WITH MODERN PHYSICS
INSTRUCTOR
SOLUTIONS
MANUAL
INSTRUCTOR
SOLUTIONS MANUAL
THIRD EDITION
FOR SCIENTISTS AND ENGINEERS
a strategic approach
randall d. knight
Larry Smith
Brett Kraabel
Snow College
PhD-Physics, University of Santa Barbara
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ISBN 13: 978-0-321-76940-4
ISBN 10: 0-321-76940-6
www.pearsonhighered.com
Contents
Preface ....................................................................................................................................... v
PART I
Newton’s Laws
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Concepts of Motion ............................................................................................. 1-1
Kinematics in One Dimension ............................................................................. 2-1
Vectors and Coordinate Systems ......................................................................... 3-1
Kinematics in Two Dimensions........................................................................... 4-1
Force and Motion................................................................................................. 5-1
Dynamics I: Motion Along a Line ....................................................................... 6-1
Newton’s Third Law ............................................................................................ 7-1
Dynamics II: Motion in a Plane ........................................................................... 8-1
PART II
Conservation Laws
Chapter 9
Chapter 10
Chapter 11
Impulse and Momentum ...................................................................................... 9-1
Energy................................................................................................................ 10-1
Work .................................................................................................................. 11-1
PART III
Applications of Newtonian Mechanics
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Rotation of a Rigid Body ................................................................................... 12-1
Newton’s Theory of Gravity.............................................................................. 13-1
Oscillations ........................................................................................................ 14-1
Fluids and Elasticity........................................................................................... 15-1
PART IV
Thermodynamics
Chapter 16
Chapter 17
Chapter 18
Chapter 19
A Macroscopic Description of Matter ............................................................... 16-1
Work, Heat, and the First Law of Thermodynamics.......................................... 17-1
The Micro/Macro Connection............................................................................ 18-1
Heat Engines and Refrigerators ......................................................................... 19-1
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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
PART V
Waves and Optics
Chapter 20
Chapter 21
Chapter 22
Chapter 23
Chapter 24
Traveling Waves ................................................................................................ 20-1
Superposition ..................................................................................................... 21-1
Wave Optics....................................................................................................... 22-1
Ray Optics.......................................................................................................... 23-1
Optical Instruments............................................................................................ 24-1
PART VI
Electricity and Magnetism
Chapter 25
Chapter 26
Chapter 27
Chapter 28
Chapter 29
Chapter 30
Chapter 31
Chapter 32
Chapter 33
Chapter 34
Chapter 35
Electric Charges and Forces............................................................................... 25-1
The Electric Field............................................................................................... 26-1
Gauss’s Law....................................................................................................... 27-1
The Electric Potential......................................................................................... 28-1
Potential and Field ............................................................................................. 29-1
Current and Resistance ...................................................................................... 30-1
Fundamentals of Circuits ................................................................................... 31-1
The Magnetic Field ............................................................................................ 32-1
Electromagnetic Induction ................................................................................. 33-1
Electromagnetic Fields and Waves.................................................................... 34-1
AC Circuits ........................................................................................................ 35-1
PART VII
Relativity and Quantum Physics
Chapter 36
Chapter 37
Chapter 38
Chapter 39
Chapter 40
Chapter 41
Chapter 42
Relativity............................................................................................................ 36-1
The Foundations of Modern Physics ................................................................. 37-1
Quantization....................................................................................................... 38-1
Wave Functions and Uncertainty....................................................................... 39-1
One-Dimensional Quantum Mechanics ............................................................. 40-1
Atomic Physics .................................................................................................. 41-1
Nuclear Physics.................................................................................................. 42-1
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Preface
This Instructor Solutions Manual has a twofold purpose. First, and most obvious, is to provide worked solutions for the use of instructors. Second, but equally important, is to provide
examples of good problem-solving techniques and strategies that will benefit your students if
you post these solutions.
Far too many solutions manuals simply plug numbers into equations, thereby reinforcing one
of the worst student habits. The solutions provided here, by contrast, attempt to:
•
•
•
•
Follow, in detail, the problem-solving strategies presented in the text.
Articulate the reasoning that must be done before computation.
Illustrate how to use drawings effectively.
Demonstrate how to utilize graphs, ratios, units, and the many other “tactics” that must be
successfully mastered and marshaled if a problem-solving strategy is to be effective.
• Show examples of assessing the reasonableness of a solution.
• Comment on the significance of a solution or on its relationship to other problems.
Most education researchers believe that it is more beneficial for students to study a smaller
number of carefully chosen problems in detail, including variations, than to race through a
larger number of poorly understood calculations. The solutions presented here are intended
to provide a basis for this practice.
So that you may readily edit and/or post these solutions, they are available for download as
editable Word documents and as pdf files via the “Resources” tab in the textbook’s Instructor Resource Center (www.pearsonhighered.com/educator/catalog/index.page) or from the
textbook’s Instructor Resource Area in MasteringPhysics® (www.masteringphysics.com).
We have made every effort to be accurate and correct in these solutions. However, if you do
find errors or ambiguities, we would be very grateful to hear from you. Please contact your
Pearson Education sales representative.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
v
vi
PREFACE
Acknowledgments for the First Edition
We are grateful for many helpful comments from Susan Cable, Randall Knight, and Steve
Stonebraker. We express appreciation to Susan Emerson, who typed the word-processing
manuscript, for her diligence in interpreting our handwritten copy. Finally, we would like to
acknowledge the support from the Addison Wesley staff in getting the work into a publishable state. Our special thanks to Liana Allday, Alice Houston, and Sue Kimber for their willingness and preparedness in providing needed help at all times.
Pawan Kahol
Missouri State University
Donald Foster
Wichita State University
Acknowledgments for the Second Edition
I would like to acknowledge the patient support of my wife, Holly, who knows what is important.
Larry Smith
Snow College
I would like to acknowledge the assistance and support of my wife, Alice Nutter, who helped
type many problems and was patient while I worked weekends.
Scott Nutter
Northern Kentucky University
Acknowledgments for the Third Edition
To Holly, Ryan, Timothy, Nathan, Tessa, and Tyler, who make it all worthwhile.
Larry Smith
Snow College
I gratefully acknowledge the assistance of the staff at Physical Sciences Communication.
Brett Kraabel
PhD-University of Santa Barbara
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CONCEPTS OF MOTION
1
Conceptual Questions
1.1.
(a) 3 significant figures.
(b) 2 significant figures. This is more clearly revealed by using scientific notation:
2 sig. figs.
P
0.53 = 5.3 × 10−1
(c) 4 significant figures. The trailing zero is significant because it indicates increased precision.
(d) 3 significant figures. The leading zeros are not significant but just locate the decimal point.
1.2.
(a) 2 significant figures. Trailing zeros in front of the decimal point merely locate the decimal point and are
not significant.
(b) 3 significant figures. Trailing zeros after the decimal point are significant because they indicate increased
precision.
(c) 4 significant figures.
(d) 3 significant figures. Trailing zeros after the decimal point are significant because they indicate increased
precision.
1.3. Without numbers on the dots we cannot tell if the particle in the figure is moving left or right, so we can’t tell
if it is speeding up or slowing down. If the particle is moving to the right it is slowing down. If it is moving to the left
it is speeding up.
1.4. Because the velocity vectors get longer for each time step, the object must be speeding up as it travels to the
left. The acceleration vector must therefore point in the same direction as the velocity, so the acceleration vector also
points to the left. Thus, a x is negative as per our convention (see Tactics Box 1.4).
1.5. Because the velocity vectors get shorter for each time step, the object must be slowing down as it travels in
the ⫺ y direction (down). The acceleration vector must therefore point in the direction opposite to the velocity;
namely, in the +y direction (up). Thus, a y is positive as per our convention (see Tactics Box 1.4).
1.6.
The particle position is to the left of zero on the x-axis, so its position is negative. The particle is moving to the
right, so its velocity is positive. The particle’s speed is increasing as it moves to the right, so its acceleration vector
points in the same direction as its velocity vector (i.e., to the right). Thus, the acceleration is also positive.
1.7.
The particle position is below zero on the y-axis, so its position is negative. The particle is moving down, so
its velocity is negative. The particle’s speed is increasing as it moves in the negative direction, so its acceleration
vector points in the same direction as its velocity vector (i.e., down). Thus, the acceleration is also negative.
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1-1
1-2
Chapter 1
1.8. The particle position is above zero on the y-axis, so its position is positive. The particle is moving down, so its
velocity is negative. The particle’s speed is increasing as it moves in the negative direction, so its acceleration vector
points in the same direction as its velocity vector (i.e., down). Thus, the acceleration is also negative.
Exercises and Problems
Section 1.1 Motion Diagrams
1.1.
Model: Imagine a car moving in the positive direction (i.e., to the right). As it skids, it covers less distance
between each movie frame (or between each snapshot).
Solve:
Assess: As we go from left to right, the distance between successive images of the car decreases. Because the time
interval between each successive image is the same, the car must be slowing down.
1.2.
Model: We have no information about the acceleration of the rocket, so we will assume that it accelerates
upward with a constant acceleration.
Solve:
Assess: Notice that the length of the velocity vectors increases each step by approximately the length of the
acceleration vector.
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Concepts of Motion
1-3
1.3.
Model: We will assume that the term “quickly” used in the problem statement means a time that is short
compared to 30 s.
Solve:
Assess: Notice that the acceleration vector points in the direction opposite to the velocity vector because the car is
decelerating.
Section 1.2 The Particle Model
1.4.
Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated
into a single point. The size and shape of the object will not be considered. This is a reasonable approximation of
reality if (i) the distance traveled by the object is large in comparison to the size of the object and (ii) rotations and
internal motions are not significant features of the object’s motion. The particle model is important in that it allows
us to simplify a problem. Complete reality—which would have to include the motion of every single atom in the
object—is too complicated to analyze. By treating an object as a particle, we can focus on the most important aspects
of its motion while neglecting minor and unobservable details.
(b) The particle model is valid for understanding the motion of a satellite or a car traveling a large distance.
(c) The particle model is not valid for understanding how a car engine operates, how a person walks, how a bird flies,
or how water flows through a pipe.
Section 1.3 Position and Time
Section 1.4 Velocity
1.5. Model: We model the ball’s motion from the instant after it is released, when it has zero velocity, to the
instant before it hits the ground, when it will have its maximum velocity.
Solve:
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1-4
Chapter 1
Assess: Notice that the “particle” we have drawn has a finite dimensions, so it appears as if the bottom half of this
“particle” has penetrated into the ground in the bottom frame. This is not really the case; our mental particle has no
size and is located at the tip of the velocity vector arrow.
1.6.
Solve: The player starts from rest and moves faster and faster.
1.7.
Solve: The player starts with an initial velocity but as he slides he moves slower and slower until coming to rest.
Section 1.5 Linear Acceleration
1.8.
G
G
Solve: (a) Let v0 be the velocity vector between points 0 and 1 and v1 be the velocity vector between points
1 and 2. Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.
(b) To find the acceleration, use the method of Tactics Box 1.3:
Assess: The acceleration vector points in the same direction as the velocity vectors, which makes sense because the
speed is increasing.
1.9.
G
G
Solve: (a) Let v0 be the velocity vector between points 0 and 1 and v1 be the velocity vector between points 1
and 2. Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.
(b) Acceleration is found by the method of Tactics Box 1.3.
Assess: The acceleration vector points in the same direction as the velocity vectors, which makes sense because the
speed is increasing.
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Concepts of Motion
1.10.
Solve:
(a)
1.11.
(b)
Solve:
(a)
1.12.
1-5
(b)
Model: Represent the car as a particle.
Visualize: The dots are equally spaced until brakes are applied to the car. Equidistant dots on a single line indicate
constant average velocity. Upon braking, the dots get closer as the average velocity decreases, and the distance
between dots changes by a constant amount because the acceleration is constant.
1.13.
Model: Represent the (child + sled) system as a particle.
Visualize: The dots in the figure are equally spaced until the sled encounters a rocky patch. Equidistant dots on a
single line indicate constant average velocity. On encountering a rocky patch, the average velocity decreases and the
sled comes to a stop. This part of the motion is indicated by a decreasing separation between the dots.
1.14.
Model: Represent the wad of paper as a particle. Ignore air resistance.
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1-6
Chapter 1
Visualize: The dots become more closely spaced because the particle experiences a downward acceleration. The
distance between dots changes by a constant amount because the acceleration is constant.
1.15.
Model: Represent the tile as a particle.
Visualize: Starting from rest, the tile’s velocity increases until it hits the water surface. This part of the motion is
represented by dots with increasing separation, indicating increasing average velocity. After the tile enters the water,
it settles to the bottom at roughly constant speed, so this part of the motion is represented by equally spaced dots.
1.16.
Model: Represent the tennis ball as a particle.
Visualize: The ball falls freely for three stories. Upon impact, it quickly decelerates to zero velocity while compressing, then accelerates rapidly while re-expanding. As vectors, both the deceleration and acceleration are an upward
vector. The downward and upward motions of the ball are shown separately in the figure. The increasing length between
the dots during downward motion indicates an increasing average velocity or downward acceleration. On the other hand,
the decreasing length between the dots during upward motion indicates acceleration in a direction opposite to the
motion, so the average velocity decreases.
Assess: For free-fall motion, acceleration due to gravity is always vertically downward. Notice that the acceleration
due to the ground is quite large (although not to scale—that would take too much space) because in a time interval
much shorter than the time interval between the points, the velocity of the ball is essentially completely reversed.
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Concepts of Motion
1-7
1.17.
Model: Represent the toy car as a particle.
Visualize: As the toy car rolls down the ramp, its speed increases. This is indicated by the increasing length of the
G
velocity arrows. That is, motion down the ramp is under a constant acceleration a. At the bottom of the ramp, the toy
car continues with a constant velocity and no acceleration.
Section 1.6 Motion in One Dimension
1.18.
(a)
Solve:
Dot
1
2
3
4
5
6
7
8
9
Time (s)
0
2
4
6
8
10
12
14
16
x (m)
0
30
95
215
400
510
600
670
720
(b)
1.19. Solve: A forgetful physics professor is walking from one class to the next. Walking at a constant speed, he
covers a distance of 100 m in 200 s. He then stops and chats with a student for 200 s. Suddenly, he realizes he is
going to be late for his next class, so the hurries on and covers the remaining 200 m in 200 s to get to class on time.
1.20. Solve: Forty miles into a car trip north from his home in El Dorado, an absent-minded English professor
stopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking that he was
supposed to go on this trip on Sunday. Absent-mindedly he missed his exit and stopped after one hour of driving at
another rest area 20 miles south of El Dorado. After waiting there for one hour, he drove back very slowly, confused
and tired as he was, and reached El Dorado two hours later.
Section 1.7 Solving Problems in Physics
Visualize: The bicycle move forward with an acceleration of 1.5 m/s 2 . Thus, the velocity will increase by
1.5 m/s each second of motion.
1.21.
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1-8
Chapter 1
G
Visualize: The rocket moves upward with a constant acceleration a. The final velocity is 200 m/s and is
reached at a height of 1.0 km.
1.22.
Section 1.8 Units and Significant Figures
1.23.
⎛ 1s ⎞
−3
Solve: (a) 6.15 ms = (6.15 ms) ⎜⎜ 3
⎟⎟ = 6.15 × 10 s
10
ms
⎝
⎠
⎛ 103 m ⎞
(b) 27.2 km = (27.2 km) ⎜
= 27.2 × 103 m
⎜ 1 km ⎟⎟
⎝
⎠
km ⎞ ⎛ 103 m ⎞ ⎛ 1 hour ⎞
⎛
(c) 112 km/hour = ⎜112
⎟⎜
⎟⎜
⎟ = 31.1 m/s
hour ⎠ ⎜⎝ 1 km ⎟⎠ ⎝ 3600 s ⎠
⎝
3
⎛ μm ⎞ ⎛ 1 m ⎞ ⎛ 10 ms ⎞
−2
(d) 72 μm/ms = ⎜ 72
⎟ = 7.2 × 10 m/s
⎟⎟ ⎜⎜
⎟ ⎜⎜ 6
⎝ ms ⎠ ⎝ 10 μm ⎠ ⎝ 1 s ⎟⎠
1.24.
−2
⎛ 2.54 cm ⎞ ⎛ 10 m ⎞
Solve: (a) 8.0 inch = (8.0 inch) ⎜
⎜
⎟ = 0.20 m
⎟
⎝ 1 inch ⎠ ⎜⎝ 1 cm ⎟⎠
1m
⎛ feet ⎞⎛ 12 inch ⎞⎛
⎞
(b) 66 feet/s = ⎜ 66
⎟⎜
⎟⎜
⎟ = 20 m/s
s ⎠⎝ 1 foot ⎠⎝ 39.37 inch ⎠
⎝
3
⎛ miles ⎞⎛ 1.609 km ⎞ ⎛ 10 m ⎞ ⎛ 1 hour ⎞
(c) 60 mph = ⎜ 60
⎟⎜
⎟⎜
⎟ ⎜⎜
⎟ = 27 m/s
⎝ hour ⎠⎝ 1 mile ⎠ ⎝ 1 km ⎟⎠ ⎝ 3600 s ⎠
2
1m
⎛
⎞
2
−3
−3
(d) 14 square inches = (14 inch 2 ) ⎜
⎟ = 9.0 × 10 m = 9.0 × 10 square meters
⎝ 39.37 inches ⎠
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Concepts of Motion
1.25.
1-9
⎛ 3600 s ⎞
4
Solve: (a) 3 hour = (3 hour) ⎜
⎟ = 10,800 s ≈ 1 × 10 s
1
hour
⎝
⎠
⎛ 24 hours ⎞ ⎛ 3600 s ⎞
5
5
(b) 2 day = (2 day) ⎜
⎟⎜
⎟ = 1.73 × 10 s ≈ 2 × 10 s
⎝ 1 day ⎠ ⎝ 1 hour ⎠
⎛ 365.25 days ⎞ ⎛ 8.64 × 104
(c) 1 year = (1 year) ⎜
⎟⎜
⎝ 1 year ⎠ ⎜⎝ 1 day
s⎞
7
7
⎟⎟ = 3.16 × 10 s ≈ 3 × 10 s
⎠
ft ⎞⎛ 12 inch ⎞⎛
1m
⎛
⎞
(d) 215 ft/s = ⎜ 215 ⎟⎜
⎟⎜
⎟ = 65.5 m/s
s
1
ft
39.37
inch
⎝
⎠⎝
⎠⎝
⎠
Assess: The results are given to appropriate number of significant figures.
1.26.
⎛1 m ⎞
Solve: (a) 20 ft ≈ (20 ft) ⎜
⎟≈7 m
⎝ 3 ft ⎠
⎛ 1 km ⎞⎛ 1000 m ⎞
(b) 60 miles ≈ (60 miles) ⎜
⎟⎜
⎟ ≈ 100,000 m
⎝ 0.6 miles ⎠⎝ 1 km ⎠
⎛ 1 m/s ⎞
(c) 60 mph ≈ (60 mph) ⎜
⎟ ≈ 30 m/s
⎝ 2 mph ⎠
−2
⎛ 1 cm ⎞ ⎛ 10 m ⎞
(d) 8 in ≈ (8 in) ⎜
⎟ ≈ 0.2 m
⎟ ⎜⎜
⎝ 1/2 in ⎠ ⎝ 1 cm ⎟⎠
1.27.
Solve:
⎛ 4 in ⎞
(a) 30 cm ≈ (30 cm) ⎜
⎟ ≈ 12 in
⎝ 10 cm ⎠
⎛ 2 mph ⎞
(b) 25 m/s ≈ (25 m/s) ⎜
⎟ ≈ 50 mph
⎝ 1 m/s ⎠
⎛ 0.6 mi ⎞
(c) 5 km ≈ (5 km) ⎜
⎟ ≈ 3 mi
⎝ 1 km ⎠
⎛ 1/2 in ⎞
(d) 0.5 cm ≈ (0.5 cm) ⎜
⎟ ≈ 0.3 in
⎝ 1 cm ⎠
Solve: (a) 33.3 × 25.4 = 846
(b) 33.3 − 25.4 = 7.9
1.28.
(c) 33.3 = 5.77
(d) 333.3 ÷ 25.4 = 13.1
Solve: (a) (12.5)3 = 1.95 × 103.
(b) 12.5 × 5.21 = 65.1
1.29.
(c)
12.5 − 1.2 = 3.54 − 1.2 = 2.3
(d) 12.5−1 = 1.00/12.5 = 0.0800 = 8.00 × 10−2
1.30.
Solve: The length of a typical car is 15 ft or
1m
⎛ 12 inch ⎞⎛
⎞
(15 ft) ⎜
⎟⎜
⎟ = 4.6 m
1
ft
39.37
inch
⎝
⎠⎝
⎠
This length of 15 ft is approximately two-and-a-half times my height.
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1-10
Chapter 1
1.31.
Solve: The height of a telephone pole is estimated to be around 50 ft or (using 1 m ~ 3 ft) about 15 m. This
height is approximately 8 times my height.
1.32.
Solve: I typically take 15 minutes in my car to cover a distance of approximately 6 miles from home to
campus. My average speed is
⎛ 0.447 m/s ⎞
⎛ 6 miles ⎞⎛ 60 min ⎞
⎟ = 11 m/s
⎜
⎟⎜
⎟ = 24 mph = (24 mph) ⎜
⎝ 15 min ⎠⎝ 1 hour ⎠
⎝ 1 mph ⎠
1.33.
Solve: My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hair
growth is
⎛ 1 inch ⎞ ⎛ 2.54 cm ⎞ ⎛ 10−2 m ⎞ ⎛ 1 month ⎞ ⎛ 1 day ⎞⎛ 1 h ⎞
−9
⎟⎜
⎜
⎟⎜
⎟⎜
⎟ ⎜⎜
⎟⎜
⎟ = 9.8 × 10 m/s
⎟
1
month
1
inch
1
cm
30
days
24
h
3600
s
⎝
⎠
⎝
⎠⎝
⎠
⎠
⎝
⎠
⎝
⎠⎝
m ⎞ ⎛ 106 μm ⎞ ⎛ 3600 s ⎞
⎛
= ⎜ 9.8 × 10−9 ⎟ ⎜
⎟⎜
⎟ ≈ 40 μm/h
s ⎠ ⎜⎝ 1 m ⎟⎠ ⎝ 1 h ⎠
⎝
1.34.
Model: Represent the Porsche as a particle for the motion diagram. Assume the car moves at a constant
speed when it coasts.
Visualize:
1.35.
Model: Represent the jet as a particle for the motion diagram.
Visualize:
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Concepts of Motion
1-11
1.36.
Model: Represent (Sam + car) as a particle for the motion diagram.
Visualize:
1.37.
Model: Represent the wad as a particle for the motion diagram.
Visualize:
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1-12
Chapter 1
1.38.
Model: Represent the speed skater as a particle for the motion diagram.
Visualize:
1.39.
Model: Represent Santa Claus as a particle for the motion diagram.
Visualize:
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Concepts of Motion
1-13
1.40.
Model: Represent the motorist as a particle for the motion diagram.
Visualize:
1.41.
Model: Represent the car as a particle for the motion diagram.
Visualize:
1.42.
Model: Represent Bruce and the puck as particles for the motion diagram.
Visualize:
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1-14
Chapter 1
1.43.
Model: Represent the cars of David and Tina and as particles for the motion diagram.
Visualize:
1.44.
Solve: Isabel is the first car in line at a stop light. When it turns green, she accelerates, hoping to make the
next stop light 100 m away before it turns red. When she’s about 30 m away, the light turns yellow, so she starts to
brake, knowing that she cannot make the light.
1.45.
Solve: A car coasts along at 30 m/s and arrives at a hill. The car decelerates as it coasts up the hill. At the
top, the road levels and the car continues coasting along the road at a reduced speed.
1.46.
Solve: A skier starts from rest down a 25° slope with very little friction. At the bottom of the 100-m slope
the terrain becomes flat and the skier continues at constant velocity.
1.47.
Solve: A ball is dropped from a height to check its rebound properties. It rebounds to 80% of its original
height.
1.48.
Solve: Two boards lean against each other at equal angles to the vertical direction. A ball rolls up the
incline, over the peak, and down the other side.
1.49.
Solve:
(a)
(b) A train moving at 100 km/hour slows down in 10 s to a speed of 60 km/hour as it enters a tunnel. The driver
maintains this constant speed for the entire length of the tunnel that takes the train a time of 20 s to traverse. Find the
length of the tunnel.
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Concepts of Motion
1-15
(c)
1.50.
Solve:
(a)
(b) Sue passes 3rd Street doing 30 km/h, slows steadily to the stop sign at 4th Street, stops for 1.0 s, then speeds up
and reaches her original speed as she passes 5th Street. If the blocks are 50 m long, how long does it take Sue to drive
from 3rd Street to 5th Street?
(c)
1.51.
Solve:
(a)
(b) Jeremy has perfected the art of steady acceleration and deceleration. From a speed of 60 mph he brakes his car to
rest in 10 s with a constant deceleration. Then he turns into an adjoining street. Starting from rest, Jeremy accelerates
with exactly the same magnitude as his earlier deceleration and reaches the same speed of 60 mph over the same
distance in exactly the same time. Find the car’s acceleration or deceleration.
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1-16
Chapter 1
(c)
1.52.
Solve:
(a)
(b) A coyote (A) sees a rabbit and begins to run toward it with an acceleration of 3.0 m/s 2 . At the same instant, the
rabbit (B) begins to run away from the coyote with an acceleration of 2.0 m/s 2 . The coyote catches the rabbit after
running 40 m. How far away was the rabbit when the coyote first saw it?
(c)
1.53. Solve: Since area equals length ⫻ width, the smallest area will correspond to the smaller length and the
smaller width. Similarly, the largest area will correspond to the larger length and the larger width. Therefore, the
smallest area is (64 m)(100 m) = 6.4 ⫻103 m 2 and the largest area is (75 m)(110 m) = 8.3 ⫻ 103 m 2 .
1.54.
Solve: (a) We need kg/m3. There are 100 cm in 1 m. If we multiply by
3
⎛ 100 cm ⎞
3
⎜
⎟ = (1)
⎝ 1m ⎠
we do not change the size of the quantity, but only the number in terms of the new unit. Thus, the mass density of
aluminum is
3
kg
⎛
−3 kg ⎞⎛ 100 cm ⎞
= 2.7 × 103 3
⎜ 2.7 × 10
3 ⎟⎜ 1 m ⎟
cm ⎠⎝
m
⎝
⎠
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Concepts of Motion
1-17
(b) Likewise, the mass density of alcohol is
3
g ⎞⎛ 100 cm ⎞ ⎛ 1 kg ⎞
kg
⎛
⎟ = 810 3
⎜ 0.81 3 ⎟⎜
⎟ ⎜
1
m
1000
g
cm ⎠⎝
m
⎝
⎠ ⎝
⎠
1.55.
Model: In the particle model, the car is represented as a dot.
Solve:
(a)
Time t (s)
0
10
20
30
40
50
60
70
80
90
Position x (m)
1200
975
825
750
700
650
600
500
300
0
(b)
1.56.
Solve: Susan enters a classroom, sees a seat 40 m directly ahead, and begins walking toward it at a constant
leisurely pace, covering the first 10 m in 10 seconds. But then Susan notices that Ella is heading toward the same
seat, so Susan walks more quickly to cover the remaining 30 m in another 10 seconds, beating Ella to the seat. Susan
stands next to the seat for 10 seconds to remove her backpack.
1.57. Solve: A crane operator starts lifting a ton of bricks off the ground. In 8 s, he has lifted them to a height of
15 m, then he takes 4 s to make a safety check. He then continues raising the bricks the remaining 15 m, which takes 4 s.
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KINEMATICS IN ONE DIMENSION
2
Conceptual Questions
2.1. It was a typical summer day on the interstate. I started 10 mi east of town and drove for 20 min at 30 mph west
to town. My stop for gas took 10 min. Then I headed back east at 60 mph before I encountered a construction zone.
Traffic was at a standstill for 10 min and then I was able to move forward (east) at 30 mph until I got to my
destination 30 mi east of town.
2.2. With a slow start out of the blocks, a super sprinter reached top speed in about 5 s, having gone only 30 m. He
was still able to finish his 100 m in only just over 9 s by running a world record pace for the rest of the race.
2.3. The baseball team is warming up. The pitcher (who is 50 feet from home plate) lobs the ball at 100 ft/s to the
second baseman who is 100 ft from home plate. The second baseman then fires the ball at 200 ft/s to the catcher at
home plate.
2.4. (a) At t = 1 s, the slope of the line for object A is greater than that for object B. Therefore, object A’s speed is
greater. (Both are positive slopes.)
(b) No, the speeds are never the same. Each has a constant speed (constant slope) and A’s speed is always greater.
2.5. (a) A’s speed is greater at t = 1 s. The slope of the tangent to B’s curve at t = 1 s is smaller than the slope of
A’s line.
(b) A and B have the same speed at just about t = 3 s. At that time, the slope of the tangent to the curve representing
B’s motion is equal to the slope of the line representing A.
2.6. (a) B. The object is still moving, but the magnitude of the slope of the position-versus-time curve is smaller than at D.
(b) D. The slope is greatest at D.
(c) At points A, C, and E the slope of the curve is zero, so the object is not moving.
(d) At point D the slope is negative, so the object is moving to the left.
2.7. (a) The slope of the position-versus-time graph is greatest at D, so the object is moving fastest at this point.
(b) The slope is negative at points C, D, and E, meaning the object is moving to the left at these points.
(c) At point C the slope is increasing in magnitude (getting more negative), meaning that the object is speeding up to
the left.
(d) At point B the object is not moving since the slope is zero. Before point B, the slope is positive, while after B it is
negative, so the object is turning around at B.
2.8. (a) The positions of the third dots of both motion diagrams are the same, as are the sixth dots of both, so cars A
and B are at the same locations at the time corresponding to dot 3 and again at that of dot 6.
(b) The spacing of dots 4 and 5 in both diagrams is the same, so the cars are traveling at the same speeds between
times corresponding to dots 4 and 5.
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2-1
2-2
Chapter 2
2.9. No, though you have the same position along the road, his velocity is greater because he is passing you. If his
velocity were not greater, then he would remain even with the front of your car.
2.10. Yes. The acceleration vector will point west when the bicycle is slowing down while traveling east.
2.11. (a) As a ball tossed upward moves upward, its vertical velocity is positive, while its vertical acceleration is
negative, opposite the velocity, causing the ball to slow down.
(b) The same ball on its way down has downward (negative) velocity. The downward negative acceleration is
pointing in the same direction as the velocity, causing the speed to increase.
2.12. For all three of these situations the acceleration is equal to g in the downward direction. The magnitude and
direction of the velocity of the ball do not matter. Gravity pulls down at constant acceleration. (Air friction is
ignored.)
2.13. (a) The magnitude of the acceleration while in free fall is equal to g at all times, independent of the initial
velocity. The acceleration only tells how the velocity is changing.
(b) The magnitude of the acceleration is still g because the rock is still in free fall. The speed is increasing at the same
rate each instant, that is, by the same Δv each second.
2.14. The ball remains in contact with the floor for a small but noticeable amount of time. It is in free fall when not
in contact with the floor. When it hits the floor, it is accelerated very rapidly in the upward direction as it bounces.
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Kinematics in One Dimension
2-3
Exercises and Problems
Section 2.1 Uniform Motion
2.1. Model: Cars will be treated by the particle model.
Visualize:
Solve: Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows:
Δx x1 − x0
x −x
v=
=
⇒ t1 = t0 + 1 0
Δt t1 − t0
v
Using the known values identified in the pictorial representation, we find:
x
−x
400 mile
tAlan 1 = tAlan 0 + Alan 1 Alan 0 = 8:00 AM +
= 8:00 AM + 8 hr = 4:00 PM
50 miles/hour
v
400 mile
x
−x
tBeth 1 = tBeth 0 + Beth 1 Beth 0 = 9:00 AM +
= 9:00 AM + 6.67 hr = 3:40 PM
60 miles/hour
v
(a) Beth arrives first.
(b) Beth has to wait tAlan 1 − tBeth 1 = 20 minutes for Alan.
Assess: Times of the order of 7 or 8 hours are reasonable in the present problem.
2.2. Model: We will consider Larry to be a particle.
Visualize:
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2-4
Chapter 2
Solve: Since Larry’s speed is constant, we can use the following equation to calculate the velocities:
s −s
vs = f i
tf − ti
(a) For the interval from the house to the lamppost:
200 m − 600 m
v1 =
= −200 m/min
9 : 07 − 9 : 05
For the interval from the lamppost to the tree:
1200 m − 200 m
v2 =
= +333 m/min
9 :10 − 9 : 07
(b) For the average velocity for the entire run:
1200 m − 600 m
vavg =
= +120 m/min
9 :10 − 9 : 05
2.3. Solve: (a) The time for each segment is Δt1 = 50 mi/40 mph = 5/4 hr and Δt2 = 50 mi/60 mph = 5/6hr. The
average speed to the house is
100 mi
= 48 mph
5/6 h + 5/4 h
(b) Julie drives the distance Δx1 in time Δt1 at 40 mph. She then drives the distance Δx2 in time Δt2 at 60 mph. She
spends the same amount of time at each speed, thus
Δt1 = Δt2 ⇒ Δx1/40 mph = Δx2 /60 mph ⇒ Δx1 = (2/3)Δx2
But Δx1 + Δx2 = 100 miles, so (2/3)Δx2 + Δx2 = 100 miles. This means Δx2 = 60 miles and Δx1 = 40 miles. Thus,
the times spent at each speed are Δt1 = 40 mi/40 mph = 1.00 h and Δt2 = 60 mi/60 mph = 1.00 h. The total time for
her return trip is Δt1 + Δt2 = 2.00 h. So, her average speed is 100 mi/2 h = 50 mph.
2.4. Model: The jogger is a particle.
Solve: The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at t = 10 s is
Δs 50 m − 25 m
=
= 1.25 m/s
v=
Δt
20 s
The slope at t = 25 s is
50 m − 50 m
v=
= 0.0 m/s
10 s
The slope at t = 35 s is
0 m − 50 m
= −5.0 m/s
v=
10 s
Section 2.2 Instantaneous Velocity
Section 2.3 Finding Position from Velocity
2.5. Solve: (a) We can obtain the values for the velocity-versus-time graph from the equation v = Δs/Δt .
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Kinematics in One Dimension
2-5
(b) There is only one turning point. At t = 1 s the velocity changes from +20 m/s to −5 m/s, thus reversing the direction
of motion. At t = 3 s, there is an abrupt change in motion from −5 m/s to rest, but there is no reversal in motion.
2.6. Visualize: Please refer to Figure EX2.6. The particle starts at x0 = 10 m at t0 = 0. Its velocity is initially in
the –x direction. The speed decreases as time increases during the first second, is zero at t = 1 s, and then increases
after the particle reverses direction.
Solve: (a) The particle reverses direction at t = 1 s, when vx changes sign.
(b) Using the equation xf = x0 + area of the velocity graph between t1 and tf ,
x2s = 10 m − (area of triangle between 0 s and 1 s) + (area of triangle between 1 s and 2 s)
1
1
= 10 m − (4 m/s)(1 s) + (4 m/s)(1 s) = 10 m
2
2
x3s = 10 m + area of trapazoid between 2 s and 3 s
1
= 10 m + (4 m/s + 8 m/s)(3 s − 2 s) = 16 m
2
x4s = x3s + area between 3 s and 4 s
1
= 16 m + (8 m/s + 12 m/s)(1 s) = 26 m
2
2.7. Model: The graph shows the assumption that the blood isn’t moving for the first 0.1 s nor at the end of the
beat.
Visualize: The graph is a graph of velocity vs. time, so the displacement is the area under the graph—that is, the
area of the triangle. The velocity of the blood increases quickly and decreases a bit more slowly.
Solve: Call the distance traveled Δy. The area of a triangle is
Δy =
1 BH .
2
1
1
BH = (0.20 s)(0.80 m/s) = 8.0 cm
2
2
Assess: This distance seems reasonable for one beat.
2.8. Solve: (a) We can calculate the position of the particle at every instant with the equation
xf = xi + area under the velocity-versus-time graph between ti and tf
The particle starts from the origin at t = 0 s, so xi = 0 m. Notice that the each square of the grid in Figure EX2.8 has
“area” (5 m/s) × (2 s) = 10 m. We can find the area under the curve, and thus determine x, by counting squares. You
can see that x = 35 m at t = 4 s because there are 3.5 squares under the curve. In addition, x = 35 m at t = 8 s
because the 5 m represented by the half square between 4 and 6 s is cancelled by the –5 m represented by the half
square between 6 and 8 s. Areas beneath the axis are negative areas. The particle passes through x = 35 m at t = 4 s
and again at t = 8 s.
(b) The particle moves to the right for 0 s ≥ t ≥ 6 s, where the velocity is positive. It reaches a turning point at
x = 40 m at t = 6 s. The motion is to the left for t > 6 s. This is shown in the motion diagram below.
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2-6
Chapter 2
Section 2.4 Motion with Constant Acceleration
2.9. Visualize: The object has a constant velocity for 2 s and then speeds up between t = 2 and t = 4.
Solve: A constant velocity from t = 0 s to t = 2 s means zero acceleration. On the other hand, a linear increase in
velocity between t = 2 s and t = 4 s implies a constant positive acceleration which is the slope of the velocity line.
2.10. Visualize: The graph is a graph of velocity vs. time, so the acceleration is the slope of the graph.
Solve: When the blood is speeding up the acceleration is
ay =
Δυ y
Δt
=
0.80 m/s
= 16 m/s 2
0.05s
When the blood is slowing down the acceleration is
ay =
Δυ y
Δt
=
−0.80 m/s
= −5.3 m/s 2
0.15 s
Assess: 16 m/s 2 is an impressive but reasonable acceleration.
2.11. Solve: (a) At t = 2.0 s, the position of the particle is
x2 s = 2.0 m + area under velocity graph from t = 0 s to t = 2.0 s
1
= 2.0 m + (4.0 m/s)(2.0 s) = 6.0 m
2
(b) From the graph itself at t = 2.0 s, v = 4 m/s.
(c) The acceleration is
ax =
Δvx vfx − vix 6 m/s − 0 m/s
=
=
= 2 m/s 2
Δt
Δt
3s
2.12. Solve: (a) Using the equation
xf = xi + area under the velocity-versus-time graph between ti and tf
we have
x (at t = 1 s) = x (at t = 0 s) + area between t = 0 s and t = 1 s
= 2.0 m + (4 m/s)(1 s) = 6 m
Reading from the velocity-versus-time graph, vx (at t = 1 s) = 4 m/s. Also, a x = slope = Δv/Δt = 0 m/s 2 .
(b) x(at t = 3.0 s) = x(at t = 0 s) + area between t = 0 s and t = 3 s
= 2.0 m + 4 m/s × 2 s + 2 m/s × 1 s + (1/2) × 2 m/s × 1 s = 13.0 m
Reading from the graph, vx (t = 3 s) = 2 m/s. The acceleration is
a x (t = 3 s) = slope =
vx (at t = 4 s) − vx (at t = 2 s)
= −2 m/s 2
2s
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Kinematics in One Dimension
2-7
2.13. Model: Represent the jet plane as a particle.
Visualize:
Solve: Since we don’t know the time of acceleration, we will use
v12 = v02 + 2a ( x1 − x0 )
⇒a=
v12 − v02 (400 m/s) 2 − (300 m/s) 2
=
= 8.75 m/s 2 ≈ 8.8 m/s 2
2 x1
2(4000 m)
The acceleration of the jet is not quite equal to g, the acceleration due to gravity; this seems reasonable for a jet.
2.14. Model: Model the air as a particle.
Visualize: Use the definition of acceleration and then convert units.
Solve:
ax =
Δvx 150 km/h ⎛ 1000 m ⎞ ⎛ 1 h ⎞ ⎛ 1 min ⎞
2
=
⎜
⎟⎜
⎟⎜
⎟ = 83 m/s
Δt
0.50 s ⎝ 1 km ⎠ ⎝ 60 min ⎠ ⎝ 60 s ⎠
Assess: 83 m/s 2 is a remarkable acceleration.
2.15. Model: We are using the particle model for the skater and the kinematics model of motion under constant
acceleration.
Solve: Since we don’t know the time of acceleration we will use
vf2 = vi2 + 2a ( xf − xi )
⇒a=
vf2 − vi2
(6.0 m/s) 2 − (8.0 m/s) 2
=
= −2.8 m/s 2
2( xf − xi )
2(5.0 m)
Assess: A deceleration of 2.8 m/s 2 is reasonable.
2.16. Model: We are assuming both cars are particles.
Visualize:
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2-8
Chapter 2
Solve: The Porsche’s time to finish the race is determined from the position equation
1
xP1 = xP0 + vP0 (tP1 − tP0 ) + aP (tP1 − tP0 ) 2
2
1
⇒ 400 m = 0 m + 0 m + (3.5 m/s 2 )(tP1 − 0 s) 2 ⇒ tP1 = 15.1 s
2
The Honda’s time to finish the race is obtained from Honda’s position equation as
1
xH1 = xH0 + vH0 (tH1 − tH0 ) + aH0 (tH1 − tH0 ) 2
2
1
2
400 m = 0 m + 0 m + (3.0 m/s )(tH1 + 1 s) 2 ⇒ tH1 = 15.3 s
2
So, the Porsche wins.
Assess: The numbers are contrived for the Porsche to win, but the time to go 400 m seems reasonable.
Section 2.5 Free Fall
2.17. Model: Represent the spherical drop of molten metal as a particle.
Visualize:
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Kinematics in One Dimension
2-9
Solve: (a) The shot is in free fall, so we can use free fall kinematics with a = − g . The height must be such that the
shot takes 4 s to fall, so we choose t1 = 4 s. Then,
1
1
1
y1 = y0 + v0 (t1 − t0 ) − g (t1 − t0 ) 2 ⇒ y0 = gt12 = (9.8 m/s 2 )(4 s) 2 = 78.4 m
2
2
2
(b) The impact velocity is v1 = v0 − g (t1 − t0 ) = − gt1 = −39.2 m/s.
G
Assess: Note the minus sign. The question asked for velocity, not speed, and the y-component of v is negative
because the vector points downward.
2.18. Model: Assume the ball undergoes free-fall acceleration and that the ball is a particle.
Visualize:
(a)
Solve: (a) We will use the kinematic equations
v = v0 + a (t − t0 ) and
1
y = y0 + v0 (t − t0 ) + a (t − t0 ) 2
2
as follows:
v(at t = 1.0 s) = 19.6 m/s + (−9.8 m/s 2 )(1.0 s − 0 s) = 9.8 m/s
y (at t = 1.0 s) = 0 m + (19.6 m/s)(1.0 s − 0 s) + 1/2(−9.8 m/s 2 )(1.0 s − 0 s)2 = 14.7 m
v(at t = 2.0 s) = 19.6 m/s + ( −9.8 m/s 2 )(2.0 s − 0 s) = 0 m/s
y (at t = 2.0 s) = 0 m + (19.6 m/s)(2.0 s − 0 s) + 1/2(−9.8 m/s 2 )(2.0 s − 0 s)2 = 19.6 m
v(at t = 3.0 s) = 19.6 m/s + ( −9.8 m/s 2 )(3 s − 0 s) = −9.8 m/s
y (at t = 3.0 s) = 0 m + (19.6 m/s)(3.0 s − 0 s) + 1/2(−9.8 m/s 2 )(3.0 s − 0 s) 2 = 14.7 m
v(at t = 4.0 s) = 19.6 m/s + ( −9.8 m/s 2 )(4.0 s − 0 s) = −19.6 m/s
y (at t = 4.0 s) = 0 m + (19.6 m/s)(4.0 s − 0 s) + 1/2(−9.8 m/s 2 )(4.0 s − 0 s) 2 = 0 m
(b)
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2-10
Chapter 2
Assess: (a) A downward acceleration of 9.8 m/s 2 on a particle that has been given an initial upward velocity of
+19.6 m/s will reduce its speed to 9.8 m/s after 1 s and then to zero after 2 s. The answers obtained in this solution
are consistent with the above logic.
(b) Velocity changes linearly with a negative uniform acceleration of 9.8 m/s 2 . The position is symmetrical in time
around the highest point which occurs at t = 2 s.
2.19. Model: We model the ball as a particle.
Visualize:
Solve: Once the ball leaves the student’s hand, the ball is in free fall and its acceleration is equal to the free-fall
acceleration g that always acts vertically downward toward the center of the earth. According to the constantacceleration kinematic equations of motion
1
y1 = y0 + v0 Δt + a Δt 2
2
Substituting the known values
−2 m = 0 m + (15 m/s)t1 + (1/2)( −9.8 m/s 2 )t12
One solution of this quadratic equation is t1 = 3.2 s. The other root of this equation yields a negative value for t1,
which is not valid for this problem.
Assess: A time of 3.2 s is reasonable.
2.20. Model: We will use the particle model and the constant-acceleration kinematic equations.
Visualize:
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Kinematics in One Dimension
2-11
Solve: (a) Substituting the known values into y1 = y0 + v0 Δt + 12 a Δt 2 , we get
1
−10 m = 0 m + 20 (m/s)t1 + (−9.8 m/s 2 )t12
2
One of the roots of this equation is negative and is not relevant physically. The other root is t1 = 4.53 s, which is the
answer to part (b). Using v1 = v0 + aΔt , we obtain
v1 = 20(m/s) + (−9.8 m/s 2 )(4.53 s) = −24 m/s
(b) The time is 4.5 s.
Assess: A time of 4.5 s is a reasonable value. The rock’s velocity as it hits the bottom of the hole has a negative sign
because of its downward direction. The magnitude of 24 m/s compared to 20 m/s, when the rock was tossed up, is
consistent with the fact that the rock travels an additional distance of 10 m into the hole.
Section 2.6 Motion on an Inclined Plane
2.21. Model: We will model the skier as a particle.
Visualize:
Note that the skier’s motion on the horizontal, frictionless snow is not of any interest to us. Also note that the
acceleration parallel to the incline is equal to g sin10°.
Solve: Using the following constant-acceleration kinematic equations,
vf2x = vi2x + 2a x ( xf − xi )
⇒ (15 m/s) 2 = (3.0 m/s) 2 + 2(9.8 m/s 2 )sin10°( x1 − 0 m) ⇒ x1 = 64 m
vfx = vix + a x (tf − ti )
⇒ (15 m/s) = (3.0 m/s) + (9.8 m/s 2 )(sin10°)t ⇒ t = 7.1 s
Assess: A time of 7.1 s to cover 64 m is a reasonable value.
2.22. Model: Represent the car as a particle.
Visualize:
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2-12
Chapter 2
Solve: Note that the problem “ends” at a turning point, where the car has an instantaneous speed of 0 m/s before
rolling back down. The rolling back motion is not part of this problem. If we assume the car rolls without friction,
then we have motion on a frictionless inclined plane with an accleration a = − g sin θ = − g sin10° = −1.7 m/s 2 .
Constant acceleration kinematics gives
v12 = v02 + 2a( x1 − x0 ) ⇒ 0 m 2 /s 2 = v02 + 2ax1 ⇒ x1 = −
v02
(30 m/s) 2
=−
= 265 m
2a
2( −1.7 m/s 2 )
Notice how the two negatives canceled to give a positive value for x1.
G
Assess: We must include the minus sign because the a vector points down the slope, which is in the negative
x-direction.
Section 2.7 Instantaneous Acceleration
2.23. Solve: x = (2t 2 − t + 1) m
(a) The position t = 2 s is x2s = [2(2) 2 − 2 + 1] m = 7 m.
(b) The velocity is the derivative v = dx/dt and the velocity at t = 2 s is calculated as follows:
v = (4t − 1) m/s ⇒ v2s = [4(2) − 1] m/s = 7 m/s
(c) The acceleration is the derivative a = dv/dt and the acceleration at t = 2 s is calculated as follows:
a = (4) m/s 2 ⇒ a2s = 4 m/s 2
2.24. Solve: The formula for the particle’s position along the x-axis is given by
tf
xf = xi + Ñvx dt
ti
Using the expression for vx we get
xf = xi + 23 [tf3 − ti3 ]
ax =
dvx d
= (2t 2 m/s) = 4t m/s 2
dt dt
(a) The particle’s position at t = 1 s is x1 s = 1 m + 23 m = 53 m.
(b) The particle’s speed at t = 1 s is v1 s = 2 m/s.
(c) The particle’s acceleration at t = 1 s is a1 s = 4 m/s 2 .
2.25. Solve: The formula for the particle’s velocity is given by
vf = vi + area under the acceleration curve between ti and tf
For t = 4 s, we get
1
v4 s = 8 m/s + (4 m/s 2 )4 s = 16 m/s
2
Assess: The acceleration is positive but decreases as a function of time. The initial velocity of 8.0 m/s will therefore
increase. A value of 16 m/s is reasonable.
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Kinematics in One Dimension
2-13
2.26. Solve: (a)
(b) To be completed by student.
dx
(c)
= vx = 2t − 4 ⇒ vx (at t = 1 s) = [2 m/s 2 (1 s) − 4 m/s] = −2 m/s
dt
(d) There is a turning point at t = 2 s. At that time x = −2 m.
(e) Using the equation in part (c),
vx = 4 m/s = (2t − 4) m/s ⇒ t = 4
Since x = (t 2 − 4t + 2) m, x = 2 m.
(f )
2.27. Solve: The graph for particle A is a straight line from t = 2 s to t = 8 s. The slope of this line is −10 m/s,
which is the velocity at t = 7.0 s. The negative sign indicates motion toward lower values on the x-axis. The velocity
of particle B at t = 7.0 s can be read directly from its graph. It is −20 m/s. The velocity of particle C can be obtained
from the equation
vf = vi + area under the acceleration curve between ti and tf
This area can be calculated by adding up three sections. The area between t = 0 s and t = 2 s is 40 m/s, the area
between t = 2 s and t = 5 s is 45 m/s, and the area between t = 5 s and t = 7 s is −20 m/s. We get (10 m/s) +
(40 m/s) + (45 m/s) − (20 m/s) = 75 m/s.
2.28. Visualize:
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2-14
Chapter 2
Solve: We will determine the object’s velocity using graphical methods first and then using calculus.
Graphically, v(t ) = v0 + area under the acceleration curve from 0 to t. In this case, v0 = 0 m/s. The area at each time
t requested is a triangle.
t =0 s
v(t = 0 s) = v0 = 0 m/s
1
v(t = 2 s) = (2 s)(5 m/s) = 5 m/s
2
1
v(t = 4 s) = (4 s)(10 m/s) = 20 m/s
2
1
v(t = 6 s) = (6 s)(10 m/s) = 30 m/s
2
v (t = 8 s) = v(t = 6 s) = 30 m/s
t=2s
t=4s
t =6 s
t =8 s
The last result arises because there is no additional area after t = 6 s. Let us now use calculus. The acceleration
function a(t) consists of three pieces and can be written:
0 s≤t ≤4 s
⎧ 2.5t
⎪
a (t ) = ⎨ −5t + 30 4 s ≤ t ≤ 6 s
⎪ 0
6 s≤t ≤8 s
⎩
These were determined by the slope and the y-intercept of each of the segments of the graph. The velocity function is
found by integration as follows: For 0 ≤ t ≤ 4 s,
t2
v(t ) = v(t = 0 s) + Ñ a(t )dt = 0 + 2.5
0
2
t
t
= 1.25t 2
0
This gives
t = 0 s v(t = 0 s) = 0 m/s
t = 2 s v(t = 2 s) = 5 m/s
t = 4 s v(t = 4 s) = 20 m/s
For 4 s ≤ t ≤ 6 s,
t
⎡ −5t 2
⎤
v(t ) = v(t = 4 s) + Ñ a (t ) dt = 20 m/s + ⎢
+ 30t ⎥ = −2.5t 2 + 30t − 60
4
⎢⎣ 2
⎥⎦ 4
t
This gives:
t = 6 s v (t = 6 s) = 30 m/s
For 6 s ≤ t ≤ 8 s,
t
v(t ) = v(t = 6 s) + Ñ a (t )dt = 30 m/s + 0 m/s = 30 m/s
6
This gives:
t = 8 s v (t = 8 s) = 30 m/s
Assess: The same velocities are found using calculus and graphs, but the graphical method is easier for simple
graphs.
2.29. Solve: (a) The velocity-versus-time graph is the derivative with respect to time of the distance-versus-time
graph. The velocity is zero when the slope of the position-versus-time graph is zero, the velocity is most positive
when the slope is most positive, and the velocity is most negative when the slope is most negative. The slope is zero
at t = 0, 1 s, 2 s, 3 s, . . . ; the slope is most positive at t = 0.5 s, 2.5 s, . . ; and the slope is most negative at t = 1.5 s,
3.5 s, . . .
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Kinematics in One Dimension
2-15
(b)
2.30. Solve: The given function for the velocity is vx = t 2 − 7t + 10.
(a) The turning points are when the velocity changes sign. Set vx = 0 and check that it actually changes sign at those
times. The function factors into the product of two binomials:
vx = (t − 2)(t − 5) ⇒ vx = 0 when t = 2 s and t = 5 s
Indeed, the function changes sign at those two times.
(b) The acceleration is given by the derivative of the velocity.
dv
a x = x = 2t − 7
dt
Plug in the times from part (a): a x (2 s) = 2(2) − 7 = −3 m/s 2 and a x (5 s) = 2(5) − 7 = 3 m/s 2
Assess: This problem does not have constant acceleration so the kinematic equations do not apply, but a = dv/dt
always applies.
