoms

Reflect: The average density of the universe is very small. Interstellar space contains a very small number of atoms per cubic meter, compared to the number of atoms per cubic meter in ordinary material on the earth, such as air.

30.52. Set Up: The half-life for this sample can be estimated from the table. The decay constant is given by λ=

ln 2 . The number of radioactive nuclei present can be estimated from the decay constant and the activity: T1/2

ΔN = λN. Δt Solve (a) A good estimate for the half-life can be obtained by noting that from during the first 1.5 hours the activity drops by just more than half (from 20,000 Bq to 9,980 Bq )—and it drops by half again during the next 1.5 hours A=

(from 9,980 Bq to 4,970 Bq ). Thus we have T1/2 = 1.5 hours. ln 2 ln 2 = = 0.46 h −1. T1/2 1.5 h (c) The initial number of radioactive nuclei can be found from the initial activity: A 20,000 Bq = 1.6 × 108. N0 = 0 = λ (0.46 h −1)(1 h/3600 s) (b) The decay constant is given by λ =

(d) After 7.0 hours the activity has dropped to 790 Bq so we have N =

A

λ

=

790 Bq −1

(0.46 h )(1 h/3600 s)

= 6.2 × 106

radioactive nuclei remaining. Note that we get a slightly different value if we use N = N 0e − λt to obtain our value. Reflect: Since A = A0e − λt we have ln( A/A0 ) = −λt. Thus we can find a more accurate value for the decay constant

by making a graph of ln( A/A0 ) versus t. The slope of the best-fit line of this graph should be a good estimate of − λ .

30.53. Set Up: All of the 400 kg + 400 kg = 800 kg of matter and antimatter is converted to energy. ΔE = ( Δm ) c 2 .

Solve: ΔE = (800 kg)(3.00 × 108 m/s)2 = 7.20 × 1019 J

30.54. Set Up: equivalent dose (rem) = RBE × absorbed dose (rad). 1 rad = 0.010 J/kg. For alpha particles, Table 30.3 says RBE = 20. The activity is 0

ΔN ln 2 0 = λN. λ = . One Δt T1/2

226

Ra nucleus has a mass of approximately

226mp . T1/2 = 1600 yr = 5.05 × 1010 s.



30-15

Solve: (a) The energy absorbed is (5.00 × 1012 )(1.50 × 106 eV)(1.6 × 10−19 J/eV) = 1.2 J. The absorbed does =

1.2 J = 0.0171 J/kg = 1.71 rad. 70.0 kg (b) equivalent dose = (20)(1.71 rad) = 34.2 rem 0.0100 × 10−3 kg 1.00 × 10−5 kg = = 2.65 × 1019. 226mp 226(1.67 × 10−27 kg)

(c) N =

λ=

ln 2 ln 2 = = 1.37 × 10−11 s −1. T1/2 5.05 × 1010 s

ΔN = λ N = (1.37 × 10−11 s −1)(2.65 × 1019 ) = 3.63 × 108 decays/s. Δt (d) t =

5.00 × 1012 decays 3.63 × 108 decays/s

= 1.38 × 104 s = 230 min

*30.55. Set Up: equivalent dose (rem) = RBE × absorbed dose (rad). 1 rad = 0.010 J/kg. 1 Ci = 3.70 × 1010 decays/s.

Solve: (a) 15.0 Ci = 5.55 × 1011 decays/s. Each decay produces 1.25 MeV = 2.00 × 10213 J. The

60

Co source

−13

11

J) = 0.111 J/s. Only half of this energy is absorbed, so the energy releases (5.55 × 10 decays/s)(2.00 × 10 absorbed per second is (0.111 J)/2 = 0.0555 J.

0.0555 J = 0.111 J/kg = 11.1 rad. 0.500 kg (c) The equivalent dose per second is (0.70)(11.1 rad) = 7.77 rem. (b) The absorbed dose per second is

200 rem = 25.7 s 7.77 rem/s Reflect: The equivalent dose is less than the absorbed dose since RBE is less than 1.00. (d) t =

30.56. Set Up: In one T1/2 the number of radioactive nuclei halves. If the number at t = 0 is N 0 , the number after nT1/ 2 is

( 12 )

n

N0 .

Solve: (a) 2.0 min is one T1/2 for

N19 =

( 12 )

4

N 0 = N 0 /16.

O, so n15 = 1. 2.0 min is four T1/2 for

19

O, so n19 = 4. N15 =

( 12 ) N0 = N0 /2. 1

N15 N 0 /2 = = 8.0. N19 N 0 /16

(b) n15 = 5. n19 = 20. N15 =

30.57. Set Up:

15

( 12 )

ΔN = λN. Δt

5

N 0 . N19 =

λ=

( 12 )

20

N0 .

N15 N 0 /25 = = 215 = 3.3 × 104. N19 N 0 /220

ln 2 . T1/ 2 = 1.28 × 108 yr = 4.04 × 1015 s. A T1/2

40

K nucleus has mass 40mp .