2.31. Solve: (a) The velocity-versus-time graph is given by the derivative with respect to time of the position
function:
dx
= (6t 2 − 18t ) m/s
dt
For vx = 0 m/s, there are two solutions to the quadratic equation: t = 0 s and t = 3 s.
vx =
(b) At the first of these solutions,
x (at t = 0 s) = 2(0 s)3 − 9(0 s) 2 + 12 = 12 m
The acceleration is the derivative of the velocity function:
dv
a x = x = (12t − 18) m/s 2 ⇒ a (at t = 0 s) = −18 m/s 2
dt
At the second solution,
x (at t = 3 s) = 2(3 s)3 − 9(3 s) 2 + 12 = −15 m
a x (at t = 3 s) = 12(3 s) − 18 = 118 m/s 2
2.32. Model: Represent the object as a particle.
Solve: (a) Known information: x0 = 0 m, v0 = 0 m/s, x1 = 40 m, v1 = 11 m/s, t1 = 5 s. If the acceleration is uniform
(constant a), then the motion must satisfy the three equations
1
x1 = at12 ⇒ a = 3.20 m/s 2 v1 = at1 ⇒ a = 2.20 m/s 2 v12 = 2ax1 ⇒ a = 1.51 m/s 2
2
But each equation gives a different value of a. Thus the motion is not uniform acceleration.
(b) We know two points on the velocity-versus-time graph, namely at t0 = 0 and t1 = 5 s. What shape does the
function have between these two points? If the acceleration was uniform, which it’s not, then the graph would be a
straight line. The area under the graph is the displacement Δ x. From the figure you can see that Δ x = 27.5 m for a
straight-line graph. But we know that, in reality, Δ x = 40 m. To get a larger Δ x, the graph must bulge upward above
the straight line. Thus the graph is curved, and it is concave downward.
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2-16
Chapter 2
2.33. Solve: The position is the integral of the velocity.
t1
t1
t1
t0
0
0
x1 = x0 + Ñ vx dt = x0 + Ñ kt 2 dt = x0 + 13 kt 3
= x0 + 13 kt13
We’re given that x0 = −9.0 m and that the particle is at x1 = 9.0 m at t1 = 3.0 s. Thus
9.0 m = (−9.0 m) + 13 k (3.0 s)3 = ( −9.0 m) + k (9.0 s3 )
Solving for k gives k = 2.0 m/s3.
2.34. Solve: (a) The velocity is the integral of the acceleration.
t1
t1
t0
0
(
v1x = v0 x + Ñ a x dt = 0 m/s + Ñ (10 − t )dt = 10t − 12 t 2
The velocity is zero when
(
) (
) 01 = 10t1 − 12 t12
t
)
v1x = 0 m/s = 10t1 − 12 t12 = 10 − 12 t1 × t1
⇒ t1 = 0 s or t1 = 20 s
The first solution is the initial condition. Thus the particle’s velocity is again 0 m/s at t1 = 20 s.
(b) Position is the integral of the velocity. At t1 = 20 s, and using x0 = 0 m at t0 = 0 s, the position is
t1
20
t0
0
x1 = x0 + Ñ vx dt = 0 m + Ñ
(10t − t ) dt = 5t
1 2
2
2 20
0
− 16 t 3
20
0
= 667 m
2.35. Model: Represent the ball as a particle.
Visualize: Please refer to Figure P2.35.
Solve: In the first and third segments the acceleration as is zero. In the second segment the acceleration is negative and
constant. This means the velocity vs will be constant in the first two segments and will decrease linearly in the third
segment. Because the velocity is constant in the first and third segments, the position s will increase linearly. In the second
segment, the position will increase parabolically rather than linearly because the velocity decreases linearly with time.
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Kinematics in One Dimension
2-17
2.36. Model: Represent the ball as a particle.
Visualize: Please refer to Figure P2.36. The ball rolls down the first short track, then up the second short track, and
then down the long track. s is the distance along the track measured from the left end (where s = 0). Label t = 0 at the
beginning, that is, when the ball starts to roll down the first short track.
Solve: Because the incline angle is the same, the magnitude of the acceleration is the same on all of the tracks.
Assess: Note that the derivative of the s versus t graph yields the vs versus t graph. And the derivative of the vs
versus t graph gives rise to the as versus t graph.
2.37. Model: Represent the ball as a particle.
Visualize: The ball moves to the right along the first track until it strikes the wall, which causes it to move to the
left on a second track. The ball then descends on a third track until it reaches the fourth track, which is horizontal.
Solve:
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2-18
Chapter 2
Assess: Note that the time derivative of the position graph yields the velocity graph, and the derivative of the
velocity graph gives the acceleration graph.
2.38. Visualize: Please refer to Figure P2.38.
Solve:
2.39. Visualize: Please refer to Figure P2.39.
Solve:
2.40. Model: The plane is a particle and the constant-acceleration kinematic equations hold.
Solve: (a) To convert 80 m/s to mph, we calculate 80 m/s × 1 mi/1609 m × 3600 s/h = 179 mph.
(b) Using as = Δv/Δt , we have,
as (t = 0 to t = 10 s) =
23 m/s − 0 m/s
= 2.3 m/s 2
10 s − 0 s
as (t = 20 s to t = 30 s) =
69 m/s − 46 m/s
= 2.3 m/s 2
30 s − 20 s
For all time intervals a is 2.3 m/s 2 .
(c) Using kinematics as follows:
vfs = vis + a (tf − ti ) ⇒ 80 m/s = 0 m/s + (2.3 m/s 2 )(tf − 0 s) ⇒ tf = 35 s
(d) Using the above values, we calculate the takeoff distance as follows:
1
1
sf = si + vis (tf − ti ) + as (tf − ti )2 = 0 m + (0 m/s)(35 s) + (2.3 m/s 2 )(35 s) 2 = 1410 m
2
2
For safety, the runway should be 3 × 1410 m = 4230 m or 2.6 mi. This is longer than the 2.4 mi long runway, so the
takeoff is not safe.
2.41. Model: Represent the car as a particle.
Solve: (a) First, we will convert units:
60
miles 1 hour 1610 m
×
×
= 27 m/s
hour 3600 s 1 mile
The motion is constant acceleration, so
v1 − v0 (27 m/s − 0 m/s)
=
= 2.7 m/s 2
Δt
10 s
(b) The fraction is a /g = 2.7/9.8 = 0.28. So a is 28% of g.
v1 = v0 + aΔt ⇒ a =
(c) The distance is calculated as follows:
1
1
x1 = x0 + v0Δt + a (Δt ) 2 = a ( Δt ) 2 = 1.3 × 102 m = 4.3 × 102 feet
2
2
2.42. Model: Represent the spaceship as a particle.
Solve: (a) The known information is: x0 = 0 m, v0 = 0 m/s, t0 = 0 s, a = g = 9.8 m/s 2 , and v1 = 3.0 × 108 m/s. Constant
acceleration kinematics gives
v1 = v0 + aΔt ⇒ Δt = t1 =
v1 − v0
= 3.06 × 107 s
a
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Kinematics in One Dimension
2-19
The problem asks for the answer in days, so we need a conversion:
1 hour
1 day
×
= 3.54 × 102 days ≈ 3.6 × 102 days
t1 = (3.06 × 107 s) ×
3600 s 24 hour
(b) The distance traveled is
1
1
x1 − x0 = v0Δt + a (Δt ) 2 = at12 = 4.6 × 1015 m
2
2
(c) The number of seconds in a year is
24 hours 3600 s
1 year = 365 days ×
×
= 3.15 × 107 s
1 day
1 hour
In one year light travels a distance
1 light year = (3.0 × 108 m/s)(3.15 × 107 s) = 9.47 × 1015 m
The distance traveled by the spaceship is 4.6 × 1015 m/9.46 × 1015 m = 0.49 of a light year.
Assess: Note that x1 gives “Where is it?” rather than “How far has it traveled?” “How far” is represented by
x1 − x0 . They happen to be the same number in this problem, but that isn’t always the case.
2.43. Model: The car is a particle and constant-acceleration kinematic equations hold.
Visualize:
Solve: (a) This is a two-part problem. During the reaction time,
x1 = x0 + v0 (t1 − t0 ) + 1/2a0 (t1 − t0 ) 2
= 0 m + (20 m/s)(0.50 s − 0 s) + 0 m = 10 m
After reacting, x2 − x1 = 110 m − 10 m = 100 m, that is, you are 100 m away from the intersection.
(b) To stop successfully,
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ (0 m/s) 2 = (20 m/s) 2 + 2a1 (100 m) ⇒ a1 = −2 m/s 2
(c) The time it takes to stop once the brakes are applied can be obtained as follows:
v2 = v1 + a1 (t2 − t1) ⇒ 0 m/s = 20 m/s + ( −2 m/s 2 )(t2 − 0.50 s) ⇒ t2 = 11 s
The total time to stop since the light turned red is 11.5 s.
2.44. Model: Model the car as a particle. Use the kinematic equations.
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2-20
Chapter 2
Visualize:
Solve:
(a) In the pictorial representation d = x2 − x0 and t1 = tR . At the end of the reaction time the position is
x1 = x0 + v0tR and v1 = v0 . We need to end up at rest with v2 = 0. During the braking time we have
v22 = v12 + 2a ( x2 − x1); solve that for a and set v2 = 0.
a=
v22 − v12
−v12
−v02
−v02
=
=
=
2( x2 − x1) 2( x2 − x1) 2( x2 − ( x0 + v0tR )) 2( d − v0tR )
The required deceleration is the absolute value of a; (the denominator is positive because we were told d > v0tR ).
a =
v02
2(d − v0tR )
(b) Given v0 = 21 m/s, d = 50 m, and tR = 0.50 s find the required deceleration using the answer from part (a).
a =
v02
(21 m/s)2
=
= 5.6 m/s 2
2( d − v0tR ) 2(50 m − (21 m/s)(0.50 s))
Your car’s maximum deceleration is greater than this, so yes, you can stop in time.
Assess: The units check out. The data given seem reasonable, and our answer is just less than the maximum
deceleration.
2.45. Model: We will use the particle model and the constant-acceleration kinematic equations.
Visualize:
Solve: (a) To find x2 , we first need to determine x1. Using x1 = x0 + v0 (t1 − t0 ), we get x1 = 0 m + (20 m/s)
(0.50 s − 0 s) = 10 m. Now,
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 = (20 m/s) 2 + 2( −10 m/s 2 )( x2 − 10 m) ⇒ x2 = 30 m
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Kinematics in One Dimension
2-21
The distance between you and the deer is ( x3 − x2 ) or (35 m − 30 m) = 5 m.
(b) Let us find v0 max such that v2 = 0 m/s at x2 = x3 = 35 m. Using the following equation,
v22 − v02 max = 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 − v02 max = 2(−10 m/s 2 )(35 m − x1 )
Also, x1 = x0 + v0 max (t1 − t0 ) = v0 max (0.50 s − 0 s) = (0.50 s)v0 max . Substituting this expression for x1 in the above
equation yields
−v02 max = (−20 m/s 2 )[35 m − (0.50 s) v0 max ] ⇒ v02
max
+ (10 m/s)v0 max − 700 m2 /s 2 = 0
The solution of this quadratic equation yields v0 max = 22 m/s. (The other root is negative and unphysical for the
present situation.)
Assess: An increase of speed from 20 m/s to 22 m/s is very reasonable for the car to cover an additional distance of
5 m with a reaction time of 0.50 s and a deceleration of 10 m/s 2 .
2.46. Model: The car is represented as a particle.
Visualize:
Solve: (a) This is a two-part problem. First, we need to use the information given to determine the acceleration
during braking. Second, we need to use that acceleration to find the stopping distance for a different initial velocity.
First, the car coasts at constant speed before braking:
x1 = x0 + v0 (t1 − t0 ) = v0t1 = (30 m/s)(0.5 s) = 15 m
Then, the car brakes to a halt. Because we don’t know the time interval, use
v22 = 0 = v12 + 2a1( x2 − x1)
v12
(30 m/s)2
=2
= −10 m/s 2
2( x2 − x1 )
2(60 m − 15 m)
G
We used v1 = v0 = 30 m/s. Note the minus sign, because a1 points to the left.
⇒ a1 = −
We can repeat these steps now with v0 = 40 m/s. The coasting distance before braking is
x1 = v0t1 = (40 m/s)(0.5 s) = 20 m
The position x2 after braking is found using
v22 = 0 = v12 + 2a1 ( x2 − x1 )
⇒ x2 = x1 −
v12
(40 m/s) 2
= 20 m −
= 100 m
2a1
2(−10 m/s 2 )
(b) The car coasts at a constant speed for 0.5 s, traveling 20 m. The graph will be a straight line with a slope of
40 m/s. For t ≥ 0.5 the graph will be a parabola until the car stops at t2 . We can find t2 from
v2 = 0 = v1 + a1(t2 − t1 ) ⇒ t2 = t1 −
v1
= 4.5 s
a1
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2-22
Chapter 2
The parabola will reach zero slope (v = 0 m/s) at t = 4.5 s. This is enough information to draw the graph shown in the
figure.
2.47. Model: Model the flea as a particle. Both the initial acceleration phase and the free-fall phase have constant
acceleration, so use the kinematic equations.
Visualize:
Solve: We can apply the kinematic equation vf2 − vi2 = 2aΔy twice, once to find the take-off speed and then again to
find the final height. In the first phase the acceleration is up (positive) and v0 = 0.
v12 = 2a0 ( y1 − y0 ) = 2(1000 m/s 2 )(0.50 × 10−3 m)v1 = 1.0 m/s
In the free fall phase the acceleration is a1 = − g and v1 = 1.0 m/s and v2 = 0 m/s.
y2 − y1 =
v22 − v12
− v12
−(1.0 m/s) 2
=
=
= 5.1 cm
2a1
2(− g ) 2(−9.8 m/s 2 )
So the final height is y2 = 5.1 cm + y1 = 5.1 cm + 0.50 mm = 5.2 cm.
Assess: This is pretty amazing–about 10–20 times the size of a typical flea.
2.48. Model: Model each of the animals as a particle and use kinematic equations. Assume that the time it takes the
cheetah to accelerate to 30 m/s is negligible.
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Kinematics in One Dimension
2-23
Visualize:
Solve: The cheetah is in uniform motion for the entire duration of the problem, so we can easily solve for its position
at t3 = 25 s:
x3C = x1C + (vx )1C Δt = 0 m + (30 m/s)(15 s) = 450 m
The gazelle’s motion has two phases: one of constant acceleration and then one of constant velocity. We can solve
for the position and the velocity at t2 , the end of the first phase.
(vx ) 2G = (vx )1G + (a x )G Δt = 0 m/s + (4.6 m/s 2 )(5.0 s) = 23 m/s
1
1
x2G = x1G + (vx )1G Δt + (a x )G (Δt ) 2 = 170 m + 0 m + (4.6 m/s 2 )(5.0 s)2 = 227.5 m
2
2
From t2 to t3 the gazelle moves at a constant speed, so we can use the equation for uniform motion to find its final
position.
x3G = x2G + (vx ) 2G Δt = 227.5 m + (23 m/s)(10.0 s) = 457.5 m ≈ 460 m
x3C is 450m; x3G is 460m. The gazelle is just a few meters ahead of the cheetah when the cheetah has to break off
the chase, so the gazelle escapes.
Assess: The numbers in the problem statement are realistic, so we expect our results to mirror real life. The speed for
the gazelle is close to that of the cheetah, which seems reasonable for two animals known for their speed. And the
result is the most common occurrence–the chase is very close, but the gazelle gets away.
2.49. Model: The rocket is represented as a particle.
Visualize:
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2-24
Chapter 2
Solve: (a) There are three parts to the motion. Both the second and third parts of the motion are free fall, with
a = − g . The maximum altitude is y2 . In the acceleration phase:
1
1
1
y1 = y0 + v0 (t1 − t0 ) + a (t1 − t0 ) 2 = at12 = (30 m/s 2 )(30 s) 2 = 13,500 m
2
2
2
v1 = v0 + a(t1 − t0 ) = at1 = (30 m/s 2 )(30 s) = 900 m/s
In the coasting phase,
v22 = 0 = v12 − 2 g ( y2 − y1 ) ⇒ y2 = y1 +
v12
(900 m/s) 2
= 13,500 m +
= 54,800 m = 54.8 km
2g
2(9.8 m/s 2 )
The maximum altitude is 54.8 km (≈ 33 miles).
(b) The rocket is in the air until time t3 . We already know t1 = 30 s. We can find t2 as follows:
v2 = 0 m/s = v1 − g (t2 − t1 ) ⇒ t2 = t1 +
v1
= 122 s
g
Then t3 is found by considering the time needed to fall 54,800 m:
1
1
2 y2
y3 = 0 m = y2 + v2 (t3 − t2 ) − g (t3 − t2 ) 2 = y2 − g (t3 − t2 ) 2 ⇒ t3 = t2 +
= 228 s
g
2
2
(c) The velocity increases linearly, with a slope of 30 (m/s)/s, for 30 s to a maximum speed of 900 m/s. It then begins
to decrease linearly with a slope of −9.8(m/s)/s. The velocity passes through zero (the turning point at y2 ) at
t2 = 122 s. The impact velocity at t3 = 228 s is calculated to be v3 = v2 − g (t3 − t2 ) = −1040 m/s.
Assess: In reality, friction due to air resistance would prevent the rocket from reaching such high speeds as it falls,
and the acceleration upward would not be constant because the mass changes as the fuel is burned, but that is a more
complicated problem.
2.50. Model: We will model the rocket as a particle. Air resistance will be neglected.
Visualize:
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Kinematics in One Dimension
2-25
Solve: (a) Using the constant-acceleration kinematic equations,
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (16 s − 0 s) = ao (16 s)
1
1
y1 = y0 + v0 (t1 − t0 ) + a0 (t1 − t0 )2 = a0 (16 s − 0 s) 2 = ao (128 s 2 )
2
2
1
y2 = y1 + v1 (t2 − t1 ) + a1(t2 − t1 )2
2
1
2
⇒ 5100 m = 128 s a0 + 16 s a0 (20 s − 16 s) + (−9.8 m/s2 )(20 s − 16 s) 2 ⇒ a0 = 27 m/s 2
2
(b) The rocket’s speed as it passes through a cloud 5100 m above the ground can be determined using the kinematic
equation:
v2 = v1 + a1 (t2 − t1) = (16 s)a0 + ( −9.8 m/s 2 )(4 s) = 390 m/s
Assess: 400 m/s ≈ 900 mph, which would be the final speed of a rocket that has been accelerating for 20 s at a rate of
approximately 20 m/s 2 or 66 ft/s 2 .
2.51. Model: We will model the lead ball as a particle and use the constant-acceleration kinematic equations.
Visualize:
Note that the particle undergoes free fall until it hits the water surface.
Solve: The kinematics equation y1 = y0 + v0 (t1 − t0 ) + 12 a0 (t1 − t0 ) 2 becomes
1
−5.0 m = 0 m + 0 m + (−9.8 m/s 2 )(t1 − 0) 2 ⇒ t1 = 1.01 s
2
Now, once again,
1
y2 = y1 + v1 (t2 − t1) + a1(t2 − t1 ) 2
2
⇒ y2 − y1 = v1(3.0 s − 1.01 s) + 0 m/s = 1.99 v1
v1 is easy to determine since the time t1 has been found. Using v1 = v0 + a0 (t1 − t0 ), we get
v1 = 0 m/s − (9.8 m/s 2 )(1.01 s − 0 s) = −9.898 m/s
With this value for v1, we go back to:
y2 − y1 = 1.99v1 = (1.99)(−9.898 m/s) = −19.7 m
y2 − y1 is the displacement of the lead ball in the lake and thus corresponds to the depth of the lake. The negative
sign shows the direction of the displacement vector.
Assess: A depth of about 60 ft for a lake is not unusual.
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2-26
Chapter 2
2.52. Model: The elevator is a particle moving under constant-acceleration kinematic equations.
Visualize:
Solve: (a) To calculate the distance to accelerate up:
(v1 )2 = v02 + 2a0 ( y0 − y0 ) ⇒ (5 m/s) 2 = (0 m/s) 2 + 2(1 m/s 2 )( y1 − 0 m) ⇒ y1 = 12.5 m
(b) To calculate the time to accelerate up:
v1 = v0 + a0 (t1 − t0 ) ⇒ 5 m/s = 0 m/s + (1 m/s 2 )(t1 − 0 s) ⇒ t1 = 5 s
To calculate the distance to decelerate at the top:
v32 = v22 + 2a2 ( y3 − y2 ) ⇒ (0 m/s) 2 = (5 m/s)2 + 2(−1 m/s 2 )( y3 − y2 ) ⇒ y3 − y2 = 12.5 m
To calculate the time to decelerate at the top:
v3 = v2 + a2 (t3 − t2 ) ⇒ 0 m/s = 5 m/s + (−1 m/s 2 )(t3 − t2 ) ⇒ t3 − t2 = 5 s
The distance moved up at 5 m/s is
y2 − y1 = ( y3 − y0 ) − ( y3 − y2 ) − ( y1 − y0 ) = 200 m − 12.5 m − 12.5 m = 175 m
The time to move up 175 m is given by
1
y2 − y1 = v1(t2 − t1 ) + a1 (t2 − t1) 2 ⇒ 175 m = (5 m/s)(t2 − t1 ) ⇒ (t2 − t1) = 35 s
2
To total time to move to the top is
(t1 − t0 ) + (t2 − t1 ) + (t3 − t2 ) = 5 s + 35 s + 5 s = 45 s
Assess: To cover a distance of 200 m at 5 m/s (ignoring acceleration and deceleration times) will require a time of
40 s. This is comparable to the time of 45 s for the entire trip as obtained above.
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Kinematics in One Dimension
2-27
2.53. Model: The car is a particle moving under constant-acceleration kinematic equations.
Visualize:
Solve: This is a three-part problem. First the car accelerates, then it moves with a constant speed, and then it
decelerates.
First, the car accelerates:
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + (4.0 m/s 2 )(6 s − 0 s) = 24 m/s
1
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + (4.0 m/s 2 )(6 s − 0 s) 2 = 72 m
2
2
Second, the car moves at v1:
1
x2 − x1 = v1 (t2 − t1 ) + a1 (t2 − t1) 2 = (24 m/s)(8 s − 6 s) + 0 m = 48 m
2
Third, the car decelerates:
v3 = v2 + a2 (t3 − t2 ) ⇒ 0 m/s = 24 m/s + (−3.0 m/s 2 )(t3 − t2 ) ⇒ (t3 − t2 ) = 8 s
1
1
x3 = x2 + v2 (t3 − t2 ) + a2 (t3 − t2 ) 2 ⇒ x3 − x2 = (24 m/s)(8 s) + (−3.0 m/s 2 )(8 s) 2 = 96 m
2
2
Thus, the total distance between stop signs is:
x3 − x0 = ( x3 − x2 ) + ( x2 − x1 ) + ( x1 − x0 ) = 96 m + 48m + 72 m = 216 m
Assess: A distance of approximately 600 ft in a time of around 10 s with an acceleration/deceleration of the order of
7 mph/s is reasonable.
2.54. Model: The car is a particle moving under constant linear acceleration.
Visualize:
Solve: Using the kinematic equation for position:
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2-28
Chapter 2
1
1
x2 = x1 + v1(t2 − t1) + a1(t2 − t1 ) 2 ⇒ x1 + 30 m = x1 + v1 (5.0 s − 4.0 s) + (2 m/s 2 )(5.0 s − 4.0 s)2
2
2
1
2
2
⇒ 30 m = v1 (1.0 s) + (2 m/s )(1.0 s) ⇒ v1 = 29 m/s
2
And 4.0 seconds before:
v1 = v0 + a0 (t1 − t0 ) ⇒ 29 m/s = v0 + (2 m/s 2 )(4.0 s − 0 s) ⇒ v0 = 21 m/s
Assess: 21 m/s ≈ 47 mph and is a reasonable value.
2.55. Model: Santa is a particle moving under constant-acceleration kinematic equations.
Visualize: Note that our x-axis is positioned along the incline.
Solve: Using the following kinematic equation,
v12 = v02 + 2a & ( x1 − x2 ) = (0 m/s) 2 + 2(4.9 m/s 2 )(10 m − 0 m) ⇒ v1 = 9.9 m/s
Assess: Santa’s speed of 20 mph as he reaches the edge is reasonable.
2.56. Model: The cars are represented as particles.
Visualize:
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Kinematics in One Dimension
2-29
Solve: (a) Ann and Carol start from different locations at different times and drive at different speeds. But at time t1
they have the same position. It is important in a problem such as this to express information in terms of positions (that is,
coordinates) rather than distances. Each drives at a constant velocity, so using constant velocity kinematics gives
xA1 = xA0 + vA (t1 − tA0 ) = vA (t1 − tA0 )
xC1 = xC0 + vC (t1 − tC0 ) = xC0 + vCt1
The critical piece of information is that Ann and Carol have the same position at t1, so xA1 = xC1. Equating these two
expressions, we can solve for the time t1 when Ann passes Carol:
vA (t1 − tA0 ) = xC0 + vCt1
⇒ (vA − vC )t1 = xC0 + vAtA0
⇒ t1 =
xC0 + vAtA0 2.4 mi + (50 mph)(0.5 h)
=
= 1.96 h ≈ 2.0 h
vA − vC
50 mph − 36 mph
(b) Their position is x1 = xA1 = xC1 = xC0 + vCt1 = 72.86 miles ≈ 73 miles
(c) Note that Ann’s graph doesn’t start until t = 0.5 hours, but her graph has a steeper slope so it intersects Carol’s
graph at t ≈ 2.0 hours.
2.57. Model: Model the ice as a particle and use the kinematic equations for constant acceleration. Model the “very
slippery block” and “smooth ramp”‘ as frictionless. Set the x-axis parallel to the ramp.
Visualize:
Note that the distance down the ramp is Δx = h / sin θ . Also a x = g sin θ down a frictionless ramp.
Solve:
(a) Use vf2 = vi2 + 2a x Δx , where vi = 0 .
vf2 = 2aΔx ⇒ vf = 2( g sin θ )
h
= 2 gh
sin θ
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2-30
Chapter 2
(b) For h = 0.30 m,
vf = 2(9.8 m/s 2 )(0.30 m) = 2.4 m/s
This is true for both angles as the answer is independent of the angle.
Assess: We will later learn how to solve this problem in an easier way with energy.
2.58. Model: We will model the toy train as a particle.
Visualize:
Solve: Using kinematics,
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 2 m + (2.0 m/s)(2.0 s − 0 s) + 0 m = 6.0 m
2
The acceleration can now be obtained as follows:
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 = (2.0 m/s) 2 + 2a1 (8.0 m − 6.0 m) ⇒ a1 = 21.0 m/s 2
The magnitude is 1.0 m/s 2 .
Assess: A deceleration of 1 m/s 2 in bringing the toy train to a halt over a distance of 2.0 m is reasonable.
2.59. Model: We will use the particle model and the kinematic equations at constant-acceleration.
Visualize:
Solve: To find x2 , let us use the kinematic equation
v22 = v12 + 2a1 ( x2 − x1 ) = (0 m/s) 2 = (50 m/s) 2 + 2(−10 m/s 2 )( x2 − x1) ⇒ x2 = x1 + 125 m
Since the nail strip is at a distance of 150 m from the origin, we need to determine x1:
x1 = x0 + v0 (t1 − t0 ) = 0 m + (50 m/s)(0.60 s − 0.0 s) = 30 m
Therefore, we can see that x2 = (30 + 125) m = 155 m. That is, he can’t stop within a distance of 150 m. He is in jail.
Assess: Bob is driving at approximately 100 mph and the stopping distance is of the correct order of magnitude.
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Kinematics in One Dimension
2-31
2.60. Model: We will use the particle model with constant-acceleration kinematic equations.
Visualize:
Solve: The acceleration, being the same along the incline, can be found as
v12 = v02 + 2a ( x1 − x0 ) ⇒ (4.0 m/s) 2 = (5.0 m/s)2 + 2a (3.0 m − 0 m) ⇒ a = −1.5 m/s2
We can also find the total time the puck takes to come to a halt as
v2 = v0 + a(t2 − t0 ) ⇒ 0 m/s = (5.0 m/s) + (−1.5 m/s 2 ) t2 ⇒ t2 = 3.3 s
Using the above obtained values of a and t2 , we can find x2 as follows:
1
1
x2 = x0 + v0 (t2 − t0 ) + a (t2 − t0 )2 = 0 m + (5.0 m/s)(3.3 s) + (−1.5 m/s 2 )(3.3 s) 2 = 8.3 m
2
2
That is, the puck goes through a displacement of 8.3 m. Since the end of the ramp is 8.5 m from the starting position
x0 and the puck stops 0.2 m or 20 cm before the ramp ends, you are not a winner.
2.61. Model: Model the ball as a particle. Ignore air resistance.
Visualize:
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2-32
Chapter 2
Solve:
(a) We can apply the kinematic equation vf2 − vi2 = 2aΔy twice, once for the launching phase and again for the free
fall phase. In the launching phase the acceleration is up (positive), v0 = 0 and Δy = y1 − y0 = d .
v12 = 2a0d
In the free fall phase the acceleration is a1 = − g , v2 = 0, and Δy = y2 − y1 = h.
−v12 = 2a1h = 2(− g )h
Cancel the negative signs and set the two expressions for v12 equal to each other.
2a0d = 2 gh
Solve for a0 .
a0 =
(b) For h = 3.2m, and d = 0.45 m we get
h
g
d
h
3.2 m
g=
(9.8 m/s 2 ) = 69.7 m/s 2 ≈ 70 m/s 2
d
0.45 m
Assess: The answer is independent of the mass of the ball. The units check out.
a0 =
2.62. Model: The ball is a particle that exhibits freely falling motion according to the constant-acceleration
kinematic equations.
Visualize:
Solve: Using the known values, we have
v12 = v02 + 2a0 ( y1 − y0 ) ⇒ (−10 m/s) 2 = v02 + 2( −9.8 m/s 2 )(5.0 m − 0 m) ⇒ v0 = 14 m/s
2.63. Model: The car is a particle that moves with constant linear acceleration.
Visualize:
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Kinematics in One Dimension
2-33
Solve: The reaction time is 1.0 s, and the motion during this time is
x1 = x0 + v0 (t1 − t0 ) = 0 m + (20 m/s)(1.0 s) = 20 m
During slowing down,
1
x2 = x1 + v1(t2 − t1 ) + a1 (t2 − t1 ) 2 = 200 m
2
1
= 20 m + (20 m/s)(15 s − 1.0 s) + a1 (15 s − 1.0 s) 2 ⇒ a1 = −1.02 m/s 2
2
The final speed v2 can now be obtained as
v2 = v1 + a1 (t2 − t1) = (20 m/s) + (−1.02 m/s 2 )(15 s − 1 s) = 5.7 m/s
2.64. Solve: (a) The quantity
2 P 2(3.6 × 104 W)
=
= 60 m 2 /s3. Thus
1200 kg
m
vx = (60 m 2 /s3 )t
At t = 10 s, vx = (60 m 2 /s3 )(10 s) = 24 m/s ( ≈ 50 mph), and at t = 20 s, vx = 35 m/s ( ≈ 75 mph).
(b) With vx =
2P 1/2
t , we have
m
ax =
(c) At t = 1 s, ax =
dvx
2 P 1 −1/2
P
=
× t
=
2mt
dt
m 2
P
(3.6 × 104 W)
=
= 3.9 m/s 2 . Similarly, at t = 10 s, ax = 1.2 m/s 2 .
2mt
2(1200 kg)(1 s)
(d) Consider the limiting case of very short times. Note that a x → ∞ as t → 0. This is physically impossible for the
Alfa Romeo.
(e) We can use the relationship that vx =
dx
2P 1/2
and integrate to find x(t ). We have vx =
and the initial
t
dt
m
condition xi = 0 at ti = 0. Thus
x
Ñ0 dx =
and x =
2 P t 1/2
t dt
m Ñ0
2 P t 3/2 2 2 P 3/2
=
t
m 3/2 3 m
(f) Time to travel a distance x is found by solving the above equation for t.
⎡3 m ⎤
t=⎢
x⎥
⎣⎢ 2 2 P ⎦⎥
2/3
For x = 402 m, t = 18.2 s.
2.65. Model: Both cars are particles that move according to the constant-acceleration kinematic equations.
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2-34
Chapter 2
Visualize:
Solve: (a) David’s and Tina’s motions are given by the following equations:
1
xD1 = xD0 + vD0 (tD1 − tD0 ) + aD (tD1 − tD0 ) 2 = vD0tD1
2
1
1
2
xT1 = xT0 + vT0 (tT1 − tT0 ) + aT (tT1 − tT0 )2 = 0 m + 0 m + aTtT1
2
2
When Tina passes David the distances are equal and tD1 = tT1, so we get
1
1
2v
2(30 m/s)
2
⇒ vD0 = aTtT1 ⇒ tT1 = D0 =
= 30 s
xD1 = xT1 ⇒ vD0tD1 = aTtT1
2
2
aT
2.0 m/s 2
Using Tina’s position equation,
1
1
2
xT1 = aTtT1
= (2.0 m/s 2 )(30 s) 2 = 900 m
2
2
(b) Tina’s speed vT1 can be obtained from
vT1 = vT0 + aT (tT1 − tT0 ) = (0 m/s) + (2.0 m/s 2 )(30 s − 0 s) = 60 m/s
Assess: This is a high speed for Tina (~134 mph) and so is David’s velocity (~67 mph). Thus the large distance for
Tina to catch up with David (~0.6 miles) is reasonable.
2.66. Model: We will represent the dog and the cat in the particle model.
Visualize:
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Kinematics in One Dimension
2-35
Solve: We will first calculate the time tC1 the cat takes to reach the window. The dog has exactly the same time to
reach the cat (or the window). Let us therefore first calculate tC1 as follows:
1
xC1 = xC0 + vC0 (tC1 − tC0 ) + aC (tC1 − tC0 ) 2
2
1
2
⇒ 3.0 m = 1.5 m + 0 m + (0.85 m/s 2 )tC1
⇒ tC1 = 1.879 s
2
In the time tD1 = 1.879 s, the dog’s position can be found as follows:
1
xD1 = xD0 + vD0 (tD1 − tD0 ) + aD (tD1 − tD0 ) 2
2
1
= 0 m + (1.50 m/s)(1.879 s) + ( −0.10 m/s 2 )(1.879 s) 2 = 2.6 m
2
That is, the dog is shy of reaching the cat by 0.4 m. The cat is safe.
2.67. Model: Jill and the grocery cart will be treated as particles that move according to the constant-acceleration
kinematic equations.
Visualize:
Solve: The final position of Jill when the cart is caught is given by
1
1
1
2
xJ1 = xJ0 + vJ0 (tJ1 − tJ0 ) + aJ0 (tJ1 − tJ0 ) 2 = 0 m + 0 m + aJ0 (tJ1 − 0 s)2 = (2.0 m/s 2 )tJ1
2
2
2
The cart’s position when it is caught is
1
1
xC1 = xC0 + vC0 (tC1 − tC0 ) + aC0 (tC1 − tC0 ) 2 = 20 m + 0 m + (0.5 m/s 2 )(tC1 − 0 s) 2
2
2
2
= 20 m + (0.25 m/s 2 )tC1
Since xJ1 = xC1 and tJ1 = tC1, we get
1
2
2
2
(2.0)tJ1
= 20 s 2 + 0.25tC1
⇒ 0.75tC1
= 20 s 2 ⇒ tC1 = 5.16 s
2
2
⇒ xC1 = 20 m + (0.25 m/s 2 )tC1
= 20 m + (0.25 m/s 2 )(5.16 s)2 = 26.7 m
So, the cart has moved 6.7 m.
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2-36
Chapter 2
2.68. Model: The watermelon and Superman will be treated as particles that move according to constantacceleration kinematic equations.
Visualize:
Solve: The watermelon’s and Superman’s position as they meet each other are
1
yW1 = yW0 + vW0 (tW1 − tW0 ) + aW0 (tW1 − tW0 ) 2
2
1
yS1 = yS0 + vS0 (tS1 − tS0 ) + aS0 (tS1 − tS0 ) 2
2
1
⇒ yW1 = 320 m + 0 m + ( −9.8 m/s 2 )(t W1 − 0 s) 2
2
⇒ yS1 = 320 m + (−35 m/s)(tS1 − 0 s) + 0 m
Because tS1 = tW1,
2
yW1 = 320 m − (4.9 m/s 2 ) tW1
yS1 = 320 m − (35 m/s) tW1
Since yW1 = yS1,
2
320 m − (4.9 m/s 2 )tW1
= 320 m − (35 m/s)tW1 ⇒ t W1 = 0 s and 7.1 s
Indeed, tW1 = 0 s corresponds to the situation when Superman arrives just as the watermelon is dropped off the
Empire State Building. The other value, t W1 = 7.1 s, is the time when the watermelon will catch up with Superman.
The speed of the watermelon as it passes Superman is
vW1 = vW0 + aW0 (t W1 − t W0 ) = 0 m/s + (−9.8 m/s 2 )(7.1 s − 0 s) = −70 m/s
Note that the negative sign implies a downward velocity.
Assess: A speed of 140 mph for the watermelon is understandable in view of the significant distance (250 m)
involved in the free fall.
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Kinematics in One Dimension
2-37
2.69. Model: Treat the car and train in the particle model and use the constant acceleration kinematics equations.
Visualize:
Solve: In the particle model the car and train have no physical size, so the car has to reach the crossing at an
infinitesimally sooner time than the train. Crossing at the same time corresponds to the minimum a1 necessary to
avoid a collision. So the problem is to find a1 such that x2 = 45 m when y2 = 60 m.
The time it takes the train to reach the intersection can be found by considering its known constant velocity.
v0 y = v2 y = 30 m/s =
y2 − y0 60 m
=
⇒ t 2 = 2.0 s
t 2 − t0
t2
Now find the distance traveled by the car during the reaction time of the driver.
x1 = x0 + v0 x (t1 − t0 ) = 0 + (20 m/s)(0.50 s) = 10 m
The kinematic equation for the final position at the intersection can be solved for the minimum acceleration a1.
1
x2 = 45 m = x1 + v1x (t2 − t1 ) + a1 (t2 − t1 ) 2
2
1
= 10 m + (20 m/s)(1.5 s) + a1 (1.5 s) 2
2
⇒ a1 = 4.4 m/s 2
Assess: The acceleration of 4.4 m/s 2 = 2.0 miles/h/s is reasonable for an automobile to achieve. However, you
should not try this yourself! Always pay attention when you drive! Train crossings are dangerous locations, and
many people lose their lives at one each year.
2.70. Model: Model the ball as a particle. Since the ball is heavy we ignore air resistance.
1
Visualize: We use the kinematic equation Δy = v0Δt + a(Δt ) 2 , but we set the origin at the ground so y0 = h and
2
y1 = 0; this means Δy = y1 − y0 = −h. We release the ball from rest so at t0 = 0 we have v0 = 0 and Δt = t. We also
note that a = − g where g is the free-fall acceleration on Planet X. Making all these substitutions leaves
⎛1 ⎞
h = ⎜ g ⎟t2
⎝2 ⎠
So we expect a graph of h vs. t 2 to produce a straight line whose slope is g/2 and whose intercept is zero. Compare
to y = mx + b where y = h, m = g /2, x = t 2 , and b = 0.
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2-38
Chapter 2
Solve: First look at a graph of height vs. fall time and notice that it is not linear. It would be difficult to analyze.
Even though the point (0,0) is not a measured data point, it is valid to add to the data table and graph because it
would take zero time to fall zero distance.
However, the theory has guided us to expect that a graph of height vs. fall time squared would be linear and the
slope would be g /2. First we use a spreadsheet to square the fall times and then graph the height vs. fall time
squared to see if it looks linear and that the intercept is close to zero.
It looks linear and R 2 = 0.996 tells us the linear fit is very good. We also see that the intercept is a very small
negative number which is close to zero, so we have confidence in our model. The fit is not perfect and the intercept is
not exactly zero probably because of uncertainties in timing the fall.
We now conclude that the slope of the best fit line m = 3.7326 is g /2 in the proper units, so g = 2 × 3.7326m/s 2 =
7.5m/s 2 on Planet X.
Assess: The free-fall acceleration on Planet X is a little bit smaller than on earth, but is reasonable. It is customary to
put the independent variable on the horizontal axis and the dependent variable along the vertical axis. Had we done
so here we would have graphed t 2 vs. h and the slope would have been 2/g. Our answer to the question would be
the same.
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Kinematics in One Dimension
2-39
2.71. Model: Model the car as a particle. Ignore air resistance. Hard braking means the wheels are locked (not
turning) and the car is in full skid. For convenience, assume the car is skidding to the right.
Visualize: We use the kinematic equation vf2 − vi2 = 2aΔx. In this case vf = 0 and a < 0, but since we want the
deceleration (which is the absolute value of the acceleration) we drop the negative signs. Relabel vi as v . We’ll call the
beginning of the skid mark the origin so that xi = 0 and the skid length is Δx = x. Making these substitutions leaves
v 2 = (2a ) x
So we expect a graph of v 2 vs. x to produce a straight line whose slope is 2a and whose intercept is zero. Compare
to y = mx + b where y = v 2 , m = 2a, x = x, and b = 0.
Solve: First look at a graph of the data of speed vs. skid length and notice that it is not linear. It would be difficult to
analyze. We added the point (0,0) to the data table and graph because we are sure that if the speed were zero the skid
length would also be zero.
However, the theory has guided us to expect that a graph of speed squared vs. skid length would be linear and the
slope would be 2a . First we use a spreadsheet to square the speed and then graph the speed squared vs. skid length
to see if it looks linear and that the intercept is close to zero. Only if it is linear is the deceleration constant,
independent of speed.
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2-40
Chapter 2
(a) It looks linear and R 2 = 0.995 tells us the linear fit is very good. This means the deceleration (involved in the
slope) is constant, independent of speed. We also see that the intercept is a very small negative number which is close
to zero, so we have confidence in our model. The fit is not perfect and the intercept is not exactly zero probably
because of uncertainties in measuring the speed.
(b) We now conclude that the slope of the best fit line m = 15.812 is 2a in the proper units, so the deceleration is
1
a = × 15.812 m/s 2 = 7.9 m/s 2 .
2
Assess: The value of 7.9 m/s 2 seems reasonable for hard braking. It is customary to put the independent variable on
the horizontal axis and the dependent variable along the vertical axis. Had we done so here we would have graphed
x vs. v 2 and the slope would have been 1/2a. Our answer to the question would be the same.
2.72. Solve: A comparison of the given equation with the constant-acceleration kinematics equation
1
x1 = x0 + v0 (t1 − t0 ) + ax (t1 − t0 ) 2
2
yields the following information: x0 = 0 m, x1 = 64 m, t0 = 0 ,t1 = 4 s, and v0 = 32 m/s.
(a) After landing on the deck of a ship at sea with a velocity of 32 m/s, a fighter plane is observed to come to a
complete stop in 4.0 seconds over a distance of 64 m. Find the plane’s deceleration.
(b)
1
(c) 64 m = 0 m + (32 m/s)(4 s − 0 s) + a x (4 s − 0 s) 2
2
64 m = 128 m + (8 s 2 ) a x ⇒ a x = −8 m/s 2
The deceleration is the absolute value of the acceleration, or 8 m/s 2 .
2.73. Solve: (a) A comparison of this equation with the constant-acceleration kinematic equation
2
(v1y ) 2 = v0y
+ 2(ay )( y1 − y0 )
yields the following information: y0 = 0 m, y1 = 10 m, a y = −9.8 m/s 2 , and v1 y = 10 m/s. It is clearly a problem of
free fall. On a romantic Valentine’s Day, John decided to surprise his girlfriend, Judy, in a special way. As he
reached her apartment building, he found her sitting in the balcony of her second floor apartment 10 m above the first
floor. John quietly armed his spring-loaded gun with a rose, and launched it straight up to catch her attention. Judy
noticed that the flower flew past her at a speed of 10 m/s. Judy is refusing to kiss John until he tells her the initial
speed of the rose as it was released by the spring-loaded gun. Can you help John on this Valentine’s Day?
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Kinematics in One Dimension
2-41
(b)
2
(c) (10 m/s)2 = v0y
− 2(9.8 m/s 2 )(10 m − 0 m) ⇒ v0y = 17.2 m/s
Assess: The initial velocity of 17.2 m/s, compared to a velocity of 10 m/s at a height of 10 m, is very reasonable.
2.74. Solve: A comparison with the constant-acceleration kinematics equation
(v1x )2 = (v0x ) 2 + 2ax ( x1 − x0 )
yields the following quantities: x0 = 0 m, v0 x = 5 m/s, v1x = 0 m/s, and a x = −(9.8 m/s 2 )sin10D.
(a) A wagon at the bottom of a frictionless 10° incline is moving up at 5 m/s. How far up the incline does it move
before reversing direction and then rolling back down?
(b)
(c) (0 m/s) 2 = (5 m/s)2 − 2(9.8 m/s 2 )sin10°( x1 − 0 m)
⇒ 25(m/s) 2 = 2(9.8 m/s 2 )(0.174) x1 ⇒ x1 = 7.3 m
2.75. Solve: (a) From the first equation, the particle starts from rest and accelerates for 5 s. The second equation
gives a position consistent with the first equation. The third equation gives a subsequent position following the
second equation with zero acceleration. A rocket sled accelerates from rest at 20 m/s 2 for 5 s and then coasts at
constant speed for an additional 5 s. Draw a graph showing the velocity of the sled as a function of time up to t = 10 s.
Also, how far does the sled move in 10 s?
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2-42
Chapter 2
(b)
1
(c) x1 = (20 m/s 2 )(5 s) 2 = 250 m
2
v1 = 20 m/s 2 (5 s) = 100 m/s
x2 = 250 m + (100 m/s)(5 s) = 750 m
2.76. Model: The masses are particles.
Visualize:
Solve: The rigid rod forms the hypotenuse of a right triangle, which defines a relationship between x2 and y1:
x22 + y12 = L2 .
Taking the time derivative of both sides yields
2 x2
We can now use v2 x =
dx2
dy
+ 2 y1 1 = 0
dt
dt
dy
dx2
and v1 y = 1 to write x2v2 x + y1v1 y = 0.
dt
dt
⎛y ⎞
y
Thus v2 x = − ⎜ 1 ⎟ v1 y . But from the figure, 1 = tan θ ⇒ v2 x = −v1 y tan θ .
x
x
⎝ 2⎠
2
Assess: As x2 decreases (v2 x < 0), y1 increases (v1 y > 0), and vice versa.
2.77. Model: The rocket and the bolt will be represented as particles to investigate their motion.
Visualize:
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Kinematics in One Dimension
2-43
The initial velocity of the bolt as it falls off the side of the rocket is the same as that of the rocket, that is,
vB0 = vR1 and it is positive since the rocket is moving upward. The bolt continues to move upward with a deceleration
equal to g = 9.8 m/s 2 before it comes to rest and begins its downward journey.
Solve: To find aR we look first at the motion of the rocket:
To find aR
1
yR1 = yR0 + vR0 (tR1 − tR0 ) + aR (tR1 − tR0 ) 2
2
1
= 0 m + 0 m/s + aR (4.0 s − 0 s)2 = 8aR
2
we must determine the magnitude of yR1 or yB0 . Let us now look at the bolt’s motion:
1
yB1 = yB0 + vB0 (tB1 − tB0 ) + aB (tB1 − tB0 )2
2
1
0 = yR1 + vR1 (6.0 s − 0 s) + (−9.8 m/s 2 )(6.0 s − 0 s) 2
2
⇒ yR1 = 176.4 m − (6.0 s) vR1
Since vR1 = vR0 + aR (tR1 − tR0 ) = 0 m/s + 4 aR = 4 aR the above equation for yR1 yields yR1 = 176.4 − 6.0(4aR ). We
know from the first part of the solution that yR1 = 8aR . Therefore, 8aR = 176.4 − 24.0aR and hence aR = 5.5 m/s 2 .
2.78. Model: The rocket car is a particle that moves according to the constant-acceleration equations of motion.
Visualize:
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2-44
Chapter 2
Solve: This is a two-part problem. For the first part,
1
1
1
2
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0 ( 9.0 s − 0 s ) = (81 s 2 )a0
2
2
2
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (9.0 s − 0 s) = (9.0 s)a0
During the second part of the problem,
1
x2 = x1 + v1(t2 − t1 ) + a1 (t2 − t1) 2
2
1
1
2
⇒ 990 m = (81 s )a0 + (9.0 s)a0 (12 s − 9.0 s) + (−5.0 m/s 2 )(12 s − 9.0 s) 2
2
2
⇒ a0 = 15 m/s 2
This leads to:
v1 = (9.0 s) a0 = (9.0 s)(15 m/s 2 ) = 135 m/s
Using this value of v1, we can now calculate v2 as follows:
v2 = v1 + a1 (t2 − t1) = (135 m/s) + (−5.0 m/s 2 )(12 s − 9.0 s) = 120 m/s
That is, the car’s speed as it passes the judges is 120 m/s.
Assess: This is a very fast motion (~250 mph), but the acceleration is large and the long burn time of 9 s yields a
high velocity.
2.79. Model: Use the particle model.
Visualize:
Solve: (a) Substituting into the constant-acceleration kinematic equation
1
10 ⎞
⎛
x2 = x1 + v1(t2 − t1) + a1(t2 − t1 ) 2 ⇒ 100 m = x1 + v1 ⎜ t2 − ⎟ + 0 m
2
3⎠
⎝
100 − x1 10
+
t2 =
v1
3
Let us now find v1 and x1 as follows:
⎛ 10
⎞
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + (3.6 m/s 2 ) ⎜
s − 0 s ⎟ = 12 m/s
3
⎝
⎠
2
1
1
⎛ 10
⎞
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 )2 = 0 m + 0 m + (3.6 m/s 2 ) ⎜
s − 0 s ⎟ = 20 m
2
2
⎝ 3
⎠
The expression for t2 can now be solved as
t2 =
100 m − 20 m 10 s
+
= 10 s
12 m/s
3
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Kinematics in One Dimension
2-45
(b) The top speed = 12 m/s which means v1 = 12 m/s. To find the acceleration so that the sprinter can run the
100-meter dash in 9.9 s, we use
v1 = v0 + a0 (t1 − t0 ) ⇒ 12 m/s = 0 m/s + a0t1 ⇒ t1 =
12 m/s
a0
1
1
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0t12 = a0t12
2
2
2
Since x2 = x1 + v1(t2 − t1) + 12 a1(t2 − t1 )2 , we get
1
100 m = a0t12 + (12 m/s) (9.9 s − t1 ) + 0 m
2
Substituting the above equation for t1 in this equation,
2
⎛
12 m/s ⎞
⎛ 1 ⎞ ⎛ 12 m/s ⎞
2
100 m = ⎜ ⎟ a0 ⎜
⎟ + (12 m/s) ⎜ 9.9 s −
⎟ ⇒ a0 = 3.8 m/s
a
⎝ 2 ⎠ ⎝ a0 ⎠
0 ⎠
⎝
(c) We see from parts (a) and (b) that the acceleration has to be increased from 3.6 m/s 2 to 3.8 m/s 2 for the sprint
time to be reduced from 10 s to 9.9 s, that is, by 1%. This decrease of time by 1% corresponds to an increase of
acceleration by
3.8 − 3.6
× 100, = 5.6,
3. 6
2.80. Solve: (a) The acceleration is the time derivative of the velocity.
ax =
dvx d
= [a (1 − e −bt )] = abe −bt
dt
dt
With a = 11.81 m/s and b = 0.6887 s −1, ax = 8.134e−0.6887t m/s 2 . At the times t = 0 s, 2 s, and 4 s, this has the
values 8.134 m/s 2 , 2.052 m/s 2 , and 0.5175 m/s 2 .
dx
dx
, the position x is the integral of the velocity. With vx =
= a − ae−bt and the initial condition that
dt
dt
xi = 0 m at ti = 0 s,
(b) Since vx =
x
t
t
2bt
Ñodx = Ñoa dt − Ñoae
dt
Thus
t
a
x = at + e2bt
b
o
t
o
a
a
= at + e2bt −
b
b
This can be written a little more neatly as
a
x = (bt + e−bt − 1)
b
= 17.15(0.6887t + e −0.6887t − 1) m
(c) By trial and error, t = 9.92 s yields x = 100.0 m.
Assess: Lewis’s actual time was 9.93 s.
2.81. Model: We will use the particle-model to represent the sprinter and the equations of kinematics.
Visualize:
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2-46
Chapter 2
Solve: Substituting into the constant-acceleration kinematic equations,
1
1
1
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0 (4 s − 0 s) 2 = a0t12 = a0 (4.0 s) 2
2
2
2
2
⇒ x1 = (8 s 2 )a0
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (4.0 s − 0 s) ⇒ v1 = (4.0 s) a0
From these two results, we find that x1 = (2 s)v1. Now,
1
x2 = x1 + v1 (t2 − t1) + a1(t2 − t1 ) 2
2
⇒ 100 m = (2 s)v1 + v1 (10 s − 4 s) + 0 m ⇒ v1 = 12.5 m/s
Assess: Using the conversion 2.24 mph = 1 m/s, v1 = 12.5 m/s = 28 mph. This speed as the sprinter reaches the finish
line is physically reasonable.
2.82. Model: The balls are particles undergoing constant acceleration.
Visualize:
Solve: (a) The positions of each of the balls at t1 is found from kinematics.
1
1
( y1 ) A = ( y0 ) A + (v0 y ) A t1 − gt12 = v0t1 − gt12
2
2
1 2
1 2
( y1 ) B = ( y0 ) B + (v0 y ) B t1 − gt1 = h − gt1
2
2
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Kinematics in One Dimension
2-47
In the particle model the balls have no physical extent, so they meet when ( y1 ) A = ( y1 ) B . This means
v0t1 −
1 2
1
h
gt1 = h − gt12 ⇒ t1 =
2
2
v0
1
gh 2
Thus the collision height is ycoll = h − gt12 = h −
.
2
2v0 2
(b) We need the collision to occur while ycoll ≥ 0. Thus
h−
So hmax =
gh 2
2v02
≥ 0 ⇒ 1≥
gh
2v0 2
⇒ h≤
2v0 2
g
2v0 2
.
g
(c) Ball A is at its highest point when its velocity (v1 y ) A = 0.
(v1 y ) A = (v0 y ) A − gt1 ⇒ 0 = v0 − gt1 ⇒ t1 =
v0
g
h
h v0
v2
. Equating these,
= ⇒h= 0 .
v0
v0 g
g
Assess: Interestingly, the height at which a collision occurs while Ball A is at its highest point is exactly half of
hmax .
In (a) we found that the collision occurs at t1 =
2.83. Model: The space ships are represented as particles.
Visualize:
Solve: The difficulty with this problem is how to describe “barely avoid.” The Klingon ship is moving with constant
speed, so its position-versus-time graph is a straight line from xK0 = 100 km. The Enterprise will be decelerating, so
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2-48
Chapter 2
its graph is a parabola with decreasing slope. The Enterprise doesn’t have to stop; it merely has to slow quickly
enough to match the Klingon ship speed at the point where it has caught up with the Klingon ship. (You do the same
thing in your car when you are coming up on a slower car; you decelerate to match its speed just as you come up on
its rear bumper.) Thus the parabola of the Enterprise will be tangent to the straight line of the Klingon ship, showing
that the two ships have the same speed (same slopes) when they are at the same position. Mathematically, we can say
that at time t1 the two ships will have the same position ( xE1 = xK1 ) and the same velocity (vE1 = vK1). Note that we
are using the particle model, so the ships have zero length. At time t1,
xK1 = xK0 + vK0t1
vK1 = vK0
1
vE1 = vE0 + at1
xE1 = vE0t1 + at12
2
Equating positions and velocities at t1 :
1
xK0 + vK0t1 = vE0t1 + at12
2
vK0 = vE0 + at1
We have two simultaneous equations in the two unknowns a and t1. From the velocity equation,
t1 = (vK0 − vE0 ) /a
Substituting into the position equation gives
2
xK0 = −(vK0 − vE0 ) ⋅
⇒ a =2
(vK0 − vE0 ) 1 ⎛ (vK0 − vE0 ) ⎞
(vK0 − vE0 )2
+ a ⋅⎜
⎟ =2
2 ⎝
2a
a
a
⎠
(vK0 − vE0 ) 2
(20,000 m/s − 50,000 m/s) 2
=−
= −4500 m/s 2
2 xK0
2(100,000 m)
The magnitude of the acceleration is 4500 m/s 2 .
Assess: The deceleration is 4500 m/s 2 , which is a rather extreme ≈ 460 g. Fortunately, the Enterprise has other
methods to keep the crew from being killed.
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VECTORS AND COORDINATE SYSTEMS
3
Conceptual Questions
3.1. The magnitude of the displacement vector is the minimum distance traveled since the displacement is the vector
sum of a number of individual movements. Thus, it is not possible for the magnitude of the displacement vector to be
more than the distance traveled. If the individual movements are all in the same direction, the total displacement and
the distance traveled are equal. However, it is possible that the total displacement is less than the distance traveled, if
the individual movements are not in the same direction.
G
G
G
G
possible that C > A + B because, if A and B point in different directions, putting them tip to tail gives a resultant
with a shorter length (see figure below).
3.2. It is possible that C = A + B only if A and B both point in the same direction as in the figure below. It is not
G
G
G
G
G
A and B are parallel but in opposite directions, C will still have a length greater than or equal to zero.
3.3. It is possible that C = 0 if A = − B. It is not possible for the length of a vector to be negative, so C ≥ 0. Even if
3.4. No, it is not possible to add a scalar to a vector, since the scalar has no direction.
G
3.5. The zero vector 0 has zero length. It does not point in any direction.
3.6. A vector can have a component that is zero and still have nonzero length only if another component is nonzero.
For example, consider the vector iˆ = (1, 0), which points along the x-axis.
3.7. If one component of a vector is nonzero then it is not possible for the vector to have zero magnitude. The
magnitude of the vector depends on the sum of the squares of the components, so any component signs do not matter.
3.8. No, it is not possible for two vectors with unequal magnitudes to add to zero. To add to zero, two vectors must
be antiparallel and of the same length (magnitude).
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3-1
3-2
Chapter 3
3.9. (a) False, because the size of a vector is fixed.
(b) False, because the direction of a vector in space is independent of any coordinate system.
(c) True, because the orientation of the vector relative to the axes can be different.
Exercises and Problems
Section 3.1 Vectors
Section 3.2 Properties of Vectors
3.1. Visualize:
G G
G
G
Solve: (a) To find A + B, we place the tail of vector B on the tip of vector A and draw an arrow from the tail of
G
G
vector A to the tip of vector B.
G G G
G
G
G
(b) Since A − B = A + (− B) , we place the tail of the vector − B on the tip of vector A and then draw an arrow from
G
G
the tail of vector A to the tip of vector − B .
3.2. Visualize:
G G
G
G
Solve: (a) To find A + B, we place the tail of vector B on the tip of vector A and then draw an arrow from vector
G,
G,
A s tail to vector B s tip.
G G G
G
G G
G
G
(b) To find A − B , we note that A − B = A + (− B ). We place the tail of vector − B on the tip of vector A and then
G,
G
draw an arrow from vector A s tail to the tip of vector − B.
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Vectors and Coordinate Systems
3-3
Section 3.3 Coordinate Systems and Vector Components
3.3. Visualize:
G
Solve: Vector E points to the left and up, so the components E x and E y are negative and positive, respectively,
according to the Tactics Box 3.1.
(a) E x = − E cosθ and E y = E sin θ .
(b) E x = − E sin ϕ and E y = E cos ϕ .
Assess: Note that the role of sine and cosine are reversed because we are using a different angle. θ and φ are
complementary angles.
3.4. Visualize: The figure shows the components vx and v y , and the angle θ .
G
Solve: We have, v y = −v sin θ where we have manually inserted the minus sign because v y points in the negative-y
direction. The x-component is vx = v cosθ . Taking the ratio vx /v y and solving for vx gives vx = − v y (tan θ ) −1 =
− (−10 m/s)(tan 40°) −1 = 12 m/s.
Assess: The x-component is positive since the position vector is in the fourth quadrant.
G
3.5. Visualize: The position vector r whose magnitude r is 10 m has an x-component of 8 m. It makes an angle θ
with the + x-axis in the first quadrant.
Solve: Using trigonometry, rx = r cosθ , or 8 m = (10 m)cosθ . This gives θ = 36.9°. Thus the y-component of the
G
position vector r is ry = r sin θ = (10 m)sin (36.9°) = 6 m.
Assess: The y-component is positive since the position vector is in the first quadrant.
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3-4
Chapter 3
3.6. Visualize:
We will follow rules in Tactics Box 3.1.
G
Solve: (a) Vector r points to the right and down, so the components rx and ry are positive and negative,
respectively:
rx = r cosθ = (100 m)cos 45° = 71 m, ry = −r sin θ = −(100 m)sin 45° = −71 m
G
(b) Vector v points to the right and up, so the components vx and v y are both positive:
vx = v cos θ = (300 m/s)cos(20°) = 280 m/s, v y = v sin θ = (300 m/s)sin(20°) = 100 m/s
G
(c) Vector a has the following components:
a x = − a cos θ = − (5.0 m/s 2 )cos90° = 0 m/s 2 , a y = − a sin θ = −(5.0 m/s 2 )sin 90° = −5.0 m/s 2
Assess: The components have same units as the vectors. Note the minus signs we have manually inserted according
to Tactics Box 3.1.
3.7. Visualize:
We will follow the rules given in Tactics Box 3.1.
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Vectors and Coordinate Systems
Solve:
(a) vx = (10 m/s)cos(90.0°) = 0 m/s
2
(b) a x = (20 m/s )cos30° = 17 m/s
2
(c) Fx = − (100 N)sin(36.9°) = −60 N
3-5
v y = −(10 cm/s)sin(90.0°) = −10 m/s
a y = −(20 m/s 2 )sin 30° = −10 m/s 2
Fy = (100 N)cos(36.9°) = 80 N
Assess: The components have the same units as the vectors. Note the minus signs we have manually inserted
according to Tactics Box 3.1.
3.8. Visualize:
G
G
The components of the vector C and D, and the angles θ are shown.
K
K
Solve: For C we have C x = − (3.15 m)cos(15°) = −3.04 m and C y = (3.15 m)sin (15°) = 0.815 m. For D we have
Dx = (25.6 m)sin (30°) = 12.8 m and Dy = −(25.67 m)cos(30°) = −22.2 m.
G
G
Assess: The components of the vectors C and D have the same units as the vectors themselves. Note the minus
signs we have manually inserted, as per the rules of Tactics Box 3.1.
3.9. Visualize:
Solve: The magnitude of the vector is B = Bx2 + By2 = (2.0 T) 2 + ( −1.0 T) 2 = 5.0 T = 2.2 T. The angle θ is
θ = tan −1
By
Bx
⎛ 1.0 T ⎞
= tan −1 ⎜
⎟ = 27°
⎝ 2.0 T ⎠
Assess: Since By < Bx , the angle θ made with the +x-axis is less than 45°. θ = 45° for B y = Bx .
Section 3.4 Vector Algebra
3.10. Visualize:
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3-6
Chapter 3
Solve: (a) Using the formulas for the magnitude and direction of a vector, we have:
⎛4⎞
B = ( −4) 2 + (4) 2 = 5.7, θ = tan −1 ⎜ ⎟ = 45°
⎝4⎠
⎛ 1.0 ⎞
(b) r = ( −2.0 cm)2 + ( −1.0 cm)2 = 2.2 cm, θ = tan −1 ⎜
⎟ = 27°
⎝ 2.0 ⎠
⎛ 100 ⎞
(c) v = (−10 m/s)2 + (−100 m/s)2 = 100 m/s, θ = tan −1 ⎜
⎟ = 84°
⎝ 10 ⎠
⎛ 10 ⎞
(d) a = (10 m/s 2 ) 2 + (20 m/s 2 ) 2 = 22 m/s 2 , θ = tan −1 ⎜ ⎟ = 27°
⎝ 20 ⎠
3.11. Visualize:
Solve: (a) Using the formulas for the magnitude and direction of a vector, we have:
⎛ ry ⎞
⎛6⎞
A = (4) 2 + ( −6) 2 = 7.2, θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = 56° below the +x-axis
r
⎝4⎠
⎝ x⎠
(b) r = (50 m)2 + (80 m) 2 = 94 m
⎛ 80 m ⎞
⎟ = 58° above the +x-axis
⎝ 50 m ⎠
θ = tan −1 ⎜
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Vectors and Coordinate Systems
3-7
⎛ 40 ⎞
⎟ = 63° above the −x-axis
⎝ 20 ⎠
θ = tan −1 ⎜
(c) v = (−20 m/s) 2 + (40 m/s) 2 = 45 m/s
(d) a = (2.0 m/s 2 ) 2 + (−6.0 m/s 2 ) 2 = 6.3 m/s 2
⎛ 2.0 ⎞
⎟ = 18° to the right of the − y -axis
⎝ 6.0 ⎠
θ = tan −1 ⎜
3.12. Visualize:
G G
G G G
G
G
K
We have C = A − B or C = A + (− B ), where − B is the same as B, but in the opposite direction. Look back at
Tactics Box 1.2, which shows how to perform vector subtraction graphically.
G
G
G
G
G
Solve: To obtain vector C from A and B, we place the tail of − B on the tip of A, and then draw a vector arrow
G
G
G
from the tail of A to the tip of − B. Reading the coordonates off the graph, we find that C = 22iˆ + 5 ˆj
G G
G
G
G
3.13. Visualize: The vectors A, B, and C = A + B are shown.
G
G G G
G
Solve: (a) We have A = 4iˆ − 2 ˆj and B = −3iˆ + 5 ˆj. Thus, C = A + B = (4iˆ − 2 ˆj ) + (−3iˆ + 5 ˆj ) = 1iˆ + 3 ˆj.
G G
G
(b) Vectors A, B, and C are shown in the figure above.
G
K
(c) Since C = 1iˆ + 3 ˆj = C xiˆ + C y ˆj , C x = 1, and C y = 3. Therefore, the magnitude and direction of C are
C = (1) 2 + (3)2 = 10 = 3.2 and θ = tan −1 (C y /C x ) = tan −1(3/1) = 72°, respectively.
G
Assess: The vector C is to the right and up, thus implying that both the x and y components are positive. Also
θ > 45° since C y > C x .
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3-8
Chapter 3
3.14. Visualize:
G
G
G G
G
K
Solve: (a) We have A = 4iˆ − 2 ˆj , B = −3iˆ + 5 ˆj , and − B = 3iˆ − 5 ˆj. Thus, D = A + (− B ) = (4 + 3)iˆ + (−2 − 5) ˆj = 7iˆ − 7 ˆj.
G G
G
(b) Vectors A, B, and D are shown in the figure above.
G
G
(c) Since D = 7iˆ − 7 ˆj = Dxiˆ + Dy ˆj , Dx = 7 and D y = −7. Therefore, the magnitude and direction of D are
D = (7) 2 + ( −7) 2 = 7 2 = 9.9
(
)
θ = tan −1 Dy /Dx = tan −1(7/7) = 45°
Assess: Since Dy = Dx , the angle θ = 45°, as expected.
3.15. Visualize:
G
G
G
G
Solve: (a) We have A = 4iˆ − 2 ˆj and B = −3iˆ + 5 ˆj. This means 4 A = 16iˆ − 8 ˆj and 2 B = 26iˆ + 10 ˆj. Thus,
G
G
G
E = 4 A + 2 B = [16 + (−6)]iˆ + [( −8) + 10] ˆj = 10iˆ + 2 ˆj.
G G
G
(b) Vectors A , B, and E are shown in the figure above.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Vectors and Coordinate Systems
3-9
G
G
(c) From the E vector, E x = 10 and E y = 2. Therefore, the magnitude and direction of E are
E = (10) 2 + (2) 2 = 104 = 10, θ = tan −1 ( E y /E x ) = tan −1 (2/10) = 11°
G
So E is 10, 11° above the +x-axis.
3.16. Visualize:
G
G
G
G G
G
Solve: (a) We have A = 4iˆ − 2 ˆj and B = −3iˆ + 5 ˆj. This means 4 B = −12iˆ + 20 ˆj. Hence, F = A − 4 B = [4 − (−12)]iˆ +
[−2 − 20] ˆj = 16iˆ − 22 ˆj = Fxiˆ + Fy ˆj , so Fx = 16 and Fy = 222.
G G
G
(b) The vectors A, B, and F are shown in the above figure.
G
(c) The magnitude and direction of F are
F = Fx2 + Fy2 = (16) 2 + (−22) 2 = 27
(
)
θ = tan −1 Fy /Fx = tan −1 (22/16) = 54°
Assess: Fy > Fx implies θ > 45°, which is consistent with the figure.
3.17. Visualize:
G
Solve: In coordinate system I, the vector B makes an angle of 60° counterclockwise from vertical, so it has an angle
G
of θ = 30° with the negative x-axis. Since B points to the left and up, it has a negative x-component and a positive
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3-10
Chapter 3
y-component. Thus, Bx = − (5.0 m)cos(30°) = −4.3 m and B y = + (5.0 m)sin (30°) = 2.5 m. Thus,
G
B = −(4.3 m)iˆ + (2.5 m) ˆj.
G
In coordinate system II, the vector B makes an angle of 30° with the +y-axis and is to the left and up. This means
we have to manually insert a minus sign for the x-component. Thus, Bx = − B sin (30°) = − (5.0 m)sin (30°) = −2.5 m,
G
and B y = + B cos(30°) = (5.0 m)cos (30°) = 4.3 m. Thus B = −(2.5 m)iˆ + (4.3 m) ˆj.
G
3.18. Visualize: Refer to Figure EX3.18. The velocity vector v points south and makes an angle of 30° with the
G
−y-axis. The vector v points to the left and down, implying that both vx and v y are negative.
Solve: We have vx = − v sin (30°) = −(100 m/s)sin (30°) = −50 m/s and
v y = − v cos(30°) = −(100 m/s)cos(30°) = −87 m/s.
G
Assess: Notice that vx and v y have the same units as v .
3.19. Visualize: (a)
G G
G
Solve: (b) The components of the vectors A, B, and C are
Ax = (3.0 m)cos(20°) = 2.8 m and Ay = − (3.0 m)sin (20°) = −1.0 m; Bx = 0 m and B y = 2.0 m;
C x = − (5.0 m)cos(70°) = −1.7 m and C y = − (5.0 m)sin (70°) = −4.7 m. This means the vectors can be written as
G
G
G
A = (2.8iˆ + 1.0 ˆj ) m, B = (2.0 ˆj ) m, C = ( − 1.7iˆ − 4.7 ˆj ) m
G G G G
(c) We have D = A + B + C = (1.1 m)iˆ − (3.7 m) ˆj. This means
D = (1.1 m)2 + (3.7 m)2 = 3.9 m
G
The direction of D is south of east, 74° below the +x-axis.
θ = tan −1(3.9/1.09) = 74°
G
G
3.20. Solve: We have E = Exiˆ + E y ˆj = 2iˆ + 3 ˆj, which means Ex = 2 and E y = 3. Also, F = Fxiˆ + Fy ˆj = 2iˆ − 2 ˆj ,
which means Fx = 2 and Fy = 22.
G
G
(a) The magnitude of E is given by E = Ex2 + E y2 = (2) 2 + (3) 2 = 3.6 and the magnitude of F is given by
F = Fx2 + Fy2 = (2) 2 + (2) 2 = 2.8.
G G
G G
(b) Since E + F = 4iˆ + 1 ˆj, the magnitude of E + F is (4) 2 + (1) 2 = 4.1.
G
G
G
G
(c) Since − E − 2 F = −(2iˆ + 3 ˆj ) − 2(2iˆ − 2 ˆj ) = −6iˆ + 1 ˆj , the magnitude of − E − 2 F is
(−6) 2 + (1) 2 = 6.1.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Vectors and Coordinate Systems
G
3-11
G
3.21. Solve: We have r = (5.0iˆ + 4.0 ˆj )t 2 m. This means that r does not change the ratio of its components as
G
G
t increases; that is, the direction of r is constant. The magnitude of r is given by
r = (5.0t 2 )2 + (4.0t 2 ) 2 m = 6.40t 2 m.
(a) The particle’s distance from the origin at t = 0 s, t = 2 s, and t = 5 s is 0 m, 26 m, and 160 m.
G
dr
dt 2
G
= (5.0iˆ + 4.0 ˆj )
m/s = (5.0iˆ + 4.0 ˆj )2t m/s = (10iˆ + 8.0 ˆj )t m/s.
(b) The particle’s velocity is v (t ) =
dt
dt
(c) The magnitude of the particle’s velocity is given by v = (10t )2 + (8.0t ) 2 = 13t m/s. The particle’s speed at t = 0 s,
t = 2 s, and t = 5 s is 0 m/s, 26 m/s, and 64 m/s.
3.22. Visualize:
G
G
G
Solve: (a) Vector C is the sum of vectors A and B, which is obtained using the tip-to-tip rule of graphical
addition. Its magnitude is measured to be 4.7 and its angle made with the +x-axis is measured to be 33°.
(b) The geometry of parallelograms shows that φ = 180° − (θ B − θ A ) = 180° − (60° − 20°) = 140°. Combining this with
the law of cosines, C 2 = A2 + B 2 − 2 AB cos φ , gives
C = (3)2 + (2) 2 − 2(3)(2)cos(140°) = 4.7
Using the law of sines:
Thus, θC = α + 20° = 36°.
(c) We have:
sin α sin140°
=
⇒ α = 16°
2
4.71
Ax = A cosθ A = 3cos 20° = 2.82
Ay = A sin θ A = 3sin 20° = 1.03
Bx = B cosθ B = 2cos60° = 1.00
By = B sin θ B = 2sin 60° = 1.73
This means: C x = Ax + Bx = 3.82 and C y = Ay + B y = 2.76. The magnitude and direction of C are given by
⎛ Cy ⎞
−1 ⎛ 2.76 ⎞
C = C x2 + C y2 = (3.82)2 + (2.76)2 = 4.7 θC = tan −1 ⎜
⎟ = tan ⎜
⎟ = 36°
⎝ 3.82 ⎠
⎝ Cx ⎠
Assess: Using the method of vector components and their algebraic addition to find the resultant vector yields the
same results as using the graphical addition of vectors.
3.23. Visualize: Refer to Figure P3.23 in your textbook.
G
G G G
G
G G
Solve: (a) We are given that A + B + C = 1 ˆj with A = 4iˆ, and C = −2 ˆj. This means A + C = 4iˆ − 2 ˆj. Thus,
G G G
G G
G
B = ( A + B + C ) − ( A + C ) = (1 ˆj ) − (4iˆ − 2 ˆj ) = −4iˆ + 3 ˆj.
G
(b) We have B = Bxiˆ + By ˆj with Bx = −4 and B y = 3. Hence, B = ( −4) 2 + (3) 2 = 5.0
θ = tan −1
By
Bx
⎛3⎞
= tan −1 ⎜ ⎟ = 37°
⎝4⎠
G
G
Since B has a negative x-component and a positive y-component, the vector B is in the second quadrant and the
G
angle θ made by B is measured above the −x-axis.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3-12
Chapter 3
Assess: Since By < Bx , θ < 45° as obtained above.
3.24. Visualize:
⎛1⎞
⎛2⎞
Solve: (a) θ E = tan −1 ⎜ ⎟ = 45°, θ F = tan −1 ⎜ ⎟ = 63°. Thus φ = 180° − θ E − θ F = 72°
1
⎝ ⎠
⎝1⎠
(b) From the figure, E = 2 and F = 5 . Using
G 2 = E 2 + F 2 − 2 EF cos ϕ = ( 2) 2 + ( 5)2 − 2( 2)( 5)cos(180° − 72°)
G=3
sin α sin (180° − 72°)
Furthermore, using
=
⇒ α = 45°
2.975
5
G
Since θ E = 45°, the angle made by the vector G with the +x-axis is θG = (α + θ E ) = 45° + 45° = 90°.
(c) We have
E x = +1.0, and E y = +1.0
Fx = −1.0, and Fy = +2.0
Gx = 0.0,
and G y = 3.0
G = (0.0) 2 + (3.0) 2 = 3.0, and
θ = tan −1
Gy
Gx
⎛ 3.0 ⎞
= tan −1 ⎜
⎟ = 90°
⎝ 0.0 ⎠
G
That is, the vector G makes an angle of 90° with the x-axis.
Assess: The graphical solution and the vector solution give the same answer within the given significance of figures.
3.25. Visualize: Refer to Figure P3.25.
Solve: From the rules of trigonometry, we have Ax = 4cos(40°) = 3.06 and Ay = 4sin (40°) = 2.57. Also,
G G G G
Bx = −2cos(10°) = −1.97 and B y = +2sin (10°) = 0.35. Since A + B + C = 0,
G
G G
G
G
C = − A − B = (− A) + (− B) = ( −3.06iˆ − 2.57 ˆj ) + (+1.97iˆ − 0.35 ˆj ) = −1iˆ − 3 ˆj.
3.26. Visualize:
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Vectors and Coordinate Systems
3-13
G
G
Solve: In the tilted coordinate system, the vectors A and B are expressed as:
G
G
A = [2sin (15°) m]iˆ + [2cos(15°) m] ˆj and B = [4cos(15°) m]iˆ − [4sin (15°) m] ˆj.
G G
G
Therefore, D = 2 A + B = (4 m)[sin (15°) + cos(15°)] iˆ + (4 m)[cos(15°) − sin (15°)] ˆj = (4.9 m) iˆ + (2.9 m) ˆj.
Assess: The magnitude of this vector is D = (4.9 m) 2 + (2.9 m) 2 = 5.7 m, and it makes an angle of
θ = tan −1 (2.9 m/4.9 m) = 31° with the +x-axis. The resultant vector can be obtained graphically by using the rule of
tail-to-tip addition.
3.27. Visualize:
G
The magnitude of the unknown vector is 1 and its direction is along iˆ + ˆj . Let A = iˆ + ˆj as shown in the diagram.
G
G
That is, A = 1iˆ + 1 ˆj and the x- and y-components of A are both unity. Since θ = tan −1 ( Ay /Ax ) = 45°, the unknown
vector must make an angle of 45° with the +x-axis and have unit magnitude.
G
Solve: Let the unknown vector be B = Bxiˆ + By ˆj where
Bx = B cos(45°) =
G
We want the magnitude of B to be 1, so we have
1
B
2
and B y = B sin (45°) =
2
1
B
2
2
⎛ 1 ⎞ ⎛ 1 ⎞
B = Bx2 + By2 = 1 ⇒ ⎜
B⎟ + ⎜
B ⎟ = 1 ⇒ B2 = 1 ⇒ B = 1
⎝ 2 ⎠ ⎝ 2 ⎠
Thus,
Bx = B y =
Finally,
1
2
G
1 ˆ 1 ˆ
B = Bxiˆ + By ˆj =
i+
j
2
2
3.28. Model: Carlos will be represented as a particle and the particle model will be used for motion under constant
acceleration kinetic equations.
Visualize:
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3-14
Chapter 3
Solve: Carlos runs at constant speed without changing direction. The total distance he travels is found from
kinematics:
r1 = r0 + v0Δt = 0.0 m + (5.0 m/s)(600 s) = 3000 m
G
This displacement is north of east, or θ = 25° from the +x-axis. Thus the position r1 becomes
G
r1 = (3000 m)[cos(25°)iˆ + sin (25°) ˆj ] = 2.7 km iˆ + 1.3 km ˆj
That is, Carlos ends up 1.3 km north of his starting position.
Assess: The choice of our coordinate system is such that the x-component of the displacement is along the east and
the y-component is along the north. The displacement of 3.0 km is reasonable for Carlos to run in 10 minutes if he is
an athlete.
3.29. Visualize: The coordinate
system (x, y, z) is shown here; +x denotes east, +y denotes north, and +z denotes
G
G
G
G
upward vertical. The vectors S morning (shortened to S m ), Safternoon (shortened to Sa ), and the total displacement
G
G
G
vector S total = Sa + S m are also shown.
G
G
Solve: S m = (2000iˆ + 3000 ˆj + 200kˆ) m, and Sa = ( −1500iˆ + 2000 ˆj − 300kˆ) m. The total displacement is the sum of
the individual displacements.
(a) The sum of the z-components of the afternoon and morning displacements is Saz + S mz = −300 m + 200 m =
−100 m; that is, 100 m lower.
G
G
G
(b) S total = Sa + S m = (500iˆ + 5000 ˆj − 100kˆ) m; that is, (500 m east) + (5000 m north) – (100 m vertical). The
magnitude of your total displacement is
S total = (500) 2 + (5000)2 + (−100) 2 m = 5.0 km
3.30. Visualize:
Only the minute hand is shown in the figure.
G
G
Solve: (a) We have S8:00 = (2.0 cm) ˆj and S8:20 = (2.0 cm)cos(30°)iˆ − (2.0 cm)sin(30°) ˆj. The displacement vector is
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Vectors and Coordinate Systems
3-15
G
G G
Δr = S8:20 − S8:00
= (2.0 cm)[cos30°iˆ − (sin 30° + 1) ˆj ]
= (2.0 cm)(0.87iˆ − 1.50 ˆj )
= (1.7 cm)iˆ − (3.0 cm) ˆj
G
G
G
G G
(b) We have S8:00 = (2.0 cm) ˆj and S9:00 = (2.0 cm) ˆj. The displacement vector is Δ r = S9:00 − S8:00 = 0.0 cm.
Assess: The displacement vector in part (a) has a positive x-component and a negative y-component. The vector thus
is to the right and points down, in quadrant IV. This is where the vector drawn from the tip of the 8:00 a.m. arm to
the tip of the 8:20 a.m. arm will point.
3.31. Visualize: (a)
Note that +x is along the east and +y is along the north.
G
Solve: (b) We are given A = − (200 m) ˆj , and can use trigonometry to obtain
G
B = − (400 m)cos(45°) − (400 m)sin (45°) = − (283 m)iˆ − (283 m) ˆj and
G
G G G G
C = (200 m)sin (30°) + (200 m)cos(30°) = (100 m)iˆ + (173 m) ˆj. We want A + B + C + D = 0 , so
G G G
G
D = −A − B − C
= (200 m) ˆj − [ −(283 m)iˆ − (283 m) ˆj ] − [(100 m)iˆ + (173 m) ˆj ] = (183 m) iˆ + (310 m) ˆj
K
The magnitude and direction of D are
⎛ Dy ⎞
−1 ⎛ 310 m ⎞
D = (183 m) 2 + (310 m) 2 = 360 m and θ = tan −1 ⎜
⎟ = tan ⎜
⎟ = 59°
D
⎝ 183 m ⎠
⎝ x⎠
G
This means D = (360 m 59° north of east).
G
(c) The measured length of the vector D on the graph (with a ruler) is approximately 1.75 times the measured length
G
of vector A . Since A = 200 m, this gives D = 1.75 × 200 m = 350 m. Similarly, the angle θ measured with the
protractor is close to 60°. These answers are in close agreement to part (b).
3.32. Visualize: (a) The figure shows Sparky’s individual displacements and his net displacement.
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3-16
Chapter 3
G
G
G
G
Solve: (b) Dnet = D1 + D2 + D3 , where individual displacements are
G
D1 = (50 m)cos(45°)iˆ + (50 m)sin (45°) ˆj = (35.4 m)iˆ + (35.4 m) ˆj
G
D2 = − (70 m)iˆ
G
D3 = − (20 m) ˆj
G
Thus, to two significant figures, Sparky’s displacement is Dnet = −(35 m)iˆ + (15 m) ˆj.
(c) As a magnitude and angle,
⎛ Dnet, y ⎞
Dnet = ( Dnet, x ) 2 + ( Dnet, y ) 2 = (−35 m) 2 + (15 m) 2 = 38 m, θ net = tan −1 ⎜
⎟ = 24°
⎜ | Dnet, x | ⎟
⎝
⎠
Sparky’s net displacement is 38 m in a direction 24° north of west.
3.33. Visualize:
G
G
G
Solve: We are given A = (5.0 m) iˆ and C = ( −1.0 m)kˆ Using trigonometry, B = (3.0 m)cos(45°) iˆ − (3.0 m)sin(45° ) ˆj
G G G G
G
The total displacement is r = A + B + C = (7.12 m)iˆ − (2.12 m) ˆj − (1.0 m) kˆ. The magnitude of r is
r = (7.12 m)2 + (2.12 m) 2 + (1.0 m) 2 = 7.5 m.
Assess: A displacement of 7.5 m is a reasonable displacement.
3.34. Visualize:
G
Solve: We have v = vxiˆ + v y ˆj = v|| iˆ + v⊥ ˆj = v cosθ iˆ + v sin θ ˆj. Thus, v|| = v cos θ = (100 m/s)cos(30°) = 87 m/s.
Assess: For the angle of 30°, 87 m/s for the horizontal component seems reasonable.
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Vectors and Coordinate Systems
3-17
3.35. Visualize:
⎛ 2.5 m/s ⎞
Solve: (a) Since vx = v cos θ , we have 2.5 m/s = (3.0 m/s)cosθ ⇒ θ = cos −1 ⎜
⎟ = 34°.
⎝ 3.0 m/s ⎠
(b) The vertical component is v y = v sin θ = (3.0 m/s)sin(34°) = 1.7 m/s.
3.36. Visualize:
The coordinate system used here is tilted so that the x-axis is along the slope.
Solve: The component of the velocity parallel to the x-axis is v|| = −v cos(70°) = −v sin (20°) = −(10 m/s)(0.34) =
−3.4 m/s. This is the speed down the slope. The component of the velocity perpendicular to the slope is
v⊥ = − v sin (70°) = −v cos(20°) = −(10 m/s)(0.94) = −9.4 m/s. This is the speed toward the ground.
Assess: A final speed of approximately 10 m/s implies a fall time of approximately 1 second under free fall. Note
that g = −9.8 m/s 2 . This time is reasonable for a drop of approximately 5 m, or 16 feet.
3.37. Visualize:
(a)
(b)
Solve: (a) The river is 100 m wide. If Mary rows due north at a constant speed of vrow = 2.0 m/s, it will take her
(100 m)/(2.0 m/s) = 50 s to row across. But while she’s doing so, the current sweeps her boat sideways at a speed
vcurrent = 1.0 m/s. In the 50 s it takes her to cross the river, the current sweeps here a distance
d || = (vcurrent × 50 s) = 1.0 m/s × 50 s = 50 m, so she lands 50 m east of the point that was directly across the river
from her when she started.
G
G
G
(b) Mary’s net displacement Dnet , her displacement Dcurrent due to the river’s current, and her displacemnt Drow
due to her rowing are shown in the figure.
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3-18
Chapter 3
G
3.38. Visualize: Establish a coordinate system with origin at theG tree Gand with the x-axis pointing east. Let A be a
displacement vector directly from the tree to the treasure. Vector A is A = (100iˆ + 500 ˆj ) paces.
This describes the displacement you would undergo by walking north 500 paces, then east 100 paces. Instead, you
follow the road for 300 paces and undergo displacement
G
B = [300sin (60°)iˆ + 300cos(60°) ˆj ] paces = (260iˆ + 150 ˆj ) paces
G
G G G
Solve: Now let C be the displacement vector from your position to the treasure. From the figure A = B + C .
G G G
So the displacement you need to reach the treasure is C = A − B = (−160iˆ + 350 ˆj ) paces.
G
If θ is the angle measured between C and the y-axis,
⎛ 160 ⎞
θ = tan −1 ⎜
⎟ = 25°
⎝ 350 ⎠
You should head 25° west of north. You need to walk distance C = C x2 + C y2 = (−160) 2 + (350)2 paces = 385
paces to get to the treasure.
3.39. Visualize: A 3% grade rises 3 m for every 100 m of horizontal distance. The angle of the ground is
thus α = tan −1 (3/100) = 1.72°.
Establish a tilted coordinate system with one axis parallel to the ground and the other axis perpendicular to the
ground.
G
Solve: From the figure, the magnitude of the component vector of v perpendicular to the ground
G
is v⊥ = v sin α = 15.0 m/s. But this is only the size. We also have to note that the direction of v⊥ is down, so the
G
component is v⊥ = −(15 m/s) ˆj.
G
3.40. Visualize: The average velocity is the net displacement Dnet divided by the total time, which are marked on
the graph. We also mark on the graph of the bacterium’s individual displacements and the time for each.
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Vectors and Coordinate Systems
3-19
Solve: The magnitude of the net displacement is found with Pythagoreum’s rule, taking the values from the graph.
We have Dnet =
2
2
2
2
Dnet,
x + Dnet, y = (40 μm) + (−20 μm ) = 45 μm. The direction of this displacement is
⎛ Dnet, y ⎞
⎟ = tan −1 ⎜⎛ 20 μm ⎟⎞ = 27°
⎜ Dnet, x ⎟
⎝ 40 μm ⎠
⎝
⎠
θ = tan −1 ⎜
The total time for the displacement is the sum of the individual times, which may be found by dividing each
individual distance by the bacterium’s constant speed of 20 μm/s. This gives
Δ t AB = DAB /(20 μm/s ) = (50 μm)2 + (10 μm)2 /(20 μm/s ) = (51.0 μm)/(20 μm/s) = 2.55 s
Δ t BC = DBC /(20 μm/s ) = (10 μm)/(20 μm/s) = 0.50 s
Δ tCD = DCD /(20 μm/s ) = (40 μm)2 + (10 μm)2 /(20 μm/s ) = (41.0 μm)/(20 μm/s) = 2.06 s
Δ t DE = DDE /(20 μm/s ) = (−50 μm)2 + (−50 μm)2 /(20 μm/s ) = (70.7 μm)/(20 μm/s ) = 3.54 s
The total time is therefore Δ tTot = 2.55 s + 0.50 s + 2.06 s + 3.54 s = 8.65 s and the magnitude of the bacterium’s net
velocity is
vnet =
Dnet
45 μm
=
= 5.2 μm/s
ΔtTot 8.65 s
3.41. Visualize:
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3-20
Chapter 3
G G
G
G
G
Solve: The resulting velocity is given by v = vfly + vwind , where vwind = (6.0 m/s)iˆ and vfly = −v sin θ iˆ − v cosθ ˆj.
G
Substituting the known values we get v = −(8.0 m/s)sin θ iˆ − (8.0 m/s)cosθ ˆj + (6.0 m/s) iˆ. We need to have vx = 0.
This means 0 = −(8.0 m/s)sin θ + (6.0 m/s), so sin θ = 86 or θ = 49°. Thus the ducks should head 49° west of south.
3.42. Model: We will treat the knot in the rope as a particle in static equilibrium.
Visualize:
G
G
Solve: Expressing the vectors in component form, we have F1 = 3.0iˆ and F2 = −5.0sin (30°)iˆ + 5.0cos(30°) ˆj. Since
G
G G
G G
G G
we must have F1 + F2 + F3 = 0 for the know to remain stationary, we can write F3 = 2F1 − F2 = −0.50 iˆ − 4.33 ˆj. The
G
G
2
magnitude of F3 is given by F3 = ( −0.50 ) + ( −4.33) 2 = 4.4 units The angle between F3 and the negative x-axis is
θ = tan −1(4.33/0.50) = 83° below the negative x-axis.
Assess: The resultant vector has both components negative, and is therefore in quadrant III. Its magnitude and
G
direction are reasonable. Note the minus sign that we have manually inserted with the force F2 .
3.43. Visualize:
Use a tilted coordinate system such that x-axis is down the slope.
G
G
Solve: Expressing all three forces in terms of unit vectors, we have F1 = −(3.0 N)iˆ, F2 = +(6.0 N) ˆj , and
G
F3 = (5.0 N)sin θ iˆ − (5.0 N)cosθ ˆj.
G
(a) The component of Fnet parallel to the floor is ( Fnet ) x = −(3.0 N) + 0 N + (5.0 N)sin (30°) = −0.50 N, or 0.50 N
up the slope.
G
(b) The component of Fnet perpendicular to the floor is ( Fnet ) y = 0 N + (6.0 N) − (5.0 N)cos(30°) = 1.67 N, or 1.7 N
to two significant figures.
G
(c) The magnitude of Fnet is Fnet = ( Fnet ) x + ( Fnet ) y = (−0.50 N) 2 + (1.67 N) 2 = 1.74 N, or 1.7 N to two
G
significant figures. The angle between Fnet and the negative x-axis is
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Vectors and Coordinate Systems
(F ) y
φ = tan −1 Gnet
G
Fnet is 73° clockwise from the −x-axis.
( Fnet ) x
3-21
⎛ 1.67 N ⎞
= tan −1 ⎜
⎟ = 73°
⎝ 0.50 N ⎠
3.44. Visualize:
Solve: Using trigonometry to calculate θ, we get θ = tan −1 (100 cm/ 141 cm) = 35.3°. Expressing the three forces
G
G
G
component form gives FB = −(3.0 N)iˆ, FC = −(6.0 N) ˆj , and FD = + (2.0 N)cos(35.3°)iˆ − (2.0 N) sin (35.3°) ˆj =
G
G
G
G
G
(1.63 N)iˆ − (1.16 N) ˆj. The total force is Fnet = FB + FC + FD = −1.37 N iˆ − 7.2 N ˆj. The magnitude of Fnet is
Fnet = (1.37 N) 2 + (7.2 N) 2 = 7.3 N.
θ net = tan −1
( Fnet ) y
( Fnet ) x
⎛ 7.2 N ⎞
= tan −1 ⎜
⎟ = 79°
⎝ 1.37 N ⎠
G
Fnet = (7.3 N, 79° below the negative x-axis in quadrant III).
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
KINEMATICS IN TWO DIMENSIONS
4
Conceptual Questions
G
4.1. (a) As shown in the figure below, the acceleration a can be divided into components perpendicular (⊥) and
G
parallel ( || ) to the velocity. a|| will slow the particle down since it is in the opposite direction to v .
G
(b) The perpendicular component of a , a⊥ , is pointing to the right, and changes the particle direction to the right.
4.2. (a) A component of the acceleration either parallel or antiparallel to the velocity would speed up the particle or
slow it down, respectively. Since there is no parallel component then the speed isn’t changing.
G
(b) The perpendicular component of a , a⊥ , is pointing down, and changes the particle direction in that direction.
4.3. Approximate Tarzan as a particle in nonuniform circular motion.
G
(a) As Tarzan just steps off the vine, his velocity is zero, but increasing along his trajectory, so a is along the
v2
= 0 because v = 0.
r
dv
v2
G
(b) At the bottom of the swing, ar = t ≠ 0, but the velocity is at a maximum, so at = t = 0, so a is not zero
dt
r
G
trajectory. The component of a that is the centripetal acceleration ar =
and points up.
4.4. A typical trajectory of a projectile is shown in the figure below. The acceleration due to gravity always points
down. The velocity changes direction from the launch angle θ = θ 0 above the + x -axis to zero at the top of the
trajectory, to θ = θ 0 below the + x -axis when it hits the ground.
v1y
G
G
(a) At no time are v and a parallel if
= tan 30°
v1x
G
G
(b) At the top of the trajectory v and a are perpendicular.
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4-1
4-2
Chapter 4
4.5. For a projectile, only vx , ax , and a y are constant during the flight. Since the acceleration a y = − g is down,
a x = 0 and vx is constant. The nonzero a y is constantly changing v y , so the total speed v = vx2 + v 2y changes as
well. The positions x and y change, so r = x 2 + y 2 changes, too. Only a x is zero throughout the flight.
4.6. (a) The ball fired upward is a projectile with a horizontal component of initial velocity equal to the cart’s speed.
Without air friction, there is no horizontal component of the acceleration, so the ball stays over the cart during the
whole flight, and lands directly back in the tube.
(b) The cart accelerates after launching the ball, the horizontal component of the ball’s velocity is less than the
velocity of the cart, so the ball will land behind the cart.
4.7. After the rock is released it is in free fall, so its acceleration is equal to g.
4.8. Anita is approaching ball 2 and moving away from where ball 1 was thrown, so ball 1 was thrown with the
greater speed. This can be determined numerically as well, treating Anita as a moving reference frame with respect to
the ground, so vAnita = vball − 5 m/s. For ball 1, Anita measures 10 m/s = v1 − 5 m/s ⇒ v1 = 15 m/s. For ball 2,
−10 m/s = v2 − 5 m/s ⇒ v2 = −5 m/s. So ball 1 was thrown with greater speed.
4.9. The ball has the same horizontal velocity as the plane whether parked or flying, as does an observer in the
plane. When the plane is moving forward at a steady speed, the ball after release appears to fall straight down,
landing on the X as it did when the plane was parked.
4.10. Zach should throw his book outward and toward the back of the car. The book has the same initial velocity as
do Zach and the car, so throw 1 or 2 will cause the book to land beyond the driveway in the same direction as the car
is traveling.
4.11. Since Zach and Yvette are traveling at the same speed they share the same reference frame, so Zach should
throw the book straight to her (throw 2.)
4.12. In uniform circular motion the tangential acceleration is zero, and the speed is constant. All vector quantities
(velocity and radial acceleration) have constant magnitudes but changing directions. Note that the tangential velocity
is the same as the instantaneous velocity in uniform circular motion.
4.13. (a) ω 1= ω 2= ω 3. All points on an object turn at the same angular rate.
(b) v3 > v1 = v2 . Since v = ω r and ω is the same for all of the rotating wheel, the speeds are ranked by how far from
the center (r ) they are.
4.14. (a) ω : + α : + Rotation is counterclockwise and increasing in the counterclockwise direction.
(b) ω : − α : + Rotation is clockwise and decreasing, so the angular acceleration is counterclockwise.
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Kinematics in Two Dimensions
4-3
(c) ω : + α : − Rotation is counterclockwise and decreasing, so the angular acceleration is clockwise.
(d) ω : − α : − Rotation is clockwise and increasing in the clockwise direction.
v
r
4.15. (a) The instantaneous speed v is zero, and so ω = = 0.
(b) Rotation is beginning in the clockwise direction, so α < 0.
Exercises and Problems
Section 4.1 Acceleration
4.1. Solve: (a)
(b) A race car slows from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn. After making the
90° turn the car accelerates back up to 100 mph in the same time it took to slow down.
4.2. Solve: (a)
(b) A ball rolls along a level table at 3 m/s. It rolls over the edge and falls 1 m to the floor. How far away from the
edge of the table does it land?
4.3. Solve: To make the particle slow down the acceleration needs to have a component that is opposite the
direction of the velocity. To make the particle curve upward the acceleration must have a component upward. So the
answer is H.
4.4. Solve: To keep a steady speed there can’t be a component of the acceleration parallel to the velocity. To make
the particle curve to the right the acceleration must have a component to the right. So the answer is C.
4.5. Solve: To make the particle speed up the acceleration needs to have a component that is in the direction of the
velocity. To make the particle curve downward the acceleration must have a component downward. So the answer is E.
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4-4
Chapter 4
Section 4.2 Two-Dimensional Kinematics
4.6. Model: The boat is treated as a particle whose motion is governed by constant-acceleration kinematic
equations in a plane.
Visualize:
Solve: Resolving the acceleration into its x and y components, we obtain
G
a = (0.80 m/s 2 )cos 40°iˆ + (0.80 m/s 2 )sin 40° ˆj = (0.613 m/s 2 )iˆ + (0.514 m/s 2 ) ˆj
G G G
From the velocity equation v1 = v0 + a (t1 − t0 ),
G
v1 = (5.0 m/s)iˆ + [(0.613 m/s 2 )iˆ + (0.514 m/s 2 ) ˆj ](6 s − 0 s) = (8.68 m/s)iˆ + (3.09 m/s) ˆj
G
The magnitude and direction of v are
v = (8.68 m/s) 2 + (3.09 m/s) 2 = 9.2 m/s
⎛ v1 y ⎞
−1 ⎛ 3.09 m/s ⎞
⎟ = tan ⎜
⎟ = 20° north of east
v
⎝ 8.68 m/s ⎠
⎝ 1x ⎠
Assess: An increase of speed from 5.0 m/s to 9.2 m/s is reasonable.
θ = tan −1 ⎜
4.7. Model: Model the rocket as a particle and assume constant acceleration in both directions (vertical and
horizontal) so use the kinematic equations in direction.
Visualize:
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Kinematics in Two Dimensions
4-5
1
Solve: Use s1 = s0 + v0 s Δt + as ( Δt )2 in each direction. Δt is the same in each direction.
2
1
1
x1 = x0 + v0 x Δt + a x (Δt ) 2 = 0 m + 0 m + (1.50 m/s 2 )(6.00 s)2 = 27 m
2
2
1
1
y1 = y0 + v0 y Δt + a y ( Δt )2 = 0 m + 0 m + (6.00 m/s 2 )(6.00 s)2 = 108 m
2
2
The distance from the launch pad at t = 6.00 s is
r = x12 + y12 = (27 m) 2 + (108 m) 2 = 111.32 m ≈ 111 m
Assess: This distance seems reasonable for a model rocket after 6.00 s.
G
4.8. Solve: (a) At t = 0 s, x = 0 m and y = 0 m, or r = (0iˆ + 0 ˆj ) m. At t = 4 s, x = 0 m and y = 0 m, or
G
r = (0iˆ + 0 ˆj ) m. In other words, the particle is at the origin at both t = 0 s and at t = 4 s. From the expressions for x
and y,
dy ˆ ⎡⎛ 3 2
⎤
G dx
⎞
v = iˆ +
j = ⎢⎜ t − 4t ⎟ iˆ + (t − 2) ˆj ⎥ m/s
⎝
⎠
dt
dt
⎣ 2
⎦
G
G
ˆ
ˆ
ˆ
At t = 0 s, v = −2 j m/s, v = 2 m/s. At t = 4 s, v = (8i + 2 j ) m/s, v = 8.3 m/s.
G
(b) At t = 0 s, v is along − ˆj , or 90° south of + x. At t = 4 s,
⎛ 2 m/s ⎞
⎟ = 14° north of +x
⎝ 8 m/s ⎠
θ = tan −1 ⎜
4.9. Model: The puck is a particle and follows the constant-acceleration kinematic equations of motion.
G
Solve: (a) At t = 2 s, the graphs give vx = 16 cm/s and v y = 30 cm/s. The angle made by the vector v with the
x-axis can thus be found as
⎛ vy ⎞
−1 ⎛ 30 cm/s ⎞
⎟ = tan ⎜
⎟ = 62° above the x-axis
v
⎝ 16 cm/s ⎠
⎝ x⎠
(b) After t = 5 s, the puck has traveled a distance given by:
θ = tan −1 ⎜
5s
x1 = x0 + Ñ vx dt = 0 m + area under vx -t curve = 12 (40 cm/s)(5 s) = 100 cm
0
5s
y1 = y0 + Ñ v y dt = 0 m + area under v y -t curve = (30 cm/s)(5 s) = 150 cm
0
⇒ r1 = x12 + y12 = (100 cm) 2 + (150 cm)2 = 180 cm
4.10. Model: Use the particle model for the puck.
Solve: Since the vx vs t and v y vs t graphs are straight lines, the puck is undergoing constant acceleration along
the x- and y- axes. The components of the puck’s acceleration are
dv
Δv
(10 m/s − ( −10 m/s) )
ax = x = x =
= 2.0 m/s 2
dt
Δt
10 s − 0 s
(10 m/s − 0 m/s)
ay =
= 1.0 m/s 2
(10 s − 0 s)
The magnitude of the acceleration is a = ax2 + a 2y = 2.2 m/s 2 .
Assess: The acceleration is constant, so the computations above apply to all times shown, not just at 5 s. The puck
turns around at t = 5 s in the x direction, and constantly accelerates in the y direction. Traveling 50 m from the
starting point in 10 s is reasonable.
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4-6
Chapter 4
Section 4.3 Projectile Motion
4.11. Model: Assume the particle model for the ball, and apply the constant-acceleration kinematic equations of
motion in a plane.
Visualize:
G
Solve: (a) We know the velocity v1 = (2.0iˆ + 2.0 ˆj ) m/s at t = 1 s. The ball is at its highest point at t = 2 s, so
G
v y = 0 m/s. The horizontal velocity is constant in projectile motion, so vx = 2.0 m/s at all times. Thus v2 = 2.0iˆ m/s
at t = 2 s. We can see that the y-component of velocity changed by Δv y = −2.0 m/s between t = 1 s and t = 2 s.
Because a y is constant, v y changes by −2.0 m/s in any 1-s interval. At t = 3 s, v y is 2.0 m/s less than its value of 0
at t = 2 s. At t = 0 s, v y must have been 2.0 m/s more than its value of 2.0 m/s at t = 1 s. Consequently, at t = 0 s,
G
v0 = (2.0iˆ + 4.0 ˆj ) m/s
At t = 1 s,
G
v (1) = (2.0iˆ + 2.0 ˆj ) m/s
At t = 2 s,
G
v (2) = (2.0iˆ + 0.0 ˆj ) m/s
At t = 3 s,
G
v (3) = (2.0iˆ − 2.0 ˆj ) m/s
(b) Because v y is changing at the rate −2.0 m/s per s, the y-component of acceleration is a y = −2.0 m/s 2 . But
a y = − g for projectile motion, so the value of g on Exidor is g = 2.0 m/s 2 .
G
(c) From part (a) the components of v0 are v0 x = 2.0 m/s and v0 y = 4.0 m/s. This means
⎛ v0 y ⎞
−1 ⎛ 4.0 m/s ⎞
⎟ = tan ⎜
⎟ = 63° above +x
v
⎝ 2.0 m/s ⎠
⎝ 0x ⎠
θ = tan −1 ⎜
Assess: The y-component of the velocity vector decreases from 2.0 m/s at t = 1 s to 0 m/s at t = 2 s. This gives an
acceleration of −2 m/s 2 . All the other values obtained above are also reasonable.
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Kinematics in Two Dimensions
4-7
4.12. Model: The ball is treated as a particle and the effect of air resistance is ignored.
Visualize:
Solve: Using x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 ,
50 m = 0 m + (25 m/s)(t1 − 0 s) + 0 m ⇒ t1 = 2.0 s
Now, using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ,
y1 = 0 m + 0 m + 12 (−9.8 m/s 2 )(2.0 s − 0 s) 2 = −19.6 m
So the ball was thrown from 19.6 m high.
Assess: The minus sign with y1 indicates that the ball’s displacement is in the negative y direction or downward.
A magnitude of 19.6 m for the height is reasonable.
4.13. Model: The bullet is treated as a particle and the effect of air resistance on the motion of the bullet is
neglected.
Visualize:
Solve: (a) Using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 , we obtain
(−2.0 × 10−2 m) = 0 m + 0 m + 12 ( −9.8 m/s 2 )(t1 − 0 s) 2 ⇒ t1 = 0.0639 s ≈ 0.064 s
(b) Using x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 ,
(50 m) = 0 m + v0 x (0.0639 s − 0 s) + 0 m ⇒ v0 x = 782 m/s ≈ 780 m/s
Assess: The bullet falls 2 cm during a horizontal displacement of 50 m. This implies a large initial velocity, and a
value of 782 m/s is understandable.
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4-8
Chapter 4
4.14. Model: We will use the particle model for the food package and the constant-acceleration kinematic
equations of motion.
Visualize:
Solve: For the horizontal motion,
x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (150 m/s)(t1 − 0 s) + 0 m = (150 m/s)t1
We will determine t1 from the vertical y-motion as follows:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2
⇒ 0 m = 100 m + 0 m + 12 (−9.8 m/s 2 )t12 ⇒ t1 =
200 m
= 4.518 s ≈ 4.5 s
9.8 m/s 2
From the above x-equation, the displacement is x1 = (150 m/s)(4.518 s) = 678 m ≈ 680 m.