1 Ci = 3.70 × 1010 decays/s. Solve: (a) λ =

ln 2 4.04 × 1015 s

= 1.72 × 10−16 s −1.

N=

6.00 × 10−9 kg 6.00 × 10−9 kg = = 8.98 × 1016. 40mp 40(1.67 × 10−27 kg)

ΔN = λ N = (1.72 × 10−16 s −1 )(8.98 × 1016 ) = 15.4 decays/s. Δt


30-16

Chapter 30

⎛ ⎞ 1 Ci −10 (b) (15.4 decays/s) ⎜⎜ Ci ⎟⎟ = 4.16 × 10 10 3 70 10 decays/s . × ⎝ ⎠ Reflect: The half-life is very long, so the number of decays per second is small.

30.58. Set Up: The activity 0 ΔN /Δt 0 = λ N . λ =

ln 2 . The mass of one T1/ 2

103

Pd nucleus is 103mp . In a time of one

half-life the number of radioactive nuclei and the activity decrease by a factor of 2. Solve: (a) λ =

ln 2 ln 2 = = 4.7 × 10−7 s −1 T1/2 (17 days)(24 h/day)(3600 s/h) N=

0.250 × 10−3 kg = 1.45 × 1021 103mp

0 ΔN /Δt 0 = (4.7 × 10−7 s −1 )(1.45 × 1021 ) = 6.8 × 1014 Bq (b) 68 days is 4T1/ 2 so the activity is (6.8 × 1014 Bq)/24 = 4.2 × 1013 Bq

*30.59. Set Up:

ΔN ln 2 . T1/2 = 5.3 yr = 1.67 × 108 s. A = λN. λ = Δt T1/ 2

60

Co nucleus has mass 60mp . 1 Ci =

3.70 × 1010 decays/s. Solve: (a) λ = (b) N = (c)

ln 2 ln 2 = = 4.15 × 10−9 s −1 T1/2 1.67 × 108 s

0.0400 × 10−3 kg 4.00 × 10−5 kg = = 3.99 × 1020 60mp 60(1.67 × 10−27 kg)

ΔN = λ N = (4.15 × 10−9 s −1 )(3.99 × 1020 ) = 1.66 × 1012 decays/s Δt

⎛ ⎞ 1 Ci (d) (1.66 × 1012 ) ⎜⎜ ⎟⎟ = 44.9 Ci 10 ⎝ 3.70 × 10 decays/s ⎠

30.60. Set Up: The tritium (H-3) decays to He-3. The ratio of the number of He-3 atoms to H-3 atoms allows us to calculate the time since the decay began, which is immediately after a nuclear explosion formed the H-3. The H-3 decay is exponential. The number of tritium (H-3) nuclei decreases exponentially as N H = N 0,H e−λt , with a half-life of 12.3 years. The amount of He-3 present after a time t is equal to the original amount of tritium minus the number of tritium nuclei that remain after time t. Solve:

The number of He-3 nuclei after time t is N He = N 0,H − N H = N 0,H − N 0,H e− λt = N 0,H (1 − e− λ t ).

Taking the ratio of the number of He-3 atoms to the number of tritium (H-3) atoms gives − λt N He N 0,H (1 − e ) 1 − e− λt = = −λt = eλt − 1. NH N 0,H e− λ t e

Solving for t gives t =

ln(1 + N He /N H )

λ

. Using the given numbers and T1/2 =

ln 2

λ

, we have λ =

ln 2 = T1/2

ln(1 + 4.3) ln 2 = 0.0563/y and t = = 30 years. 0.0563/y 12.3y Reflect: One limitation on this method would be that after many years the ratio of H to He would be too small to measure accurately. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


30-17

*30.61. Set Up: The number of radioactive nuclei left after time t is given by N = N0e− λt . The decay constant for 14

C can be calculated from its known half-life of 5730 years (Example 30.4): λ = ln 2/T1/2 . The problem says

N /N 0 = 0.21; solve for t. Solve:

0.21 = e − λ t so ln(0.21) = − λt and t = − ln(0.21)/λ . We can calculate the decay constant for

known half-life: λ =

less than

()

14

C is 5730 y, so our calculated t is more than two half-lives, so the fraction remaining is

= 14 .

30.62. Set Up: The particle emitted in β − decay is an electron, radioactive nuclei decreases by a factor of 2. 6.25, = Solve: (a)

C from its

− ln(0.21) ln 2 = 1.21× 10−4 y −1 . Thus t = = 1.3 × 104 y. 5730 y 1.21 × 10−4 y −1

Reflect: The half-life of 1 2 2

14

90 38Sr

→

0 −1e

+ 90 39Y The daughter nucleus is

0 −1 e.

In a time of one half-life the number of

1 = 2−4 16 90 39 Y.