Assess: The horizontal distance of 678 m covered by a freely falling object from a height of 100 m and with an
initial horizontal velocity of 150 m/s (≈ 335 mph) is reasonable.
Section 4.4 Relative Motion
4.15. Model: Assume motion along the x-direction (downstream to the right). Call the speed of the boat with
respect to the water (vx ) BW , the speed of the water with respect to the Earth (vx ) WE , and the speed of the boat with
respect to the Earth (vx ) BE .
Solve: We seek (vx ) WE .
30 km
= 10 km/h
3.0 h
30 km
Upstream:
= −6.0 km/h
(vx )BE = −(vx ) BW + (vx ) WE =
5.0 h
Add the two equations to get 2(vx ) WE = 4.0 km/h, so the river flows at 2.0 m/s.
Downstream:
(vx )BE = (vx ) BW + (vx ) WE =
Assess: This means that the boat goes at 8.0 m/s relative to the water. Both these numbers sound reasonable.
4.16. Model: Assume motion along the x-direction. Let Δx = x1 − x0 be the distance between the gate and the
baggage claim. Call your walking speed (vx ) YS , the speed of the moving sidewalk with respect to the floor (vx )SF ,
and the speed of you with respect to the floor (vx ) YF while walking and riding.
Solve: We seek Δt , the time it takes to go Δx while walking on the moving sidewalk.
Δx
(vx )YS =
Walking alone:
50 s
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Kinematics in Two Dimensions
(vx )SF =
Standing while riding:
Walking while riding:
Cancel Δx and solve for Δt :
(vx )YF =
4-9
Δx
75 s
Δx
Δx
Δx
= (vx ) YS + (vx )SF =
+
50 s 75 s
Δt
Δx Δx
Δx
=
+
⇒ Δt = 30 s
Δt 50 s 75 s
Assess: A Δt smaller than 50 s was expected.
4.17. Model: Let the x-direction be east and the y-direction be north. Use subscripts M, W, and E for Mary, the
water, and the Earth, respectively. Let the origin be Mary’s starting point on the south bank.
Visualize: In the reference frame of the water Mary has no east-west motion; in that frame she travels 100 m across
the river at 2.0 m/s so Δt = 50 s.
Solve:
(a)
G
G
G
rME = rMW + rWE
G
G
= vMW Δt + vWE Δt
= (2.0 m/s) ˆj (50 s) + (1.0 m/s)ιˆ(50 s)
= (50 m)iˆ + (100 m) ˆj
So she lands 50 m east (downstream) from where she intended.
(b)
vME = (vx ) 2 + (v y )2 = (1.0 m/s) 2 + (2.0 m/s) 2 = 2.236 m/s ≈ 2.2 m/s
Assess: Most of Mary’s speed with respect to the shore is due to her rowing rather than the current.
4.18. Model: Let the x-direction be east and the y-direction be north. Use subscripts S, T, and G for Susan, Trent,
and Ground respectively.
G
G
G
G
G
Visualize: vTS = vTG + vGS where vGS = −vSG = ( −60 mph) ˆj.
Solve:
vTS = (vx ) 2 + (v y ) 2 = (45 mph) 2 + (−60 mph) 2 = 75 mph
Assess: We expected the relative speed between Trent and Susan to be greater than either of their speeds relative to
the ground.
Section 4.5 Uniform Circular Motion
4.19. Visualize: The angular velocity is the slope of the angular position graph.
Solve:
(a) The slope of the graph at t = 1 s is 0 rad/s.
−2π rad
= −π / 2 rad/s.
(b) The slope of the graph at t = 4s is
4s
6π rad
(c) The slope of the graph at t = 7s is
= 3π rad/s.
2s
4.20. Solve: Since ω = ( dθ /dt ) we have
θ f = θi + area under the ω -versus-t graph between ti and tf
From t = 0 s to t = 2 s, the area is (20 rad/s)(2 s) = 40 rad. From t = 2 s to t = 4 s, the area is
(10 rad/s)(2 s) = 20 rad. Thus, the area under the ω -versus-t graph during the total time interval of 4 s is 60 rad or
(60 rad) × (1 rev/2π rad) = 9.55 revolutions ≈ 9.5 revolutions.
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4-10
Chapter 4
4.21. Solve: The angular position graph is the area under the angular velocity graph. At t = 4 s the area is 80 rad.
Between 4 s and 6 s the angular velocity is zero so the angular position doesn’t change. Between 6 s and 8 s the area
is 20 rad, but it is below the axis, so we subtract it. The area under the ω versus t graph during the total time interval
of 8 s is 80 rad −20 rad = 60 rad. This is where we end up on the θ axis at 8.s
4.22. Model: Treat the record on a turntable as a particle rotating at 45 rpm.
Solve: (a) The angular velocity is
ω = 45 rpm ×
(b) The period is
T=
1 min 2π rad
×
= 1.5π rad/s ≈ 4.7 rad/s
60 s
1 rev
2π rad
=
2π rad
= 1.33 s ≈ 1.3 s
1.5π rad/s
ω
4.23. Model: The airplane is to be treated as a particle.
Visualize:
Solve: The angle you turn through is
s 5000 miles
180°
= 1.2500 rad = 1.2500 rad ×
= 71.62°
θ 2 − θ1 = =
π rad
r 4000 miles
The plane’s angular velocity is
θ − θ 71.62°
= 8.0° / h
ω= 2 1=
t2 − t1
9h
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Kinematics in Two Dimensions
4-11
Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the
separation between Kampala and Singapore is approximately one-fifth of the earth’s circumference.
Section 4.6 Velocity and Acceleration in Uniform Circular Motion
4.24. Model: Assume the beach is at sea level so that rS = 6400 km for the surfer and rC = 6403 km for the
climber. The angular velocity for each of them is 2π rad/24 h = 7.27 × 10−5 rad/s.
Visualize: v = rω.
Solve:
Δv = vC − vS = rCω − rSω = ( rC − rS )ω = (3000 m)(7.27 × 10−5 rad/s) = 0.22 m/s = 22 cm/s
Assess: The difference in speed is small because ω is small.
4.25. Solve: The plane must fly as fast as the earth’s surface moves, but in the opposite direction. That is, the plane
must fly from east to west. The speed is
km
km
1 mile
⎛ 2π rad ⎞
3
v =ωr =⎜
= 1680
×
= 1040 mph
⎟ (6.4 × 10 km) = 1680
h
h 1.609 km
⎝ 24 h ⎠
4.26. Model: The rider is assumed to be a particle.
Solve: Since ar = v 2 /r , we have
v 2 = ar r = (98 m/s 2 )(12 m) ⇒ v = 34 m/s
Assess: 34 m/s ≈ 70 mph is a large yet understandable speed.
4.27. Model: The earth is a particle orbiting around the sun.
Solve: (a) The magnitude of the earth’s velocity is displacement divided by time:
2π r
2π (1.5 × 1011 m)
v=
=
= 3.0 × 104 m/s
24 h 3600 s
T
365 days ×
×
1 day
1h
(b) Since v = rω , the angular velocity is
ω=
v 3.0 × 104 m/s
=
= 2.0 × 10−7 rad/s
r 1.5 × 1011 m
(c) The centripetal acceleration is
ar =
v 2 (3.0 × 104 m/s)2
=
= 6.0 × 10−3 m/s 2
11
r
1.5 × 10 m
Assess: A tangential velocity of 3.0 × 104 m/s or 30 km/s is large, but needed for the earth to go through a
displacement of 2π (1.5 × 1011 m) ≈ 9.4 × 108 km in 1 year.
4.28. Solve: The pebble’s angular velocity ω = (3.0 rev/s)(2π rad/rev) = 18.9 rad/s. The speed of the pebble as it
moves around a circle of radius r = 30 cm = 0.30 m is
v = ω r = (18.9 rad/s)(0.30 m) = 5.7 m/s
The radial acceleration is
v 2 (5.7 m/s)2
=
= 108 m/s 2
r
0.30 m
4.29. Model: The crankshaft is a rotating rigid body.
Visualize: The angular acceleration is the slope of the angular velocity graph.
Solve:
−200 rad/s
(a) The slope of the graph at t = 1 s is
= −100 rad/s 2 .
2s
ar =
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4-12
Chapter 4
(b) The slope of the graph at t = 3 s is 0 rad/s 2 .
150 rad/s
(c) The slope of the graph at t = 1 s is
= 50 rad/s 2 .
3s
4.30. Model: The turntable is a rotating rigid body.
Visualize: The angular velocity is the area under the α vs. t graph. Use the formula for the area of a trapezoid.
Solve: Because the turntable starts from rest ω 0= 0.
(a) The area under the graph from t = 0 s to t = 1 s is (5 rad/s 2 + 2.5 rad/s 2 )(1 s)/2 = 3.75 rad/s
(b) The area under the graph from t = 0 s to t = 2 s is (5 rad/s 2 + 0 rad/s 2 )(2 s)/2 = 5.0 rad/s
(c) The area under the graph from is not increasing after t = 2 s so ω stays the same 5.0 rad/s.
4.31. Visualize: The angular position is the slope of the area under the ω vs. t graph.
Solve: The area under the graph is 20 rad + 40 rad = 60 rad. Convert to revolutions. 60 rad(1 rev/2π rad) = 9.5 rev.
4.32. Model: Model the child on the merry-go-round as a particle in nonuniform circular motion.
Visualize:
Solve: (a) The speed of the child is v0 = rω = (2.5 m)(1.57 rad/s) = 3.9 m/s.
(b) The merry-go-round slows from 1.57 rad/s to 0 in 20 s. Thus
a
rω
(2.5 m)(1.57 rad/s)
= −0.197 m/s 2
ω 1= 0 = ω 0+ t t1 ⇒ at = − 0 = −
r
t1
20 s
During these 20 s, the wheel turns through angle
a
0.197 m/s 2
(20 s) 2 = 15.6 rad
θ1 = θ0 + ω 0t1 + t t12 = 0 + (1.57 rad/s)(20 s) −
2r
2(2.5 m)
In terms of revolutions, θ1 = (15.6 rad)(1 rev/2π rad) = 2.5 rev.
4.33. Model: The fan is in nonuniform circular motion.
Visualize:
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Kinematics in Two Dimensions
4-13
⎛ min ⎞
2
Solve: Note 1800 rev/min ⎜
⎟ = 30 rev/s. Thus ω f = ωi + αΔt ⇒ 30 rev/s = 0 rev/s + α (4.0 s) ⇒ α = 7.5 rev/s .
60
s
⎝
⎠
⎛ 2π rad ⎞
2
This can be expressed as (7.5 rev/s) ⎜
⎟ = 47 rad/s .
⎝ rev ⎠
Assess: An increase in the angular velocity of a fan blade by 7.5 rev/s each second seems reasonable.
4.34. Model: The wheel is in nonuniform circular motion.
Visualize:
Solve: (a) Express ω i in rad/s:
⎛ min ⎞⎛ 2π rad ⎞
⎟⎜
⎟ ⇒ 5.2 rad/s
⎝ 60 s ⎠⎝ rev ⎠
ω i = (50 rev/min) ⎜
After 10 s, ω f = ω i + α Δt ⇒ ω f = 5.2 rad/s + (0.50 rad/s 2 )(10 s) 2 = 55 rad/s. Converting to rpm,
⎛ 60 s ⎞⎛ rev ⎞
(55 rad/s) ⎜
⎟⎜
⎟ = 53 rpm
⎝ min ⎠⎝ 2π rad ⎠
(b) In 10 s, the wheel has turned a number of radians
θf = θi + ω 0Δt + 12 α Δt 2 ⇒ θf = 0 rad/sec + (5.2 rad/s)(10 s) + 12 (0.50 rad/s 2 )(10 s)2 = 77 radians.
Converting, 77 rad = 12.3 revolutions.
Assess: Making a bicycle wheel turn just over 12 revolutions in 10 s when it is initially turning almost one
revolution per second to begin with seems attainable by a cyclist.
4.35. Model: Model the particle on the crankshaft as being in nonuniform circular motion.
Visualize:
Solve: (a) The initial angular velocity is ω 0= 2500 rpm × (1 min/60 s) × (2π rad/rev) = 261.8 rad/s. The crankshaft
slows from 261.8 rad/s to 0 in 1.5 s. Thus
a
rω
(0.015 m)(261.8 rad/s)
= −2.618 m/s 2 = −2.6 m/s 2
ω 1= 0 = ω 0+ t t1 ⇒ at = − 0 = −
r
t1
1.5 s
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4-14
Chapter 4
(b) During these 1.5 s, the crankshaft turns through angle
a
2.618 m/s 2
(1.5 s) 2 = 196 rad
θ1 = θ0 + ω 0t1 + t t12 = 0 + (261.8 rad/s)(1.5 s) −
2r
2(0.015 m)
In terms of revolutions, θ1 = (196 rad)(1 rev/πrad) = 31.2 rev.
4.36. Model: We will assume that constant-acceleration kinematic equations in a plane apply.
Visualize:
G
Solve: (a) The particle’s position r = (9.0 ˆj ) m implies that at t0 the particle’s coordinates are x0 = 0 m and
G
y0 = 9.0 m. The particle’s position r1 = (20iˆ) m at time t1 implies that x1 = 20 m and y1 = 0 m. This is the position
where the wire hoop is located. Let us find the time t1 when the particle crosses the hoop at x1 = 20 m. From the
vx -versus-t curve and using the relation x1 = x0 + area of the vx -t graph, we get
20 m = 0 m + area of the vx -t graph = area of the vx -t graph
From Figure P4.36 we see that the area of the vx -t graph equals 20 m when t = t1 = 3 s.
(b) We can now look at the y-motion to find a y . Note that the slope of the vx -t graph (that is, a y ) is negative and
constant, and we can determine a y by substituting into y3 = y0 + v0 y (t3 − t0 ) + 12 a y (t3 − t0 ) 2:
0 m = 9 m + 0 m + 12 a y (3 s − 0 s) 2 ⇒ a y = −2 m/s 2
Therefore,
v4y = v0y + a y (t4 − t0 ) = 0 m/s + (−2 m/s 2 )(4 s − 0 s) = −8 m/s
4.37. Model: Assume the spaceship is a particle. The acceleration is constant, so we can use the kinematic
equations.
1
Visualize: We apply the kinematic equation sf = si + v0Δt + a(Δt ) 2 in each direction. Δt = 35 min = 2100 s.
2
Solve:
1
xf = 6.0 × 105 km + (9.5 km/s)(2100 s) + (0.040 km/s 2 )(2100 s) 2 s = 708000 km
2
1
5
yf = −4.0 × 10 km + (0 km/s)(2100 s) + (0 km/s 2 )(2100 s) 2 s = −400000 km
2
1
zf = 2.0 × 105 km + (0 km/s)(2100 s) + (−.020 km/s 2 )(2100 s)2 s = −156000 km
2
G
Rounding to two sig figs gives r = (710iˆ − 400 ˆj − 160kˆ) × 103 km.
f
Assess: The y-component didn’t change because there was no velocity or acceleration in the y-direction.
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Kinematics in Two Dimensions
4-15
4.38. Solve: From the expression for R, Rmax = v02 /g . Therefore,
Rmax v02 sin 2θ
1
=
⇒ sin 2θ = ⇒ θ = 15° and 75°
2
g
2
Assess: The discussion in the chapter explains why launch angles θ and (90° − θ ) give the same range.
R=
4.39. Model: Assume particle motion in a plane and constant-acceleration kinematics for the projectile.
Visualize:
Solve: (a) We know that v0 y = v0 sin θ , a y = − g , and v1 y = 0 m/s. Using v12y = v02y + 2a y ( y1 − y0 ),
v02 sin 2 θ
2g
0 m 2 /s 2 = v02 sin 2 θ + 2( − g )h ⇒ h =
(b) Using the equation for range and the above expression for θ = 30.0°:
h=
( x2 − x0 ) =
(33.6 m/s)2 sin 2 30.0°
2(9.8 m/s 2 )
= 14.4 m
v02 sin 2θ (33.6 m/s)2 sin(2 × 30.0°)
=
= 99.8 m
g
(9.8 m/s 2 )
For θ = 45.0°:
h=
(33.6 m/s) 2 sin 2 45.0°
2(9.8 m/s 2 )
( x2 − x0 ) =
= 28.8 m
(33.6 m/s) 2 sin(2 × 45.0°)
(9.8 m/s 2 )
= 115.2 m
For θ = 60.0°:
h=
(33.6 m/s) 2 sin 2 60.0°
( x2 − x0 ) =
2(9.8 m/s 2 )
= 43.2 m
(33.6 m/s) 2 sin(2 × 60.0°)
2(9.8 m/s 2 )
= 99.8 m
Assess: The projectile’s range, being proportional to sin(2θ ), is maximum at a launch angle of 45°, but the
maximum height reached is proportional to sin 2 (θ ). These dependencies are seen in this problem.
4.40. Model: Treat the kangaroo as a particle.
Visualize: We are asked to find the take-off speed and horizontal speed of the kangaroo given its initial angle, 20D ,
and its range. Since the horizontal speed is given by vx = v0 cosθ and the time of flight is given by Δt = 2v0 sin θ /g ,
the range of the kangaroo is given by the product of these: Δx = 2v0 sin θ cosθ /g .
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4-16
Chapter 4
Solve: (a) We can solve the above formula for v0 and then plug in the range and angle to find the take-off speed:
v0 = g Δx/(2sin θ cosθ ) = (9.8 m/s 2 )(10 m)/(2 sin20° cos 20°) = 12.3 m/s
Its take-off speed is 12 m/s, to two significant figures.
(b) h =
v02 sin 2 θ (12.3 m/s) 2 sin 2 20°
=
= 0.90 m
2g
2(9.8 m/s 2 )
Assess: These numbers seem reasonable for a kangaroo.
4.41. Model: Assume the particle model for the projectile and motion in a plane.
Visualize:
Solve: (a) Using y2 = y0 + v0 y (t2 − t0 ) + 12 a y (t2 − t0 ) 2 ,
y2 = 0 m + (30 m/s)sin 60°(7.5 s − 0 s) + 12 (−9.8 m/s 2 )(7.5 s − 0 s) 2 = −80.8 m
Thus the launch point is 81 m higher than where the projectile hits the ground.
(b) Using v12y = v02y + 2a y ( y1 − y0 ),
0 m 2 /s 2 = (30sin 60° m/s) 2 + 2( −9.8 m/s 2 )( y1 − 0 m) ⇒ y1 = 34.4 m, or y1 = 34 m
Assess: The projectile hits the ground at an angle of 73°.
4.42. Model: Assume the particle model and motion under constant-acceleration kinematic equations in a plane.
Visualize:
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Kinematics in Two Dimensions
4-17
Solve: (a) Using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ,
0 m = 1.80 m + v0 sin 40°(t1 − 0 s) + 12 ( −9.8 m/s 2 )(t1 − 0 s) 2
= 1.80 m + (7.713 m/s)t1 − (4.9 m/s 2 )t12 ⇒ t1 = −0.206 s and 1.780 s
The negative value of t1 is unphysical for the current situation. Using t1 = 1.780 s and x1 = x0 + v0 x (t1 − t0 ), we get
x1 = 0 + (v0 cos 40° m/s)(1.780 s − 0 s) = (12.0 m/s)cos 40°(1.78 s) = 16.36 m ≈ 16.4 m
(b) We can repeat the calculation for each angle. A general result for the flight time at angle θ is
(
)
t1 = 12sin θ + 144sin 2 θ + 35.28 /9.8 s
and the distance traveled is x1 = (12.0)cosθ × t1. We can put the results in a table.
θ
t1
x1
40.0°
1.780 s
16.36 m
42.5°
1.853 s
16.39 m
45.0°
1.923 s
16.31 m
47.5°
1.990 s
16.13 m
Maximum distance is achieved at θ ≈ 42.5°.
Assess: The well-known “fact” that maximum distance is achieved at 45° is true only when the projectile is launched
and lands at the same height. That isn’t true here. The extra 0.03 m = 3 cm obtained by increasing the angle from
40.0° to 42.5° could easily mean the difference between first and second place in a world-class meet.
4.43. Model: The golf ball is a particle following projectile motion.
Visualize:
(a) The distance traveled is x1 = v0 xt1 = v0 cosθ × t1. The flight time is found from the y-equation, using the fact that
the ball starts and ends at y = 0:
y1 − y0 = 0 = v0 sin θ t1 − 12 gt12 = (v0 sin θ − 12 gt1 ) t1 ⇒ t1 =
2v0 sin θ
g
Thus the distance traveled is
x1 = v0 cosθ ×
2v0 sin θ 2v02 sin θ cosθ
=
g
g
For θ = 30°, the distances are
( x1 )earth =
2v02 sin θ cosθ 2(25 m/s) 2 sin30°cos30°
=
= 55.2 m
gearth
9.80 m/s 2
( x1 ) moon =
2v02 sin θ cosθ 2v02 sin θ cosθ
2v02 sin θ cosθ
=
=
×
= 6( x1 )earth = 331.2 m
6
1g
g moon
gearth
6 earth
The golf ball travels 331.2 m − 55.2 m = 276 m farther on the moon than on earth.
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4-18
Chapter 4
(b) The flight times are
(t1 )earth =
2v0 sin θ
= 2.55 s
g earth
(t1) moon =
2v0 sin θ 2v0 sin θ
= 1
= 6(t1 )earth = 15.30 s
g moon
g
6 earth
The ball spends 15.30 s − 2.55 s = 12.75 s longer in flight on the moon.
4.44. Model: The particle model for the ball and the constant-acceleration equations of motion are assumed.
Visualize:
Solve: (a) Using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ,
h = 0 m + (30 m/s)sin 60°(4 s − 0 s) + 12 (−9.8 m/s 2 )(4 s − 0 s) 2 = 25.5 m
The height of the cliff is 26 m.
(b) Using (v 2y ) top = v 2y + 2a y ( ytop − y0 ),
0 m 2 /s 2 = (v0 sin θ ) 2 + 2( − g )( ytop ) ⇒ ytop =
(v0 sin θ ) 2 [(30 m/s)sin 60°]2
=
= 34.4 m
2g
2(9.8 m/s 2 )
The maximum height of the ball is 34 m.
(c) The x and y components are
v1 y = v0 y + a y (t1 − t0 ) = v0 sin θ − gt1 = (30 m/s)sin 60° − (9.8 m/s 2 ) × (4.0 s) = −13.22 m/s
v1x = v0 y = v0 cos60° = (30 m/s)cos60° = 15.0 m/s
⇒ v1 = v12x + v12y = 20.0 m/s
The impact speed is 20 m/s.
Assess: Compared to a maximum height of 34.4 m, a height of 25.5 for the cliff is reasonable.
4.45. Model: The particle model for the ball and the constant-acceleration equations of motion in a plane are
assumed.
Visualize:
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Kinematics in Two Dimensions
4-19
Solve: The initial velocity is
v0 x = v0 cos5.0° = (20 m/s)cos5.0° = 19.92 m/s
v0 y = v0 sin 5.0° = (20 m/s)sin 5.0° = 1.743 m/s
The time it takes for the ball to reach the net is
x1 = x0 + v0 x (t1 − t0 ) ⇒ 7.0 m = 0 m + (19.92 m/s)(t1 − 0 s) ⇒ t = 0.351 s
G G G
The vertical position at v = v′ + V is
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2
= (2.0 m) + (1.743 m/s)(0.351 s − 0 s) + 12 (−9.8 m/s 2 )(0.351 s − 0 s) 2 = 2.01 m
Thus the ball clears the net by 1.01 m ≈ 1.0 m.
Assess: The vertical free fall of the ball, with zero initial velocity, in 0.351 s is 0.6 m. The ball will clear by
approximately 0.4 m if the ball is thrown horizontally. The initial launch angle of 5° provides some initial vertical
velocity and the ball clears by a larger distance. The above result is reasonable.
4.46. Model: The particle model for the ball and the constant-acceleration equations of motion in a plane are
assumed.
Visualize:
Solve: (a) The time for the ball to fall is calculated as follows:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2
⇒ 0 m = 4 m + 0 m + 12 (−9.8 m/s 2 )(t1 − 0 s) 2 ⇒ t1 = 0.9035 s
Using this result for the horizontal velocity:
x1 = x0 + v0 x (t1 − t0 ) ⇒ 25 m = 0 m + v0 x (0.9035 s − 0 s) ⇒ v0 x = 27.7 m/s
The friend’s pitching speed is 28 m/s.
(b) We have v0 y = ±v0 sin θ , where we will use the plus sign for up 5° and the minus sign for down 5°. We can write
y1 = y0 ± v0 sin θ (t1 − t0 ) −
g
g
(t1 − t0 ) 2 ⇒ 0 m = 4 m ± v0 sin θ t1 − t12
2
2
Let us first find t1 from x1 = x0 + v0 x (t1 − t0 ):
25 m = 0 m + v0 cosθ t1 ⇒ t1 =
25 m
v0 cosθ
Now substituting t1 into the y-equation above yields
⎛ 25 m ⎞ g ⎛ 25 m ⎞
0 m = 4 m ± v0 sin θ ⎜
⎟− ⎜
⎟
⎝ v0 cosθ ⎠ 2 ⎝ v0 cosθ ⎠
2
⎫
g (25 m) 2 ⎧
1
⎨
⎬ = 22.3 m/s and 44.2 m/s
2
2cos θ ⎩ 4 m ± (25 m) tan θ ⎭
The range of speeds is 22 m/s to 44 m/s, which is the same as 50 mph to 92 mph.
Assess: These are reasonable speeds for baseball pitchers.
⇒ v02 =
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4-20
Chapter 4
4.47. Model: We will use the particle model and the constant-acceleration kinematic equations in a plane.
Visualize:
Solve: The x-and y-equations of the ball are
x1B = x0B + (v0B ) x (t1B − t0B ) + 12 (aB ) x (t1B − t0B ) 2 ⇒ 65 m = 0 m + (v0B cos30°)t1B + 0 m
2
y1B = y0B + (v0B ) y (t1B − t0B ) + 12 (aB ) y (t1B − t0B ) 2 ⇒ 0 m = 0 m + (v0B sin 30°)t1B + 12 (− g )t1B
From the y-equation,
v0B =
gt1B
(2sin 30°)
Substituting this into the x-equation yields
2
g cos30° t1B
2sin 30°
⇒ t1B = 2.77 s
65 m =
For the runner:
t1R =
20 m
= 2.50 s
8.0 m/s
Thus, the throw is too late by 0.27 s.
Assess: The times involved in running the bases are small, and a time of 2.5 s is reasonable.
4.48. Model: Use the particle model for the ball and the constant-acceleration kinematic equations.
Visualize:
Solve: (a) The distance from the ground to the peak of the house is 6.0 m. From the throw position this distance is
5.0 m. Using the kinematic equation v12y = v02y + 2a y ( y1 − y0 ),
0 m 2 /s 2 = v02y + 2(−9.8 m/s 2 )(5.0 m − 0 m) ⇒ v0 y = 9.899 m/s
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Kinematics in Two Dimensions
4-21
The time for up and down motion is calculated as follows:
y2 = y0 + v0 y (t2 − t0 ) + 12 a y (t2 − t0 ) 2 ⇒ 0 m = 0 m + (9.899 m/s)t2 − 12 (9.8 m/s 2 )t22 ⇒ t2 = 0 s and 2.02 s
The zero solution is not of interest. Having found the time t2 = 2.02 s, we can now find the horizontal velocity
needed to cover a displacement of 18.0 m:
x2 = x0 + v0 x (t2 − t0 ) ⇒ 18.0 m = 0 m + v0 x (2.02 s − 0 s) ⇒ v0 x = 8.911 m/s
⇒ v0 = (8.911 m/s)2 + (9.899 m/s)2 = 13.3 m/s ≈ 13 m/s
G
(b) The direction of v0 is given by
θ = tan −1
v0 y
v0 x
= tan −1
9.899
= 48°
8.911
Assess: Since the maximum range corresponds to an angle of 45°, the value of 48° corresponding to a range of 18 m
and at a modest speed of 13.3 m/s is reasonable.
4.49. Model: We will assume a particle model for the sand, and use the constant-acceleration kinematic equations.
Visualize:
Solve: Using the equation x1 = x0 + v0 x (t1 − t0 ) + 12 a x (t1 − t0 ) 2 ,
x1 = 0 m + (v0 cos15°)(t1 − 0 s) + 0 m = (60 m/s)(cos15°)t1
We can find t1 from the y-equation, but note that v0 y = −v0 sin15° because the sand is launched at an angle below
horizontal.
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ 0 m = 3.0 m − (v0 sin15°)t1 − 12 gt12
= 3.0 m − (6.0 m/s)(sin15°)t1 − 12 (9.8 m/s 2 )t12
⇒ 4.9t12 + 1.55t1 − 3.0 = 0 ⇒ t1 = 0.6399 s and − 0.956 s ( unphysical )
Substituting this value of t1 in the x-equation gives the distance
d = x1 = (6.0 m/s)cos15°(0.6399 s) = 3.71 m ≈ 3.7 m
4.50. Model: We will use the particle model and the constant-acceleration kinematic equations for the car.
Visualize:
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4-22
Chapter 4
Solve: (a) The initial velocity is
v0 x = v0 cosθ = (20 m/s)cos 20° = 18.79 m/s
v0 y = v0 sin θ = (20 m/s)sin 20° = 6.840 m/s
Using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ,
0 m = 30 m + (6.840 m/s)(t1 − 0 s) + 12 (−9.8 m/s 2 )(t1 − 0 s)2 ⇒ 4.9t12 − 6.840t1 − 30 = 0
The positive root to this equation is t1 = 3.269 s. The negative root is physically unreasonable in the present case.
Using x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 , we get
x1 = 0 m + (18.79 m/s)(3.269 s − 0 s) + 0 = 61.4 m
The car lands 61 m from the base of the cliff.
(b) The components of the final velocity are v1x = v0 x = 18.79 m/s and
v1 y = v0 y + a y (t1 − t0 ) = 6.840 m/s − (9.8 m/s 2 )(3.269 s − 0 s) = −25.2 m/s
⇒ v = (18.79 m/s) + ( −25.2 m/s) 2 = 31.4 m/s
The car’s impact speed is 31 m/s.
Assess: A car traveling at 45 mph and being driven off a 30-m high cliff will land at a distance of approximately
200 feet (61.4 m). This distance is reasonable.
4.51. Model: Assuming constant acceleration allows us to use the kinematic equations.
Visualize: We apply the kinematic equations during the free-fall flight to find the velocity as the javelin left the
hand. Then use vf2 = vi2 + 2as Δs where Δs = 0.70m.
Solve: The range is Δx = 62m.
Δx = (v0 ) x Δt = v0 cosθΔt ⇒ Δt =
Δx
v0 cosθ
1
yf = yi + (v0 sin θ )Δt + a y (Δt ) 2
2
Insert our new expression for Δt .
Δy = (v0 sin θ )
⎛ Δx ⎞
Δx
1
+ (− g ) ⎜
⎟
v0 cosθ 2
⎝ v0 cosθ ⎠
2
Solve for v0 .
⎛ Δx ⎞
1
Δy = (tan θ ) Δx + (− g ) ⎜
⎟
2
⎝ v0 cosθ ⎠
2
2
⎛ Δx ⎞
1
(g)⎜
⎟ = (tan θ ) Δx − Δy
2
⎝ v0 cosθ ⎠
v02 =
2
⎞
1
g ⎛ Δx ⎞ ⎛
⎟
⎜
⎟ ⎜
2 ⎝ cosθ ⎠ ⎝ (tan θ ) Δx − Δy ⎠
2
v0 =
⎞
1
g ⎛ Δx ⎞ ⎛
⎟
⎜
⎟ ⎜
2 ⎝ cosθ ⎠ ⎝ (tan θ ) Δx − Δy ⎠
2
=
⎞
9.8 m/s 2 ⎛ 62 m ⎞ ⎛
1
⎟ = 25.78 m/s
⎜
⎟ ⎜
°
°
−
−
2
⎝ cos30 ⎠ ⎝ (tan 30 )(62 m) ( 2 m) ⎠
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Kinematics in Two Dimensions
4-23
This is the speed as the javelin leaves the hand. It now becomes vf as we consider the time during the throw (as the
hand accelerates it from rest).
vf2 − vi2 (25.78 m/s) 2 − (0 m/s)2
=
= 470 m/s 2
2Δs
2(0.70 m)
Assess: This is a healthy acceleration, but what is required for a good throw.
a=
4.52. Model: Both ships have a common origin at t = 0 s. Use subscripts A, B, and E for the ships and the Earth.
G
G
G
rAB = rAE + rEB
Solve: (a) The velocity vectors of the two ships are:
G
vAE = (20 mph)[cos30°iˆ − sin 30° ˆj ] = (17.32 mph)iˆ − (10.0 mph) ˆj
G
vBE = (25 mph)[cos 20°iˆ + sin 20° ˆj ] = (23.49 mph)iˆ + (8.55 mph) ˆj
G G
Since r = v Δt ,
G
G
rAE = vAE (2 h) = (34.64 miles)iˆ − (20.0 miles) ˆj
G
G
rBE = vBE (2 h) = (46.98 miles)iˆ + (17.10 miles) ˆj
G
G
G
G
G
rAB = rAE + rEB = rAE − rBE = (−12.34 miles)iˆ − (37.10 miles) ˆj ⇒ R = 39.1 miles
The distance between the ships two hours after they depart is 39 miles.
G
G
G
(b) Because vAB = vAE + vEB
G
G
G
G
G
vAB = vAE + vEB = vAE − vBE = −(6.17 mph)iˆ − (18.55 mph) ˆj ⇒ vAB = 19.5 mph ≈ 20 mph
The speed of ship A as seen by ship B is 19.5 mph.
Assess: The value of the speed is reasonable.
4.53. Model: We define the x-axis along the direction of east and the y-axis along the direction of north.
Solve: (a) The kayaker’s speed of 3.0 m/s is relative to the water. Since he’s being swept toward the east, he needs to
point at angle θ west of north. His velocity with respect to the water is
G
vKW = (3.0 m/s, θ west of north) = ( −3.0sin θ m/s)iˆ + (3.0cosθ m/s) ˆj
G
G
G
G
We can find his velocity with respect to the earth vKE = vKW + vWE , with vWE = (2.0 m/s)iˆ. Thus
G
vKE = ((−3.0sin θ + 2.0) m/s)iˆ + (3.0cosθ m/s) ˆj
In order to go straight north in the earth frame, the kayaker needs (vx ) KE = 0. This will be true if
2.0
⎛ 2.0 ⎞
⇒ θ = sin −1 ⎜
⎟ = 41.8°
3.0
⎝ 3.0 ⎠
Thus he must paddle in a direction 42° west of north.
(b) His northward speed is v y = 3.0 cos(41.8°) m/s = 2.236 m/s. The time to cross is
sin θ =
t=
100 m
= 44.7 s
2.236 m/s
The kayaker takes 45 s to cross.
4.54. Model: Mike and Nancy coincide at t = 0 s. Use subscripts B, M, N for the ball, Mike, and Nancy
respectively.
G
G
G
Solve: (a) According to the Galilean transformation of velocity vBN = vBM + vMN . Mike throws the ball with velocity
G
G
vBM = (22 m/s)cos63°iˆ + (22 m/s)sin63° ˆj , and vNM = (30 m/s)iˆ. Thus with respect to Nancy
G
G
G
vBN = vBM − vNM = (22cos63° − 30)iˆ m/s + (22sin 63°) ˆj m/s = (−20.0iˆ + 19.6 ˆj ) m/s
θ = tan −1
v′y
| v′x |
= tan −1
19.6 m/s
= 44.4°
20.0 m/s
The direction of the angle is 44.4° above the − x′ axis (in the second quadrant).
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4-24
Chapter 4
(b) With respect to Nancy
xBN = −(20.0 m/s)t
and
yBN = 0 m + (19.6 m/s)t − 12 (9.8 m/s 2 )t 2 = (19.6 m/s)t − (4.9 m/s 2 )t 2
4.55. Model: Use subscripts C, R, and G for car, rain, and ground respectively.
G
G
G
G
Solve: The Galilean transformation of velocity is vRG = vRC + vCG . While driving north, vCG = (25 m/s)iˆ and
vRG = −vR cosθ ˆj − vR sin θ iˆ. Thus,
G
G
G
vRC = vRG − vCG = (−vR sin θ − 25 m/s)iˆ − vR cosθ ˆj
Since the observer in the car finds the raindrops making an angle of 38° with the vertical, we have
G
While driving south, vCG
vR sin θ + 25 m/s
= tan 38°
vR cosθ
G
= −(25 m/s)iˆ, and v = −vR cosθ ˆj − vR sin θ iˆ. Thus,
G
vRG = ( −vR sin θ + 25 m/s)iˆ − vR cosθ ˆj
Since the observer in the car finds the raindrops falling vertically straight, we have
−vR sin θ + 25 m/s
= tan 0° = 0 ⇒ vR sin θ = 25 m/s
vR cosθ
Substituting this value of vR sin θ into the expression obtained for driving north yields:
25 m/s + 25 m/s
50 m/s
= tan 38° ⇒ vR cosθ =
= 64.0 m/s
vR cosθ
tan 38°
Therefore, we have for the velocity of the raindrops:
(vR sin θ ) 2 + (vR cosθ ) 2 = (25 m/s) 2 + (64.0 m/s) 2 ⇒ vR2 = 4721(m/s) 2 ⇒ vR = 68.7 m/s
tan θ =
vR sin θ 25 m/s
=
⇒ θ = 21.3°
vR cosθ 64 m/s
The raindrops fall at 69 m/s while making an angle of 21° with the vertical.
4.56. Model: Model the shaft as a rotating rigid body in the counterclockwise direction. Use the kinematic
equations for constant angular acceleration.
Visualize: First graph the data in a spreadsheet and see if it looks linear.
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Kinematics in Two Dimensions
4-25
Solve: The data looks moderately linear, but when the spreadsheet program puts on a second order polynomial
trendline the fit ( R 2 ) is much better and the intercept is closer to zero. We conclude that the graph isn’t linear, that
the circular motion is not uniform. The angular velocity is the slope of this graph, and the slope is decreasing, so ω
is decreasing. This means the angular acceleration α is negative.
4.57. Model: Model the DVD as a rotating rigid body.
Visualize: The formula for centripetal acceleration is a = ω 2r. Use ratios so that all the quantities that don’t change
cancel out.
Solve:
(a) The angular velocity ω doesn’t change in this part.
a2 ω 2r2 ω 2 (2r1)
=
=
= 2 ⇒ a2 = 2a1 = 2(20 m/s 2 ) = 40 m/s 2
a1 ω 2r1
ω 2 r1
(b) Call a3 the acceleration of the first speck when ω is doubled. ω 3= 2ω 1. The distance r from the center doesn’t
change in this part.
a3 ω 32 r (2ω 1) 2 r
=
=
= 22 = 4 ⇒ a3 = 4a1 = 4(20 m/s 2 ) = 80 m/s 2
a1 ω 12 r
ω 12 r
Assess: This ratio technique is very powerful; it’s harder to make mistakes and the ratios reveal relationships
between quantities.
4.58. Model: We will use the particle model for the test tube which is in nonuniform circular motion.
Solve: (a) The radial acceleration is
2
rev 1 min 2π rad ⎞
⎛
4
2
ar = rω 2 = (0.1 m) ⎜ 4000
×
×
⎟ = 1.75 × 10 m/s
min 60 s
1 rev ⎠
⎝
(b) An object falling 1 meter has a speed calculated as follows:
v12 = v02 + 2a y ( y1 − y0 ) = 0 m + 2( −9.8 m/s 2 )(−1.0 m) ⇒ v1 = 4.43 m/s
When this object is stopped in 1 × 10−3 s upon hitting the floor,
v2 = v1 + a y (t2 − t1 ) ⇒ 0 m/s = −4.43 m/s + a y (1 × 10−3 s) ⇒ a y = 4.4 × 103 m/s 2
This result is one-fourth of the above radial acceleration.
Assess: The radial acceleration of the centrifuge is large, but it is also true that falling objects are subjected to large
accelerations when they are stopped by hard surfaces.
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4-26
Chapter 4
4.59. Model: We will use the particle model for the astronaut undergoing nonuniform circular motion.
Solve: (a) The initial conditions are ω 0= 0 rad/s, θ 0 = 0 rad, t0 = 0 s, and r = 6.0 m. After 30 s,
1 rev 1 rev 2π rad
=
×
×
= 4.83 rad/s
ω 1=
1.3 s
1.3
s
rev
Using these values at t1 = 30 s,
ω 1= ω 0+ (at /r )(t1 − t0 ) = 0 + (at /r )t1
⎛ 1 ⎞
2
⇒ at = (6.0 m)(4.83 rad/s) ⎜
⎟ = 0.97 m/s
⎝ 30 s ⎠
(b) The radial acceleration is
g
= 14.3 g ≈ 14 g
(9.8 m/s 2 )
Assess: The above acceleration is typical of what astronauts experience during liftoff.
ar = rω 12= (6.0 m)(4.83 rad/s) 2
v2
.
r
Solve: The centripetal acceleration is given as 1.5 times the acceleration of gravity, so
a = (1.5)(9.80 m/s 2 ) = 14.7 m/s 2
4.60. Visualize: The magnitude of centripetal acceleration is given by a =
The radius of the turn is given by
r=
v 2 (25 m/s) 2
=
= 43 m
a 14.7 m/s 2
Assess: This seems reasonable.
4.61. Model: The earth is a rigid, rotating, and spherical body.
Visualize:
Solve: At a latitude of θ degrees, the radius is r = Re cos θ with Re = 6400 km = 6.400 × 106 m.
(a) In Miami θ = 26°, and we have r = (6.400 × 106 m)(cos 26°) = 5.752 × 106 m. The angular velocity of the earth is
ω=
2π
2π
=
= 7.272 × 10−5 rad/s
T
24 × 3600 s
Thus, vstudent = rω = (5.752 × 106 m)(7.272 × 10−5 rad/s) = 418 m/s ≈ 420 m/s.
(b) In Fairbanks θ = 65°, so r = (6.400 × 106 m)cos 65° = 2.705 × 106 m and vstudent = rω = (2.705 × 106 m)
(7.272 × 10−5 rad/s) = 197 m/s ≈ 200 m/s.
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Kinematics in Two Dimensions
4-27
4.62. Model: The satellite is a particle in uniform circular motion.
Visualize:
Solve: (a) The satellite makes one complete revolution in 24 h about the center of the earth. The radius of the motion
of the satellite is
r = 6.37 × 106 m + 3.58 × 107 m = 4.22 × 107 m
(distance traveled) 2π r
The speed of the satellite is v =
=
= 3.07 × 103 m/s.
(time taken)
24 h
(b) The acceleration of the satellite is centripetal, with magnitude
2
v 2 (3.07 × 103 m/s)
=
= 0.223 m/s 2
r
4.22 × 107 m
Assess: The small centripetal acceleration makes sense when realized it is for an object traveling in a circle with
radius ≈ 26,400 miles.
ar =
4.63. Model: The magnetic computer disk is a rigid rotating body.
Visualize:
Solve: (a) Using the rotational kinematic equation ω f = ω i + α Δt , we get
ω 1= 0 rad + (600 rad/s 2 )(0.5 s − 0 s) = 300 rad/s
ω 2= (300 rad/s) + (0 rad/s 2 )(1.0 s − 0.5 s) = 300 rad/s
The speed of the painted dot v2 = rω 2= (0.04 m)(300 rad/s) = 12 m/s.
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4-28
Chapter 4
(b) The number of revolutions during the time interval t0 to t2 is
1
2
1
2
θ1 = θ 0 + ω 0(t1 − t0 ) + α 0 (t1 − t0 ) 2 = 0 rad + 0 rad + (600 rad/s2 )(0.5 s − 0 s)2 = 75 rad
1
2
θ 2 = θ1 + ω 1(t2 − t1 ) + α1(t2 − t1 ) 2
⎛ 1 rev ⎞
= 75 rad + (300 rad/s)(1.0 s − 0.5 s) + 0 rad = 225 rad = (225 rad) ⎜
⎟ = 36 rev
⎝ 2π rad ⎠
4.64. Model: Model the turbine as a rotating rigid body. Assume the angular acceleration is constant. Δt = T .
Δω
1
Visualize: First find α from
and then use Δθ = ω0T + α T 2 .
2
T
Solve:
(a)
α=
Δω 0 − ω0 −ω0
=
=
T
T
T
1
1 ⎛ −ω ⎞
1
Δθ = ω 0T + α T 2 = ω0T + ⎜ 0 ⎟ T 2 = ω 0T
2
2⎝ T ⎠
2
(b) While we were thinking of SI units in the first part, any set of consistent units will do. We want the answer in
revolutions, so we’ll use the data in the units given.
1
1⎛
rev ⎞
Δθ = ω 0T = ⎜ 3800
⎟ (10 min) = 19000 rev
2
2⎝
min ⎠
Assess: That is a lot of revolutions, but the turbine was spinning fast and it took a long time to slow down.
4.65. Model: The drill is a rigid rotating body.
Visualize:
The figure shows the drill’s motion from the top.
Solve: (a) The kinematic equation ω f = ω i + α (tf − ti ) becomes, after using
ω i = 2400 rpm = (2400)(2π )/60 = 251.3 rad/s, tf − ti = 2.5 s − 0 s = 2.5 s, and wf = 0 rad/s,
0 rad = 251.3 rad/s + α (2.5 s) ⇒ α = −100 rad/s 2
(b) Applying the kinematic equation for angular position yields:
1
2
θ f = θi + ω i(tf − ti ) + α (tf − ti )2
1
= 0 rad + (251.3 rad/s)(2.5 s − 0 s) + ( −100 rad/s 2 )(2.5 s − 0 s) 2
2
= 3.2 × 102 rad = 50 rev
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Kinematics in Two Dimensions
4-29
4.66. Visualize: Please refer to Figure P4.66.
Solve: Since ω f = ω i + (area under α vs t curve), at t = 3 s, the angular velocity is
ω f = 60 rpm + (4.0 rad/s 2 )(2 s − 1 s)
⎛ 20 rev 60 s ⎞
= 60 rpm + (4 rad/s) ⎜
×
⎟
⎝ 2π rad 1 min ⎠
= 60 rpm + 38 rpm = 98 rpm
4.67. Model: Model the tire as a rotating rigid body. Assume the angular acceleration is constant. The radius of the
tire is 32 cm.
Visualize: ω i = 3.5 rev/s = 22 rad/s; ω f = 6.0 rev/s = 37.7 rad/s.
Solve: Δθ =
Δx
.
r
α=
ω f2− ω i2 ω f2− ω i2 (37.7 rad/s) 2 − (22 rad/s) 2
=
=
= 0.75 rad/s 2
Δx
200 m
2Δθ
2
2
r
0.32 m
Assess: The units all check out.
4.68. Model: Model the motor as a rotating rigid body. The angular acceleration is not constant, but we still know
that the angular acceleration is the derivative of the angular velocity and that the change in angle is the integral of the
angular velocity.
Visualize: The area under the ω vs. t graph is shown in the accompanying figure.
Solve:
(a) The motor reverses direction at a turning point, when ω = 0.
1
20 rad/s = t 2 rad/s ⇒ t = 40 s = 6.325 s ≈ 6.3 s
2
(b) The angular position is the area under the angular velocity graph, but we need calculus to do this for a non-linear graph.
Δθ = Ñ
6.325
0
6.325
1
1 ⎤
⎡
20 − t 2 dt = ⎢ 20t − t 3 ⎥
2
6 ⎦0
⎣
= 84 rad
Assess: Δθ seems reasonable given the angular speed and the time.
4.69. Model: Model the rider as a particle and the wheel as a rotating rigid body. Let t = 0 as the wheel starts from
rest. Assume α is constant.
Visualize: We integrate the angular acceleration to get the angular velocity and integrate that to get the angular position.
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4-30
Chapter 4
Solve:
(a) We want the answers expressed in terms of α and Δθ .
ω = ∫ α dt = αt + ω 0
But ω 0= 0. Now integrate again.
1
2
θ = ∫ α t dt = α t 2 + θ 0
1
2 Δθ
Δθ = α t 2 ⇒ t 2 =
α
2
ω = αt = α
2 Δθ
α
= 2αΔθ
v = ω R = 2αΔθ R
(b) Centripetal acceleration is a = ω 2 R.
a = ω 2 R = 2αΔθ R
Assess: The units check out.
4.70. Model: Model the gear as a rotating rigid body. The angular acceleration is not constant, but we still know
that the angular acceleration is the derivative of the angular velocity.
Visualize: The area under the ω vs. t graph is shown in the accompanying figure.
Solve:
(a) Take the derivative of ω (t ).
α=
dω d ⎛
1 ⎞
= ⎜ 2.0 + t 2 ⎟ = t
dt dt ⎝
2 ⎠
α (t = 4 s) = 4 rad/s 2
(b)
at = α r = (4 rad/s 2 )(0.060 m) = 0.24 m/s 2
Assess: We don’t need to know ω 0 or θ 0 because we don’t need those constants of integration when we are taking
derivatives.
4.71. Model: Model the shaft as a rotating rigid body. Assume the angular acceleration is constant.
Visualize: The data gives angular velocity as a function of time. If the angular acceleration is constant it will be the
slope of the best-fit straight line through the data. First convert the data from rpm to rad/s.
t (s)
0
1
2
3
4
5
6
ω (rpm)
ω (rad/s)
3010
2810
2450
2250
1940
1810
1510
315.21
294.26
256.56
235.62
203.16
189.54
158.13
Solve: The slope of the linear regression line is the angular acceleration.
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Kinematics in Two Dimensions
4-31
The spreadsheet says the slope is α = −26.22 rad/s 2 ≈ −26 rad/s 2 . The magnitude is 26 rad/s 2 .
Assess: It’s hard to have an intuitive feel for reasonable values of α , but our answer doesn’t seem ridiculous.
4.72. Model: Model the car as a particle in nonuniform circular motion.
Visualize:
Note that the tangential acceleration stays the same at 1.0 m/s 2 . As the tangential velocity increases, the radial
acceleration increases as well. After a time t1 , as the car goes through an angle θ1 − θ 0 , the total acceleration will
increase to 2.0 m/s 2 . Our objective is to find this angle.
Solve: Using v1 = v0 + at (t1 − t0 ), we get
v1 = 0 m/s + (1.0 m/s 2 )(t1 − 0 s) = (1.0 m/s 2 ) t1
⇒ ar =
rv12
r
2
=
(1.0 m/s 2 ) 2 t12
t2
= 1 (m/s 4 )
120 m
120
2
⎡ t2
⎤
⇒ atotal = 2.0 m/s 2 = at2 + ar2 = (1.0 m/s 2 )2 + ⎢ 1 (m/s 4 ) ⎥ ⇒ t1 = 14.4 s
⎢⎣120
⎥⎦
We can now determine the angle θ1 using
⎛ at ⎞
(t − t ) 2
⎝ r ⎟⎠ 1 0
θ1 = θ 0 + ω 0(t1 − t0 ) + 12 ⎜
1 (1.0 m/s 2 )
(14.4 s)2 = 0.864 rad = 49.5°
2 (120 m)
The car will have traveled through an angle of 50°.
= 0 rad + 0 rad +
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4-32
Chapter 4
4.73. Model: The string is wrapped around the spool in such a way that it does not pile up on itself, and unwinds
without slipping.
Visualize:
Solve: Since the string unwinds without slipping, the angular distance the spool turns as the string is pulled 1.0 m is
Δx
1.0 m
Δθ =
=
= 33 radians.
r 3.0 × 10−2 m
The angular acceleration of the spool due to the pull on the string is
a
1.5 m/s 2
α= t =
= 50 rad/s 2
r 3.0 × 10−2 m
The angular velocity of the spool after pulling the string is found with kinematics.
ω f2= ω i2+ 2αΔθ ⇒ ω f2= 0 rad 2 /s 2 + 2(50 rad/s 2 )(33 rad)
⇒ ωf2 = 57 rad/s
Converting to revolutions per minute,
⎛ rev ⎞⎛ 60 s ⎞
2
(57 rad/s) ⎜
⎟⎜
⎟ = 5.5 × 10 rpm
⎝ 2π rad ⎠⎝ min ⎠
Assess: The angular speed of 57 rad/s ≈ 9 rev/s is reasonable for a medium-sized spool.
4.74. Solve: (a) A golfer hits an iron shot with a new club as she approaches the green. She is pretty sure, based on
past experience, that she hit the ball with a speed of 50 m/s, but she is not sure at what angle the golf ball took flight.
She observed that the ball traveled 100 m before hitting the ground. What angle did she hit the ball?
(b) From the second equation,
(50 m/s)sin θ
(4.9 m/s 2 )t12 − (50sin θ m/s) t1 = 0 ⇒ t1 = 0 s and t1 =
4.9 m/s 2
Using the above value for t1 in the first equation yields:
100 m =
(50cosθ )(50sin θ ) m 2 /s 2
4.9 m/s 2
9.8
= 0.392 ⇒ 2θ = 23.1 ⇒ θ = 11.5°
25
Assess: Although the original speed is reasonably high (50 m/s = 112 mph), the ball travels a distance of only
⇒ 2cosθ sin θ = sin 2θ =
100 m, implying either a small launch angle around 10° or an angle closer to 80°. The calculated angle of 11.5° is
thus pretty reasonable.
4.75. Solve: (a) A submarine moving east at 3.0 m/s sees an enemy ship 100 m north of its path. The submarine’s
torpedo tube happens to be stuck in a position pointing 45° west of north. The tube fires a torpedo with a speed of
6.0 m/s relative to the submarine. How far east or west of the ship should the sub be when it fires?
(b) Relative to the water, the torpedo will have velocity components
vx = −6.0cos 45° m/s + 3.0 m/s = −4.24 m/s + 3 m/s = −1.24 m/s
v y = +6.0cos 45° m/s = +4.2 m/s
The time to travel north to the ship is
100 m = (4.2 m/s) t1 ⇒ t1 = 24 s
Thus, x = (1.24 m/s)(24 s) = −30 m. That is, the ship should be 30 m west of the submarine.
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Kinematics in Two Dimensions
4-33
4.76. Solve: (a) A 1000 kg race car enters a 50 m radius curve and accelerates around the curve for 10.0 s. The
forward force provided by the car’s wheels is 1500 N. After 10.0 s the car has moved 125 m around the track. Find
the initial and final angular velocities.
(b) From Newton’s second law,
Ft = mat ⇒ 1500 N = (1000 kg) at ⇒ at = 1.5 m/s 2
θ f = θ i + ω it +
Δθ =
Δs 125 m
=
= 2.5 rad
r
50 m
at 2
1.5 m/s 2
(10 s) 2 ⇒ ω i = 0.10 rad/s
t ⇒ 2.5 rad = 0 rad + ω i (10 s) +
2r
2(50 m)
ω f = ω i+
at
1.5 m/s 2
(10 s) = 0.40 rad/s
t = 0.1 rad/s +
50 m
r
4.77. Model: The ions are particles that move in a plane. They have vertical acceleration while between the
acceleration plates, and they move with constant velocity from the plates to the tumor. The flight time will be so
small, because of the large speeds, that we’ll ignore any deflection due to gravity.
Visualize:
Solve: There’s never a horizontal acceleration, so the horizontal motion is constant velocity motion at
vx = 5.0 × 106 m/s. The times to pass between the 5.0-cm-long acceleration plates and from the plates to the tumor
are
0.050 m
= 1.00 × 10−8 s
5.0 × 106 m/s
1.50 m
t2 − t1 =
= 3.00 × 10−7 s
5.0 × 106 m/s
t1 − t0 = t1 =
Upon leaving the acceleration plates, the ion has been deflected sideways to position y1 and has velocity v1 y . These are
y1 = y0 + v0 yt1 + 12 a yt12 = 12 a yt12
v1 y = v0 y + a yt1 = a yt1
In traveling from the plates to the tumor, with no vertical acceleration, the ion reaches position
y2 = y1 + v1 y (t2 − t1) = 12 a yt12 + (a yt1)(t2 − t1 ) = ( 12 t12 + t1 (t2 − t1 )) a y
We know y2 = 2.0 cm = 0.020 m, so we can solve for the acceleration a y that the ion had while between the plates:
ay =
1 t2
2 1
y2
+ t1 (t2 − t1 )
=
0.020 m
1 (1.00 × 10−8
2
2
s) + (1.00 × 10
−8
s)(3.00 × 10
−7
s)
= 6.6 × 1012 m/s 2
Assess: This acceleration is roughly 1012 times larger than the acceleration due to gravity. This justifies our
assumption that the acceleration due to gravity can be neglected.
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4-34
Chapter 4
4.78. Model: We will use the particle model for the ball’s motion under constant-acceleration kinematic equations.
Note that the ball’s motion on the smooth, flat board is a y = − g sin 20° = −3.352 m/s 2 .
Visualize:
Solve: The ball’s initial velocity is
v0 x = v0 cosθ = (3.0 m/s)cosθ
Using x1 = x0 + v0 x (t1 − t0 ) +
v0 y = v0 sin θ = (3.0 m/s)sin θ
1 a (t − t ) 2 ,
0
2 x 1
2.5 m = 0 m + (3.0 m/s)cosθ (t1 − 0 s) + 0 m ⇒ t1 =
(2.5 m)
0.833 s
=
(3.0 m/s)cosθ
cosθ
Using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 and the above equation for t1,
2
⎛ 0.833 s ⎞ 1
2 (0.833 s)
0 m = 0 m + (3.0 m/s)sin θ ⎜
(3.352
m/s
)
−
⎟
cos 2 θ
⎝ cosθ ⎠ 2
sin θ 1.164
⇒ (2.5 m)
=
⇒ 2.5sin θ cosθ = 1.164 ⇒ 2θ = 68.6° ⇒ θ = 34.3°
cosθ cos 2 θ
4.79. Model: Use the particle model for the arrow and the constant-acceleration kinematic equations.
Visualize:
Solve: Using v1 y = v0 y + a y (t1 − t0 ), we get
v1 y = 0 m/s − gt1 ⇒ v1 y = − gt1
Also using x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 ,
60 m = 0 m + v0 xt1 + 0 m ⇒ v0 x =
60 m
= v1x
t1
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Kinematics in Two Dimensions
4-35
Since v1 y /v1x = − tan 3.0° = −0.0524, using the components of v0 gives
− gt1
(0.0524)(60 m)
= −0.0524 ⇒ t1 =
= 0.566 s
(60 m/t1 )
(9.8 m/s 2 )
Having found t1, we can go back to the x-equation to obtain v0 x = 60 m/0.566 s = 106 m/s ≈ 110 m/s
Assess: In view of the fact that the arrow took only 0.566 s to cover a horizontal distance of 60 m, a speed of
106 m/s or 237 mph for the arrow is understandable.
4.80. Model: Use the particle model for the arrow and the constant-acceleration kinematic equations. We will
assume that the archer shoots from 1.75 m above the slope (about 5′ 9′′).
Visualize:
Solve: For the y-motion:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ y1 = 1.75 m + (v0 sin 20°)t1 − 12 gt12
⇒ y1 = 1.75 m + (50 m/s)sin 20°t1 − 12 gt12
For the x-motion:
x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (v0 cos 20°)t1 + 0 m = (50 m/s)(cos 20°)t1
Because y1/x1 = − tan15° = −0.268,
1.75 m + (50 m/s)(sin 20°)t1 − 12 gt12
(50 m/s)(cos 20°)t1
= −0.268 ⇒ t1 = 6.12 s and −0.058 s (unphysical)
Using t1 = 6.12 s in the x- and y-equations above, we get y1 = −77.0 m and x1 = 287 m. This means the distance
down the slope is
x12 + y12 = (287 m) 2 + (−77.0 m) 2 = 297 m.
Assess: With an initial speed of 112 mph (50 m/s) for the arrow, which is shot from a 15° slope at an angle of 20°
above the horizontal, a horizontal distance of 287 m and a vertical distance of 77.0 m are reasonable numbers.
4.81. Model: Treat the ball as a particle and apply the constant-acceleration equations of kinematics.
Visualize:
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4-36
Chapter 4
Solve: After the first bounce, the ball leaves the surface at 40° relative to the vertical or 50° relative to the
horizontal. We first calculate the time t1 between the second bounce and the first bounce as follows:
x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 ⇒ 3.0 m = 0 m + (v0 cos50°)t1 + 0 m ⇒ t1 =
3.0 m
v0 cos50°
In this time, the ball undergoes a vertical displacement of y1 − y0 = −(3.0 m) tan 20° = −1.092 m. Substituting these
values in the equation for the vertical displacement yields:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2
⎛ 3.0 m ⎞ 1
2 ⎛ 3.0 m ⎞
−1.092 m = 0 m + (v0 sin 50°)t1 − 12 gt12 = (v0 sin 50°) ⎜
⎟ − 2 (9.8 m/s ) ⎜
⎟
⎝ v0 cos50° ⎠
⎝ v0 cos50° ⎠
⇒ −1.092 m − 3.575 m =
2
−106.73 m3/s 2
⇒ v0 = 4.78 m/s, or v0 = 4.8 m/s
v02
Assess: A speed of 4.8 m/s or 10.7 mph on the first bounce is reasonable.
4.82. Model: Treat the skateboarder as a particle.
Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for
the second. The skateboarder’s final velocity at the top of the ramp is her initial velocity as she becomes airborne.
Solve: Without friction, the skateboarder’s acceleration on the ramp is a0 = − g sin 30° = −4.90 m/s 2 . The length of
the ramp is s1 = (1.0 m)/ sin 30° = 2.0 m. We can use kinematics to find her speed at the top of the ramp:
v12 = v02 + 2a0 ( s1 − s0 ) = v02 + 2a0 s1
⇒ v1 = (7.0 m/s) 2 + 2(−4.90 m/s 2 )(2.0 m) = 5.4 m/s
This is the skateboarder’s initial speed into the air, giving her velocity components v1x = v1 cos30° = 4.7 m/s and
v1 y = v1 cos30° = 2.7 m/s. We can use the y-equation of projectile motion to find her time in the air:
y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 1.0 m + (2.7 m/s)t2 − (4.90 m/s 2 )t22
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Kinematics in Two Dimensions
4-37
This quadratic equation has roots t2 = −0.253 s (unphysical) and t2 = 0.805 s. The x-equation of motion is thus
x2 = x1 + v1xt2 = 0 m + (4.7 m/s)t2 = 3.8 m
She touches down 3.8 m from the end of the ramp.
4.83. Model: Use the particle model for the motorcycle daredevil and apply the kinematic equations of motion.
Visualize:
Solve: We need to find the coordinates of the landing ramp ( x1, y1 ). We have
x1 = x0 + v0 x (t1 − t0 ) = 0 m + (40 m/s)t1
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 = 0 m + 0 m + 12 (−9.8 m/s 2 )t12 = −(4.9 m/s 2 )t12
G
This means we must find t1. Since v1 makes an angle of 20° below the horizontal we can find v1y as follows:
| v1 y |
v1x
= tan 20° ⇒ v1 y = −v1x tan 20° = −(40 m/s) tan 20° = −14.56 m/s
We now use this value of v1y , a y = −9.8 m/s 2 , and v0y = 0 m/s in the following equation to obtain t1:
v1 y = v0 y + a y (t1 − t0 ) ⇒ −14.56 m/s = 0 m/s − (9.8 m/s 2 )t1 ⇒ t1 = 1.486 s
Now, we are able to obtain x1 and y1 using the above x- and y-equations:
x1 = (40 m/s)(1.486 s) = 59.4 m
y1 = −(4.9 m/s 2 )(1.486 s) 2 = −10.82 m
That is, the landing ramp should be placed 11 m lower and 59 m away from the edge of the horizontal platform.
4.84. Model: The train and projectile are treated in the particle model. The height of the cannon above the tracks is
ignored.
Visualize:
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4-38
Chapter 4
Solve: In the ground reference frame, the projectile is launched with velocity components
(v0 x ) P = v0 cosθ + vtrain
(v0 y ) P = v0 sin θ
While the projectile is in free fall, vfy = viy − g Δt . The time for the projectile to rise to the highest point is
(with vfy = 0) Δt =
viy
g
. So the time to vertically rise and fall is
(v0 y ) P
2v
= 0 sin θ
g
g
During this time the projectile travels a horizontal distance
2v 2
2v v
( x1 ) P = (v0 x ) P t1 = 0 sin θ cosθ + 0 train sin θ
g
g
During the same time, the train travels a horizontal distance
1
2v v
2v 2a
( x1 )T = (v0 x )T t1 + at12 = 0 train sin θ + 02 sin 2 θ
2
g
g
t1 = 2Δt = 2
The range R is the difference between the two horizontal distances:
⎞
2v 2 ⎛
a
R = ( x1 )P − ( x1 )T = 0 ⎜ sin θ cosθ − sin 2 θ ⎟
g ⎝
g
⎠
Note that the range is independent of vtrain the train’s steady motion. This makes sense, since the train and projectile
share that motion when the projectile is launched.
2v 2
dR
= 0. Thus (ignoring the constant 0 )
Maximizing the range R requires
g
dθ
dR
a
a
= cos 2 θ − sin 2 θ − (2sin θ cosθ ) = cos 2θ − sin 2θ = 0
dθ
g
g
Solving for θ ,
sin 2θ
g
1
⎛g⎞
= tan 2θ = ⇒ θ = tan −1 ⎜ ⎟
cos 2θ
a
2
⎝a⎠
Note that
and
a > 0 (train speeding up) gives θ < 45°
a < 0 (train slowing down) gives θ > 45°
⎛g⎞
since tan −1 ⎜ ⎟ will be in the 2nd quadrant.
⎝a⎠
Assess: As a check, see what the angle θ is for the limiting case in which the train does not accelerate:
1
1
a = 0 ⇒ θ = tan −1 (∞) = × 90° = 45°
2
2
This is the expected answer.
4.85. Model: Assume the river flows toward the east. Use subscripts B, W, E for the boat, water, and earth
respectively. The child is at rest with respect to the water; this means they are in the same reference frame.
Visualize:
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Kinematics in Two Dimensions
4-39
Solve: The boat can go directly to the child at angle θ = tan −1 (200/1500) = 7.595°. The boat’s speed is 8.0 m/s, so
the components of the boat’s velocity in with respect to the water are
(vx )BW = −(8.0 m/s)cos7.595° = −7.93 m/s
(v y ) BW = (8.0 m/s)sin 7.595° = 1.06 m/s
The river flows with velocity (vx ) WE = 2.0iˆ m/s relative to the earth. In the earth’s frame, which is also the frame of
the riverbank and the boat dock, the boat’s velocity is
(vx ) BE = −5.93 m/s and (v y ) BW = 1.06 m/s
Thus the boat’s angle with respect to the riverbank is θ = tan −1 (5.93/1.06) = 10.1° ≈ 10°
Assess: The boat, like the child, is being swept downstream. This moves the boat’s angle away from the shore.
4.86. Model: The ball is a particle launched into projectile motion by the wheel.
Visualize:
Solve: The initial velocity of the projectile is the tangential velocity at the point of release, and the direction is
tangential to the wheel. The strategy is first to find the required projectile launch velocity v0 . The launch velocity is
the end of an angular acceleration through Δθ = 11/12 rev = 330°, and it’s related to the final angular velocity by
v0 = ωf r. Finally, use ωf to find the required angular acceleration α .
Since the release point is Δθ = 11/12 rev = 330°, φ = 30°. In the coordinate system of the figure,
x0 = −(20 cm)sin30° = −0.100 m
y0 = (20 cm)cos30° = 0.173 m
x1 = 100 cm = 1.00 m
y1 = 0 m
ay = −g
v0 x = v0 cosφ
v0 y = v0 sinφ
ax = 0
The launch is tangent to the circle, so the launch angle is also φ = 30°.
We will use Equations 4.17 for the position of a projectile as a function of time, and use the fact that the time to
travel the horizontal distance to the cup is the same for the vertical motion. First, find the flight time required for the
ball to hit the cup from the horizontal motion:
1.00 m = −0.100 m + (v0cos30°) t ⇒ t =
1.10 m
1.27 m
=
v0cos30°
v0
Substitute this time in the equation for the vertical motion:
⎛ 1.27 m ⎞ 1 ⎛ 1.27 m ⎞
0 m = 0.173 m + v0 sin 30° ⎜
⎟ − g⎜
⎟
⎝ v0 ⎠ 2 ⎝ v0 ⎠
2
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4-40
Chapter 4
Solving for the required initial velocity, v0 = (1.27 m) 2 g/2(0.808 m) = 3.128 m/s.
Now it is time to consider the angular motion. The required final angular velocity
ωf =
v0 3.128 m/s
=
= 15.64 rad/s
r
0.20 m
Using ωf2 = ωi2 + 2α Δθ , with Δθ = 330° = 5.75 rad,
α=
ωf2 − ωi2 (15.64 rad/s) 2 − 0
=
= 21.2 rad/s 2
2Δθ
2(5.76 rad)
Assess: An angular acceleration of 21.2 rad/s 2 ≈ 3.3 rev/s 2 seems reasonable for a spring-loaded wheel.
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FORCE AND MOTION
5
Conceptual Questions
5.1.
G
G
Two forces are present, tension T in the cable and gravitational force FG as seen in the figure.
5.2.
G
G
Four forces act on the block: the push of the spring Fsp , gravitational force FG , a normal force from the table top
G
G
n , and a kinetic frictional force due to the rough table surface f k
5.3.
G
G
Two forces act on the brick. Air resistance, or drag D, may be present, and the noncontact force due to gravity FG .
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5-1
5-2
Chapter 5
G
G
G
5.4. Four forces act on the block: the string tension T , normal force n and friction f with the surface, and
G
gravitational force FG .
G
5.5. (a) Two forces act on the ball after it leaves your hand: the long-range gravitational force FG and the contact
G
force of air resistance or drag D.
G
G
(b) FG is due to an interaction between the ball and earth, so earth is the agent. Drag D is due to an interaction
between the ball and the air, so the air is the agent. Some students are tempted to list a force due to the hand acting on
the ball. As can be seen in the figure, the hand is not touching the ball, and therefore it no longer applies a force to the
ball.
5.6. (a) The object with the largest mass accelerates the slowest, since a = mF . Thus B has the largest mass.
(b) The object with the smallest mass accelerates the fastest, so C has the smallest mass.
(c) Since the same force is applied to both blocks,
m
a
3
FA = FB ⇒ mA aA = mBaB ⇒ A = B =
mB aA 5
5.7. A force F causes an object of mass m to accelerate at a = mF = 10 m/s 2 . Let a′ be the new acceleration.
(a) If the force is doubled, (2 F ) = ma′ ⇒ a′ = 2
( mF ) = 2a = 20 m/s2. The acceleration is proportional to the force, so
if the force is doubled, the acceleration is also doubled.
(b) Doubling the mass means that F = (2m) a′ ⇒ a′ = 12
( mF ) = a2 = 5.0 m/s2.
(c) If the force and mass are both doubled, then (2 F ) = (2m) a′ ⇒ a′ =
F
m
= a = 10 m/s 2 .
5.8. A force F causes an object of mass m to accelerate at a = mF = 8 m/s 2 . Let a′ be the new acceleration.
(a) If the force is halved,
( F2 ) = ma′ ⇒ a′ = ( 2Fm ) = a2 = 4 m/s2. The acceleration is proportional to the force, so if the
force is halved, the acceleration is also halved.
F = 2a = 16 m/s 2 .
(b) If the mass is halved, F = m2 a′ ⇒ a′ = 2 m
( )
( )
F = a = 8 m/s 2 .
(c) If the force and mass are both halved, then ( F2 ) = ( m2 ) a′ ⇒ a′ = m
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Force and Motion
5-3
5.9. No. If an object is at rest its acceleration is zero, so you can only conclude that the net force is zero. Thus there
may be several forces acting on the object, but they all must sum to zero.
5.10. Yes—other applied forces may act to cancel the first force, so that the net force may be zero.
5.11. False. An object will accelerate in the direction of the net force. Its initial velocity may be in any direction.
G
G
5.12. No. Newton’s second law relates the applied net force Fnet to the resulting change in motion a. So the
G
G
quantity ma is better understood as related to a change in motion. Note, however, that ma has units that are the
same as force.
5.13.
Yes, it is possible for the friction force on an object to be in the same direction as the object’s motion. Consider the
case shown in the figure, in which a box is dropped onto a moving conveyer belt. The box is pushed horizontally in
the same direction as its motion. While an observer standing next to the conveyer belt sees the box move to the right
and eventually reach a constant speed (same as the conveyer belt), an observer standing on the conveyer belt would
see the box slide to the left and eventually come to a stop. The direction of the kinetic friction force is opposite to the
relative direction of motion between the two adjacent surfaces. In the example above, the box is moving to the left in
the reference frame of the conveyer belt and, as expected, the kinetic friction force is to the right.
5.14. The friction force between two surfaces is parallel to the surfaces. Since the wall is vertical, the friction force
is also vertical. To determine whether the static friction force is up or down, imagine that friction slowly disappears.
You can imagine the book beginning to slide downwards as no vertical forces are left to counteract the gravitational
force (the normal force is perpendicular to the surface, so is horizontal in this problem.) Thus static friction must be
up to balance the gravitational force downward, so the response is c.
5.15. The ball follows path C as it emerges from the tube. The centripetal force keeping it moving in a circle within
the tube is the normal force exerted by the tube wall on the ball. When that force is removed, the ball moves in a
straight line in accordance with Newton’s first law with the velocity it had when the acceleration went to zero, which
was tangential to the circle.
5.16.
G
G
(a) Basketball A is not in equilibrium because | F2 | > | F1 |, so there is a net downward force on A.
G G
G
(b) Basketball B is in equilibrium because the vector sum of the three forces is zero: F1 + F2 + F3 = 0.
5.17. Only a and b are inertial reference frames because the car is at constant velocity in only those two cases.
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5-4
Chapter 5
Exercises and Problems
Section 5.3 Identifying Forces
5.1. Visualize:
5.2 Visualize:
5.3. Model: Assume friction is negligible compared to other forces.
Visualize:
5.4. Visualize:
5.5. Visualize:
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Force and Motion
5-5
Section 5.4 What Do Forces Do? A Virtual Experiment
5.6. Solve: Let the object have mass m and each rubber band exert a force F. For two rubber bands to accelerate the
object with acceleration a, we must have a =
1 m.
2
2F .
m
We will need N rubber bands to give acceleration 3a to a mass
Find N:
3a =
NF
⎛ 2 F ⎞ 2 NF
⇒ 3⎜
⇒ N =3
⎟=
m/2
m
⎝ m ⎠
Three rubber bands are required.
5.7. Model: An object’s acceleration is linearly proportional to the net force.
Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F ∝ a and two
rubber bands (force 2F ) produce an acceleration of 1.2 m/s 2 , four rubber bands will produce an acceleration of 2.4 m/s 2 .
(b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). From part (a), we know that
2F /m = 1.2 m/s 2 , we have
2 F = (2m) a ⇒ a = F/m = 0.60 m/s 2
5.8. Visualize: Please refer to Figure EX5.8.
Solve: Newton’s second law is F = ma. Applying this to curves 1 and 2 at the point F = 2 rubber bands gives
2 F = m1(5a1) ⎫
5m1
2
2
⇒ m1 = m2 = (0.20 kg) = 0.080 kg
⎬1 =
2 F = m2 (2a1) ⎭ 2m2
5
5
Repeating the calculation for curves 2 and 3 at the point F = 5 rubber bands gives
5 F = m2 (5a1 ) ⎫ 5m2
5
5
⇒ m3 = m2 = (0.20 kg) = 0.50 kg
⎬1 =
5F = m3 (2a1 ) ⎭ 2m3
2
2
Assess: The line with the steepest slope should have the smallest mass, so we expect m1 < m2 < m3 , which is
consistent with our calculation.
5.9. Visualize: Please refer to Figure EX5.9.
Solve: Newton’s second law is F = ma. Applying this to curves 1 at the point F = 3 rubber bands and to curve 2 at
the point F = 5 rubber bands gives
3F = m1 (5a1 ) ⎫ 3 5m1
m1 12
⇒
=
⎬ =
5 F = m2 (4a1 ) ⎭ 5 4m2
m2 25
Assess: The line with the steepest slope should have the smallest mass, so we expect m1 < m2 , which is consistent
with our calculation.
5.10. Solve: Use proportional reasoning. Given that distance traveled is proportional to the square of the time,
d ∝ t 2 , so
d
t2
should be constant. We have
2.0 furlongs
=
x
2
⇒x=
(4.0 s)2
(2.0 furlongs) = 8.0 furlongs
(2.0 s)
(4.0 s)
(2.0 s)2
Thus the distance traveled by the object in 4.0 s is x = 8.0 furlongs.
Assess: A longer time should result in a longer distance traveled.
2
5.11 Solve: Use proportional reasoning. Let T = period of the pendulum, L = length of pendulum. We are given
T ∝ L , so T/ L should be constant. We have
3.0 s
x
3.0 m
=
⇒ x=
(3.0 s) = 3.7 s
2.0 m
3.0 m
2.0 m
Solving, the period of the 3.0 m long pendulum is x = 3.7 s.
Assess: Increasing the length increases the period, as expected.
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5-6
Chapter 5
Section 5.5 Newton’s Second Law
5.12. Visualize:
Solve: (a) Newton’s second law is F = ma. When F = 2 N, we have 2 N = (0.5 kg) a, so a = 4 m/s 2 .
(b) When F = 1 N, we have 1 N = (0.5 kg) a, so a = 2 m/s 2 .
After repeating this procedure at various points, the above graph is obtained.
5.13. Solve: Newton’s second law tells us that F = ma. Compute F for each case:
(a) F = (0.200 kg)(5 m/s 2 ) = 1 N.
(b) F = (0.200 kg)(10 m/s 2 ) = 2 N.
Assess: To double the acceleration we must double the force, as expected.
5.14. Visualize: Please refer to Figure EX5.14.
Solve: Newton’s second law is F = ma. The graph tells us the acceleration as a function of force. Knowing the force
and acceleration for any given point, we can find the mass. We chose the F = 1.0 N, which gives a = 4.0 m/s 2 .
Newton’s second law yields m = F/a = (1.0 N)/(4.0 m/s 2 ) = 0.25 kg.
Assess: To double-check the result insert m = 0.25 kg into Newton’s law for F = 0.50 kg. This gives a = 2.0 m/s 2 ,
which is consistent with the graph.
5.15. Visualize: Please refer to Figure EX5.15.
Solve: Newton’s second law is F = ma. The graph tells us the acceleration as a function of mass. Knowing the mass
and acceleration for any given point, we can find the force. We chose the m = 600g = 0.60 kg, which gives a = 6.0 m/s2.
Newton’s second law yields F = ma = (0.60 kg)(6.0 m/s 2 ) = 3.6 N
Assess: To double-check the result insert F = 3.6 N into Newton’s law for m = 200 g = 0.20 kg. This gives
a = F m = (3.6 N)/(0.2 kg) = 18 m/s 2 , which is consistent with the graph.
5.16. Solve: (a) This problem calls for an estimate, so we are looking for an approximate answer. Table 5.1 gives
us no information on laptops, but does give the weight of a one-pound object. Place a pound weight in one hand and
the laptop on the other. The sensation on your hand is the weight of the object. The sensation from the laptop is about
five times the sensation from the pound weight. So we conclude the weight of the laptop is about five times the
weight of the one-pound object or about 25 N.
(b) According to Table 5.1, the propulsion force on a car is 5000 N. A bicycle (including the rider) is about 100 kg.
This is about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of a
bicycle is somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as follows:
1
1
5000 N
F (bicycle) = (mass of car) × (acceleration of car) =
= 100 N
10
5
50
So we would roughly estimate the propulsion force of a bicycle to be 100 N.
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Force and Motion
5-7
5.17. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 5.1 gives
us no information on pencils, but does give us the weight of the U.S. quarter. Put the quarter on one hand and a pencil
on the other hand. The sensation on your hand is the weight of the object. The sensation from the quarter is about the
same as the sensation from the pencil, so they both have about the same weight. We can estimate the weight of the
pencil to be 0.05 N.
(b) According to Table 5.1, the propulsion force on a car is 5000 N. The mass of a sprinter is about 100 kg. This is
about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of a sprinter is
somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as follows:
1
1
5000 N
F (sprinter) = (mass of car) × (acceleration of car) =
= 100 N
10
5
50
So, we would roughly estimate the propulsion force of a sprinter to be 100 N.
Assess: This is the same estimated number as we obtained in Exercise 5.16. This is reasonable because, in both the
cases, the propulsion force comes from the human body and it the manner in which the force is delivered is not very
significant.
Section 5.6 Newton’s First Law
5.18. Visualize:
G
G G
Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction so
that the sum of all the three forces is zero.
5.19. Visualize:
G
G G
Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction so
that the sum of all three forces is zero.
5.20. Visualize:
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5-8
Chapter 5
G
G G
Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction so
that the sum of all the three forces is zero.
Section 5.7 Free-Body Diagrams
5.21. Visualize:
Solve: The free-body diagram shows two equal and opposite forces such that the net force is zero. The force directed
down is labeled as a gravitational force, and the force directed up is labeled as a tension. With zero net force the
acceleration is zero. So, a possible description is “an object hangs from a rope and is at rest” or “an object hanging
from a rope is moving up or down with a constant speed.”
5.22. Visualize:
Solve: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward. There is a
G
G
G
force labeled FG directed down, a force Fthrust directed up, and a force D directed down. So a possible description
is: “A rocket accelerates upward.”
5.23. Visualize:
G
Solve: The free-body diagram shows three forces. There is a gravitational force FG , which is down. There is a
G
G
G
normal force labeled n, which is up. The forces FG and n are shown with vectors of the same length so they are
equal in magnitude and the net vertical force is zero. So we have an object on a surface and which is not moving
G
vertically. The only horizontal force is a kinetic friction force f k that acts to the left, so the velocity of the object
must be to the right because friction always acts against the velocity. This means there is a net force to the left
producing an acceleration to the left. A possible description is “a baseball player sliding into second base.”
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Force and Motion
5-9
5.24. Visualize:
Assess: The cat is stationary, so there is no frictional force.
5.25. Visualize:
Assess: The problem says that there is no friction and it tells you nothing about any drag; so we do not include either
of these forces. The only remaining forces are the weight and the normal force.
5.26. Visualize:
Assess: The problem uses the word “sliding.” Any real situation involves friction with the surface. Since we are not
told to neglect it, we show that force.
5.27. Visualize:
Assess: Since the velocity is constant, the acceleration is zero, and the net force is zero.
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5-10
Chapter 5
5.28. Visualize:
Figure (a) shows velocity as downward, so the object is moving down. The length of the vector increases with each
step showing that the speed is increasing (like a dropped ball). Thus, the acceleration is directed down, as shown.
G
G
Since F = ma the force is in the same direction as the acceleration and must be directed down.
Figure (b), however, shows the velocity as upward, so the object is moving upward. But the length of the vector
decreases with each step showing that the speed is decreasing (like a ball thrown up). Thus, the acceleration is also
directed down, as shown. As in part (a), the net force must be directed down.
5.29. Visualize:
The velocity vector in figure (a) is shown downward and to the left, so movement is downward and to the left. The
velocity vectors get successively longer, which means the speed is increasing. Therefore the acceleration is
G
G
downward and to the left, as shown. By Newton’s second law F = ma , the net force must be in the same direction as
the acceleration. Thus, the net force is downward and to the left.
The velocity vector in (b) is shown to be upward and to the right. So movement is upward and to the right. The
velocity vector gets successively shorter, which means the speed is decreasing. Therefore the acceleration is
downward and to the left, as shown From Newton’s second law, the net force must be in the direction of the
acceleration, so it is directed downward and to the left.
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Force and Motion
5-11
5.30. Visualize:
Solve: According to Newton’s second law F = ma, the force at any time is found simply by multiplying the value of
the acceleration by the mass of the object. Thus, for example, the point at (2 s, 3 m/s2) become (2 s, 2 m/s2 × 2.0 kg) =
(2 s, 6 N).
5.31. Visualize:
Solve: According to Newton’s second law F = ma, the force at any time is found simply by multiplying the value of
the acceleration by the mass of the object. Thus, for example, the point at (1 s, 1 m/s2) become (1 s, 1 m/s2 × 0.5 kg) =
(0 s, 0.5 N).
5.32. Visualize:
Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing the
value of the force by the mass of the object. Thus, for example, the point at (0 s, 3 N) become [0 s, (3 N)/(2.0 kg)] =
(0 s, 1.5 m/s2).
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5-12
Chapter 5
5.33. Visualize:
Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing the
value of the force by the mass of the object. Thus, for example, the point at (2 s, −0.5 N) become [2 s, (−0.5 N)/
(0.5 kg)] = (0 s, −1 m/s2).
5.34. Model: Use the particle model for the object.
Solve: (a) We are given that, for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ), the
acceleration of the mass is a0 = 8.0 m/s 2 . According to Newton’s second law, F0 = m0 a0 so F0 /m0 = a0 . If the force
is doubled to F ′ = 2 F0 , Newton’s second law gives
F ′ = m0 a′
a′ = F ′/m0 = 2 F0 /m0 = 2a0 = 16 m/s 2
(b) The force is F0 and the mass is now m′ = 2m0 . Newton’s second law gives
F0 = m′a′
F0 = 2m0a′ ⇒ a′ = F0 /(2m0 ) = a0 /2 = 4.0 m/s 2
(c) The force is F ′ = 2 F0 and the mass is m′ = 2m0 . By inspection of Newton’s second law, it is evident that the
acceleration stays the same, so a′ = 8.0 m/s 2 .
(d) The force is F ′ = 2 F0 and the mass is m′ = m0 /2. Newton’s second law gives
F ′ = m′a′
2 F0 = (m0 /2)a′ ⇒ a′ = 4 F0 /m0 = 4a0 = 32 m/s 2
5.35. Model: Use the particle model for the object.
Solve: (a) We are told that, for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ), the
acceleration of the mass is a0 = 10 m/s 2 . According to Newton’s second law, F0 = m0a0 so F0 /m0 = a0 = 10 m/s 2 . If
the force becomes F ′ = 12 F0 , Newton’s second law gives
F ′ = m0a′
F0 /2 = m0a′ ⇒ a′ = F0 /(2m0 ) = 12 a0 = 5.0 m/s 2
(b) The force is F0 and the mass is now m′ = 12 m0 . Newton’s second law gives
F0 = m′a′
F0 = 12 m0a′ ⇒ a′ = 2 F0 /m0 = 2a0 = 20 m/s 2
(c) The force is F ′ = 12 F0 and the mass is m′ = 12 m0 . By inspection of Newton’s second law, it is evident that the
acceleration stays the same, so a′ = 10 m/s 2 .
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Force and Motion
5-13
(d) The force is F ′ = 12 F0 and the mass is m′ = 2m0 Newton’s second law gives
F ′ = m′a′
1F
2 0
= 2m0a′ ⇒ a′ = 14 F0 /m0 = 14 a0 = 2.5 m/s 2
5.36. Visualize:
Solve: (d) There is a normal force and a gravitational force that are equal and opposite, so this is an object on a
horizontal surface, or at least balanced in the vertical direction. The velocity vector is most likely in the same
direction as the thrust. The description of this free-body diagram could be “a jet-powered race car driving at constant
speed.”
5.37. Visualize:
Solve: (d) There are a normal force and a gravitational force that are equal and opposite, so this is an object on a
horizontal surface. The object is being pulled to the left with a nonzero net force, so it is accelerating to the left.
Because the friction force is kinetic, we know that the velocity is nonzero, and is most likely in the direction of the
net force. The description could be “a tow truck pulls a car out of the mud.”
5.38. Visualize:
Solve: (d) There is only a single force, which is the force due to gravity. We are unable to determine the direction of
motion. The description could be “Galileo has dropped a ball from the Leaning Tower of Pisa.”
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5-14
Chapter 5
5.39. Visualize:
Solve: (d) This is an object on a horizontal surface because FG = n. It must be moving to the right because the
kinetic friction is to the left. It is experiencing a net force to the right so it is accelerating to the right. The description
of the free-body diagram could be “a compressed spring is pushing a wooden block to the right over a table top.”
5.40. Visualize:
Solve: (d) There is an object on an inclined surface with a tension force pulling down the surface. There is a small
kinetic frictional force directed up the surface, which implies that the object is sliding down the slope. A description
could be “a box is being pulled down a slope with a rope that is parallel to the slope.”
5.41. Visualize:
Solve: (d) There is an object on an inclined surface. The net force is down the surface so the acceleration is down
the surface. The net force includes both a kinetic frictional force and a component of the gravitational force. Because
the kinetic frictional force is pointing down the slope, the object must be moving upward. The description could be “a
car is skidding up an embankment.”
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Force and Motion
5-15
5.42. Visualize:
Solve: (a) Newton’s second law is F = ma, so if we plot force as a function of acceleration (i.e., force is on vertical
axis, acceleration on horizontal axis), the slope of the curve should be the acceleration (see figure above).
(b) It would be reasonable to add the point (0 m/s2, 0 N) because Newton’s second law tells us that zero force
corresponds to an acceleration of zero.
(c) From the figure, we can estimate the mass from the slope. The result is m = 0.57 kg.
5.43. Visualize:
Tension is the only contact force. The downward acceleration implies that FG > T .
5.44. Visualize:
5.45. Visualize:
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5-16
Chapter 5
The normal and lift forces are perpendicular to the ground. The thrust force is parallel to the ground and in the
direction of acceleration. The drag force is opposite to the direction of motion.
5.46. Visualize:
The normal force is perpendicular to the hill. The frictional force is parallel to the hill.
5.47. Visualize:
The normal force is perpendicular to the hill. The kinetic frictional force is parallel to the hill and directed upward
opposite to the direction of motion. The wind force is given as horizontal. Since the skier stays on the slope (that is,
there is no acceleration away from the slope) the net force must be parallel to the slope.
5.48. Visualize:
As the rock slides there is kinetic friction between it and the rough concrete sidewalk. Since the rock stays on the
level surface, the net force must be along that surface, and is equal to the kinetic friction.
5.49. Visualize:
The drag force due to air is directed opposite to the motion.
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Force and Motion
5-17
5.50. Visualize:
The ball rests on the floor of the barrel because the gravitational force is equal to the normal force. The force of the
spring pushes to the right and causes an acceleration to the right.
5.51. Visualize:
There are no contact forces on the rock. The gravitational force is the only force acting on the rock.
5.52. Visualize:
The gymnast experiences the long range force of gravity. There is also a contact force from the trampoline that is the
normal force of the trampoline on the gymnast. The gymnast is moving downward and the trampoline is decreasing
her speed, so the acceleration is upward and there is a net force upward. Thus, the normal force must be larger than
the gravitational force. The actual behavior of the normal force over time will be complicated as it involves the
dynamic stretching of the trampoline.
5.53. Visualize:
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5-18
Chapter 5
Solve: (b) While the leaf hopper is in the act of jumping, it experiences an upward acceleration of 4 m/s 2 , so the
net force acting on it must be upward. Because only the normal force and the force due to gravity are acting in the
vertical direction, the normal force from the ground must be greater than the force due to gravity.
5.54. Visualize:
You can see from the motion diagram that the box accelerates to the right along with the truck. According to
G
G
Newton’s second law, F = ma , there must be a force to the right acting on the box. This is friction, but not kinetic
friction. The box is not sliding against the truck. Instead, it is static friction, the force that prevents slipping. Were it
not for static friction, the box would slip off the back of the truck. Static friction acts in the direction needed to
prevent slipping. In this case, friction must act in the forward (toward the right) direction.
5.55. Visualize:
You can see from the motion diagram that the bag accelerates to the left along with the car as the car slows down.
G
G
According to Newton’s second law, F = ma , there must be a force to the left acting on the bag. This is friction, but
not kinetic friction. The bag is not sliding across the seat. Instead, it is static friction, the force that prevents slipping.
Were it not for static friction, the bag would slide off the seat as the car stops. Static friction acts in the direction
needed to prevent slipping. In this case, friction must act in the backward (toward the left) direction.
5.56. Visualize: (a)
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Force and Motion
5-19
(b)
(c)
(d) The ball accelerates downward until the instant it makes contact with the ground. Once it makes contact, it
begins to compress and slow down. The compression covers a short but nonzero distance, as shown in the motion
diagram above. The point of maximum compression is the turning point, where the ball has an instantaneous speed of
v = 0 m/s and reverses direction. The ball then expands and speeds up until it loses contact with the ground. The
G
motion diagram shows that the acceleration vector a points upward the entire time that the ball is in contact with the
G
ground. An upward acceleration implies that there is a net upward force Fnet on the ball. The only two forces on the
ball are the gravitational force downward and the normal force of the ground upward. To have a net force upward
requires n > FG , so the ball bounces because the normal force of the ground exceeds the gravitational force, causing
a net upward force during the entire time that the ball is in contact with the ground. This net upward force accelerates
the ball in the upward direction to first slow it down, then reverse its velocity, and finally accelerate it upward until it
loses contact with the ground. Once contact with the ground is lost, the normal force vanishes and the ball is simply
in free fall.
5.57. Visualize: (a)
You are sitting on a bench driving along to the right. Both you and the bench are moving with a constant speed, so
there are no horizontal forces. There is a force on you due to gravity, which is directed down. There is a contact force
between you and the bench, which is directed up. Since you are not accelerating up or down the net vertical force on
you is zero, which means the two vertical forces are equal in magnitude. The statement of the problem gives no
indication of any other contact forces. Specifically, we are told that the bench is very slippery. We can take this to
mean there is no frictional force. So our force diagram includes only the normal force up, the gravitational force
down, and no horizontal force.
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5-20
Chapter 5
(b) The above considerations lead to the free-body diagram that is shown.
(c) The car (and therefore the bench) slows down. Does this create any new force on you? No. The forces remain the
same. This means the pictorial representation and the free-body diagram are unchanged.
(d) The car slows down because of some new contact force on the car (maybe the brakes lock the wheels and the
road exerts a force on the tires). But there is no new contact force on you. So the force diagram for you remains
unchanged. There are no horizontal forces on you. You do not slow down and you continue at constant velocity until
something in the picture changes for you (for example, you fall off the bench or hit the windshield).
(e) The net force on you has remained zero because the net vertical force is zero and there are no horizontal forces at
all. According to Newton’s first law, if the net force on you is zero, then you continue to move in a straight line with
a constant velocity. That is what happens to you when the car slows down. You continue to move forward with a
constant velocity. The statement that you are “thrown forward” is misleading and incorrect. To be “thrown” there
would need to be a net force on you and there is none. It might be correct to say that the car has been “thrown
backward” leaving you to continue onward (until you part company with the bench).
(f) We are now asked to consider what happens if the bench is NOT slippery. That implies there is a frictional force
between the bench and you. This force is certainly horizontal (parallel to the surface of the bench). Is the frictional
force directed forward (in the direction of motion) or backward? The car is slowing down and you are staying on the
bench. That means you are slowing down with the bench. Your velocity to the right is decreasing (you are moving
right and slowing down) so you are accelerating to the left. By Newton’s second law that means the force producing
the acceleration must be to the left. That force is the force of static friction and it is shown on the free-body diagram
below. Of course, when the car accelerates (increases in speed to the right) and you accelerate with it, then your
acceleration is to the right and the frictional force must be to the right.
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DYNAMICS I: MOTION ALONG A LINE
6
Conceptual Questions
6.1. (a) Static equilibrium. The barbell is not accelerating and has a velocity of zero.
(b) Dynamic equilibrium. The girder is not accelerating but has a nonzero constant velocity.
(c) Not in equilibrium. Slowing down means the acceleration is not zero.
(d) Dynamic equilibrium. The plane is not accelerating but has a nonzero constant velocity.
(e) Not in equilibrium. The box slows down with the truck, so has a nonzero acceleration.
6.2. No. The ball is still changing its speed, and just momentarily has zero velocity.
6.3. Kat is closest to the correct statement, which should read “Gravity pulls down on it, but the table pushes it up so
that the net force on the book is zero.”
6.4. No, because the net force is not necessarily in the same direction as the motion. For example, a car using its
brakes to slow its forward motion has a net force opposite its direction of motion.
6.5.
Equal. The tension in the cable is equal to the force of gravity, since the net force must be zero in order for the
elevator to move with constant speed.
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6-1
6-2
Chapter 6
6.6.
Greater. Since the elevator is slowing down as it moves downward, it has an upward net force, so the tension must be
greater than the gravitational force.
6.7. (a) False. The mass of an object is a measure of its inertia, which is the same regardless of location.
(b) True. The weight of an object is a measurement of how much force an object presses down on a surface with,
and varies depending on location and whether the object is accelerating.
(c) False. Mass and weight describe very different things, as pointed out in parts (a) and (b).
6.8. Yes, the scale shows the astronaut’s weight on the moon, since it shows how hard the astronaut is pressing down
on the surface of the moon. His weight is different on the earth, of course.
6.9. d > b = c > a. The net force on each ball is the gravitational force mg, so the net force on the balls is ranked by mass.
6.10. Correct. There will be a correct amount of salt in the pan balance since a pan balance measures mass, which is
independent of any gravitational force or acceleration present.
6.11. Zero. The passenger’s weight is zero once the box is launched since the passenger is in free fall (ignoring any
air friction). While gravity still pulls the passenger down, a scale placed under his feet would not register any weight
without a support under it.
6.12. The ball filled with lead is more massive. Since the balls are weightless, the astronaut must measure their
inertia (mass) directly. One easy option is to move each ball side to side in turn. More force is required to change the
more massive lead-filled ball’s direction of motion.
6.13. Larger. A free-body diagram for the book is shown in the figure. The normal force of the table on the book is
larger than its weight, since the net force is zero.
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Dynamics I: Motion Along a Line
6-3
6.14. (a) 2(Δt ). Your constant push Fx provides the net force, so the puck accelerates with constant acceleration
ax =
Fx
. From kinematics, with vix = 0,
m
⎛F ⎞
v = a x Δt = ⎜ x ⎟ Δt
⎝m⎠
If the mass is doubled, the time must also be doubled to reach the same speed, so you must push for a time 2(Δt ).
(b)
2( Δt ). From kinematics, with xi = 0 and vix = 0,
1
1⎛ F ⎞
d = a x ( Δt )2 = ⎜ x ⎟ (Δt ) 2
2
2⎝ m ⎠
If m is doubled, then (Δt ) 2 must be doubled, which means Δt is increased by a factor of
a time
2. So you must push for
2( Δt ).
6.15.
(a) d. Kinetic friction f k = μ k n determines the horizontal acceleration a =
− fk
, which slows down the block. From
m
−( μ k mg )
= − μ k g , which is independent of mass. Note that changing
m
the mass has no effect on the distance the block slides.
(b) 4d. From kinematics, with vfx = 0,
the free body diagram, n = FG = mg . So a =
2
⎛ m⎞
2
⎜ a ⎟ = v0 x − 2( μ k g ) Δx
s
⎝
⎠
⇒ Δx =
v02x
2μk g
Thus Δx ∝ v02x , and we use proportional reasoning:
d
v02x
=
Δx
(2v0 x ) 2
⇒ Δx = 4 d
6.16. Yes, the friction force on a crate dropped on a conveyor belt speeds the crate up to the belt’s speed.
6.17. North. The friction force on the crate is the only horizontal force and is responsible for speeding the crate up
along with the truck. Therefore the friction force points in the same direction as the motion of the crate.
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6-4
Chapter 6
6.18. ae > aa = ab > ad > ac . The balls all have the same cross-sectional area A. All of the balls have gravity pulling
down, resulting in an acceleration g. The drag force D = 14 Av 2 results in an addition or subtraction to g of
1 D.
m
For
ball e, the drag force adds to gravity, resulting in a higher acceleration. D = 0 for balls a and b. The drag force
opposes gravity for balls c and d, and since md > mc , ad > ac .
Exercises and Problems
Section 6.1 Equilibrium
6.1. Model: We can assume that the ring is a single massless particle in static equilibrium.
Visualize:
Solve: Written in component form, Newton’s first law is
( Fnet ) x = ∑ Fx = T1x + T2 x + T3 x = 0 N ( Fnet ) y = ∑ Fy = T1 y + T2 y + T3 y = 0 N
Evaluating the components of the force vectors from the free-body diagram:
T1x = −T1 T2 x = 0 N T3 x = T3 cos30°
T1 y = 0 N T2 y = T2 T3 y = −T3 sin 30°
Using Newton’s first law:
−T1 + T3 cos30° = 0 N T2 − T3 sin 30° = 0 N
Rearranging:
T1 = T3 cos30° = (100 N)(0.8666) = 86.7 N T2 = T3 sin 30° = (100 N)(0.5) = 50.0 N
G
Assess: Since T3 acts closer to the x-axis than to the y-axis, it makes sense that T1 > T2 .
6.2. Model: We can assume that the ring is a particle.
Visualize:
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Dynamics I: Motion Along a Line
6-5
This is a static equilibrium problem. We will ignore the weight of the ring, because it is “very light,” so the only three
forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle θ .
G
Solve: Because the ring is in equilibrium it must obey Fnet = 0 N. This is a vector equation, so it has both x- and
y-components:
( Fnet ) x = T3 cosθ − T2 = 0 N ⇒ T3 cosθ = T2
( Fnet ) y = T1 − T3 sin θ = 0 N ⇒ T3 sin θ = T1
We have two equations in the two unknowns T3 and θ . Divide the y-equation by the x-equation:
T3 sin θ
T 80 N
= tan θ = 1 =
= 1.6 ⇒ θ = tan −1 (1.6) = 58°
T3 cosθ
T2 50 N
Now we can use the x-equation to find
T3 =
50 N
T2
=
= 94 N
cosθ cos58°
The tension in the third rope is 94 N directed 58° below the horizontal.
6.3. Model: We assume the speaker is a particle in static equilibrium under the influence of three forces: gravity and
the tensions in the two cables.
Visualize:
Solve: From the lengths of the cables and the distance below the ceiling we can calculate θ as follows:
sin θ =
2m
= 0.677 ⇒ θ = sin −1 0.667 = 41.8°
3m
Newton’s first law for this situation is
( Fnet ) x = ∑ Fx = T1x + T2 x = 0 N ⇒ −T1 cosθ + T2 cosθ = 0 N
( Fnet ) y = ∑ Fy = T1 y + T2 y + wy = 0 N ⇒ T1 sin θ + T2 sin θ − w = 0 N
The x-component equation means T1 = T2 . From the y-component equation:
2T1 sin θ = w ⇒ T1 =
w
mg
(20 kg)(9.8 m/s 2 ) 196 N
=
=
=
= 147 N
2sin θ 2sin θ
2sin 41.8°
1.333
Assess: It’s to be expected that the two tensions are equal, since the speaker is suspended symmetrically from the
two cables. That the two tensions add to considerably more than the weight of the speaker reflects the relatively large
angle of suspension.
6.4. Model: We can assume that the coach and his sled are a particle being towed at a constant velocity by the two
ropes, with friction providing the force that resists the pullers.
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6-6
Chapter 6
Visualize:
Solve: Since the sled is not accelerating, it is in dynamic equilibrium and Newton’s first law applies:
( Fnet ) x = ∑ Fx = T1x + T2 x + f kx = 0 N ( Fnet ) y = ∑ Fy = T1 y + T2 y + f ky = 0 N
From the free-body diagram:
⎛1 ⎞
⎛1 ⎞
T1 cos ⎜ θ ⎟ + T2 cos ⎜ θ ⎟ − f k = 0 N
2
⎝
⎠
⎝2 ⎠
⎛1 ⎞
⎛1 ⎞
T1 sin ⎜ θ ⎟ − T2 sin ⎜ θ ⎟ + 0 N = 0 N
2
⎝
⎠
⎝2 ⎠
From the second of these equations T1 = T2 . Then from the first:
1000 N 1000 N
=
= 508 N ≈ 510 N
2cos10° 1.970
Assess: The two tensions are equal, as expected, since the two players are pulling at the same angle. The two add up
to only slightly more than 1000 N, which makes sense because the angle at which the two players are pulling is
small.
2T1 cos10° = 1000 N ⇒ T1 =
6.5. Model: Model the worker as a particle.
Visualize: In equilibrium the net force is zero in both directions. There must be a static friction force to keep her
from sliding off.
Solve: We only need to examine the y-direction.
(∑ F ) y = n − mgcosθ = 0 ⇒ n = mgcosθ = (850 N)(cos 20°) = 799 N ≈ 800 N
Assess: A good way to assess solutions like this is to consider what happens in the limit as θ → 0 and as θ → 90D.
In the first case n → mg and in the second n → 0 as expected.
Section 6.2 Using Newton’s Second Law
6.6. Solve: (a) Applying Newton’s second law to the diagram,
( Fnet ) y 3.0 N − 3.0 N
(F )
2.0 N − 4.0 N
ax = net x =
= 21.0 m/s 2 a y =
=
= 0 m/s 2
m
2.0 kg
m
2.0 kg
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Dynamics I: Motion Along a Line
(b) Applying Newton’s second law to the diagram,
(F )
4 N−2 N
a x = net x =
= 1.0 m/s 2
m
2 kg
ay =
( Fnet ) y
m
=
6-7
3 N −1 N − 2 N
= 0 m/s 2
2 kg
6.7. Solve: (a) For the diagram on the left, three of the vectors lie along the axes of the tilted coordinate system.
Notice that the angle between the 3 N force and the –y-axis is the same 20° by which the coordinates are tilted.
Applying Newton’s second law,
(F )
5.0 N − 1.0 N − (3.0sin 20°) N
a x = net x =
= 1.49 m/s 2 ≈ 1.5 m/s 2
m
2.0 kg
( Fnet ) y 2.82 N − (3.0cos 20°) N
ay =
=
= 0 m/s 2
m
2.0 kg
(b) For the diagram on the right, the 2-newton force in the first quadrant makes an angle of 15° with the positive
x-axis. The other 2-newton force makes an angle of 15° with the negative y-axis. The accelerations are
( Fnet ) x (2.0cos15°) N + (2.0sin15°) N − 3.0 N
=
= −0.28 m/s 2
m
2.0 kg
( Fnet ) y 1.414 N + (2.0sin15°) N − (2.0cos15°) N
ay =
=
= 0 m/s 2
m
2.0 kg
ax =
6.8. Solve: We can use the constant slopes of the three segments of the graph to calculate the three accelerations.
For t between 0 s and 2 s,
Δv 12 m/s − 0 s
ax = x =
= 6 m/s 2
2s
Δt
For t between 3 s and 6 s, Δvx = 0 m/s, so a x = 0 m/s 2 . For t between 6 s and 8 s,
Δvx 0 m/s − 12 m/s
=
= 24 m/s 2
Δt
3s
From Newton’s second law, at t = 1 s we have
ax =
Fnet = ma x = (2.0 kg)(6 m/s 2 ) = 12 N
At t = 4 s, a x = 0 m/s 2 , so Fnet = 0 N. At t = 7 s,
Fnet = max = (2.0 kg)(−4.0 m/s 2 ) = −8 N
Assess: The magnitudes of the forces look reasonable, given the small mass of the object. The positive and negative
signs are appropriate for an object first speeding up, then slowing down.
6.9. Visualize: Assuming the positive direction is to the right, positive forces result in the object accelerating to the
right and negative forces result in the object accelerating to the left. The final segment of zero force is a period of
constant speed.