(b) 56 yr is 2T1/2 so N = N 0 /22 = N 0 /4; 25% is left. (c)

N N 1 = 2− n ; = 6.25, = = 2−4 so t = 4T1/ 2 = 112 yr N0 N0 16

30.63. Set Up: The reaction energy is Q = ( M A + M B − M C − M D )c 2 , where A = 12H, B = 147N, C = 36Li and D = 105 B. If neutral atom masses from Table 30.2 are used, the electron masses cancel. If Q > 0, energy is liberated in the reaction. If Q < 0, energy is absorbed in the reaction. Solve: M A + M B − M C − M D = 2.014101 u + 14.003074 u − 6.015123 u − 10.012937 u = −0.01088 u Q = (−0.01088 u)(931.5 MeV/u) = −10.1 MeV.

10.1 MeV is absorbed. Reflect: Kinetic energy of the reactants is converted into binding energy in the products.

30.64. Set Up: The energy released in the decay is the energy equivalent of the mass decrease. If we use atomic masses we must be careful to correctly for the masses of the electrons. 56 Solve: (a) The mass must decrease in the decay, so 56 27 Co decays into 26 Fe. (b) The nucleon number doesn’t change and the atomic number decreases by 1, so it is β + (positron) decay. 56 27 Co

→ e+ + 56 26Fe.

(c) The nuclear mass of

56 27 Co

(

56 27 Co

) − 27m , e

where m

(

56 27 Co

) is the atomic mass of the neutral atom. The

) − 26m . Using the nuclear masses, the mass decrease is Δm = [m ( Co ) − 27 m ] − [m ( Fe) − 26m ] − m = m ( Co ) − m ( Fe) − 2m .

nuclear mass of 56 27

56 26 Fe

is m e

(

is m

56 26 Fe

e

56 26

e

e

56 27

56 26

e

Δm = 55.939847 u − 55.934939 u − 2(0.000549 u) = 3.81 × 10−3 u. The energy equivalent of the mass decrease is (3.81 × 10−3 u)(931.5 Mev/u) = 355 MeV.

*30.65. Set Up: Table 30.5 gives m(K + ) = 493.7 MeV/c 2 , m(π 0 ) = 135.0 MeV/c 2 , and m(π ± ) = 139.6 MeV/c 2 . Solve: (a) Charge must be conserved, so K + → π 0 + π + is the only possible decay. (b) The mass decrease is m(K + ) − m(π 0 ) − m(π + ) = 493.7 MeV/c 2 − 135.0 MeV/c 2 − 139.6 MeV/c 2 = 219.1 MeV/c 2 .

The energy released is 219.1 MeV. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30-18

Chapter 30

30.66. Set Up: Take the minimum uncertainty product, so ΔE Δt = Solve: The lifetime is Δt =

h 2π

h (4.136 × 10−15 eV ⋅ s) = = 1.6 × 10−22 s. 2π ΔE 2π (4.0 × 106 eV)

*30.67. Set Up: Table 30.6 lists the properties of quarks. The values of Q, B, and S for an antiquark are opposite those given for the corresponding quark. Solve: (a) S = 1 indicates the presence of one s antiquark and no s quark. To have baryon number 0 there can be only one other quark, and to have net charge +e that quark must be a u, and the quark content is us . (b) The particle has an s antiquark, and for a baryon number of –1 the particle must consist of three antiquarks. For a net charge of –e, the quark content must be dd s . (c) Since S = −2 we know that there are two s quarks, and for baryon number 1 there must be one more quark. For a charge of 0 the third quark must be a u quark and the quark content is uss. Reflect: From Table 30.5 we discover that the particle described in part (a) is a K + (us ). The other combinations are not listed.

Solutions to age Problems 30.68. Set Up: We will apply E = mc 2 , which gives the rest energy of a particle in of its mass. Solve: We are told that the electron and positron are moving slowly—this means that the particles have negligible kinetic energy. Thus, the total energy liberated when they annihilate is essentially equal to Etotal = mec 2 + mec 2 = 2mec 2 . Since this energy produces a pair of photons, each photon gets half of this energy, which is

Ephoton = mec 2 . The correct answer is D.

*30.69. Set Up: We know that the energy of a photon is given by E = hf =

hc

λ

, thus λ =

hc . E

Solve: Using the result of the previous exercise we know that the energy of each photon is Ephoton = mec 2 . Thus we

have λ =

hc hc h = = . The correct answer is C. E mec 2 mec

30.70. Set Up: We are told that the index of refraction of biological tissue is equal to 1; thus, the speed of light inside the tissue will be c. Solve: The event is 3 cm closer to one detector than the midpoint and 3 cm farther from the other detector than the midpoint. This means that the path difference for the two photons is 6 cm. This corresponds to a time difference of 6 × 10−2 m Δt = = 2 × 10−10 s = 0.2 ns. Since this value is less than the given threshold of 10 ns, the two photons will 8 3 × 10 m/s be counted as arriving simultaneously. The correct answer is D.


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