Solve: We have the mass and net force for all the three segments. This means we can use Newton’s second law to
calculate the accelerations. The acceleration from t = 0 s to t = 3 s is
F
4N
ax = x =
= 2 m/s 2
m 2.0 kg
The acceleration from t = 3 s to t = 5 s is
F
−2 N
ax = x =
= −1 m/s 2
m 2.0 kg
The acceleration from t = 5 s to 8 s is ax = 0 m/s 2 . In particular, ax (at t = 6 s) = 0 m/s 2 .
We can now use one-dimensional kinematics to calculate v at t = 6 s as follows:
v = v0 + a1 (t1 − t0 ) + a2 (t2 − t0 )
= 0 + (2 m/s 2 )(3 s) + ( − 1 m/s 2 )(2 s) = 6 m/s − 2 m/s = 4 m/s
G
Assess: The positive final velocity makes sense, given the greater magnitude and longer duration of the positive F1.
A velocity of 4 m/s also seems reasonable, given the magnitudes and directions of the forces and the mass involved.
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6-8
Chapter 6
6.10. Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless, the
tension in the rope is the only horizontal force.
Visualize:
Solve: (a) Since the box is at rest, a x = 0 m/s 2 , and the net force on the box must be zero. Therefore, according to
Newton’s first law, the tension in the rope must be zero.
(b) For this situation again, a x = 0 m/s 2 , so Fnet = T = 0 N.
(c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since
a x = 5.0 m/s 2 ,
Fnet = T = ma x = (50 kg)(5.0 m/s 2 ) = 250 N
Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is exerting no horizontal force
on the box. For part (c), the 250 N force (the equivalent of about half the weight of a small person) seems reasonable
to accelerate a box of this mass at 5.0 m/s 2 .
6.11. Model: We assume that the box is a point particle that is acted on only by the tension in the rope and the pull
of gravity. Both the forces act along the same vertical line.
Visualize:
Solve: (a) Since the box is at rest, a y = 0 m/s 2 and the net force on it must be zero:
Fnet = T − FG = 0 N ⇒ T = FG = mg = (50 kg)(9.8 m/s 2 ) = 490 N
(b) Since the box is rising at a constant speed, again a y = 0 m/s 2 , Fnet = 0 N, and T = FG = 490 N.
(c) The velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since
a y = 5.0 m/s 2 ,
Fnet = T − FG = ma y = (50 kg)(5.0 m/s2 ) = 250 N
⇒ T = 250 N + w = 250 N + 490 N = 740 N
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Dynamics I: Motion Along a Line
6-9
(d) The situation is the same as in part (c), except that the rising box is slowing down. Thus a y = −5.0 m/s 2 and we
have instead
Fnet = T − FG = ma y = (50 kg)( −5.0 m/s 2 ) = −250 N
⇒ T = −250 N + FG = −250 N + 490 N = 240 N
Assess: For parts (a) and (b) the zero accelerations immediately imply that the gravitational force on the box must be
exactly balanced by the upward tension in the rope. For part (c) the tension not only has to support the gravitational
force on the box but must also accelerate it upward, hence, T must be greater than FG . When the box accelerates
downward, the rope need not support the entire gravitational force, hence, T is less than FG .
6.12. Model: We assume that the block is a point particle that is acted on only the force shown.
Visualize: We apply Newton’s second law in both parts.
Solve: (a) Since the net force is to the right the block is accelerating to the right, so it is speeding up in this case.
The answer is A. (b) The net force, while decreasing, is still to the right, so the block continues to accelerate to the
right and in this case continues to speed up. The answer is A.
Section 6.3 Mass, Weight, and Gravity
6.13. Model: Use the particle model for the woman.
Solve: (a) The woman’s weight on the earth is
wearth = mg earth = (55 kg)(9.80 m/s 2 ) = 540 N
(b) Since mass is a measure of the amount of matter, the woman’s mass is the same on Mars as on the earth. Her
weight on Mars is
wMars = mg Mars = (55 kg)(3.76 m/s 2 ) = 210 N
Assess: The smaller acceleration due to gravity on Mars reveals that objects are less strongly attracted to Mars than
to the earth. Thus the woman’s smaller weight on Mars makes sense.
6.14. Model: We assume that the passenger is a particle subject to two vertical forces: the downward pull of gravity
and the upward push of the elevator floor. We can use one-dimensional kinematics and Equation 6.10.
Visualize:
Solve: (a) The weight is
⎛ ay ⎞
⎛
0⎞
w = mg ⎜ 1 + ⎟ = mg ⎜1 + ⎟ = mg = (60 kg)(9.80 m/s 2 ) = 590 N
g ⎠
⎝ g⎠
⎝
(b) The elevator speeds up from v0y = 0 m/s to its cruising speed at v y = 10 m/s. We need its acceleration before we
can find the apparent weight:
ay =
Δv 10 m/s − 0 m/s
=
= 2.5 m/s 2
Δt
4.0 s
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6-10
Chapter 6
The passenger’s weight is
⎛
⎛ ay ⎞
2.5 m/s 2 ⎞
w = mg ⎜ 1 + ⎟ = (590 N) ⎜1 +
= (590 N)(1.26) = 740 N
⎜ 9.80 m/s 2 ⎟⎟
g ⎠
⎝
⎝
⎠
(c) The passenger is no longer accelerating since the elevator has reached its cruising speed. Thus, w = mg = 590 N
as in part (a).
Assess: The passenger’s weight is the gravitational force on the passenger in parts (a) and (c), since there is no
acceleration. In part (b), the elevator must not only support the gravitational force but must also accelerate him
upward, so it’s reasonable that the floor will have to push up harder on him, increasing his weight.
6.15. Model: We assume that the passenger is a particle acted on by only two vertical forces: the downward pull of
gravity and the upward force of the elevator floor.
Visualize: The graph has three segments corresponding to different conditions: (1) increasing velocity, meaning an
upward acceleration; (2) a period of constant upward velocity; and (3) decreasing velocity, indicating a period of
deceleration (negative acceleration).
Solve: Given the assumptions of our model, we can calculate the acceleration for each segment of the graph and then
apply Equation 6.10. The acceleration for the first segment is
v −v
8 m/s − 0 m/s
ay = 1 0 =
= 4 m/s 2
t1 − t0
2 s−0 s
⎛
⎛ ay ⎞
4 m/s 2 ⎞
4 ⎞
⎛
⇒ w = mg ⎜1 + ⎟ = mg ⎜1 +
= (75 kg)(9.80 m/s 2 ) ⎜1 +
⎟ = 1035 N
⎜ 9.80 m/s 2 ⎟⎟
9.80
g
⎝
⎠
⎝
⎠
⎝
⎠
For the second segment, a y = 0 m/s 2 and the weight is
⎛ 0 m/s 2 ⎞
2
w = mg ⎜1 +
⎟⎟ = mg = (75 kg)(9.80 m/s ) = 740 N
⎜
g
⎝
⎠
For the third segment,
ay =
v3 − v2 0 m/s − 8 m/s
=
= −2 m/s 2
t3 − t2
10 s − 6 s
⎛
−2 m/s 2 ⎞
⇒ w = mg ⎜1 +
= (75 kg)(9.80 m/s 2 )(1 − 0.2) = 590 N
⎜ 9.80 m/s 2 ⎟⎟
⎝
⎠
Assess: As expected, the weight is greater than the gravitational force on the passenger when the elevator is
accelerating upward and lower than normal when the acceleration is downward. When there is no acceleration the
weight is the gravitational force. In all three cases the magnitudes are reasonable, given the mass of the passenger and
the accelerations of the elevator.
6.16. Model: We assume the rocket is a particle moving in a vertical straight line under the influence of only two
forces: gravity and its own thrust.
Visualize:
Solve: (a) Using Newton’s second law and reading the forces from the free-body diagram,
Fthrust − FG = ma ⇒ Fthrust = ma + mg Earth = (0.200 kg)(10 m/s 2 + 9.80 m/s 2 ) = 3.96 N
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Dynamics I: Motion Along a Line
6-11
(b) Likewise, the thrust on the moon is (0.200 kg)(10 m/s 2 + 1.62 m/s 2 ) = 2.32 N.
Assess: The thrust required is smaller on the moon, as it should be, given the moon’s weaker gravitational pull. The
magnitude of a few newtons seems reasonable for a small model rocket.
Section 6.4 Friction
6.17. Model: We assume that the safe is a particle moving only in the x-direction. Since it is sliding during the
entire problem, we can use the model of kinetic friction.
Visualize:
Solve: The safe is in equilibrium, since it’s not accelerating. Thus we can apply Newton’s first law in the vertical
and horizontal directions:
( Fnet ) x = ∑ Fx = FB + FC − f k = 0 N ⇒ f k = FB + FC = 350 N + 385 N = 735 N
(Fnet ) y = ∑ Fy = n − FG = 0 N ⇒ n = FG = mg = (300 kg)(9.80 m/s 2 ) = 2.94 × 103 N
Then, for kinetic friction:
fk
735 N
=
= 0.250
n 2.94 × 103 N
Assess: The value of μ k = 0.250 is hard to evaluate without knowing the material the floor is made of, but it seems
f k = μk n ⇒ μk =
reasonable.
6.18. Model: We assume that the mule is a particle acted on by two opposing forces in a single line: the farmer’s
pull and friction. The mule will be subject to static friction until (and if!) it begins to move; after that it will be
subject to kinetic friction.
Visualize:
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6-12
Chapter 6
Solve: Since the mule does not accelerate in the vertical direction, the free-body diagram shows that n = FG = mg .
The maximum friction force is
fsmax = μs mg = (0.8)(120 kg)(9.80 m/s 2 ) = 940 N
The maximum static friction force is greater than the farmer’s maximum pull of 800 N; thus, the farmer will not be
able to budge the mule.
Assess: Maybe the farmer could put something smoother under the mule.
6.19. Model: We will represent the crate as a particle.
Visualize:
G G
Solve: (a) When the belt runs at constant speed, the crate has an acceleration a = 0 m/s 2 and is in dynamic
G
G
equilibrium. Thus Fnet = 0. It is tempting to think that the belt exerts a friction force on the crate. But if it did, there
would be a net force because there are no other possible horizontal forces to balance a friction force. Because there is
no net force, there cannot be a friction force. The only forces are the upward normal force and the gravitational force
on the crate. (A friction force would have been needed to get the crate moving initially, but no horizontal force is
needed to keep it moving once it is moving with the same constant speed as the belt.)
(b) If the belt accelerates gently, the crate speeds up without slipping on the belt. Because it is accelerating, the crate
must have a net horizontal force. So now there is a friction force, and the force points in the direction of the crate’s
motion. Is it static friction or kinetic friction? Although the crate is moving, there is no motion of the crate relative to
the belt. Thus, it is a static friction force that accelerates the crate so that it moves without slipping on the belt.
(c) The static friction force has a maximum possible value ( fs )max = μs n. The maximum possible acceleration of the
crate is
( fs )max μs n
=
m
m
If the belt accelerates more rapidly than this, the crate will not be able to keep up and will slip. It is clear from the
free-body diagram that n = FG = mg . Thus,
amax =
amax = μs g = (0.5)(9.80 m/s 2 ) = 4.9 m/s 2
6.20. Model: Model the cabinet as a particle.
Visualize: In equilibrium the net force is zero.
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Dynamics I: Motion Along a Line
6-13
Solve: The cabinet is in static equilibrium, so the static frictional force must have the same magnitude as Bob’s
pulling force: 200 N.
Assess: A possible misconception is that fs = μ n always. That value is the maximum possible value. If Bob pulled
harder and harder and got up to μ n = 235 N then the cabinet would move. But the static frictional force can easily be
less than this value.
6.21. Model: We assume that the truck is a particle in equilibrium, and use the model of static friction.
Visualize:
Solve: The truck is not accelerating, so it is in equilibrium, and we can apply Newton’s first law. The normal force
has no component in the x-direction, so we can ignore it here. For the other two forces:
( Fnet ) x = ∑ Fx = fs − ( FG ) x = 0 N ⇒ fs = ( FG ) x = mg sinθ = (4000 kg)(9.8 m/s 2 )(sin15°) = 10,145 N ≈ 10,000 N
Assess: The truck’s weight (mg) is roughly 40,000 N. A friction force that is ≈ 25, of the truck’s weight seems
reasonable.
6.22. Model: The car is a particle subject to Newton’s laws and kinematics.
Visualize:
Solve: Kinetic friction provides a horizontal acceleration which stops the car. From the figure, applying Newton’s
first and second laws gives
∑ Fx = − f k = ma x
∑ Fy = n − FG = 0 ⇒ n = FG = mg
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6-14
Chapter 6
Combining these two equations with f k = μk n yields
a x = − μk g = −(0.50)(9.80 m/s 2 ) = −4.9 m/s
Kinematics can be used to determine the initial velocity.
vf2 = vi2 + 2aΔx ⇒ vi2 = −2a x Δx
Thus
vi = −2( −4.9 m/s 2 )(65 m − 0 m) = 25 m/s
Assess: The initial speed of 25 m/s ≈ 56 mph is a reasonable speed to have initially for a vehicle to leave 65-meter-
long skid marks.
6.23. Model: We treat the train as a particle subject to rolling friction but not to drag (because of its slow speed and
large mass). We can use the one-dimensional kinematic equations. Look up the coefficient of rolling friction in the
table.
Visualize:
Solve: The locomotive is not accelerating in the vertical direction, so the free-body diagram shows us that
n = FG = mg . Thus,
f r = μr mg = (0.002)(50,000 kg)(9.80 m/s 2 ) = 980 N
From Newton’s second law for the decelerating locomotive,
−f
−980 N
ax = r =
= −0.01960 m/s 2
m
50,000 kg
Since we’re looking for the distance the train rolls, but we don’t have the time:
v 2 − v02 (0 m/s) 2 − (10 m/s)2
v12 − v02 = 2ax ( Δx) ⇒ Δx = 1
=
= 2.55 × 103 m ≈ 2.6 × 103 m
2
2a x
2( −0.01960 m/s )
Assess: The locomotive’s enormous inertia (mass) and the small coefficient of rolling friction make this long
stopping distance seem reasonable.
Section 6.5 Drag
6.24. Model: We assume that the skydiver is shaped like a box and is a particle. But we will also model the diver as
a cylinder falling end down to use C = 0.8.
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Dynamics I: Motion Along a Line
6-15
Visualize:
The skydiver falls straight down toward the earth’s surface, that is, the direction of fall is vertical. Since the skydiver
falls feet first, the surface perpendicular to the drag has the cross-sectional area A = 20 cm × 40 cm. The physical
conditions needed to use Equation 6.16 for the drag force are satisfied. The terminal speed corresponds to the
situation when the net force acting on the skydiver becomes zero.
Solve: The expression for the magnitude of the drag with v in m/s is
D ≈ 12 C ρ Av 2 = 0.5(0.8)(1.2 kg/m3 )(0.20 × 0.40)v 2 N = 0.038v 2 N
The gravitational force on the skydiver is FG = mg = (75 kg)(9.8 m/s 2 ) = 735 N. The mathematical form of the
condition defining dynamical equilibrium for the skydiver and the terminal speed is
G
G
G
Fnet = FG + D = 0 N
735
≈ 140 m/s
0.038
Assess: The result of the above simplified physical modeling approach and subsequent calculation, even if
approximate, shows that the terminal velocity is very high. This result implies that the skydiver will be very badly
hurt at landing if the parachute does not open in time.
2
⇒ 0.038vterm
N − 735 N = 0 N ⇒ vterm =
6.25. Model: We will represent the tennis ball as a particle. The drag coefficient is 0.5.
Visualize:
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6-16
Chapter 6
The tennis ball falls straight down toward the earth’s surface. The ball is subject to a net force that is the resultant of
the gravitational and drag force vectors acting vertically, in the downward and upward directions, respectively. Once
the net force acting on the ball becomes zero, the terminal velocity is reached and remains constant for the rest of the
motion.
Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is
G
G
G G
Fnet = FG + D = 0 N
Since only the vertical direction matters, one can write:
∑ Fy = 0 N ⇒ Fnet = D − FG = 0 N
When this condition is satisfied, the speed of the ball becomes the constant terminal speed v = vterm . The magnitudes
of the gravitational and drag forces acting on the ball are:
FG = mg = m(9.80 m/s 2 )
1
2
2
D ≈ (C ρ Avterm
) = 0.5(0.5)(1.2 kg/m3 )(π R 2 )vterm
= (0.3π )(0.0325 m)2 (26 m/s)2 = 0.67 N
2
The condition for dynamic equilibrium becomes:
0.67 N
(9.80 m/s 2 )m − 0.67 N = 0 N ⇒ m =
= 69 g
9.80 m/s 2
Assess: The value of the mass of the tennis ball obtained above seems reasonable.
6.26. Visualize:
We used the force-versus-time graph to draw the acceleration-versus-time graph. The peak acceleration was
calculated as follows:
F
10 N
amax = max =
= 2 m/s 2
m
5 kg
Solve: The acceleration is not constant, so we cannot use constant acceleration kinematics. Instead, we use the more
general result that
v(t ) = v0 + area under the acceleration curve from 0 s to t
The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 6 s is
1 (4
2
s)(2 m/s 2 ) = 4.0 m/s. We’ve used the fact that the area between 4 s and 6 s is zero. Thus, at t = 6 s,
vx = 4.0 m/s.
6.27. Visualize:
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Dynamics I: Motion Along a Line
6-17
The acceleration is a x = Fx /m, so the acceleration-versus-time graph has exactly the same shape as the force-versustime graph. The maximum acceleration is amax = Fmax /m = (6 N)/(2 kg) = 3 m/s 2 .
Solve: The acceleration is not constant, so we cannot use constant-acceleration kinematics. Instead, we use the more
general result that
v(t ) = v0 + area under the acceleration curve from 0 s to t
The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 4 s is a rectangle
(3 m/s 2 × 2 s = 6 m/s) plus a triangle
(
1 ×3
2
)
m/s 2 × 2 s = 3 m/s . Thus vx = 9 m/s at t = 4 s.
6.28. Model: You can model the beam as a particle in static equilibrium.
Visualize:
Solve: Using Newton’s first law, the equilibrium equations in vector and component form are:
G
G
G G G
Fnet = T1 + T2 + FG = 0 N
( Fnet ) x = T1x + T2 x + FGx = 0 N
( Fnet ) y = T1 y + T2 y + FGy = 0 N
Using the free-body diagram yields:
−T1 sin θ1 + T2 sin θ 2 = 0 N
T1 cosθ1 + T2 cosθ 2 − FG = 0 N
The mathematical model is reduced to a simple algebraic system of two equations with two unknowns, T1 and T2 .
Substituting θ1 = 20°, θ 2 = 30°, and FG = mg = 9800 N, the simultaneous equations become
−T1 sin 20° + T2 sin 30° = 0 N
T1 cos 20° + T2 cos30° = 9800 N
You can solve this system of equations by simple substitution. The result is T1 = 6397 N ≈ 6400 N and T2 = 4376 N ≈
4380 N.
Assess: The above approach and result seem reasonable. Intuition indicates there is more tension in the left rope than
in the right rope.
6.29. Model: The plastic ball is represented as a particle in static equilibrium.
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6-18
Chapter 6
Visualize:
Solve: (a) The electric force, like the weight, is a long-range force. So the ball experiences the contact force of the
string’s tension plus two long-range forces. The equilibrium condition is
( Fnet ) x = Tx + ( Felec ) x = T sin θ − Felec = 0 N
( Fnet ) y = Ty + ( FG ) y = T cosθ − mg = 0 N
We can solve the y-equation to get
mg
(0.001 kg)(9.8 m/s 2 )
=
= 0.0104 N
cosθ
cos 20°
Substituting this value into the x-equation,
Felec = T sin θ = (1.04 × 1022 N)sin 20° = 0.0036 N
(b) The tension in the string is 0.0104 N ≈ 0.010 N.
T=
6.30. Model: The piano is in static equilibrium and is to be treated as a particle.
Visualize:
Solve: (a) Based on the free-body diagram, Newton’s second law is
( Fnet ) x = 0 N = T1x + T2 x = T2 cosθ 2 − T1 cosθ1
( Fnet ) y = 0 N = T1 y + T2 y + T3 y + FGy = T3 − T1 sin θ1 − T2 sin θ 2 − mg
Notice how the force components all appear in the second law with plus signs because we are adding forces. The
negative signs appear only when we evaluate the various components. These are two simultaneous equations in the
two unknowns T2 and T3 . From the x-equation we find
T2 =
T1 cosθ 1 (500 N)cos15°
=
= 533 N
cosθ 2
cos 25°
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Dynamics I: Motion Along a Line
6-19
(b) Now we can use the y-equation to find
T3 = T1 sin θ1 + T2 sin θ 2 + mg = 5.25 × 103 N
6.31. Model: We will represent Henry as a particle. His motion is governed by constant-acceleration kinematic
equations.
Solve: (a) Henry undergoes an acceleration from 0 s to 2.0 s, constant velocity motion from 2.0 s to 10.0 s, and
another acceleration as the elevator brakes from 10.0 s to 12.0 s. The weight is the same as the gravitational force
during constant velocity motion, so Henry’s weight w = FG = mg is 750 N. His weight is less than the gravitational
force on him during the initial acceleration, so the acceleration is in a downward direction (negative a). Thus, the
elevator’s initial motion is down.
(b) Because the gravitational force on Henry is 750 N, his mass is m = FG /g = 76.5 kg ≈ 77 kg.
(c) The apparent weight during vertical motion is given by
⎛ w
⎞
⎛
a⎞
− 1⎟
w = mg ⎜ 1 + ⎟ ⇒ a = g ⎜
g
F
⎝
⎠
⎝ G
⎠
During the interval 0 s ≤ t ≤ 2 s, the elevator’s acceleration is
⎛ 600 N ⎞
− 1⎟ = −1.96 m/s 2
a = g⎜
⎝ 750 N ⎠
At t = 2 s, Henry’s position is
1
1
y1 = y0 + v0Δt0 + a(Δt0 )2 = a (Δt0 ) 2 = −3.92 m
2
2
and his velocity is
v1 = v0 + aΔt0 = aΔt0 = −3.92 m/s
During the interval 2 s ≤ t ≤ 10 s, a = 0 m/s 2 . This means Henry travels with a constant velocity v1 = −3.92 m/s. At
t = 10 s he is at position
y2 = y1 + v1Δt1 = −35.3 m
and he has a velocity v2 = v1 = −3.92 m/s. During the interval 10 s ≤ t ≤ 12.0 s, the elevator’s acceleration is
⎛ 900 N ⎞
− 1⎟ = +1.96 m/s 2
a = g⎜
750
N
⎝
⎠
The upward acceleration vector slows the elevator and Henry feels heavier than normal. At t = 12.0 s Henry is at
position
1
y3 = y2 + v2 ( Δt2 ) + a (Δt2 )2 = −39.2 m
2
Thus Henry has traveled distance 39.2 m ≈ 39 m.
6.32. Model: We’ll assume Zach is a particle moving under the effect of two forces acting in a single vertical line:
gravity and the supporting force of the elevator.
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6-20
Chapter 6
Visualize:
Solve: (a) Before the elevator starts braking, Zach is not accelerating. His weight is
⎛ 0 m/s 2 ⎞
⎛ a⎞
2
w = mg ⎜1 + ⎟ = mg ⎜1 +
⎟⎟ = mg = (80 kg)(9.80 m/s ) = 784 N
⎜
g
g
⎝
⎠
⎝
⎠
Zach’s weight is 7.8 × 102 N.
(b) Using the definition of acceleration,
a=
Δv v1 − v0 0 − ( −10) m/s
=
=
= 3.33 m/s 2
Δt t1 − t0
3.0 s
⎛ 3.33 m/s 2 ⎞
⎛
a⎞
⇒ w = mg ⎜1 + ⎟ = (80 kg)(9.80 m/s 2 ) ⎜1 +
= (784 N)(1 + 0.340) = 1050 N
2⎟
⎜
⎟
g⎠
⎝
⎝ 9.80 m/s ⎠
Now Zach’s weight is 1.05 × 103 N ≈ 1.1 kN.
Assess: While the elevator is braking, it not only must support the gravitational force on Zach but must also push
upward on him to decelerate him, so his weight is greater than the gravitational force.
6.33. Model: We can assume the foot is a single particle in equilibrium under the combined effects of gravity, the
tensions in the upper and lower sections of the traction rope, and the opposing traction force of the leg itself. We can
also treat the hanging mass as a particle in equilibrium. Since the pulleys are frictionless, the tension is the same
everywhere in the rope. Because all pulleys are in equilibrium, their net force is zero. So they do not contribute to T.
Visualize:
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Dynamics I: Motion Along a Line
6-21
Solve: (a) From the free-body diagram for the mass, the tension in the rope is
T = FG = mg = (6 kg)(9.80 m/s 2 ) = 58.8 N ≈ 59 N
(b) Using Newton’s first law for the vertical direction on the pulley attached to the foot,
( Fnet ) y = ∑ Fy = T sin θ − T sin15° − ( FG )foot = 0 N
⇒ sin θ =
T sin15° + ( FG )foot
m g
(4 kg)(9.80 m/s 2 )
= sin15° + foot = 0.259 +
= 0.259 + 0.667 = 0.926
T
T
58.8 N
⇒ θ = sin −1 0.926 = 67.8° ≈ 68°
(c) Using Newton’s first law for the horizontal direction,
( Fnet ) x = ∑ Fx = T cosθ + T cos15° − Ftraction = 0 N
⇒ Ftraction = T cosθ + T cos15° = T (cos67.8° + cos15°)
= (58.8 N)(0.3778 + 0.9659) = (58.8 N)(1.344) = 79 N
Assess: Since the tension in the upper segment of the rope must support the foot and counteract the downward pull
of the lower segment of the rope, it makes sense that its angle is larger (a more direct upward pull). The magnitude of
the traction force, roughly one-tenth of the gravitational force on a human body, seems reasonable.
6.34. Model: We can assume the person is a particle moving in a straight line under the influence of the combined
decelerating forces of the air bag and seat belt or, in the absence of restraints, the dashboard or windshield.
Visualize:
Solve: (a) In order to use Newton’s second law for the passenger, we’ll need the acceleration. Since we don’t have
the stopping time:
v 2 − v02
0 m 2 /s 2 − (15 m/s) 2
=
= −112.5 m/s 2
v12 = v02 + 2a( x1 − x0 ) ⇒ a = 1
2( x1 − x0 )
2(1 m − 0 m)
⇒ Fnet = F = ma = (60 kg)(−112.5 m/s 2 ) = −6750 N
The net force is 6750 N to the left.
(b) Using the same approach as in part (a),
v 2 − v02
0 m 2 /s 2 − (15 m/s) 2
= (60 kg)
= −1,350,000 N
F = ma = m 1
2( x1 − x0 )
2(0.005 m)
The net force is 1,350,000 N to the left.
(c) The passenger’s weight is mg = (60 kg)(9.8 m/s 2 ) = 588 N. The force in part (a) is 11.5 times the passenger’s
weight. The force in part (b) is 2300 times the passenger’s weight.
Assess: An acceleration of 11.5g is well within the capability of the human body to withstand. A force of 2300 times
the passenger’s weight, on the other hand, would surely be catastrophic.
6.35. Visualize: All the motion is in the horizontal (i.e., x ) direction. Acceleration is the second derivative of
position.
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6-22
Chapter 6
Solve: The first derivative is v =
dx
dv
= (6t 2 − 6t ) m/s. The second derivative is a =
= (12t − 6) m/s 2 . Apply
dt
dt
Newton’s second law: F = ma = (2.0 kg)((12t − 6) m/s 2 ). Plug in the two values for t.
(a)
F |0s = (2.0 kg)((12(0 s) − 6) m/s 2 ) = −12 N
(b)
F |1s = (2.0 kg)((12(1 s) − 6) m/s 2 ) = 12 N
Assess: The net force changed direction between t = 0 s and t = 1 s.
6.36. Visualize: We’ll use vf2 = vi2 + 2aΔs to find the acceleration of the balls, which will be inversely proportional
to the mass of the balls. Δs = 15 cm and vi = 0 in each case.
Solve: Newton’s second law relates mass, acceleration, and net force: a = F
1
1
then the slope
. If we graph a vs.
m
m
of the straight line should be the size of the piston’s force.
We see that the linear fit is very good. The slope is 59.12 N ≈ 59 N; this is the size of the piston’s force.
Assess: We are glad to see that the intercept of our line looks very small, even though we don’t have a ball the
inverse of whose mass is zero.
6.37. Model: The ball is represented as a particle that obeys constant-acceleration kinematic equations.
Visualize:
Solve: This is a two-part problem. During part 1 the ball accelerates upward in the tube. During part 2 the ball
undergoes free fall (a = − g ). The initial velocity for part 2 is the final velocity of part 1, as the ball emerges from the
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Dynamics I: Motion Along a Line
6-23
tube. The free-body diagram for part 1 shows two forces: the air pressure force and the gravitational force. We need
only the y-component of Newton’s second law:
( Fnet ) y Fair − FG Fair
2N
=
=
−g=
− 9.80 m/s 2 = 30.2 m/s 2
ay = a =
0.05 kg
m
m
m
We can use kinematics to find the velocity v1 as the ball leaves the tube:
v12 = v02 + 2a ( y1 − y0 ) ⇒ v1 = 2ay1 = 2(30.2 m/s2 )(1 m) = 7.77 m/s
For part 2, free-fall kinematics v22 = v12 − 2 g ( y2 − y1 ) gives
y2 − y1 =
v12
= 3.1 m
2g
6.38. Model: Model the rocket as a particle. Assume the mass of the rocket is constant so the acceleration is constant.
Assume the rocket starts from rest. Neglect air resistance.
Visualize: We’ll use vf2 = vi2 + 2aΔy to find the speed of the rocket. The net force is Fthrust − mg.
Solve:
(a) Δy = h,
vi = 0,
a = Fnet /m
− mg ⎞
⎛F
vf2 = vi2 + 2aΔy = 2 ⎜ thurst
⎟h
m
⎝
⎠
For v as a function of h we have:
⎛F
⎞
v(h) = vf = 2 ⎜ thurst − g ⎟ h
⎝ m
⎠
(b) For h = 85 m
⎛ 9.5 N
⎞
v = 2⎜
− 9.8 m/s 2 ⎟ (85 m) = 54 m/s
⎝ 0.35 kg
⎠
Assess: This 54 m/s speed seems reasonable for a model rocket.
Some of our assumptions would not be good approximations for large fast rockets that go very high: air resistance
wouldn’t be negligible, and the fuel expended reduces the mass of the rocket which increases the acceleration. At
very high altitudes (where air resistance no longer has an effect) even g decreases slightly.
6.39. Model: We will represent the bullet as a particle.
Visualize:
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6-24
Chapter 6
Solve: (a) We have enough information to use kinematics to find the acceleration of the bullet as it stops. Then we
can relate the acceleration to the force with Newton’s second law. (Note that the barrel length is not relevant to the
problem.) The kinematic equation is
v2
(400 m/s) 2
v12 = v02 + 2aΔx ⇒ a = − 0 = −
= −6.67 × 105 m/s 2
2Δx
2(0.12 m)
G
Notice that a is negative, in agreement with the vector a in the motion diagram. Turning to forces, the wood exerts
two forces on the bullet. First, an upward normal force that keeps the bullet from “falling” through the wood. Second,
G
G
a retarding frictional force f k that stops the bullet. The only horizontal force is f k , which points to the left and thus
has a negative x-component. The x-component of Newton’s second law is
( Fnet ) x = − f k = ma ⇒ f k = −ma = −(0.01 kg)(−6.67 × 105 m/s 2 ) = 6670 N ≈ 6700 N
Notice how the signs worked together to give a positive value of the magnitude of the force.
(b) The time to stop is found from v1 = v0 + aΔt as follows:
Δt = −
v0
= 6.00 × 10−4 s = 600 μs
a
(c)
Using the above kinematic equation, we can find the velocity as a function of t. For example at t = 60 μs,
vx = 400 m/s + (− 6.667 × 105 m/s 2 )(60 × 10−6 s) = 360 m/s
6.40. Model: Represent the rocket as a particle that follows Newton’s second law.
Visualize:
Solve: (a) The y-component of Newton’s second law is
ay = a =
( Fnet ) y
m
=
Fthrust − mg 3.0 × 105 N
=
− 9.80 m/s 2 = 5.2 m/s 2
m
20,000 kg
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Dynamics I: Motion Along a Line
6-25
(b) At 5000 m the acceleration has increased because the rocket mass has decreased. Solving the equation of part (a)
for m gives
m5000 m =
Fthrust
3.0 × 105 N
=
= 1.9 × 104 kg
a5000 m + g 6.0 m/s 2 + 9.80 m/s 2
The mass of fuel burned is mfuel = minitial − m5000 m = 1.0 × 103 kg.
6.41. Model: Model the object as a particle. Neglect air resistance.
Visualize: We’ll use v12 = v02 + 2aΔx to find the speed of the object. Since v0 = 0, v1 = 2ax L .
We’ll also use Newton’s second law in both directions in order to find ax .
Solve:
(a)
∑ Fy = n − mgcosθ = 0 ⇒ n = mgcosθ
∑ Fx = mgsinθ − f k = max
mgsinθ − μk n = max
mgsinθ − μk mgcosθ = ma x
Cancel the m.
ax = g (sinθ − μk cosθ )
Now put this back in to the equation for v1.
v1 = 2a x L = 2 [ g (sinθ − μ k cosθ ) ] L = 2g (h − μ k L2 − h 2 )
(b) For h = 12 m, L = 100 m, μ k = 0.07 we have
(
)
v1 = 2(9.8 m/s 2 ) (12 m − 0.07) (100 m) 2 − (12 m) 2 = 9.949 m/s ≈ 9.9 m/s
Assess: Sam’s mass was extra unneeded information because m cancels out of the equation for v1. Any skier,
regardless of their mass, would achieve the same speed at the bottom of the same hill with the same μ k .
6.42. Model: We assume that Sam is a particle moving in a straight horizontal line under the influence of two
forces: the thrust of his jet skis and the resisting force of friction on the skis. We can use one-dimensional kinematics.
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6-26
Chapter 6
Visualize:
Solve: (a) The friction force of the snow can be found from the free-body diagram and Newton’s first law, since
there’s no acceleration in the vertical direction:
n = FG = mg = (75 kg)(9.80 m/s 2 ) = 735 N ⇒ f k = μk n = (0.10)(735 N) = 73.5 N
Then, from Newton’s second law:
( Fnet ) x = Fthrust − f k = ma0 ⇒ a0 =
Fthrust − f k 200 N − 73.5 N
=
= 1.687 m/s 2
m
75 kg
From kinematics:
v1 = v0 + a0t1 = 0 m/s + (1.687 m/s 2 )(10 s) = 16.9 m/s
(b) During the acceleration, Sam travels to
1
1
x1 = x0 + v0t1 + a0t12 = (1.687 m/s 2 )(10 s) 2 = 84 m
2
2
After the skis run out of fuel, Sam’s acceleration can again be found from Newton’s second law:
( Fnet ) x = − f k = −73.5 N ⇒ a1 =
Fnet −73.5 N
=
= −0.98 m/s 2
m
75 kg
Since we don’t know how much time it takes Sam to stop:
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ x2 − x1 =
v22 − v12 0 m 2 /s 2 − (16.9 m/s) 2
=
= 145 m
2a1
2(−0.98 m/s 2 )
The total distance traveled is ( x2 − x1 ) + x1 = 145 m + 84 m = 229 m.
Assess: A top speed of 16.9 m/s (roughly 40 mph) seems quite reasonable for this acceleration, and a coasting
distance of nearly 150 m also seems possible, starting from a high speed, given that we’re neglecting air resistance.
6.43. Model: We assume Sam is a particle moving in a straight line down the slope under the influence of gravity,
the thrust of his jet skis, and the resisting force of friction on the snow.
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Dynamics I: Motion Along a Line
6-27
Visualize:
Solve: From the height of the slope and its angle, we can calculate its length:
h
h
50 m
= sin θ ⇒ x1 − x0 =
=
= 288 m
x1 − x0
sin θ sin10°
Since Sam is not accelerating in the y-direction, we can use Newton’s first law to calculate the normal force:
( Fnet ) y = ∑ Fy = n − FG cosθ = 0 N ⇒ n = FG cosθ = mg cosθ = (75 kg)(9.80 m/s 2 )(cos10°) = 724 N
One-dimensional kinematics gives us Sam’s acceleration:
v12 = v02 + 2a x ( x − x0 ) ⇒ a x =
v12 − v02
(40 m/s) 2 − 0 m 2 /s 2
=
= 2.78 m/s 2
2( x1 − x2 )
2(288 m)
Then, from Newton’s second law and the equation f k = μ k n:
( Fnet ) x = ∑ Fx = FG sin θ + Fthrust − f k = ma x
⇒ μk =
mg sin θ + Fthrust − ma (75 kg)(9.80 m/s 2 )(sin10°) + 200 N − (75 kg)(2.78 m/s 2 )
=
= 0.165
n
724 N
Assess: This coefficient seems a bit high for skis on snow, but not impossible.
6.44. Model: We assume the suitcase is a particle accelerating horizontally under the influence of friction only.
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6-28
Chapter 6
Visualize:
Solve: Because the conveyor belt is already moving, friction drags your suitcase to the right. It will accelerate until it
matches the speed of the belt. We need to know the horizontal acceleration. Since there’s no acceleration in the
vertical direction, we can apply Newton’s first law to find the normal force:
n = FG = mg = (10 kg)(9.80 m/s 2 ) = 98.0 N
The suitcase is accelerating, so we use μ k to find the friction force
f k = μ k mg = (0.3)(98.0 N) = 29.4 N
We can find the horizontal acceleration from Newton’s second law:
f
29.4 N
( Fnet ) x = ∑ Fx = f k = ma ⇒ a = k =
= 2.94 m/s 2
m 10 kg
From one of the kinematic equations:
v 2 − v02 (2.0 m/s) 2 − (0 m/s) 2
v12 = v02 + 2a ( x1 − x0 ) ⇒ x1 − x0 = 1
=
= 0.68 m
2a
2(2.94 m/s 2 )
The suitcase travels 0.68 m before catching up with the belt and riding smoothly.
Assess: If we imagine throwing a suitcase at a speed of 2.0 m/s onto a motionless surface, 0.68 m seems a
reasonable distance for it to slide before stopping.
6.45. Model: The box of shingles is a particle subject to Newton’s laws and kinematics.
Visualize:
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Dynamics I: Motion Along a Line
6-29
Solve: Newton’s laws can be used in the coordinate system in which the direction of motion of the box of shingles
JG
defines the +x-axis. The angle that F G makes with the −y-axis is 25°.
(∑ F ) x = FGsin25° − f k = ma
(∑ F ) y = n − FG cos25° = 0 ⇒ n = FG cos25°
We have used the observation that the shingles do not leap off the roof, so the acceleration in the y-direction is zero.
Combining these equations with f k = μ k n and FG = mg yields
mg sin 25° − μ k mg cos 25° = ma
⇒ a = (sin25° − μ k cos25°) g = −0.743 m/s 2
where the minus sign indicates the acceleration is directed up the incline. The required initial speed to have the box
come to rest after 5.0 m is found from kinematics.
vf2 = vi2 + 2aΔx ⇒ vi2 = −2( −0.743 m/s 2 )(5.0 m) ⇒ vi = 2.7 m/s
Assess: To give the shingles an initial speed of 2.7 m/s requires a strong, determined push, but is not beyond
reasonable.
6.46. Model: We will model the box as a particle, and use the models of kinetic and static friction.
Visualize:
The pushing force is along the +x-axis, but the force of friction acts along the −x-axis. A component of the
gravitational force on the box acts along the −x-axis as well. The box will move up if the pushing force is at least
equal to the sum of the friction force and the component of the gravitational force in the x-direction.
Solve: Let’s determine how much pushing force you would need to keep the box moving up the ramp at steady
speed. Newton’s second law for the box in dynamic equilibrium is
( Fnet ) x = ∑ Fx = nx + ( FG ) x + ( f k ) x + ( Fpush ) x = 0 N − mg sinθ − f k + Fpush = 0 N
( Fnet ) y = ∑ Fy = n y + ( FG ) y + ( f k ) y + ( Fpush ) y = n − mg cosθ + 0 N + 0 N = 0 N
The x-component equation and the model of kinetic friction yield:
Fpush = mg sinθ + f k = mg sinθ + μk n
Let us obtain n from the y-component equation as n = mg cosθ , and substitute it in the above equation to get
Fpush = mg sinθ + μ k mg cosθ = mg (sin θ + μ k cosθ )
= (100 kg)(9.80 m/s 2 )(sin 20° + 0.60 cos 20°) = 888 N
The force is less than your maximum pushing force of 1000 N. That is, once in motion, the box could be kept moving
up the ramp. However, if you stop on the ramp and want to start the box from rest, the model of static friction
applies. The analysis is the same except that the coefficient of static friction is used and we use the maximum value
of the force of static friction. Therefore, we have
Fpush = mg (sinθ + μs cosθ ) = (100 kg)(9.80 m/s 2 )(sin 20° + 0.90 cos 20°) = 1160 N
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6-30
Chapter 6
Since you can push with a force of only 1000 N, you can’t get the box started. The big static friction force and the
weight are too much to overcome.
6.47. Model: We assume that the plane is a particle accelerating in a straight line under the influence of two forces:
the thrust of its engines and the rolling friction of the wheels on the runway. We can use one-dimensional kinematics.
Visualize:
Solve: We can use the definition of acceleration to find a, and then apply Newton’s second law. We obtain:
a=
Δv 82 m/s − 0 m/s
=
= 2.34 m/s 2
Δt
35 s
( Fnet ) = ∑ Fx = Fthrust − f r = ma ⇒ Fthrust = f r + ma
For rubber rolling on concrete, μ r = 0.02 (Table 6.1), and since the runway is horizontal, n = FG = mg . Thus:
Fthrust = μ r FG + ma = μ r mg + ma = m( μr g + a )
= (75,000 kg)[(0.02)(9.8 m/s 2 ) + 2.34 m/s 2 ] = 190,000 N
Assess: It’s hard to evaluate such an enormous thrust, but comparison with the plane’s mass suggests that 190,000 N
is enough to produce the required acceleration.
6.48. Model: We will represent the wood block as a particle, and use the model of kinetic friction and kinematics.
Assume w sin θ > fs , so it does not hang up at the top.
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Dynamics I: Motion Along a Line
6-31
Visualize:
The block ends where it starts, so x2 = x0 = 0 m. We expect v2 to be negative, because the block will be moving in
the −x-direction, so we’ll want to take | v2 | as the final speed. Because of friction, we expect to find | v2 |< v0 .
G
G
Solve: (a) The friction force is opposite to v , so f k points down the slope during the first half of the motion and up
G
G
the slope during the second half. FG and n are the only other forces. Newton’s second law for the upward motion is
( Fnet ) x − FG sin θ − f k − mg sin θ − f k
=
=
m
m
m
(
F
)
n
−
F
cos
θ
n
−
mg
cosθ
net y
G
a y = 0 m/s 2 =
=
=
m
m
m
a x = a0 =
The friction model is f k = μ k n. First solve the y-equation to give n = mg cosθ . Use this in the friction model to
get f k = μ k mg cosθ . Now substitute this result for f k into the x-equation:
− mg sin θ − μ k mg cosθ
= − g (sin θ + μ k cosθ ) = −(9.8 m/s 2 )(sin 30° + 0.20cos30°) = −6.60 m/s 2
m
Kinematics now gives
v 2 − v02 0 m 2 /s 2 − (10 m/s) 2
=
= 7.6 m
v12 = v02 + 2a0 ( x1 − x0 ) ⇒ x1 = 1
2a0
2(−6.60 m/s 2 )
a0 =
The block’s height is then h = x1 sinθ = (7.6 m)sin 30° = 3.8 m.
G
(b) For the return trip, f k points up the slope, so the x-component of the second law is
( Fnet ) x − FG sin θ + f k − mg sin θ + f k
=
=
m
m
m
Note the sign change. The y-equation and the friction model are unchanged, so we have
a x = a1 =
a1 = − g (sin θ − μ k cosθ ) = −3.20 m/s 2
The kinematics for the return trip are
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ v2 = −2a1x1 = 2(−3.20 m/s 2 )(−7.6 m) = −7.0 m/s
Notice that we used the negative square root because v2 is a velocity with the vector pointing in the –x-direction.
The final speed is | v2 |= 7.0 m/s.
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6-32
Chapter 6
6.49. Model: We will model the sled and friend as a particle, and use the model of kinetic friction because the sled
is in motion.
Visualize:
The net force on the sled is zero (note the constant speed of the sled). That means the component of the pulling force
along the +x-direction is equal to the magnitude of the kinetic force of friction in the −x-direction. Also note that
( Fnet ) y = 0 N, since the sled is not moving along the y-axis.
Solve: Newton’s second law is
( Fnet ) x = ∑ Fx = nx + ( FG ) x + ( f k ) x + ( Fpull ) x = 0 N + 0 N − f k + Fpull cosθ = 0 N
( Fnet ) y = ∑ Fy = n y + ( FG ) y + ( f k ) y + ( Fpull ) y = n − mg + 0 N + Fpull sin θ = 0 N
The x-component equation using the kinetic friction model f k = μ k n reduces to
μk n = Fpull cosθ
The y-component equation gives
n = mg − Fpull sin θ
We see that the normal force is smaller than the gravitational force because Fpull has a component in a direction
opposite to the direction of the gravitational force. In other words, Fpull is partly lifting the sled. From the
x-component equation, μ k can now be obtained as
μk =
Fpull cosθ
mg − Fpull sin θ
=
(75 N)(cos30°)
(60 kg)(9.80 m/s 2 ) − (75 N)(sin 30°)
= 0.12
Assess: A quick glance at the various μk values in Table 6.1 suggests that a value of 0.12 for μ k is reasonable.
6.50. Model: Model the small box as a particle and use the model of static friction. The acceleration of the small
box must be the same as the acceleration of the large box in order for it not to slip.
Visualize: First use Newton’s second law in both directions on the small box. The force that is responsible for the
small box’s acceleration is the static friction force. We use this to determine amax . Then we use Newton’s second
law on the the two-box system.
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Dynamics I: Motion Along a Line
6-33
Solve:
(a)
∑ Fy = n − mg = 0 ⇒ n = mg
∑ Fx = fs = ma x
( fs ) max = μs n = μs mg = mamax
amax = μs g
Now consider the two-box system.
∑ Fx = Tmax = ( M + m)amax
Put these together to arrive at
Tmax = ( M + m) μs g
(b) Insert the known values for M and m, and look up μs for wood on wood in the table.
Tmax = (10 kg + 5 kg)(0.5)(9.8 m/s 2 ) = 73.5 N ≈ 74 N
Assess: Check the reasonableness of our answer by examining the dependence of Tmax on μs : if the small box were
glued to the large box ( μs → ∞) then one could pull on the rope with any tension desired; if the friction between the
two boxes were zero then one could not pull at all without causing the small box to slip. We expect a similar
dependence on g.
6.51. Model: Model the steel cabinet as a particle. It touches the truck’s bed, so only the steel bed can exert contact
forces on the cabinet. As long as the cabinet does not slide, the acceleration a of the cabinet is equal to the
acceleration of the truck.
Visualize: First use Newton’s second law in both directions on the cabinet. The force that is responsible for the
small box’s acceleration is the static friction force. We use this to determine a.
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6-34
Chapter 6
Solve:
(a)
∑ Fy = n − mg = 0 ⇒ n = mg
∑ Fx = 2 fs = ma x
−( fs ) max = − μs n = − μs mg = max
a x = − μs g
Now use the kinematic equation
v12
= v02
+ 2aΔx where Δx = d min and v1 = 0.
d min =
v12 − v02
−v02
v02
=
=
2a x
2( − μs g ) 2( μs g )
(b) Insert the known value for v0 and look up μs for steel on steel in the table.
d min =
(15 m/s) 2
2(0.80)(9.8 m/s 2 )
= 14.35 m ≈ 14 m
Assess: Check the reasonableness of our answer by examining the dependence of Tmax on μs : if the cabinet were
glued to the truck ( μs → ∞) then one could stop in an arbitrarily small distance without the cabinet slipping; if the
friction between the cabinet and truck were zero then d min = ∞ and there is no minimum stopping distance without
causing the cabinet to slip.
6.52. Model: Model the block as a particle.
Visualize: First use Newton’s second law in both directions on the block. The force that is responsible for the small
box’s acceleration is the kinetic friction force. We use this to determine a x .
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Dynamics I: Motion Along a Line
6-35
Solve:
∑ Fy = n − mg = 0 ⇒ n = mg
∑ Fx = − f k = ma x
−( f k ) = − μ k n = − μk mg = ma x
ax = − μk g
Now use the kinematic equation
v12
= v02
+ 2aΔx where v1 = 0 and Δx is the sliding distance.
v02 = −2a x Δx = −2(− μk g )Δx = 2 μ k g Δx
This says that a graph of v02 vs. Δx would be a straight line with a slope of 2 μ k g .
We see that the linear fit is very good and that the slope is 4.82 m/s 2 .
2 μ k g = 4.82 m/s 2 ⇒ μ k =
slope 4.82 m/s 2
=
= 0.246 ≈ 0.25
2g
2(9.8 m/s 2 )
Assess: Our answer for wood on smooth metal is higher than we expected because the table gives μ k for wood on
wood as 0.20. We expected the intercept of our graph to be small; in fact, we included (0,0) in the data table. The
mass of the block canceled out and so was unnecessary information.
6.53. Model: The antiques (mass = m) in the back of your pickup (mass = M ) will be treated as a particle. The
antiques touch the truck’s steel bed, so only the steel bed can exert contact forces on the antiques. The pickupantiques system will also be treated as a particle, and the contact force on this particle will be due to the road.
Visualize:
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6-36
Chapter 6
Solve: (a) We will find the smallest coefficient of friction that allows the truck to stop in 55 m, then compare that to
the known coefficients for rubber on concrete. For the pickup-antiques system, with mass m + M , Newton’s second
law is
( Fnet ) x = ∑ Fx = N x + (( FG )PA ) x + ( f ) x = 0 N + 0 N − f = (m + M )a x = (m + M )a
( Fnet ) y = ∑ Fy = N y + (( FG ) PA ) y + ( f ) y = N − (m + M ) g + 0 N = 0 N
The model of static friction is f = μ N , where μ is the coefficient of friction between the tires and the road. These
equations can be combined to yield a = − μ g . Since constant-acceleration kinematics gives v12 = v02 + 2a ( x1 + x0 ), we
find
a=
v12 − v02
v02
(25 m/s) 2
⇒ μ min =
=
= 0.58
2( x1 − x0 )
2 g ( x1 − x0 ) (2)(9.8 m/s 2 )(55 m)
The truck cannot stop if μ is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and 0.80
respectively (see Table 6.1), are larger. So the truck can stop.
(b) The analysis of the pickup-antiques system applies to the antiques, and it gives the same value of 0.58 for μ min .
This value is smaller than the given coefficient of static friction ( μs = 0.60) between the antiques and the truck bed.
Therefore, the antiques will not slide as the truck is stopped over a distance of 55 m.
Assess: The analysis of parts (a) and (b) are the same because mass cancels out of the calculations. According to the
California Highway Patrol Web site, the stopping distance (with zero reaction time) for a passenger vehicle traveling
at 25 m/s or 82 ft/s is approximately 43 m. This is smaller than the 55 m over which you are asked to stop the truck.
6.54. Model: The box will be treated as a particle. Because the box slides down a vertical wood wall, we will also
use the model of kinetic friction.
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Dynamics I: Motion Along a Line
6-37
Visualize:
Solve: The normal force due to the wall, which is perpendicular to the wall, is here to the right. The box slides down
JG
G
G G
the wall at constant speed, so a = 0 and the box is in dynamic equilibrium. Thus, F net = 0. Newton’s second law for
this equilibrium situation is
( Fnet ) x = 0 N = n − Fpush cos 45°
( Fnet ) y = 0 N = f k + Fpush sin 45° − FG = f k + Fpush sin 45° − mg
The friction force is f k = μ k n. Using the x-equation to get an expression for n, we see that f k = μk Fpush cos 45°.
Substituting this into the y-equation and using Table 6.1 to find μ k = 0.20 gives,
μk Fpush cos 45° + Fpush sin 45° − mg = 0 N
⇒ Fpush =
mg
(2.0 kg)(9.80 m/s 2 )
=
= 23 N
μk cos 45° + sin 45° 0.20cos 45° + sin 45°
6.55. Model: Use the particle model for the block and the model of static friction.
Visualize:
Solve: The block is initially at rest, so initially the friction force is static friction. If the 12 N push is too strong, the
box will begin to move up the wall. If it is too weak, the box will begin to slide down the wall. And if the pushing
force is within the proper range, the box will remain stuck in place. First, let’s evaluate the sum of all the forces
except friction:
∑ Fx = n − Fpush cos30° = 0 N ⇒ n = Fpush cos30°
∑ Fy = Fpush sin 30° − FG = Fpush sin 30° − mg = (12 N)sin 30° − (1 kg)(9.8 m/s 2 ) = −3.8 N
In the first equation we utilized the fact that any motion is parallel to the wall, so a x = 0 m/s 2 . These three forces add
up to −3.8 ˆj N. This means the static friction force will be able to prevent the box from moving if fs = +3.8 ˆj N.
Using the x-equation and the friction model we get
( fs ) max = μs n = μs Fpush cos30° = 5.2 N
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6-38
Chapter 6
G
where we used μs = 0.5 for wood on wood. The static friction force fs needed to keep the box from moving is less
than ( fs ) max . Thus the box will stay at rest.
6.56. Visualize: The book is in static equilibrium so the net force is zero. The maximum static frictional force the
person can exert will determine the heaviest book he can hold.
Solve: Consider the free-body diagram below. The force of the fingers on the book is the reaction force to the
normal force of the book on the fingers, so is exactly equal and opposite the normal force on the fingers.
The maximal static friction force will be equal to fs max = μs n = (0.80)(6.0 N) = 4.8 N. The frictional force is exerted
on both sides of the book. Considering the forces in the y-direction, we have that the weight supported by the
maximal frictional force is
w = fs max + fs max = 2 fs max = 9.6 N
Assess: Note that the force on both sides of the book are exactly equal also because the book is in equilibrium.
6.57. Model: We will model the skier along with the wooden skis as a particle of mass m. The snow exerts a contact
force and the wind exerts a drag force on the skier. We will therefore use the models of kinetic friction and drag.
Assume the skier is a cylinder end-forward so that C = 0.8.
Visualize:
We choose a coordinate system such that the skier’s motion is along the +x-direction. While the forces of kinetic
G
G
friction f k and drag D act along the −x-direction opposing the motion of the skier, the gravitational force on the
skier has a component in the +x-direction. At the terminal speed, the net force on the skier is zero as the forces along
the +x-direction cancel out the forces along the −x-direction.
Solve: Newton’s second law and the models of kinetic friction and drag are
1
( Fnet ) x = ∑ Fx = +( FG ) x + ( f k ) x + ( D) x = mg sin θ − f k − C ρ Av 2 = max = 0 N
2
( Fnet ) y = ∑ Fy = n y + ( FG ) y = n − mg cosθ = 0 N
f k = μk n
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Dynamics I: Motion Along a Line
6-39
These three equations can be combined together as follows:
(1/2)C ρ Av 2 = mg sin θ − f k = mg sin θ − μ k n = mg sin θ − μ k mg cosθ
⇒ vterm
⎛
sin θ − μk cosθ
= ⎜ mg
1 Cρ A
⎜
⎝
2
1/2
⎞
⎟
⎟
⎠
Using μk = 0.06 and A = 1.8 m × 0.40 m = 0.72 m 2 , we find
1/2
⎡
⎛
⎞⎤
sin 40° − 0.06cos 40°
⎟ ⎥ = 37 m/s
vterm = ⎢(80 kg)(9.8 m/s 2 ) ⎜
⎜ 1 (0.8)(1.2 kg/m3 )(0.72 m 2 ) ⎟ ⎥
⎢
⎝2
⎠⎦
⎣
Assess: A terminal speed of 37 m/s corresponds to a speed of ≈ 82 mph. This speed is reasonable but high due to
the steep slope angle of 40° and a small coefficient of friction.
6.58. Model: The ball is a particle experiencing a drag force and traveling at twice its terminal velocity.
Visualize:
Solve: (a) An object falling at greater than its terminal velocity will slow down to its terminal velocity. Thus the
drag force is greater than the force of gravity, as shown in the free-body diagrams. When the ball is shot straight up,
1
1
⎛
⎞
(∑ F ) y = ma = −( FG + D) = − ⎜ mg + C ρ Av 2 ⎟ = − mg − C ρ A(2vterm ) 2 =
2
2
⎝
⎠
⎛ 2mg ⎞
1
−mg − C ρ A ⎜ 4
⎟ = − mg − (4mg ) = −5mg
2
⎝ Cρ A ⎠
Thus a = −5 g , where the minus sign indicates the downward direction. We have used Equations 6.16 for the drag
force and 6.19 for the terminal velocity.
(b) When the ball is shot straight down,
⎛ 2mg ⎞
1
1
(∑ F ) y = ma = D − FG = C ρ A(2vterm ) 2 − mg = C ρ A ⎜ 4
⎟ − mg = 3mg
2
2
⎝ Cρ A ⎠
Thus a = 3 g , this time directed upward.
(c)
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6-40
Chapter 6
The ball will slow down to its terminal velocity, slowing quickly at first, and more slowly as it gets closer to the
terminal velocity because the drag force decreases as the ball slows.
6.59. Model: We will model the sculpture as a particle of mass m. The ropes that support the sculpture will be
assumed to have zero mass.
Visualize:
Solve: Newton’s first law in component form is
( Fnet ) x = ∑ Fx = T1x + T2 x + FGx = −T1 sin 30° + T2 sin 60° + 0 N = 0 N
( Fnet ) y = ∑ Fy = T1 y + T2 y + FGy = −T1 cos30° + T2 cos60° − FG = 0 N
Using the x-component equation to obtain an expression for T1 and substituting into the y-component equation
yields:
T2 =
FG
500 lbs
=
= 250 lbs
(sin 60°)(cos30°)
2
+ cos 60°
sin 30°
Substituting this value of T2 back into the x-component equation,
T1 = T2
sin 60°
sin 60°
= 250 lbs
= 433 lbs
sin 30°
sin 30°
We will now find a rope size for a tension force of 433 lbs, that is, the diameter of a rope with a safety rating of
433 lbs. Since the cross-sectional area of the rope is 14 π d 2 , we have
1/2
⎡
⎤
4(433 lbs)
d =⎢
2 ⎥
⎣ π (4000 lbs/inch ) ⎦
= 0.371 inch
Any diameter larger than 0.371 inch will ensure a safety rating of at least 433 lbs. The rope size corresponding to a
diameter of 3/8 of an inch will therefore be appropriate.
Assess: If only a single rope were used to hang the sculpture, the rope would have to support a gravitational force of
500 lbs. The diameter of the rope for a safety rating of 500 lbs is 0.399 inches, and the rope size jumps from a
diameter of 3/8 to 4/8 of an inch. Also note that the gravitational force on the sculpture is distributed in the two ropes.
It is the sum of the y-components of the tensions in the ropes that will equal the gravitational force on the sculpture.
6.60. Model: We will model the skier as a particle, and use the model of kinetic friction.
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Dynamics I: Motion Along a Line
6-41
Visualize:
G
G
G G
Solve: Your best strategy, if it’s possible, is to travel at a very slow constant speed (a = 0 so Fnet = 0). Alternatively,
you want the smallest positive a x . A negative a x would cause you to slow and stop. Let’s find the value of μ k that
G
G
gives Fnet = 0.
Newton’s second law for the skier and the model of kinetic friction are
(Fnet ) x = ∑ Fx = nx + (FG ) x + ( f k ) x + (D) x = 0 + mgsinθ − f k − Dcosθ = 0 N
(Fnet ) y = ∑ Fy = n y + (FG ) y + ( f k ) y + (D) y = n − mgcosθ + 0 N − Dsinθ = 0 N
f k = μk n
The x- and y-component equations are
f k = + mgsinθ − Dcosθ n = mgcosθ + Dsinθ
From the model of kinetic friction,
μk =
f k mgsinθ − Dcosθ 82 kg(9.8 m/s 2 )sin15° − (50 N)cos15°
=
=
= 0.20
n mgcosθ + Dsinθ 82 kg(9.8 m/s 2 )cos15° + (50 N)sin15°
Yellow wax with μ k = 0.20 is perfect.
6.61. Model: The astronaut is a particle oscillating on a spring.
Solve: (a) The position versus time function x(t) can be used to find the velocity versus time function v(t ) =
dx
. We
dt
have
d
{(0.30 m)sin((π rad/s)t )} = (0.30π m/s)cos((π rad/s)t )
dt
dv
This can then be used to find the acceleration a (t ) = .
dt
v(t ) =
dv
= −(0.30π 2 m/s 2 )sin((π rad/s)t )
dt
Newton’s second law yields a general expression for the force on the astronaut.
a (t ) =
Fnet (t ) = ma(t ) = −(75 kg)(0.30π 2 m/s 2 )sin((π rad/s)t )
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6-42
Chapter 6
Evaluating this at t = 1.0 s gives Fnet (1.0 s) = 0 N, since sin(π ) = 0.
(b) Evaluating at t = 1.5 s,
⎛ 3π
Fnet = −22.5π 2 N sin ⎜
⎝ 2
⎞
2
⎟ = 2.2 × 10 N
⎠
Assess: The force of 220 N is only one-third of the astronaut’s weight on earth, so is easy for her to withstand.
6.62. Solve: Using a x =
dvx
, we express Newton’s second law as a differential equation, which we then use to
dt
solve for vx .
Fx = m
dvx
F
ct
⇒ dvx = x dt = dt
dt
m
m
Integrating from the initial to final conditions for each variable of integration,
vx
c
t
ct 2
Ñ dvx = m Ñt dt ⇒ vx − v0 x = 2m
v0 x
Thus
0
vx = v0 x +
ct 2
2m
6.63. Model: Model the object as a particle. The acceleration is not constant so we can’t use the kinematic
equations. All the motion is in the x-direction.
Visualize: Divide F by m to get a and then integrate twice. The constants of integration are both zero because of the
initial conditions.
Solve:
F
F ⎛
t⎞
a x (t ) = x = 0 ⎜ 1 − ⎟
m m⎝ T⎠
(a)
F ⎛
t⎞
F ⎛
t2 ⎞
F ⎛
t2 ⎞
vx (t ) = ∫ a x dt = 0 ∫ ⎜1 − ⎟ dt = 0 ⎜ t −
+ v0 = 0 ⎜ t −
⎟
⎟
m ⎝ T⎠
m ⎜⎝ 2T ⎟⎠
m ⎜⎝ 2T ⎟⎠
F ⎛
T 2 ⎞ F0 T
vx (T ) = 0 ⎜ T −
⎟=
m ⎜⎝
2T ⎟⎠ m 2
(b)
F ⎛
t2 ⎞
F ⎛ t 2 t3 ⎞
F ⎛ t 2 t3 ⎞
x (t ) = ∫ vx dt = 0 ∫ ⎜ t −
dt = 0 ⎜ −
+ x0 = 0 ⎜ −
⎟
⎟
⎟
m ⎜⎝ 2T ⎟⎠
m ⎜⎝ 2 6T ⎟⎠
m ⎜⎝ 2 6T ⎟⎠
F ⎛ T 2 T 3 ⎞ F0 T 2
x (T ) = 0 ⎜
−
⎟=
m ⎜⎝ 2 6T ⎟⎠ m 3
Assess: It seems reasonable that the velocity after time T would increase with T and that the position at time T would
increase with T 2 .
6.64. Model: Model the object as a particle. The acceleration is not constant so we can’t use the kinematic
equations. All the motion is in the x-direction.
Visualize: Divide F by m to get a and then integrate twice. The constants of integration are both zero because of the
initial conditions.
Solve:
F
F
a x (t ) = x = 0 (e −t/T )
m m
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Dynamics I: Motion Along a Line
6-43
(a)
t
⎛ −t ⎞
F0 ⎛⎜ − T ⎞⎟
F0
⎜e T ⎟ + C
=
−
(
)
e
dt
T
⎟
⎜
⎟
m ∫⎜
m
⎝
⎠
⎝
⎠
F0
The constant of integration is not zero. v(0) = 0 ⇒ C = (T )
m
t
t
⎛ − ⎞ F
⎛
− ⎞
F
F
vx (t ) = 0 ( −T ) ⎜ e T ⎟ + 0 (T ) = 0 (T ) ⎜1 − e T ⎟
⎜
⎟ m
⎜
⎟
m
m
⎝
⎠
⎝
⎠
vx (t ) = ∫ a x dt =
F0
T.
m
Assess: It seems reasonable that the velocity after time T would increase with T and that the position at time T would
increase with T 2 .
(b) After a very long time the decaying exponential term is close to zero so vx (t ) →
6.65. Model: Use the linear model of drag. Assume the microorganisms are swimming in water at 20°C.
Visualize: The viscosity of water is η = 1.0 × 10−3 N ⋅ s/m 2 at 20°C.
Solve:
(a)
G G
G
∑ F = Fprop − D = 0 ⇒ Fprop = 6πη Rv
For a paramecium
Fprop = 6π (1.0 × 10−3 N ⋅ s/m 2 )(50 × 10−6 m)(0.0010 m/s) = 9.4 × 10−10 N
For an E. coli bacterium
Fprop = 6π (1.0 × 10−3 N ⋅ s/m 2 )(1.0 × 10−6 m)(30 × 10−6 m/s) = 5.7 × 10−13 N
(b)
a=
Fprop
m
=
Fprop
ρV
=
Fprop
4
ρ π R2
3
For a paramecium
a=
9.4 × 10−10 N
= 1.8 m/s 2
3 4
3
−6
(1000 kg/m ) π (50 × 10 m)
3
For an E. coli bacterium
5.7 × 10−13 N
= 135 m/s 2
3 4
3
−6
(1000 kg/m ) π (1.0 × 10 m)
3
Assess: The two accelerations are within a factor of two of each other.
a=
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6-44
Chapter 6
6.66. Model: Use the linear model of drag.
Visualize: The viscosity of air is η = 2.0 × 10−5 N ⋅ s/m 2 at 25°C. The density of dust is ρ = 2700 kg/m3.
Solve:
(a) At terminal speed the net force is zero.
∑ Fy = D − mg = 0 ⇒ 6πη Rvterm = mg
vterm =
(b)
vterm
mg
6πη R
(
)
3
4
2
ρVg ρ 3 π R g ρ 43 R g
=
=
=
=
6πη R
6πη R
6η
(2700 kg/m3 ) 43 (25 × 10−6 m) 2 (9.8 m/s 2 )
6(2.0 × 10−5 N ⋅ s/m 2 )
Δt =
= 0.18375 m/s
Δy
300 m
=
= 1633 s = 27 min
vterm 0.18375 m/s
Assess: 27 min sounds like a long time, but isn’t too surprising for dust 300 m in the air.
6.67. Solve: (a) A 1.0 kg block is pulled across a level surface by a string, starting from rest. The string has a
tension of 20 N, and the block’s coefficient of kinetic friction is 0.50. How long does it take the block to move
1.0 m?
(b) Newton’s second law for the block is
( Fnet ) y n − FG n − mg
(F )
T − f k T − μk n
=
=
=
a x = a = net x =
a y = 0 m/s 2 =
m
m
m
m
m
m
where we have incorporated the friction model into the first equation. The second equation gives n = mg . Sub-
stituting this into the first equation gives
a=
T − μk mg 20 N − 4.9 N
=
= 15.1 m/s 2
m
1.0 kg
Constant acceleration kinematics gives
1
1
2 x1
2(1.0 m)
x1 = x0 + v0Δt + a(Δt ) 2 = a (Δt ) 2 ⇒ Δt =
=
= 0.36 s
a
2
2
15.1 m/s 2
6.68. Solve: (a) A 15,000 N truck starts from rest and moves down a 15° hill with the engine providing a 12,000 N
force in the direction of the motion. Assume the frictional force between the truck and the road is very small. If the
hill is 50 m long, what will be the speed of the truck at the bottom of the hill?
(b) Newton’s second law is
∑ Fy = n y + FGy + f y + E y = ma y = 0
∑ Fx = nx + FGx + f x + Ex = max ⇒ 0 N + FG sin θ + 0 N + 12 ,000 N = ma
⇒a=
mg sinθ + 12,000 N (15,000 N)sin15° + 12,000 N
=
= 10.4 m/s 2
2
m
(15,000 N/9.8 m/s )
where we have calculated the mass of the truck from the gravitational force on it. Using the constant-acceleration
kinematic equation vx2 − v02 = 2ax,
vx2 = 2a x x = 2(10.4 m/s 2 )(50 m) ⇒ vx = 32 m/s
6.69. Solve: (a) A driver traveling at 40 m/s in her 1500 kg auto slams on the brakes and skids to rest. How far does
the auto slide before coming to rest?
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Dynamics I: Motion Along a Line
6-45
(b)
(c) Newton’s second law is
∑ Fy = n y + ( FG ) y = n − mg = ma y = 0 N
∑ Fx = −0.80n = ma x
2
The y-component equation gives n = mg = (1500 kg)(9.8 m/s ). Substituting this into the x-component equation
yields
(1500 kg)a x = −0.80(1500 kg)(9.8 m/s 2 ) ⇒ a x = (−0.80)(9.8 m/s 2 ) = −7.8 m/s 2
Using the constant-acceleration kinematic equation v12 = v02 + 2aΔx, we find
Δx = −
v02
(40 m/s) 2
=−
= 102 m
2a
2(−7.8 m/s 2 )
6.70. Solve: (a) A 20.0 kg wooden crate is being pulled up a 20° wooden incline by a rope that is connected to an
electric motor. The crate’s acceleration is measured to be 2.0 m/s 2 . The coefficient of kinetic friction between the
crate and the incline is 0.20. Find the tension T in the rope.
(b)
(c) Newton’s second law for this problem in the component form is
( Fnet ) x = ∑ Fx = T − 0.20n − (20 kg)(9.80 m/s 2 )sin 20° = (20 kg)(2.0 m/s 2 )
( Fnet ) y = ∑ Fy = n − (20 kg)(9.80 m/s 2 )cos 20° = 0 N
Solving the y-component equation, n = 184.18 N. Substituting this value for n in the x-component equation yields
T = 144 N.
6.71. Solve: (a) You wish to pull a 20 kg wooden crate across a wood floor ( μ k = 0.20) by pulling on a rope
attached to the crate. Your pull is 100 N at an angle of 30° above the horizontal. What will be the acceleration of the
crate?
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6-46
Chapter 6
(b)
(c) Newton’s equations and the model of kinetic friction are
∑ Fx = nx + Px + ( FG ) x + f x = 0 N + (100 N)cos30° + 0 N − f k = (100 N)cos30° − f k = max
∑ Fy = n y + Py + ( FG ) y + f y = n + (100 N)sin 30° − mg − 0 N = ma y = 0 N
f k = μk n
From the y-component equation, n = 150 N. From the x-component equation and using the model of kinetic friction
with μk = 0.20,
(100 N)cos30° − (0.20)(150 N) = (20 kg) a x ⇒ ax = 2.8 m/s 2
6.72. Model: The acceleration of the block is not constant before it gets to L; it increases until L and is then constant
(with increasing v ).
Visualize: Since the coefficient of friction is a function of the roughness of the two surfaces, it is understandable that
it could be a function of x and not t.
Solve:
(a) Use the chain rule.
dv
dv dx
dv
ax = x = x
= vx x
dt
dx dt
dx
(b)
x⎞
⎛
∑ Fx = F0 − f k = F0 − μk mg = F0 − μ0 ⎜1 − ⎟ mg = max
L
⎝
⎠
ax =
F0
⎛x ⎞
+ μ0 g ⎜ − 1 ⎟
m
⎝L ⎠
Now examine the result in part (a).
∫ ax dx =∫ vx dvx
F
⎛x
⎞
∫ m0 + μ0 g ⎜⎝ L − 1⎟⎠ dx = ∫ vx dvx
⎛ x2
⎞ 1
F0
x + μ0 g ⎜
− x ⎟ = vx2 + C
⎜ 2L
⎟ 2
m
⎝
⎠
The constant of integration C is zero because vx = 0 at x = 0.
⎡F
⎛ x2
⎞⎤
v x ( x ) = 2 ⎢ 0 x + μ0 g ⎜
− x ⎟⎥
⎜
⎟⎥
⎝ 2L
⎠⎦
⎣⎢ m
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Dynamics I: Motion Along a Line
6-47
⎡F
⎛ L2
⎞⎤
⎡F
⎛ L ⎞⎤
⎛ 2F
⎞
v x ( L ) = 2 ⎢ 0 L + μ0 g ⎜
− L ⎟ ⎥ = 2 ⎢ 0 L + μ0 g ⎜ − ⎟ ⎥ = L ⎜ 0 − μ0 g ⎟
⎜
⎟
m
2
L
m
2
m
⎝
⎠⎦
⎝
⎠
⎣
⎢⎣
⎝
⎠ ⎥⎦
Assess: Check dependencies; we expect vx ( L ) to increase with L and decrease with increasing m, μ0 , and g.
6.73. Model: We will model the shuttle as a particle and assume the elastic cord to be massless. We will also use
the model of kinetic friction for the motion of the shuttle along the square steel rail.
Visualize:
Solve: The upward tension component Ty = T sin 45° = 14.1 N is larger than the gravitational force on the shuttle.
Consequently, the elastic cord pulls the shuttle up against the rail and the rail’s normal force pushes downward.
Newton’s second law in component form is
( Fnet ) x = ∑ Fx = Tx + ( f k ) x + (n) x + ( FG ) x = T cos 45° − f k + 0 N + 0 N = ma x = max
( Fnet ) y = ∑ Fy = Ty + ( f k ) y + ( n) y + ( FG ) y = T sin 45° + 0 N − n − mg = ma y = 0 N
The model of kinetic friction is f k = μ k n. We use the y-component equation to get an expression for n and hence
f k . Substituting into the x-component equation and using the value of μ k in Table 6.1 gives us
ax =
=
T cos 45° − μ k (T sin 45° − mg )
m
(20 N)cos 45° − (0.60)[+(20 N)sin 45° − (0.800 kg)(9.80 m/s 2 )]
= 13 m/s 2
0.800 kg
Assess: The x-component of the tension force is 14.1 N. On the other hand, the net force on the shuttle in the
x-direction is ma x = (0.800 kg)(13.0 m/s 2 ) = 10.4 N. This value for ma is reasonable since a part of the 14.1 N
tension force is used up to overcome the force of kinetic friction.
6.74. Model: Assume the ball is a particle on a slope, and that the slope increases as the x-displacement increases.
Assume that there is no friction and that the ball is being accelerated to the right so that it remains at rest on the slope.
Visualize: Although the ball is on a slope, it is accelerating to the right. Thus we’ll use a coordinate system with
horizontal and vertical axes.
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6-48
Chapter 6
Solve: Newton’s second law is
∑ Fx = n sin θ = max
∑ Fy = n cosθ − FG = ma y = 0 N
Combining the two equations, we get
ma x =
FG
sin θ = mg tan θ ⇒ a x = g tan θ
cosθ
The curve is described by y = x 2 . Its slope a position x is tanθ, which is also the derivative of the curve. Hence,
dy
= tan θ = 2 x ⇒ a x = (2 x) g
dx
(b) The acceleration at x = 0.20 m is a x = (2)(0.20)(9.8 m/s 2 ) = 3.9 m/s 2 .
6.75. Visualize:
Solve: (a) The horizontal velocity as a function of time is determined by the horizontal net force. Newton’s second
law as the x-direction gives
( Fnet ) x = max = − D cosθ = −bv cosθ = −bvx
JJG
G
Note that D points opposite to v , so the angle θ with the x-axis is the same for both vectors, and the x components
of both vectors have the same cosθ term. As the particle changes direction as it falls, the evolution of the horizontal
motion depends only on the horizontal component of the velocity.
Thus
dv
m x = −bvx
dt
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Dynamics I: Motion Along a Line
vx ( t )
Separating and integrating,
Ñ
v0
6-49
t
dvx
b
= 2 Ñdt
vx
m0
⎛ v (t ) ⎞
b
⇒ ln ⎜ x ⎟ = − t
m
⎝ v0 ⎠
Solving,
vx (t ) = v0e
−
bt
m
= v0e
−
6πη Rt
m
1
(b) The time to reach v(t ) = v0 is found by solving for the time when
2
−
1
v0 = v0e
2
6πη Rt
m
Hence
t=
m ln(2)
6πη R
With η = 1.0 × 10−3 Ns/m 2 , R = 2.0 × 10−2 m, and m = 0.033 kg, we get t = 61 s.
6πη R
v = (1.1 × 10−2 s −1)vx . This is a small fraction of the
m
velocity, so a time of about one minute to slow to half the initial speed is reasonable.
Assess: The magnitude of the acceleration is a x =
6.76. Visualize:
dvx ⎛ dvx ⎞⎛ dx ⎞
dvx
=⎜
.
⎟⎜ ⎟ = vx
dt ⎝ dx ⎠⎝ dt ⎠
dx
(b) The horizontal motion is determined by using Newton’s second law in the horizontal direction. Using the freebody diagram at a later time t,
( Fnet ) x = ma x = − D cosθ = −bv cosθ = −bvx
JJG
G
Note that since D points opposite to v , the angle θ with the x-axis is the same for both vectors, and the
x-components of both vectors have the same cosθ term. Thus
dv
ma x = mvx x = −bvx
dx
Solve: (a) Using the chain rule, a x =
vx ( x )
Separating and integrating,
Ñ
v0
dvx = −
b
m
x (t )
Ñ dx
x0
⇒ vx ( x ) − v0 = −
b
( x (t ) − x0 )
m
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6-50
Chapter 6
Solving with x0 = 0,
b
6πη R
x = v0 −
x
m
m
(c) The marble stops after traveling a distance d when vx (d ) = 0.
vx ( x ) = v0 −
6πη R
d
m
mv0
⇒d =
6πη R
v0 =
Hence
Using v0 = 10 cm/s, R = 0.50 cm, m = 1.0 × 10 −3 kg, and using η = 1.0 × 10−3 Ns/m,
(1.0 × 10−3 kg)(0.10 m/s)
d=
= 1.1 m
6π (1.0 × 10−3 Ns/m 2 )(5.0 × 10−3 m)
Assess: The equation for d indicates that a marble with a faster initial velocity travels a farther distance.
6.77. Model: We will model the object as a particle, and use the model of drag.
Visualize:
Solve: (a) We cannot use the constant-acceleration kinematic equations since the drag force causes the acceleration to
change with time. Instead, we must use a x = dvx /dt and integrate to find vx . Newton’s second law for the object is
1
dv
( Fnet ) x = ∑ Fx = D = − C ρ Avx2 = ma x = m x
2
dt
This can be written
dvx
vx2
=
Cρ A
dt
2m
We can integrate this from the start (v0 x at t = 0) to the end (vx at t ):
vx
Ñv
0x
dvx
vx2
=
Cρ A t
1
1
Cρ A
=
dt ⇒ − +
t
2m Ñ0
vx v0 x
2m
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Dynamics I: Motion Along a Line
6-51
Solving for vx gives
vx =
v0 x
1 + C ρ Av0 xt/2m
(b) Using A = (1.6 m)(1.4 m) = 2.24 m 2 , v0 x = 20 m/s, and m = 1500 kg, we get
20 m/s
vx =
3
2
=
20
1
⎛
⎞ ⎛ 20 m/s ⎞
⇒t =⎜
− 1⎟
⎟⎜
1 + 0.006272t
⎝ 0.006272 ⎠ ⎝ vx
⎠
(0.35)(1.2 kg/m )(2.24 m )t (20 m/s)
2 × 1500
where t is in seconds. We can now obtain the time t for v = 10 m/s:
1
⎛
⎞⎛ 20 m/s ⎞
⎛ 20 ⎞
− 1⎟ = 159.44 ⎜ − 1⎟ = 160 s
t =⎜
⎟⎜
⎝ 0.006272 ⎠⎝ 10 m/s ⎠
⎝ 10 ⎠
1+
When vx = 5 m/s, then t = 480 s.
(c) If the only force acting on the object was kinetic friction with, say, μ k = 0.05, that force would be (0.05)
(1500 kg) (9.8 m/s 2 ) = 735 N. The drag force at an average speed of 10 m/s is D = 14 (2.24)(10) 2 N = 56 N. We
conclude that it is not reasonable to neglect the kinetic friction force.
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NEWTON’S THIRD LAW
7
Conceptual Questions
7.1. If you were to throw the rocks in the opposite direction you wanted to go, you would be pushed by the rocks in
the right direction. Throwing the rocks requires a force to accelerate them (Newton’s second law). So you exert a
force on the rock in one direction and the rock exerts an equal force on you in the opposite direction (Newton’s third
law). This force will cause you to slide along the ice in the opposite direction that you threw the rock. Note that you
will move most efficiently when you use a horizontal force, which means that you throw the rock horizontally.
7.2. The paddle, you, and the canoe can be treated as a single object. You can push backward on the water with
the paddle so that the water pushes forward on the paddle. The figure shows how the backward force of the paddle on
the water and the forward force of the water on the paddle are action/reaction pairs. Since you hold the paddle while
sitting in the canoe, the force of the water on the paddle causes the paddle-person-canoe object to move forward. The
vertical force between the canoe and water we label as a normal force here but will later identify as the buoyant force.
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7-1
7-2
Chapter 7
7.3. The rocket forces the exhaust gases down, and the hot gases push up on the rocket. The two forces are a
Newton’s third law pair. The rocket accelerates upward because the force of the exhaust gases on the rocket is greater
than the force of gravity.
7.4. The player pushes down on the floor, which pushes back up on him. The player accelerates upward because the
force of his push is greater than the force of gravity.
7.5. Newton’s third law tells us that the force of the mosquito on the car has the same magnitude as the force of the
car on the mosquito.
7.6. The mosquito has a much smaller mass than the car, so the magnitude of the interaction force between the car
and mosquito, although equal on each, causes the mosquito to have a much larger acceleration. In fact, the
acceleration is usually fatal to the mosquito.
7.7. Newton’s third law tells us that the magnitude of the forces are equal. The acceleration of the truck and car are
determined by the net force on each.
7.8. The force of the wagon on the girl acts on the girl, whereas the force of the girl on the wagon acts on the wagon.
The wagon’s motion is determined by the net force acting on it, so if the girl pulls hard enough to overcome any
other opposing forces acting on the wagon, the wagon will move forward. So try saying, “But, my dear, the net force
on the wagon determines if it will move forward. The forces you mention act on different objects, and so cannot
cancel.”
7.9. The net force on each team determines that team’s motion. The net horizontal force on each team is the
difference between the rope’s pull and friction with the ground. So the team that wins the tug-of-war is not the team
that pulls harder, but the team that is best able to keep from sliding along the ground.
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Newton’s Third Law
7-3
7.10. This technique will not work because the magnet is part of the cart, not external to it. The forces between the
magnet and cart have the same magnitude but act in the opposite directions. Therefore, although the two objects may
accelerate toward each other, the cart-magnet system as a whole will not move. (Actually, it would be more precise to
say that the center of mass of the cart-magnet system does not move.)
7.11. The scale reads 5 kg. The left-hand mass performs a function no different than the ceiling would if the rope
were attached to the ceiling (i.e., both pull upward with the force required to suspend 5 kg). The force of gravity
acting on the right-hand mass provides the downward force on the spring scale that the spring scale converts to mass
in its display.
7.12. The scale reads 5 kg. The left-hand mass performs a function no different than the wall would if the rope were
attached to the wall (i.e., both pull leftward with the force required to suspend 5 kg). The right-hand mass provides
the rightward force on the spring scale that the spring scale converts to mass and displays.
7.13.
The figure shows the horizontal forces on blocks B and A using the massless-string approximation in the absence of
friction. The hand must accelerate both blocks A and B, so more force is required to accelerate the greater mass. Thus
the force of the string on B is smaller than the force of the hand on A.
7.14.
The pulley will not rotate. As shown in the free-body diagrams above, the force of gravity pulls down equally on
both blocks so the tension forces, which act as if they were a Newton third law pair, pull up equally on each with the
same magnitude force as the force of gravity. The net force on each block is therefore zero, so they do not move and
the pulley does not rotate.
7.15. Block A’s acceleration is greater in case b. In case a, the hanging 10 N must accelerate both the mass of A and its
own mass, leading to a smaller acceleration than case b, where the entire 10 N force accelerates the mass of block A.
Case a
10 N = ( M A + M10 N ) a
10 N
a=
( M A + M10 N )
Case b
10 N = M A a
10 N
a=
MA
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7-4
Chapter 7
Exercises and Problems
Section 7.2 Analyzing Interacting Objects
7.1. Visualize:
Solve: (a) The weight lifter is holding the barbell in dynamic equilibrium as he stands up, so the net force on the
barbell and on the weight lifter must be zero. The barbells have an upward contact force from the weight lifter and
the gravitational force downward. The weight lifter has a downward contact force from the barbells and an upward
one from the surface. Gravity also acts on the weight lifter.
(b) The system is the weight lifter and barbell, as indicated in the figure.
(c)
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Newton’s Third Law
7-5
7.2. Visualize:
Solve: (a) Both the bowling ball and the soccer ball have a normal force from the surface and gravitational force on
them. The interaction forces between the two are equal and opposite.
(b) The system consists of the soccer ball and bowling ball, as indicated in the figure.
(c)
Assess: Even though the soccer ball bounces back more than the bowling ball, the forces that each exerts on
the other are part of an action/reaction pair, and therefore have equal magnitudes. Each ball’s acceleration due to the
forces on it is determined by Newton’s second law, a = Fnet /m, which depends on the mass. Since the masses of the
balls are different, their accelerations are different.
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7-6
Chapter 7
7.3. Visualize:
Solve: (a) Both the mountain climber and bag of supplies have a normal force from the surface on them, as well as a
gravitational force vertically downward. The rope has gravity acting on it, along with pulls on each end from the
mountain climber and supply bag. The mountain climber experiences static friction with the surface, whereas the bag
experiences kinetic friction with the surface.
(b) The system consists of the mountain climber, rope, and bag of supplies, as indicated in the figure.
(c)
Assess: Since the motion is along the surface, it is convenient to choose the x-coordinate axis along the surface. The
free-body diagram of the rope shows pulls that are slightly off the x-axis since the rope is not massless.
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Newton’s Third Law
7-7
7.4. Visualize:
Solve: (a) The car and rabbit both experience a normal force and friction from the floor and a gravitational force
from the Earth. The push that each exerts on the other is a Newton’s third law force pair.
(b) The system consists of the car and stuffed rabbit, as indicated in the figure.
(c)
7.5. Visualize: Please refer to Figure EX7.5.
Solve: (a) Gravity acts on both blocks. Block A is in contact with the floor and experiences a normal force and friction.
The string tension is the same on both blocks since the rope and pulley are massless and the pulley is frictionless. There are
two third law pair of forces at the surface where the two blocks touch. Block B pushes against Block A with a normal force,
while Block A pushes back against Block B. There is also friction between the two blocks at the surface.
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7-8
Chapter 7
(b) A string that will not stretch constrains the two blocks to accelerate at the same rate but in opposite directions.
Block A accelerates down the incline with an acceleration equal in magnitude to the acceleration of Block B up the
incline. The system consists of the two blocks, as indicated in the figure above.
(c)
Assess: The inclined coordinate systems allows the acceleration a to be purely along the x-axis. This is convenient
because it simplifyies the mathematical expression of Newton’s second law.
7.6. Visualize: Please refer to Figure EX7.6.
Solve: (a) For each block, there is a gravitational force due to the Earth, a normal force and kinetic friction due to
the surface, and a tension force due to the rope.
(b) The tension in the massless ropes over the frictionless pulley is the same on both blocks. Block A accelerates
down the incline with the same magnitude acceleration that Block B has up the incline. The system consists of the
two blocks, as indicated in the figure.
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Newton’s Third Law
7-9
(c)
Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is convenient
since then one component of a is zero, simplifying the mathematical expression of Newton’s second law.
Section 7.3 Newton’s Third Law
7.7. Model: We will model the astronaut and the chair as particles. The astronaut and the chair will be denoted by
A and C, respectively, and they are separate systems. The launch pad is a part of the environment.
Visualize:
Solve: (a) Newton’s second law for the astronaut is
∑( Fon A ) y = nC on A − ( FG ) A = mA aA = 0 N ⇒ nC on A = ( FG ) A = mA g
By Newton’s third law, the astronaut’s force on the chair is
nA on C = nC on A = mA g = (80 kg)(9.8 m/s 2 ) = 7.8 × 102 N
(b) Newton’s second law for the astronaut is:
∑( Fon A ) y = nC on A − ( FG ) A = mA aA
⇒ nC on A = ( FG ) A + mA aA = mA ( g + aA )
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7-10
Chapter 7
By Newton’s third law, the astronaut’s force on the chair is
nA on C = nC on A = mA ( g + aA ) = (80 kg)(9.8 m/s 2 + 10 m/s 2 ) = 1.6 × 103 N
Assess: This is a reasonable value because the astronaut’s acceleration is greater than g.
7.8. Visualize: Please refer to Figure EX7.8.
Solve: Since the ropes are massless we can treat the tension force they transmit as a Newton’s third law force pair on
the blocks. The connection shown in Figure EX7.8 has the same effect as a frictionless pulley on these massless
ropes. The blocks are in equilibrium as the mass of A is increased until block B slides, which occurs when the static
friction on B is at its maximum value. Applying Newton’s first law to the vertical forces on block B gives
nB = ( FG ) B = mB g . The static friction force on B is thus
( fs ) B = μs nB = μs mB g .
Applying Newton’s first law to the horizontal forces on B gives ( fs ) B = TA on B , and the same analysis of the vertical
forces on A gives TB on A = ( FG ) A = mA g . Since TA on B = TB on A , we have ( fs ) B = mA g , so
μs mB g = mA g ⇒ mA = μs mB = (0.60)(20 kg) = 12 kg
7.9. Model: Model the car and the truck as particles denoted by the symbols C and T, respectively. Denote the
surface of the ground by the symbol S.
Visualize:
Solve: (a) The x-component of Newton’s second law for the car gives
∑( Fon C ) x = FS on C − FT on C = mC aC
The x-component of Newton’s second law for the truck gives
∑( Fon T ) x = FC on T = mT aT
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Newton’s Third Law
7-11
Using aC = aT ≡ a and FT on C = FC on T , we get
⎛ 1 ⎞
⎛ 1 ⎞
( FC on S − FC on T ) ⎜
⎟ = a and ( FC on T ) ⎜
⎟=a
m
⎝ mT ⎠
⎝ C⎠
Combining these two equations,
⎛ 1 ⎞
⎛ 1
⎛ 1 ⎞
⎛ 1 ⎞
1 ⎞
( FC on S − FC on T ) ⎜
+
⎟ = ( FC on T ) ⎜
⎟ = ( FC on S ) ⎜
⎟
⎟ ⇒ FC on T ⎜
⎝ mT ⎠
⎝ mC ⎠
⎝ mC mT ⎠
⎝ mC ⎠
⎛ mT ⎞
⎛
⎞
2000 kg
FC on T = ( FC on S ) ⎜
⎟ = (4500 N) ⎜
⎟ = 3000 N
⎝ 1000 kg + 2000 kg ⎠
⎝ mC + mT ⎠
(b) Due to Newton’s third law, FT on C = 3000 N.
7.10. Model: The blocks are to be modeled as particles and denoted as 1, 2, and 3. The surface is frictionless and
along with the earth it is a part of the environment. The three blocks are our three systems of interest.
Visualize:
The force applied on block 1 is FA on 1 = 12 N. The acceleration for all the blocks is the same and is denoted by a.
Solve: (a) Newton’s second law for the three blocks along the x-direction is
∑( Fon 1) x = FA on 1 − F2 on 1 = m1a, ∑( Fon 2 ) x = F1 on 2 − F3 on 2 = m2a, ∑( Fon 3 ) x = F2 on 3 = m3a
Summing these three equations and using Newton’s third law ( F2 on 1 = F1 on 2 and F3 on 2 = F2 on 3 ), we get
FA on 1 = (m1 + m2 + m3 )a ⇒ (12 N) = (1 kg + 2 kg + 3 kg) a ⇒ a = 2 m/s 2
Using this value of a, the force equation for block 3 gives
F2 on 3 = m3a = (3 kg)(2 m/s 2 ) = 6 N
(b) Substituting into the force equation on block 1,
12 N − F2 on 1 = (1 kg)(2 m/s 2 ) ⇒
F2 on 1 = 10 N
Assess: Because all three blocks are pushed forward by a force of 12 N, the value of 10 N for the force that the 2 kg
block exerts on the 1 kg block seems reasonable.
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7-12
Chapter 7
7.11. Model: We treat the two objects of interest, the block (B) and steel cable (C), like particles. The motion of
these objects is governed by the constant-acceleration kinematic equations. The horizontal component of the external
force is 100 N.
Visualize:
Solve: Using v12x = v02x + 2a x ( x1 − x0 ), we find
(4.0 m/s) 2 = 0 m 2 /s 2 + 2ax (2.0 m) ⇒ a x = 4.0 m/s 2
From the free-body diagram on the block:
∑( Fon B ) x = ( FC on B ) x = mBax
⇒ ( FC on B ) x = (20 kg)(4.0 m/s 2 ) = 80 N
Also, according to Newton’s third law ( FB on C ) x = ( FC on B ) x = 80 N. Applying Newton’s second law to the cable gives
∑( Fon C ) x = ( Fext ) x − ( FB on C ) x = mCa x
⇒ 100 N − 80 N = mC (4.0 m/s 2 ) ⇒ mC = 5.0 kg
Section 7.4 Ropes and Pulleys
7.12. Model: The man (M) and the block (B) are interacting with each other through a rope. We will assume the
pulley to be frictionless, which implies that the tension in the rope is the same on both sides of the pulley. The system
is the man and the block.
Visualize:
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Newton’s Third Law
7-13
Solve: Clearly the entire system remains in equilibrium since mB > mM . The block would move downward but it is
already on the ground. From the free-body diagrams, we can write Newton’s second law in the vertical direction as
∑( Fon M ) y = TR on M − ( FG ) M = 0 N ⇒ TR on M = ( FG ) M = (60 kg)(9.8 m/s 2 ) = 590 N
Since the tension is the same on both sides, TB on R = TM on R = T = 590 N.
7.13. Model: The two ropes and the two blocks (A and B) will be treated as particles.
Visualize:
Solve: (a) The two blocks and two ropes form a combined system of total mass M = 2.5 kg. This combined system
is accelerating upward at a = 3.0 m/s 2 under the influence of a force F and the gravitational force − Mg ˆj. Newton’s
second law applied to the combined system gives
( Fnet ) y = F − Mg = Ma ⇒ F = M (a + g ) = (2.5 kg)(3.0 m/s 2 + 9.8 m/s 2 ) = 32 N
(b) The ropes are not massless. We must consider both the blocks and the ropes as systems. The force F acts only on
block A because it does not contact the other objects. We can proceed to apply the y-component of Newton’s second
law to each system, starting at the top. Each object accelerates upward at a = 3.0 m/s 2 . For block A,
( Fnet on A ) y = F − mA g − T1 on A = mA a ⇒ T1 on A = F − mA ( a + g ) = 19 N
(c) Applying Newton’s second law to rope 1 gives
( Fnet on 1) y = TA on 1 − m1g − TB on 1 = m1a
G
G
G
where TA on 1 and T1 on A are an action/reaction pair. But, because the rope has mass, the two tension forces TA on 1
G
and TB on 1 are not the same. The tension at the lower end of rope 1, where it connects to B, is
TB on 1 = TA on 1 − m1 (a + g ) = 16 N
(d) We can continue to repeat this procedure, noting from Newton’s third law that
T1 on B = TB on 1 and T2 on B = TB on 2
Newton’s second law applied to block B is
( Fnet on B ) y = T1 on B − mB g − T2 on B = mBa ⇒ T2 on B = T1 on B − mB ( a + g ) = 3.2 N
7.14. Model: Together the carp (C) and the trout (T) make up the system that will be represented through the
particle model. The fishing rod line (R) is assumed to be massless.
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7-14
Chapter 7
Visualize:
Solve: Jimmy’s pull T2 = 60 N is larger than the total weight of the fish, so they accelerate upward. They are tied
together, so each fish has the same acceleration a. Applying Newton’s second law along the y-direction for the carp
and the trout gives
∑( Fon C ) y = T2 − T1 − ( FG )C = mCa ⇒ ∑( Fon T ) y = T1 − ( FG )T = mT a
Adding these two equations gives
T − ( FG )C − ( FG )T 60 N − (1.5 kg)(9.8 m/s 2 ) − (3.0 kg)(9.8 m/s 2 )
=
= 3.533 m/s 2
a= 2
( mC + mT )
1.5 kg + 3.0 kg
Substituting this value of acceleration back into the force equation for the trout, we find that
T1 = mT (a + g ) = (3.0 kg)(3.533 m/s 2 + 9.8 m/s 2 ) = 40 N
( FG )T = mT g = (3.0 kg)(9.8 m/s 2 ) = 29 N ⇒ ( FG )C = mC g = (1.5 kg)(9.8 m/s 2 ) = 15 N
Thus, T2 > T1 > ( FG )T > ( FG )C .
7.15. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We must therefore
consider both the block of ice and the rope as objects in the system.
Visualize:
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Newton’s Third Law
7-15
G
Solve: (a) The force Fext acts only on the rope. Since the rope and the ice block move together, they have the same
acceleration. Also because the rope has mass, Fext on the front end of the rope is not the same as FI on R that acts on
the rear end of the rope. Applying Newton’s second law along the x-axis to the ice block and the rope gives
∑( Fon I ) x = ( FR on I ) x = mI a = (10 kg)(2.0 m/s 2 ) = 20 N
(b) Applying Newton’s second law to the rope gives
∑( Fon R ) x = ( Fext ) x − ( FI on R ) x = mR a ⇒ ( Fext ) x = ( FR on I ) x + mR a = 20 N + (0.500 kg)(2.0 m/s 2 ) = 21 N
7.16. Model: The hanging block and the rail car are objects in the systems.
Visualize:
Solve: The mass of the rope is very small in comparison to the 2000-kg block, so we assume a massless rope. In this case,
G
G
G
the forces T1 and T1′ act as if they are an action/reaction pair. The hanging block is in static equilibrium, with Fnet = 0 N,
so T1′ = mblock g = (2000 kg)(9.8 m/s 2 ) = 19,600 N. The rail car with the pulley is also in static equilibrium, so
T2 + T3 − T1 = 0 N. Notice how the tension force in the cable pulls both the top and bottom of the pulley to the right.
Now, T1 = T1′ = 19,600 N by Newton’s third law. Also, the cable tension is T2 = T3 = T . Thus, T = 12 T1′ = 9800 N.
7.17. Visualize:
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7-16
Chapter 7
Solve: The rope is treated as two 1.0-kg interacting objects. At the midpoint of the rope, the rope has a tension
TB on T = TT on B ≡ T . Apply Newton’s first law to the bottom half of the rope to find T.
( Fnet ) y = 0 = T − ( FG ) B
⇒ T = mB g = (1.0 kg)(9.8 m/s 2 ) = 9.8 N
Assess: 9.8 N is half the gravitational force on the whole rope. This is reasonable since the top half is holding up the
bottom half of the rope against gravity.
7.18. Model: The cat and dog are modeled as two blocks in the pictorial representation below. The rope is assumed
to be massless. The two points (knots) where the blocks are attached to the rope and the two hanging blocks form a
system. These four objects are treated at particles, form the system, and are in static equilibrium.
Visualize:
Solve: (a) We consider both the two hanging blocks and the two knots. The blocks are in static equilibrium with
G
G
Fnet = 0 N. Note that there are three action/reaction pairs. For Block 1 and Block 2, Fnet = 0 N and we have
T4′ = ( FG )1 = m1g , T5′ = ( FG ) 2 = m2 g
By Newton’s third law:
T4 = T4′ = m1g , T5 = T5′ = m2 g
The knots are also in equilibrium. Newton’s law applied to the left knot is
( Fnet ) x = T2 − T1 cosθ1 = 0 N ⇒ ( Fnet ) y = T1 sin θ1 − T4 = T1 sin θ1 − m1g = 0 N
The y-equation gives T1 = m1g/ sin θ1. Substitute this into the x-equation to find
T2 =
m1g cosθ1 m1g
=
sin θ1
tan θ1
Newton’s law applied to the right knot is
( Fnet ) x = T3 cosθ3 − T2′ = 0 N ⇒ ( Fnet ) y = T3 sin θ3 − T5 = T3 sin θ3 − m2 g = 0 N
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Newton’s Third Law
7-17
These can be combined just like the equations for the left knot to give
m g cosθ3 m2 g
=
T2′ = 2
sin θ3
tan θ3
G
G
But the forces T2 and T2′ are an action/reaction pair, so T2 = T2′ . Therefore,
m1g
m g
= 2
tan θ1 tan θ3
⇒ tan θ3 =
m2
tan θ1 ⇒ θ3 = tan −1[2 tan(20°)] = 36°
m1
We can now use the y-equation for the right knot to find T3 = m2 g/ sin θ3 = 67 N.
7.19. (a) Visualize: The upper magnet is labeled U and the lower magnet L. Each magnet exerts a long-range
magnetic force on the other. Each magnet and the table exert a contact force (normal force) on each other. In
addition, the table experiences a normal force due to the surface.
(b) Solve: Each object is in static equilibrium with Fnet = 0. Start with the lower magnet. Because
FU on L = 3( FG ) L = 6.0 N, equilibrium requires nT on L = 4.0 N. For the upper magnet, FL on U = FU on L = 6.0 N
because these are an action/reaction pair. Equilibrium for the upper magnet requires nT on U = 8.0 N. For the table,
the action/reaction pairs are nL on T = nT on L = 4.0 N and nU on T = nT on U = 8.0 N . The table’s gravitational force is
( FG )T = 20 N, so nS on T = 24 N for the table to be in equilibrium. Summarizing, we have
Upper magnet
( FG ) U = 2.0 N
Table
( FG )T = 20 N
Lower magnet
( FG ) L = 2.0 N
nT on U = 8.0 N
nU on T = 8.0 N
nT on L = 4.0 N
FL on U = 6.0 N
nL on T = 4.0 N
FU on L = 6.0 N
nS on T = 24 N
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7-18
Chapter 7
Assess: The result nS on T = 24 N makes sense. The combined gravitational force on the table and two magnets is
24 N. Because the table is in equilibrium, the upward normal force of the surface has to exactly balance the total
gravitational force on the table and magnets.
7.20. Model: The astronaut and the satellite, the two objects in our system, will be treated as particles.
Visualize:
Solve: The astronaut and the satellite accelerate in opposite directions for 0.50 s. The force on the satellite and the
force on the astronaut are an action/reaction pair, so both have a magnitude of 100 N. Newton’s second law for the
satellite along the x-direction gives
FA on S −(100 N)
∑( Fon S ) x = FA on S = mSaS ⇒ aS =
=
= −0.156 m/s 2
mS
640 kg
Newton’s second law for the astronaut along the x-direction is
FS on A FA on S 100 N
∑( Fon A ) x = FS on A = mA aA ⇒ aA =
=
=
= 1.25 m/s 2
mA
mA
80 kg
Let us first calculate the positions and velocities of the astronaut and the satellite at t1 = 0.50 s under the
accelerations aA and aS:
x1A = x0A + v0A (t1 − t0 ) + 12 aA (t1 − t0 ) 2 = 0 m + 0 m + 12 (1.25 m/s 2 )(0.50 s − 0.00 s) 2 = 0.156 m
x1S = x0S + v0S (t1 − t0 ) + 12 aS (t1 − t0 ) 2 = 0 m + 0 m + 12 (−0.156 m/s 2 )(0.50 s − 0.00 s) 2 = −0.020 m
v1A = v0A + aA (t1 − t0 ) = 0 m/s + (1.25 m/s 2 )(0.50 s − 0.00 s) = 0.625 m/s
v1S = v0S + aS (t1 − t0 ) = 0 m/s + ( −0.156 m/s 2 )(0.5 s − 0.00 s) = −0.078 m/s
With x1A and x1S as initial positions, v1A and v1S as initial velocities, and zero accelerations, we can now obtain
the new positions at (t2 − t1 ) = 59.5 s:
x2A = x1A + v1A (t2 − t1 ) = 0.156 m + (0.625 m/s)(59.5 s) = 37.34 m
x2S = x1S + v1S (t2 − t1) = −0.02 m + (−0.078 m/s)(59.5 s) = −4.66 m
Thus the astronaut and the satellite are x2A − x2S = (37.34 m) − (−4.66 m) = 42 m apart.
7.21. Model: The block (B) and the steel cable (C), the two objects in the system, are modeled as particles and their
motion is determined by the constant-acceleration kinematic equations.
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Newton’s Third Law
7-19
Visualize:
Solve: Using v1x = v0 x + a x (t1 − t0 ),
4.0 m/s = 0 m/s + a x (2.0 s − 0.0 s) ⇒ ax = 2.0 m/s 2
Newton’s second law along the x-direction for the block gives
∑( Fon B ) x = ( FC on B ) x = mBa x = (20 kg)(2.0 m/s 2 ) = 40 N
( Fext ) x acts on the right end of the cable and ( FB on C ) x acts on the left end. According to Newton’s third law,
( FB on C ) x = ( FC on B ) x = 40 N. The difference in the horizontal component of the tension between the two ends of
the cable is thus
( Fext ) x − ( FB on C ) x = 100 N − 40 N = 60 N
7.22. Model: The gliders and spring form the system and are modeled as particles. Because the spring is
compressed, it may be modeled as a rigid rod, so the three objects are constrained to have the same acceleration.
Visualize:
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7-20
Chapter 7
Solve: By Newton’s third law, we know the action/reaction forces are of equal magnitude but point in the opposite
direction, so they cancel when considering the entire system. Using Newton’s second law in the x-direction, the
acceleration of the system is
Fext
6.0 N
∑( F ) x = Fext = (mA + mSp + mB ) a ⇒ a =
=
= 5.0 m/s 2
mA + mSp + mB 0.40 kg + 0.20 kg + 0.60 kg
Applying Newton’s second law to glider A gives
G
G
∑( F ) x = Fext − ( FSp on A ) x = mA a ⇒ ( FSp on A ) x = Fext − mA a = 6.0 N − (0.40 kg)(5.0 m/s2 ) = 4.0 N
Applying Newton’s second law to glider B gives
G
∑( F ) x = ( FSp on B ) x = mBa = (0.60 kg)(5.0 m/s 2 ) = 3.0 N
G
G
Applying Newton’s second law to the spring in the y-direction, and using the fact that ( FSp on A ) y = ( FSp on B ) y by
symmetry, we find
G
G
G
∑( F ) y = ( FSp on A ) y + ( FSp on B ) y − ( FG )Sp = 0 ⇒ 2( FSp on B ) y = mSp (9.8 m/s 2 )
G
G
( FSp on A ) y = ( FSp on B ) y = 12 (0.20 kg)(9.8 m/s 2 ) = 0.98 N
Adding the x- and y-components in quadrature gives the total force exerted by the spring on each block:
glider A : FSp on A = (4.0 N) 2 + (0.98 N)2 = 4.1 N
glider B : FSp on B = (3.0 N) 2 + (0.98 N) 2 = 3.2 N
Assess: The result seems reasonable because more force is exerted on glider A by the spring than on glider B, as
expected. The force exerted on glider A by the spring is, by Newton’s third law, the force that must accelerate the
spring + glider B, whereas the force exerted by the spring on glider B only has to accelerate glider B.
7.23. Model: Sled A, sled B, and the dog (D) are treated like particles in the model. The rope is considered to be
massless.
Visualize:
Solve: The acceleration constraint is (aA ) x = (aB ) x = a x . Newton’s second law applied to sled A gives
G
∑( Fon A ) y = nA − ( FG ) A = 0 N ⇒ nA = ( FG ) A = mA g
G
∑( Fon A ) x = T1 on A − f A = mA ax
Using f A = μk nA , the x-equation yields
T 1 on A − μk nA = mA ax
⇒ 150 N − (0.10)(100 kg)(9.8 m/s 2 ) = (100 kg)ax
⇒ ax = 0.52 m/s 2
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Newton’s Third Law
7-21
Newton’s second law applied to sled B gives
G
∑( Fon B ) y = nB − ( FG ) B = 0 N ⇒ nB = ( FG ) B = mB g
G
∑( Fon B ) x = T2 − T1 on B − f B = mBax
T 1 on B and T 1 on A act as if they are an action/reaction pair, so T 1 on B = 150 N. Using f B = μk nB = (0.10)(80 kg)
(9.8 m/s 2 ) = 78.4 N, we find
T2 − 150 N − 78.4 N = (80 kg)(0.52 m/s 2 ) ⇒ T2 = 270 N
Thus the tension T2 = 2.7 × 102 N.
7.24. Model: Consider an element of the rope dm = ρ dy, where ρ = m/L is the mass density of the rope. Model
this element as a particle.
Visualize:
Solve: The rope is stationary, so Newton’s second law applied to the particle gives
∑( Fdm ) y = T ( y ) − FG ( y ) = 0 ⇒ T ( y ) = FG ( y )
The force FG ( y ) due to gravity is the weight of the rope below the point y, which is
FG ( y ) = yρg = y ( m/L) g
Inserting this into the expression above gives the tension: T ( y ) = ymg/L.
Assess: The result seems reasonable because T(y) = 0 at the bottom of the rope (y = 0) and T(y) = mg at the top of the
rope (y = L).
7.25. Model: The coffee mug (M) is the only object in the system, and it will be treated as a particle. The model of
friction and the constant-acceleration kinematic equations will also be used.
Visualize:
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7-22
Chapter 7
Solve: The mug and the car have the same velocity. If the mug does not slip, their accelerations will also be the
same. Using v12x = v02x + 2a x ( x1 − x0 ), we get
0 m 2 /s 2 = (20 m/s) 2 + 2ax (50 m) ⇒ a x = −4.0 m/s 2
The static force needed to stop the mug is
( Fnet ) x = − fs = ma x = (0.5 kg)(−4.0 m/s 2 ) = −2.0 N ⇒ fs = 2.0 N
The maximum force of static friction is
( fs ) max = μs n = μs FG = μs mg = (0.50)(0.50 kg)(9.8 m/s 2 ) = 2.5 N
Since ( fs ) max < ( fs )max , the mug does not slide.
7.26. Model: For car tires on dry concrete, the coefficient μs of static friction is typically about 0.80 (see Table 6.1).
Visualize: The car and the ground are denoted by C and S, respectively.
Solve: The car presses down against the ground with both the drive wheels (assumed to be the front wheels F,
although this is not critical) and the nondrive wheels. For this car, two-thirds of the gravitational force rests on the
G
G
front wheels. Physically, force FS on C is a static friction force, so its maximum value is ( FS on C ) max = ( fs ) max = μs n.
The maximum acceleration of the car on the ground (or concrete surface) occurs when the static friction reaches this
maximum possible value:
( F S on C ) max = ( fs ) max = μs nF = μs ( FG ) F = μs 23 mg = (0.80) 23 (1500 kg)(9.8 m/s 2 ) = 7840 N
(
)
()
Use this force in Newton’s second law to find the acceleration:
( FS on C ) max 7840 N
amax =
=
= 5.2 m/s 2
m
1500 kg
7.27. Model: The starship and the shuttlecraft will be denoted as M and m, respectively, and both will be treated as
particles. We will also use the constant-acceleration kinematic equations.
Visualize:
G
Solve: (a) The tractor beam is some kind of long-range force FM on m . Regardless of what kind of force it is, by
G
Newton’s third law there must be a reaction force Fm on M on the starship. As a result, both the shuttlecraft and the
starship move toward each other (rather than the starship remaining at rest as it pulls the shuttlecraft in). However,
the very different masses of the two crafts means that the distances they each move will also be very different. The
pictorial representation shows that they meet at time t1 when xM1 = xm1. There’s only one force on each craft, so
Newton’s second law is very simple. Furthermore, because the forces are an action/reaction pair,
FM on m = Fm on M = Ftractor beam = 4.0 × 104 N
The accelerations of the two craft are
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Newton’s Third Law
aM =
Fm on M
M
=
4.0 × 104 N
2.0 × 106 kg
2
= 0.020 m/s and am =
G
FM on m
m
=
−4.0 × 104 N
2.0 × 104 kg
7-23
= −2.0 m/s 2
Acceleration am is negative because the force and acceleration vectors point in the negative x-direction. Now we
have a constant-acceleration problem in kinematics. At a later time t1 the positions of the crafts are
xM1 = xM0 + vM0 (t1 − t0 ) + 12 aM (t1 − t0 ) 2 = 12 aMt12
xm1 = xm0 + vm0 (t1 − t0 ) + 12 am (t1 − t0 ) 2 = xm0 + 12 amt12
The craft meet when xM1 = xm1, so
1 a t2
2 M1
= xm0 + 12 amt12
⇒ t1 =
2 xm0
2 xm0
2(10,000 m)
=
=
= 99.5 s
aM − am
aM + | am |
2.02 m/s 2
Knowing t1, we can now find the starship’s position as it meets the shuttlecraft:
xM1 = 12 aMt12 = 99 m
The starship moves 99 m as it pulls in the shuttlecraft from 10 km away.
7.28. Model: We shall only consider horizontal forces. The head and the baseball are the two objects in our system
and are treated as particles. We will also use the constant-acceleration kinematic equations.
Visualize:
Solve: (a) The ball experiences an average acceleration of
v −v
−30 m/s
aB = B1 B0 =
= −20,000 m/s 2
−3
t
1.5 × 10 s
Insert this into Newton’s second law to find the force on the baseball:
FH on B = mBaB = (0.14 kg) −20,000 m/s 2 = 2800 N
(b) By Newton’s third law, the magnitude of the force exerted by the ball on the head is the same as that exerted by
the head on the ball. Thus, FB on H = 2800 N.
(c) Because 2800 N < 6000 N, the ball will not fracture your forehead, but will fracture your cheekbone because
2800 N > 1300 N.
Assess: A 90 mph fastball travels at (90 mph)(1609.3 m/mile)(1 h/3600 s) = 40 m/s, so it will not fracture your
forehead, but it will fracture your cheekbone. This explains why baseball helmets protect the cheekbone.
7.29. Model: The rock (R) and Bob (b) are the two objects in our system, and will be treated as particles. We will
also use the constant-acceleration kinematic equations.
Visualize:
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7-24
Chapter 7
G
Solve: (a) Bob exerts a forward force FB on R on the rock to accelerate it forward. The rock’s acceleration is
calculated as follows:
2
2
v1R
= v0R
+ 2a0R Δx ⇒ aR =
2
v1R
(30 m/s) 2
=
= 450 m/s 2
2Δx 2(1.0 m)
The force is calculated from Newton’s second law:
FB on R = mR aR = (0.500 kg)(450 m/s 2 ) = 225 N
Bob exerts a force of 2.3 × 102 N on the rock.
G
G
G
(b) Because Bob pushes on the rock, the rock pushes back on Bob with a force FR on B . Forces FR on B and FB on R
are an action/reaction pair, so FR on B = FB on R = 225 N. The force causes Bob to accelerate backward with an
acceleration of
( Fnet on B ) x
FR on B
225 N
= −3.0 m/s 2
75 kg
mB
mB
This is a rather large acceleration, but it lasts only until Bob releases the rock. We can determine the time interval by
returning to the kinematics of the rock:
v
v1R = v0R + aR Δt = aR Δt ⇒ Δt = 1R = 0.0667 s
aR
At the end of this interval, Bob’s velocity is
v1B = v0B + aBΔt = aBΔt = −0.20 m/s
aB =
=−
=−
Thus his recoil speed is 0.20 m/s.
7.30. Model: The boy (B) and the crate (C) are the two objects in our system, and they will be treated in the particle
model. We will also use the static and kinetic friction models.
Visualize:
Solve: The fact that the boy’s feet occasionally slip means that the maximum force of static friction must exist
between the boy’s feet and the sidewalk. That is, fsB = μsBnB . Also f kC = μ kC nC .
Newton’s second law applied to the crate gives
∑( Fon C ) y = nC − ( FG )C = 0 N ⇒ nC = mC g
∑( Fon C ) x = FB on C − f kC = 0 N ⇒
FB on C = f kC = μkC nC = μ kC mC g
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Newton’s Third Law
7-25
Newton’s second law for the boy is
∑( Fon B ) y = nB − ( FG ) B = 0 N ⇒ nB = mB g
∑( Fon B ) x = fsB − FC on B = 0 N ⇒
G
G
FC on B and FB on C are an action/reaction pair, so
FC on B = FB on C
⇒ μsBmB g = μkC mC g
FC on B = fsB = μsBnB = μsBmB g
⇒ mC =
μsBmB (0.8)(50 kg)
=
= 2 × 102 kg
μkC
(0.2)
7.31. Model: Assume package A and package B are particles. Use the model of kinetic friction and the constantacceleration kinematic equations.
Visualize:
Solve: Package B has a smaller coefficient of friction, so its acceleration down the ramp is greater than that of
package A. It will therefore push against package A and, by Newton’s third law, package A will push back on B. The
acceleration constraint is aA = aB ≡ a.
Newton’s second law applied to each package gives
∑( Fon A ) x = FB on A + ( FG )A sin θ − f kA = mA a
FB on A + mA g sin θ − μ kA (mA g cosθ ) = mA a
∑( Fon B ) x = − FA on B − f kB + ( FG ) B sin θ = mBa
− FA on B − μ kB (mB g cosθ ) + mB g sin θ = mBa
where we have used nA = mA cosθ g and nB = mB cos θ g . Adding the two force equations, and using FA on B = FB on A
because they are an action/reaction pair, we get
( μ m + μkBmB )( g cosθ ) [(020)(5.0 kg) + (0.15)(10 kg)](9.8 m/s 2 )cos(20°)
a = g sin θ − kA A
=
= 1.82 m/s 2
mA + mB
5.0 kg + 10 kg
Finally, using x1 = x0 + v0 (t1 − t0 ) + 12 a (t1 − t0 ) 2 , we find
2.0 m = 0 m + 0 m + 12 (1.82 m/s 2 )(t1 − 0 s)2
⇒ t1 = 2(2.0 m)/(1.82 m/s 2 ) = 1.5 s
7.32. Model: The two blocks form a system of interacting objects. We shall treat them as particles.
Visualize: Please refer to Figure P7.32.
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7-26
Chapter 7
Solve: It is possible that the left-hand block (Block L) is accelerating down the slope faster than the right-hand block
(Block R), causing the string to be slack (zero tension). If that were the case, we would get a zero or negative answer
for the tension in the string. Newton’s first law applied in the y-direction on Block L yields
(∑ FL ) y = 0 = nL − ( FG ) L cos(20°) ⇒ nL = mL g cos(20°)
Therefore
( f k ) L = ( μ k ) L mL g cos(20°) = (0.20)(1.0 kg)(9.80 m/s 2 )cos(20°) = 1.84 N
A similar analysis of the forces in the y-direction on Block R gives ( f k ) R = 1.84 N as well. Using Newton’s second
law in the x-direction for Block L gives
(∑ FL ) x = mL a = TR on L − ( f k ) L + ( FG ) L sin(20°) ⇒ mL a = TR on L − 1.84 N + mL g sin(20°)
For Block R,
(∑ FR ) x = mR a = ( FG ) R sin(20°) − 1.84 N − TL on R
⇒ mR a = mR g sin(20°) − 1.84 N − TL on R
Solving these two equations in the two unknowns a and TL on R = TR on L ≡ T , we obtain a = 2.12 m/s 2 and T = 0.61 N.
Assess: The tension in the string is positive, and is about 1/3 of the kinetic friction force on each of the blocks,
which is reasonable.
7.33. Model: The two blocks (1 and 2) form the system of interest and will be treated as particles. The ropes are
assumed to be massless, and the model of kinetic friction will be used.
Visualize:
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Newton’s Third Law
7-27
Solve: (a) The separate free-body diagrams for the two blocks show that there are two action/reaction pairs. Notice how
G
G
block 1 both pushes down on block 2(force n1′ )and exerts a retarding friction force f 2 top on the top surface of block 2.
Block 1 is in static equilibrium (a1 = 0 m/s 2 ) but block 2 is accelerating to the right. Newton’s second law for block 1 is
( Fnet on 1) x = f1 − Trope = 0 N ⇒ Trope = f1
( Fnet on 1) y = n1 − m1g = 0 N ⇒ n1 = m1g
Although block 1 is stationary, there is a kinetic force of friction because there is motion between block 1 and block 2.
The friction model means f1 = μ k n1 = μ k m1g . Substitute this result into the x-equation to get the tension in the rope:
Trope = f1 = μ k m1g = (0.40)(1.0 kg)(9.8 m/s 2 ) = 3.9 N
(b) Newton’s second law for block 2 is
ax ≡ a =
( Fnet on 2 ) x
m2
a y = 0 m/s 2 =
=
Tpull − f 2 top − f 2 bot
( Fnet on 2 ) y
m2
=
n2 − n1′ − m2 g
m2
m2
G
G
Forces n1 and n1′ are an action/reaction pair, so n1′ = n1 = m1g . Substituting into the y-equation gives n2 = ( m1 + m2 ) g .
This is not surprising because the combined weight of both objects presses down on the surface. The kinetic friction
on the bottom surface of block 2 is then
f 2 bot = μ k n2 = μ k ( m1 + m2 ) g
G
G
The forces f1 and f 2 top are an action/reaction pair, so f 2 bot = f1 = μ k m1g . Inserting these friction results into the
x-equation gives
a=
=
( Fnet on 2 ) x
m2
=
Tpull − μk m1g − μk (m1 + m2 ) g
m2
20 N − (0.40)(1.0 kg)(9.8 m/s 2 ) − (0.40)(1.0 kg + 2.0 kg)(9.8 m/s 2 )
= 2.2 m/s 2
2.0 kg
7.34. Model: The 3-kg and 4-kg blocks constitute the system and are to be treated as particles. The models of
kinetic and static friction and the constant-acceleration kinematic equations will be used.
Visualize:
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7-28
Chapter 7
Solve: The minimum time will be achieved when static friction is at its maximum possible value. Newton’s second
law for the 4-kg block is
∑( Fon 4 ) y = n3 on 4 − ( FG ) 4 = 0 N ⇒ n3 on 4 = ( FG ) 4 = m4 g = (4.0 kg)(9.8 m/s 2 ) = 39.2 N
fs4 = ( fs )max = μs n3 on 4 = (0.60)(39.2 N) = 23.5 N
Newton’s second law for the 3-kg block is
∑( Fon 3 ) y = n3 − n4 on 3 − ( FG )3 = 0 N ⇒ n3 = n4 on 3 + ( FG )3 = 39.2 N + (3.0 kg)(9.8 m/s 2 ) = 68.6 N
Friction forces f and fs4 are an action/reaction pair. Thus
∑( Fon 3 ) x = fs3 − f k3 = m3a3
⇒
fs4 − μ k n3 = m3a3 ⇒ 23.5 N − (0.20)(68.6 N) = (3.0 kg) a3
a3 = 3.27 m/s 2
Since block 3 does not slip, this is also the acceleration of block 4. The time is calculated as follows:
x1 − x0 + v0 x (t1 − t0 ) + 12 a (t1 − t0 ) 2 ⇒ 5.0 m = 0 m + 0 m + 12 (3.27 m/s 2 )(t1 − 0 s) 2 ⇒ t1 = 1.8 s
7.35. Model: Blocks 1 and 2 make up the system of interest and will be treated as particles. Assume a massless
rope and frictionless pulley.
Visualize:
Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration constraint is
a2 = a = − a1, where a will have a positive value. There are two real action/reaction pairs. The two tension forces will act as if
they are action/reaction pairs because we are assuming a massless rope and a frictionless pulley. Make sure you understand
G
why the friction forces point in the directions shown in the free-body diagrams, especially force f1′ exerted on block 2 by block
1. We have quite a few pieces of information to include. First, Newton’s second law applied to blocks 1 and 2 gives
G
( Fnet on 1) x = f1 − T1 = μ k n1 − T1 = m1a1 = − m1a
( Fnet on 1) y = n1 − m1g = 0 N ⇒ n1 = m1g
( Fnet on 2 ) x = Tpull − f1′ − f 2 − T2 = Tpull − f1′ − μk n2 − T2 = m2a2 = m2a
( Fnet on 2 ) y = n2 − n1′ − m2 g = 0 N ⇒ n2 = n1′ + m2 g
We’ve already used the kinetic friction model in both x-equations. Next, Newton’s third law gives
n1′ = n1 = m1g f1′ = f1 = μk n1 = μk m1g T1 = T2 = T
Knowing n1′ , we can now use the y-equation of block 2 to find n2 . Substitute all these pieces into the two
x-equations, and we end up with two equations with two unknowns:
μk m1g − T = −m1a Tpull − T − μk m1g − μk (m1 + m2 ) g = m2a
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Newton’s Third Law
7-29
Subtract the first equation from the second to get
Tpull − μ k (3m1 + m2 ) g = (m1 + m2 )a
a=
Tpull − μ k (3m1 + m2 ) g
m1 + m2
=
20 N − (0.30)[3(1.0 kg) + 2.0 kg](9.8 m/s 2 )
= 1.8 m/s 2
1.0 kg + 2.0 kg
7.36. Model: Blocks 1 and 2 make up the system of interest and will be treated as particles. We shall use the kinetic
friction model.
Visualize:
Notice that the coordinate system of for block B is rotated so that the motion in the positive x-direction is consistent
between the two free-body diagrams.
Solve: The blocks are constrained to have the same magnitude acceleration. Applying Newton’s second law to block B gives
∑( F ) y = −T + ( FG ) B = ma ⇒ T − mg = − ma
Applying Newton’s second law in both the x-and y-directions to the block A gives
∑( F ) y = n − ( FG ) A = 0 ⇒ n = Mg
∑( F ) x = T = Ma ⇒ T = Ma
Using the first equation to eliminate the acceleration a gives the tension:
mMg
m+M
Assess: The result is positive, as it should be for our choice of coordinate system. Consider m = 0. In this case,
T = 0, as expected. For m M , the tension is independent of the mass of the hanging block because its acceleration
will be g, as we can see by solving for the acceleration:
T
Mg
a=− +g =g−
→ g for m M
m
m+M
T = Ma = M ( g − T/m) ⇒ T =
7.37. Model: The sled (S) and the box (B) will be treated in the particle model, and the model of friction will be
used. Refer to Table 6.1 for the required friction coefficients.
Visualize:
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7-30
Chapter 7
In the sled’s free-body diagram nS is the normal (contact) force on the sled due to the snow. Similarly f kS is the force
of kinetic friction on the sled due to snow.
Solve: Newton’s second law on the box in the y-direction is
nS on B − ( FG )B cos(20°) = 0 N ⇒ nS on B = (10 kg)(9.8 m/s 2 )cos(20°) = 92.1 N
G
The static friction force fS on B accelerates the box. The maximum acceleration occurs when static friction reaches its
maximum possible value.
( fs ) max = μSnS on B = (0.50)(92.1 N) = 46.1 N
Newton’s second law along the x-direction thus gives the maximum acceleration
fS on B − ( FG ) B sin(20°) = mBa ⇒ 46.1 N − (10 kg)(9.8 m/s2 )sin(20°) = (10 kg)a ⇒ a = 1.25 m/s 2
Newton’s second law for the sled along the y-direction is
nS − nB on S − ( FG )S cos(20°) = 0 N
nS = nB on S + mS g cos(20°) = (92.1 N) + (20 kg)(9.8 m/s 2 )cos(20°) = 276.3 N
Therefore, the force of friction on the sled by the snow is
f kS = ( μ k )nS = (0.06)(276.3 N) = 16.6 N
Newton’s second law along the x-direction is
Tpull − wS sin(20°) − f kS − f B on S = mSa
The friction force f B on S = fS on B because these are an action/reaction pair. We’re using the maximum acceleration,
so the maximum tension is
Tmax − (20 kg)(9.8 m/s 2 )sin(20°) − 16.6 N − 46.1 N = (20 kg)(1.25 m/s 2 ) = 160 N
7.38. Model: The masses m and M are to be treated in the particle model. We will also assume a massless rope and
frictionless pulley, and use the constant-acceleration kinematic equations for m and M.
Visualize:
Solve: Using y1 = y0 + v0 y (t1 − t0 ) + 12 aM (t1 − t0 ) 2 ,
(−1.0 m) = 0 m + 0 m + 12 aM (6.0 s − 0 s) 2
Newton’s second law for m and M gives
∑( Fon m ) y = TR on m − ( FG ) m = mam
⇒ aM = −0.0556 m/s 2
∑( Fon M ) y = TR on M − ( FG ) M = MaM
The acceleration constraint is am = −aM . Also, the tensions are an pseudo-action/reaction pair, so TR on m = TR on M .
With these, the second-law equations become
TR on M − Mg = MaM
TR on M − mg = −maM
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Newton’s Third Law
7-31
Subtracting the second from the first gives
− Mg + mg = MaM + maM
⎡ g + aM ⎤
m=M⎢
⎥
⎣ g − aM ⎦
⎡ 9.8 m/s 2 − 0.556 m/s 2 ⎤
= (100 kg) ⎢
= 99 kg
2
2⎥
⎢⎣ 9.8 m/s + 0.556 m/s ⎥⎦
Assess: Note that am = − aM = 0.0556 m/s 2 . For such a small acceleration, the 1% mass difference seems reasonable.
7.39. Model: Use the particle model for the block of mass M and the two massless pulleys.
Additionally, the rope is
G
massless and the pulleys are frictionless. The block is kept in place by an applied force F.
Visualize:
Solve: Since there is no friction on the pulleys, T2 = T3 = T5 . Newton’s second law for mass M gives
T1 − FG = 0 N ⇒ T1 = Mg = (10.2 kg)(9.8 m/s 2 ) = 100 N
Newton’s second law for the small pulley is
T2 + T3 − T1 = 0 N ⇒ T2 = T3 =
T1
= 50 N = T5 = F
2
Newton’s second law for the large pulley is
T4 − T2 − T3 − T5 = 0 N ⇒ T4 = T2 + T3 + T5 = 150 N
7.40. Model: Assume the particle model for m1, m2 , and m3 , and the model of kinetic friction. Assume the ropes to
be massless, and the pulleys to be frictionless and massless.
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7-32
Chapter 7
Visualize:
Solve: Newton’s second law for m1 gives T1 − ( FG )1 = m1a1. Newton’s second law for m2 gives
∑( Fon m2 ) y = n2 − ( FG ) 2 = 0 N ⇒ n2 = m2 g
∑( Fon m2 ) x = T2 − f k2 − T = m2 a2
⇒ T2 − μ k n2 − T1 = m2 a2
Newton’s second law for m3 gives T2 − ( FG )3 = m3a3. Since m1, m2 , and m3 move together, a1 = a2 = −a3 ≡ a.
The equations for the three masses thus become
T1 − ( FG )1 = m1a T2 − μ k n2 − T1 = m2a T2 − ( FG )3 = − m3a
Subtracting the third equation from the sum of the first two equations yields:
−( FG )1 − μk n2 + ( FG )3 = − m1g − μ k m2 g + m3 g = (m1 + m2 + m3 )a
a=
− m1g − μ k m2 g + m3 g −1.0 kg − (0.30)(2.0 kg) + 3.0 kg
=
(9.8 m/s 2 ) = 2.3 m/s 2
(m1 + m2 + m3 )
1.0 kg + 2.0 kg + 3.0 kg
7.41. Model: Assume the particle model for the two blocks, and the model of kinetic and static friction.
Visualize:
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Newton’s Third Law
7-33
Solve: (a) If the mass m is too small, the hanging 2.0 kg mass will pull it up the slope. We want to find the smallest
mass that will stick as a result of friction. The smallest mass will be the one for which the force of static friction is at
its maximum possible value: f s = ( fs ) max = μs n. As long as the mass m is stuck, both blocks are at rest with
G
Fnet = 0 N. In this situation, Newton’s second law for the hanging mass M gives
( Fnet ) x = −TM + Mg = 0 N ⇒ TM = Mg = (2.0 kg)(9.8 m/s 2 ) = 19.6 N
For the smaller mass m,
( Fnet ) x = Tm − fs − mg sin θ = 0 N ( Fnet ) y = n − mg cosθ ⇒ n = mg cosθ
G
G
For a massless string and frictionless pulley, forces Tm and TM act as if they are an action/reaction pair. Thus Tm = TM . Mass
m is a minimum when fs = ( fs )max = μs n = μs mg cosθ . Substituting these expressions into the x-equation for m gives
TM − μs mg cosθ − mg sin θ = 0 N
TM
19.6 N
=
= 1.83 kg
( μs cosθ + sin θ ) g [(0.80)cos(20°) + sin(20°)](9.8 m/s 2 )
or 1.8 kg to two significant figures.
(b) Because μ k < μS the 1.8 kg block will begin to slide up the ramp and the 2.0 kg mass will begin to fall if
m=
the block is nudged ever so slightly. In this case, the net force and the acceleration are not zero. Notice how, in the
pictorial representation, we chose different coordinate systems for the two masses. The magnitudes of the
accelerations are the same because the blocks are tied together. Thus, the acceleration constraint is am = aM ≡ a,
where a will have a positive value. Newton’s second law for block M gives
( Fnet ) x = −T + Mg = MaM = Ma
For block m we have
( Fnet ) x = T − f k − mg sin θ = T − μ k mg cosθ − mg sin θ = mam = ma
In writing these equations, we used Newton’s third law to obtain Tm = TM = T . Also, notice that the x-equation and
the friction model for block m don’t change, except for μs becoming μ k , so we already know the expression for f k
from part (a). Notice that the tension in the string is not the gravitational force Mg. We have two equations with the
two unknowns T and a:
Mg − T = Ma T − ( μ k cosθ + sin θ )mg = ma
Adding the two equations to eliminate T gives
Mg − ( μ k cosθ + sin θ ) mg = Ma + ma
M − ( μ k cosθ + sin θ )m
M +m
2 2.0 kg − [(0.50)cos(20°) + sin(20°)](1.83 kg)
= (9.8 m/s )
1.3 m/s 2
2.0 kg + 1.83 kg
a=g
7.42. Model: Assume the particle model for the two blocks and use the friction model.
Visualize:
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7-34
Chapter 7
Solve: (a) The slope is frictionless, so the blocks stay in place only if held. Once m is released, the blocks will move
G
one way or the other. As long as m is held, the blocks are in static equilibrium with Fnet = 0 N. In this case,
Newton’s second law for the hanging block M is
( Fnet on M ) y = TM − Mg = 0 N ⇒ TM = Mg = 19.6 N
Because the string is massless and the pulley is frictionless, TM = Tm = T = 20 N (to two significant figures).
(b) The free-body diagram shows box m after it is released. Whether it moves up or down the slope depends on
whether the acceleration a is positive or negative. The acceleration constraint is (am ) x − (aM ) y ≡ a Newton’s second
law for each system gives
( Fnet on m ) x = T − mg sin θ = m(am ) x = ma ( Fnet on M ) y = T − Mg = M (aM ) y = − Ma
We have two equations in two unknowns. Subtract the second from the first to eliminate T:
M − m sin θ
2.0 kg − (4.0 kg)sin (35°)
g=
− mg sin θ + Mg = ( m + M )a ⇒ a =
= −0.48 m/s 2
M +m
2.0 kg + 4.0 kg
Since a < 0 m/s 2 , the box accelerates down the slope.
(c) It is now straightforward to compute T = Mg − Ma = 21 N. Notice how the tension is larger than when the blocks
were motionless.
7.43. Model: Use the particle model for the book (B) and the coffee cup (C), the models of kinetic and static
friction, and the constant-acceleration kinematic equations.
Visualize:
Solve: (a) Using v12x = v02x + 2a ( x1 − x0 ), we find
0 m 2 /s 2 = (3.0 m/s) 2 + 2a ( x1 ) ⇒ ax1 = −4.5 m 2 /s 2
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Newton’s Third Law
7-35
To find x1, we must first find a. Newton’s second law applied to the book and the coffee cup gives
∑( Fon B ) y = nB − ( FG ) B cos(20°) = 0 N ⇒ nB = (1.0 kg)(9.8 m/s 2 )cos(20°) = 9.21 N
∑( Fon B ) x = −T − f k − ( FG ) B sin(20°) = mBaB
∑( Fon C ) y = T − ( FG )C = mCaC
The last two equations can be rewritten, using aC = aB = a, as
−T − μ k nB − mB g sin(20°) = mBa T − mC g = mCa
Adding the two equations gives
a (mC + mB ) = − g[mC + mB sin(20°)] − μk (9.21 N)
(1.5 kg)a = −(9.8 m/s 2 )[0.500 kg + (1.0 kg)sin 20°] − (0.20)(9.21 N) ⇒ a = −6.73 m/s 2
Using this value for a, we can now find x1 as follows:
−4.5 m 2 /s 2 −4.5 m 2 /s 2
=
= 0.67 m
a
−6.73 m/s 2
(b) The maximum static friction force is ( fs ) max = μs nB = (0.50)(9.21 N) = 4.60 N. We’ll see if the force fs needed
x1 =
to keep the book in place is larger or smaller than ( fs ) max . When the cup is at rest, the string tension is T = mC g .
Newton’s first law for the book is
∑( Fon B ) x = fs − T − wB sin(20°) = fs − mC g − mB g sin(20°) = 0
fs = ( M C + M B sin 20°) g = 8.25 N
Because fs > ( fs ) max , the book slides back down.
7.44. Model: Use the particle model for the cable car and the counterweight. Assume a massless cable.
Visualize:
Solve: (a) Notice the separate coordinate systems for the cable car (object 1) and the counterweight (object 2).
G
G
G
G
Forces T1 and T2 act as if they are an action/reaction pair. The braking force FB works with the cable tension T1 to
G
allow the cable car to descend at a constant speed. Constant speed means dynamic equilibrium, so Fnet = 0 N for both
systems. Newton’s second law applied to the cable car gives
( Fnet on 1) x = T1 + FB − m1g sin θ1 = 0 N ( Fnet on 1 ) y = n1 − m1g cosθ1 = 0 N
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7-36
Chapter 7
Newton’s second law applied to the counterweight gives
( Fnet on 2 ) x = m2 g sin θ 2 − T2 = 0 N ( Fnet on 2 ) y = n2 − m2 g cosθ 2 = 0 N
From the x-equation for the counterweight, T2 = m2 g sin θ 2 . Because we can neglect the pulley’s friction and the
cable is assumed to be massless, T1 = T2 . Thus the x-equation for the cable car then becomes
FB = m1g sin θ1 − T1 = m1g sin θ1 − m2 g sin θ 2 = 3770 N = 3.8 kN
(b) If the brakes fail, then FB = 0 N. The car will accelerate down the hill on one side while the counterweight
accelerates up the hill on the other side. Both will have negative accelerations because of the direction of the
acceleration vectors. The constraint is a1x = a2 x = a, where a will have a negative value. Using T1 = T2 = T , the two
x-equations are
( Fnet on 1 ) x = T − m1g sin θ1 = m1a1x = m1a ( Fnet on 2 ) x = m2 g sin θ 2 − T = m2a2 x = m2a
Note that the y-equations aren’t needed in this problem. Add the two equations to eliminate T:
m sin θ1 − m2 sin θ 2
− m1g sin θ1 + m2 g sin θ = (m1 + m2 )a ⇒ a = − 1
g = −0.991 m/s 2
m1 + m2
Now we have a problem in kinematics. The speed at the bottom is calculated as follows:
v12 = v02 + 2a ( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2(−0.991 m/s 2 )(−400 m) = 28 m/s
Assess: A speed of approximately 60 mph as the cable car travels a distance of 2000 m along a frictionless slope of
30° is reasonable.
7.45. Model: Assume the cable mass is negligible compared to the car mass and that the pulley is frictionless. Use
the particle model for the two cars.
Visualize: Please refer to Figure P7.45.
Solve: (a) The cars are moving at constant speed, so they are in dynamic equilibrium. Consider the descending car
D. We can find the rolling friction force on car D, and then find the cable tension by applying Newton’s first law. In
the y-direction for car D,
( Fnet ) y = 0 = nD − ( FG ) D cos(35°)
nD = mD g cos(35°)
So the rolling friction force on car D is
( f R ) D = μ R nD = μ R mD g cos(35°)
Applying Newton’s first law to car D in the x-direction gives
( Fnet ) x = TA on D + ( f R ) D − ( FG ) D sin(35°) = 0
Thus,
TA on D = mD g sin(35°) − μ R mD g cos(35°)
= (1500 kg)(9.80 m/s 2 )[sin(35°) − (0.020)cos(35°)]
= 8.2 × 103 N
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Newton’s Third Law
7-37
(b) Similarly, we find that for car A, ( f R )A = μ R mA g cos(35°). In the x-direction for car A,
( Fnet ) x = Tmotor + TD on A − ( f R ) A − ( FG ) A sin(35°) = 0
Tmotor = mA g sin(35°) + μR mA g cos(35°) − mD g sin(35°) + μR mD g cos(35°)
Here, we have used TA on D = TD on A . If we also use m A = m D , then
Tmotor = 2 μ R mA g cos(35°) = 4.8 × 102 N.
Assess: Careful examination of the free-body diagrams for cars D and A yields the observation that Tmotor = 2( FR ) A
in order for the cars to be in dynamic equilibrium. It is a tribute to the design that the motor must only provide such a
small force compared to the tension in the cable connecting the two cars.
7.46. Model: The painter and the chair are treated as a single object and represented as a particle. We assume that
the rope is massless and that the pulley is massless and frictionless.
Visualize:
Solve: If the painter pulls down on the rope with force F, Newton’s third law requires the rope to pull up on the
painter with force F. This is just the tension in the rope. With our model of the rope and pulley, the same tension
force F also pulls up on the painter’s chair. Newton’s second law for (painter + chair) gives
2 F − FG = (mP + mC ) a
( 12 )[(mP + mC )a + (mP + mC ) g ] = 12 (mP + mC )(a + g )
= ( 12 ) (70 kg + 10 kg)(0.20 m/s 2 + 9.8 m/s 2 ) = 4.0 × 102 N
F=
Assess: A force of 400 N, which is approximately one-half the total gravitational force, is reasonable since the
upward acceleration is small.
7.47. Model: Model Jorge as a particle and use the friction model.
Visualize:
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7-38
Chapter 7
Solve: If the Jorge pulls on the rope with force F, Newton’s third law requires the rope to pull up on Jorge with force
F. This is just the tension in the rope (i.e., F = T). With our model of the rope and pulley, the same tension pulls at
Jorge’s waist where the rope is tied. Applying Newton’s second law to Jorge in the y-direction gives
∑( F ) y = n − FG = 0 ⇒ n = FG = mg
From the friction model, we have f r = μr n = μ r mg. Using this in the equation below, which is derived by using
Newton’s second law applied in the x-direction, gives
∑( F ) x = F + T − f r = ma
2 F − μr mg r = ma
a=
2F
− μr g
m
7.48. Model: Use the particle model for the tightrope walker and the rope. The rope is assumed to be massless, so
the tension in the rope is uniform.
Visualize:
Solve: Newton’s second law applied to the tightrope walker gives
FR on W − FG = ma ⇒ FR on W = m( a + g ) = (70 kg)(8.0 m/s 2 + 9.8 m/s 2 ) = 1.25 × 103 N
Newton’s second law applied to the rope gives
∑( Fon R ) y = T sin θ + T sin θ − FW on R = 0 N ⇒ T =
FW on R
2sin(10°)
=
FR on W
2sin(10°)
=
1.25 × 103 N
= 3.6 × 103 N
2sin(10°)
We used FW on R = FR on W because they are an action/reaction pair.
7.49. Model: Use the particle model for the wedge and the block.
Visualize:
The block will not slip relative to the wedge if they both have the same horizontal acceleration a. Note that n1 on 2
and n2 on 1 form a third law pair, so n1 on 2 = n2 on 1.
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Newton’s Third Law
7-39
Solve: Newton’s second law applied to block m2 in the y-direction gives
∑( Fon 2 ) y = n1 on 2 cosθ − ( FG ) 2 = 0 N ⇒ n1 on 2 =
m2 g
cosθ
Combining this equation with the x-component of Newton’s second law yields:
n
sin θ
∑( Fon 2 ) x = n1 on 2 sin θ = m2a ⇒ a = 1 on 2
= g tan θ
m2
Newton’s second law applied to the wedge gives
∑( Fon 1 ) x = F − n2 on 1 sin θ = m1a
F = m1a + n2 on 1 sin θ = m1a + m2a = (m1 + m2 ) a = ( m1 + m2 ) g tan θ
7.50. Model: Treat the basketball player (P) as a particle, and use the constant-acceleration kinematic equations.
Visualize:
Solve: (a) While in the process of jumping, the basketball player is pressing down on the floor as he straightens his
G
legs. He exerts a force FP on F on the floor. The player experiences a gravitational force ( FG ) P as well as a normal
G
G
force from the floor nF on P . The floor experiences the force FP on F exerted by the player.
G
G
(b) The player standing at rest exerts a force FP on F on the floor. The normal force nF on P is the reaction force to
G
G
FP on F . But nF on P = FP on F , so Fnet = 0 N. When the basketball player accelerates upward by straightening his
legs, his speed has to increase from zero to v1 y , which is the speed with which he leaves the floor. Thus, according
to Newton’s second law, there must be a net upward force on him during this time. This can be true only if
nF on P > ( FG ) P . In other words, the player presses on the floor with a force FP on F larger than the gravitational force
G
on him, which is equal to his weight. The reaction force nF on P then exceeds his weight and accelerates him upward
until his feet leave the floor.
(c) The height of 80 cm = 0.80 m is sufficient to determine the speed v1y with which he leaves the floor. Once his
feet are off the floor, he is simply in free fall, with a1 = − g . From kinematics,
v22 y = v12y + 2a1 ( y2 − y1 ) ⇒ 0 m 2 /s 2 = v12y + 2(− g )(0.80 m) ⇒ v1 y = (2 g )(0.80 m) = 3.96 m/s
The basketball player reaches v1 y = 4.0 m/s by accelerating from rest through a distance of 0.60 m.
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7-40
Chapter 7
(d) Assuming a0 to be constant during the jump, we find
v12y = v02 y + 2a0 ( y1 − y0 ) = 0 m 2 /s 2 + 2a0 ( y1 − 0 m) ⇒ a0 =
v12y
2 y1
=
(3.96 m/s) 2
= 13 m/s2
2(0.60 m)
(e) The scale reads the value of nF on P , which is the force exerted by the scale on the player. Before jumping,
nF on P − ( FG ) P = 0 N ⇒ nF on P = ( FG )P = mg = (100 kg)(9.8 m/s 2 ) = 980 N
While accelerating upward,
⎛ a ⎞
⎛ 13.1 ⎞
⇒ nF on P = ma0 + mg = mg ⎜1 + 0 ⎟ = (980 N) ⎜1 +
⎟ = 2.3 kN
g ⎠
9.8 ⎠
⎝
⎝
= 0 N because there is no contact with the scale.
nF on P − mg = ma0
After leaving the scale, nF on P
7.51. A 1.0 kg wood block is placed on top of a 2.0 kg wood block. A horizontal rope pulls the 2.0 kg block across a
frictionless floor with a force of 21.0 N. Does the 1.0 kg block on top slide?
Visualize:
Solve: The 1.0 kg block is accelerated by static friction. It moves smoothly with the lower block if fs < ( fs ) max . It
slides if the force that would be needed to keep it in place exceeds ( fs ) max . Begin by assuming that the blocks move
together with a common acceleration a. Newton’s second law gives
Top block: ∑( Fon 1 ) x = fs = m1a
Bottom block: ∑( Fon 2 ) x = Tpull − fs = m2 a
Adding these two equations gives Tpull = (m1 + m2 )a, or a = (21.0 N)/(1.0 kg + 2.0 kg) = 7.0 m/s 2 . The static friction
force needed to accelerate the top block at 7.0 m/s2 is
fs m1a = (1.0 kg)(7.0 m/s 2 ) = 7.0 N
To find the maximum possible static friction force ( fs ) max = μs n1, the y-equation of Newton’s second law for the top
block shows that n1 = m1g . Thus
( fs ) max = μs m1g = (0.50)(1.0 kg)(9.8 m/s 2 ) = 4.9 N
Because 7.0 N > 4.9 N, static friction is not sufficient to accelerate the top block, so it slides.
7.52. A 1.0 kg wood block is placed behind a 2.0 kg wood block on a horizontal table. The coefficients of kinetic
friction with the table are 0.30 for the 1.0 kg block and 0.50 for the 2.0 kg block. The 1.0 kg block is pushed forward,
against the 2.0 block, and released with a speed of 2.0 m/s. How far do the blocks travel before stopping?
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Newton’s Third Law
7-41
Visualize:
Solve: The 2.0 kg block in front has a larger coefficient of friction. Thus the 1.0 kg block pushes against the rear of
the 2.0 kg block and, in reaction, the 2.0 kg block pushes back against the 1.0 kg block. There’s no vertical
acceleration, so n1 = m1g and n2 = m2 g , leading to f1 = μ1m1g and f 2 = μ2 m2 g . Applying Newton’s second law
along the x-axis gives
1 kg block: ∑( Fon 1 ) x = − F2 on 1 − f1 = − F2 on 1 − μ1m1g = m1a
2 kg block: ∑ ( Fon 2 ) x = F1 on 2 − f 2 = F2 on 1 − μ 2 m2 g = m2 a
where we used a1 = a2 = a. Also, F1 on 2 = F2 on 1 because they are an third-law action/reaction pair. Adding these
two equations gives
−( μ1m1 + μ2 m2 ) g = (m1 + m2 )a
μ1m1 + μ2m2
(0.30)(1.0 kg) + (0.50)(2.0 kg)
(9.8 m/s 2 ) = −4.25 m/s 2
m1 + m2
1.0 kg + 2.0 kg
We can now use constant-acceleration kinematics to find
v2
(2.0 m/s)2
v12x = 0 = v02x + 2a( x1 − x0 ) ⇒ x1 = − 0 x = −
= 0.47 m
2a
2( −4.25 m/s 2 )
a=−
g =−
7.53. Model: Treat the ball of clay and the block as particles.
Visualize:
G
G
Solve: (a) Forces FC on B and FB on C are an action/reaction pair, so FB on C = FC on B . Note that aB ≠ aC because
the clay is decelerating while the block is accelerating. Newton’s second law applied in the x-direction gives
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7-42
Chapter 7
Clay: ∑( Fon C ) x = − FB on C = mC aC
Block: ∑( Fon B ) x = FC on B = FB on C = mBaB
Equating the two expressions for FB on C gives
aC = −
mB
aB
mC
Turning to kinematics, the velocity of each after Δt is
(vC )1 = (vC )0 + aC Δt
(vB )1 = (vB )0 + aBΔt = aBΔt
But (vC )1 = (vB )1 because the clay and the block are moving together after Δt has elapsed. Equating these two
expressions gives (vC )0 + aC Δt = aBΔt , from which we find
aC = aB −
(vC )0
Δt
We can now equate the two expressions for aC :
−
mB
(v )
aB = aB − C 0
mC
Δt
⇒ aB =
(vC )0 /Δt (10 m/s)/(0.010 s)
=
= 100 m/s 2
1 + mB /mC 1 + (900 g)/100 g
Then aC = −9aB = −900 m/s 2. With the acceleration now known, we can use either kinematic equation to find
(vC )1 = (vB )1 = (100 m/s 2 )(0.010 s) = 1.0 m/s
(b) FC on B = mBaB = (0.90 kg)(100 m/s 2 ) = 90 N.
(c) FB on C = mC aC = (0.10 kg)(900 m/s 2 ) = 90 N.
Assess: The two forces are of equal magnitude, as expected from Newton’s third law.
7.54. Model: Use the particle model for the two blocks. Assume a massless rope and massless, frictionless pulleys.
Visualize:
Note that for every meter block 1 moves forward, one meter is provided to block 2. So each rope on m2 has to be
lengthened by one-half meter. Thus, the acceleration constraint is a2 = − 12 a1.
Solve: Newton’s second law applied to block 1 gives T = m1a1. Newton’s second law applied to block 2 gives
2T − ( FG ) 2 = m2 a2 . Combining these two equations gives
2( m1a1 ) − m2 g = m2 (− 12 a1 ) ⇒ a1 (4m1 + m2 ) = 2m2 g
⇒ a1 =
2m2 g
4m1 + m2
where we have used a2 = − 12 a1.
Assess: If m1 = 0 kg, then a2 = − g . This is what is expected for a freely falling object.
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Newton’s Third Law
7-43
7.55. Model: Use the particle model for the two blocks. Assume a massless rope and massless, frictionless pulleys.
Visualize:
For every one meter that the 1.0-kg block goes down, each rope on the 2.0-kg block will be shortened by one-half
meter. Thus, the acceleration constraint is a1 = −2a2 .
Solve: Newton’s second law applied to the two blocks gives
2T = m2 a2 T − ( FG )1 = m1a1
Since a1 = −2a2 , the above equations become
2T = m2a2 T − m1g = m1( −2a2 )
a
m2 2 + m1(2a2 ) = m1g
2
a2 =
2m1g
2(1.0 kg)(9.8 m/s 2 )
=
= 3.3 m/s 2
(2.0 kg + 4.0 kg)
m2 + 4m1
Assess: If m1 = 0 kg, then a2 = 0 m/s 2 , which is expected.
7.56. Model: The hamster of mass m and the wedge with mass M will be treated as objects 1 and 2, respectively.
They will be treated as particles.
Visualize:
The scale is denoted by the letter s.
G
Solve: (a) The reading of the scale is the magnitude of the force n2 that the scale exerts upward. There are two
G
action/reaction pairs. Initially the hamster of mass m is stuck in place and is in static equilibrium with Fnet = 0 N.
G
Because of the shape of the blocks, it is not clear whether the scale has to exert a horizontal friction force fs on 2 to
prevent horizontal motion. We’ve included one just in case. Newton’s second law for the hamster is
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7-44
Chapter 7
( Fnet on 1) x = mg sin θ − f 2 on 1 = 0 N ⇒
f 2 on 1 = mg sin θ
( Fnet on 1) y = n1 − mg cosθ = 0 N ⇒ n1 = mg cosθ
For the wedge, we see from Newton’s third law that n1′ = n1 = mg cosθ and that f 2 on 1 = f1 on 2 = mg sin θ . Using
these equations, Newton’s second law for the wedge is
( Fnet on 2 ) x = f1 on 2 cosθ + fs on 2 − n1′ sin θ = mg sin θ cosθ + fs on 2 − mg cosθ sin θ = 0 N ⇒
fs on 2 = 0 N
( Fnet on 2 ) y = n2 − n1′ cosθ − f1 on 2 sin θ − Mg = n2 − mg cos 2 θ − mg sin 2 θ − Mg = 0 N
n2 = mg (cos 2 θ + sin 2 θ ) + Mg = ( M + m) g = (0.800 kg + 0.200 kg)(9.8 m/s 2 ) = 9.8 N
First we find that fs on 2 = 0 N, so no horizontal static friction is needed to prevent motion. More interesting, the
scale reading is ( M + m) g which is the total gravitational force resting on the scale. This is the expected result.
(b) Now suppose that the hamster is accelerating down the wedge. The total mass is still M + m, but is the reading
still ( M + m) g ? The frictional forces between the systems 1 and 2 have now vanished, and system 1 now has an
acceleration. However, the acceleration is along the hamster’s x-axis, so a1 y = 0 m/s 2 . The hamster’s y-equation is still
( Fnet on 1 ) y = n1 − mg cosθ = 0 N ⇒ n1 = mg cosθ
We still have n1′ = n1 = mg cosθ , so the y-equation for block 2 (with a2 y = 0 m/s 2 ) is
( Fnet on 2 ) y = n2 − n1′ cosθ − Mg = n2 − mg cos 2 θ − Mg = 0 N
n2 = mg cos 2 θ + Mg = ( M + m cos 2 θ ) g = 9.0 N
Assess: The scale reads less than it did when the hamster was at rest. This makes sense if you consider the limit
θ → 90°, in which case cosθ → 0. If the face of the wedge is vertical, then the hamster is simply in free fall and can
have no effect on the scale (at least until impact!). So for θ = 90° we expect the scale to record Mg only, and that is
indeed what the expression for n2 gives.
7.57. Model: The hanging masses m1, m2 , and m3 are modeled as particles. Pulleys A and B are massless and
frictionless. The strings are massless.
Visualize:
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Newton’s Third Law
7-45
Solve: (a) The length of the string over pulley B is constant. Therefore,
( yB − y3 ) + ( yB − yA ) = LB ⇒ yA = 2 yB − y3 − LB
The length of the string over pulley A is constant. Thus,
( yA − y2 ) + ( yA − y1 ) = LA = 2 yA − y1 − y2
2(2 yB − y3 − LB ) − y1 − y2 = LA ⇒ 2 y3 + y2 + y1 = constant
This constraint implies that
dy
dy
dy
2 3 + 2 + 1 = 0 m/s = 2v3 y + v2 y + v1 y
dt
dt
dt
Also by differentiation, 2a3 y + a2 y + a1 y = 0 m/s 2 .
(b) Applying Newton’s second law to the masses m3 , m2 , m1, and pulley A gives
TB − m3 g = m3a3 y
TA − m2 g = m2a2 y
TA − m1g = m1a1 y
TB − 2TA = 0 N
The pulley equation is zero because the pulley is massless. These four equations plus the acceleration constraint
consitute five equations with five unknowns (two tensions and three accelerations). To solve for TA , multiply the
m3 equation by 2, substitute 2TB = 4TA , then divide each of the mass equations by the mass. This gives the three
equations
4TA /m3 − 2 g = 2a3 y
TA /m2 − g = a2 y
TA /m1 − g = a1 y
If these three equations are added, the right side adds to zero because of the acceleration constraint. Thus
4g
(4/m3 + 1/m2 + 1/m2 )TA − 4 g = 0 ⇒ TA =
(4/m3 + 1/m2 + 1/m2 )
(c) Using numerical values, we find TA = 18.97 N. Then
a1 y = TA /m1 − g = −2.2 m/s 2
a2 y = TA /m2 − g = 2.9 m/s 2
a3 y = 2TA /m3 − g = −0.32 m/s 2
(d) m3 = m1 + m2 , so it appears at first as if m3 should hang in equilibrium. For this to be so, tension TB would need to
equal m3 g. However, TB is not (m1 + m2 ) g because masses m1 and m2 are accelerating rather than hanging at rest.
Consequently, tension TB is not able to balance the weight of m3 .
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DYNAMICS II: MOTION IN A PLANE
8
Conceptual Questions
8.1. For an object in uniform circular motion the speed is constant (that’s what uniform means), the angular velocity
is constant, and the magnitude of the net force is constant. The velocity and centripetal acceleration are both vectors
whose magnitude is constant but whose direction changes.
8.2. The free-body diagram (a) is correct. The forces acting on the car at the bottom of the hill are the downward
gravitational force and an upward normal force. The car can be considered to be in circular motion about a point
above the bottom center of the valley, which requires a net force toward the center of the circle. In this case, the
circle center is above the car, so the normal force is greater than the gravitational force.
8.3. Tc > Ta = Td > Tb . Use T =
For (d), Td =
mv 2
mv 2
mv 2 1
(2m)v 2
. For (a), Ta =
. For (b), Tb =
= Ta . For (c), Tc =
= 2Ta .
r
r
2r
2
r
(2m)v 2
= Ta .
2r
8.4. The tension in the vine at the lowest part of Tarzan’s swing is greater than the gravitational force on Tarzan. If
Tarzan is at rest on the vine, just hanging, the tension in the vine is equal to the gravitational force on Tarzan. But
when Tarzan is swinging he is in circular motion, with the center of the circle at the top end of the vine, and the vine
must provide the additional centripetal force necessary to move him in a circle.
mv 2
. Since the
r
velocity v is the same for both, the greater radius for B means that the tension in case A is greater than for case B.
(b) In this case, we use ( Fr ) net = mω 2r . The angular velocity ω is the same for both A and B, so the larger radius for
8.5. (a) The difference in the tension between A and B is due to the centripetal force ( Fr )net =
B means that the tension is case A is less than the tension for case B.
8.6. Neither Sally nor Raymond is completely correct. Both have partially correct descriptions, but are missing key
points. In order to speed up, there must be a nonzero acceleration parallel to the track. In order to move in a circle,
there must be a nonzero centripetal acceleration. Since both of these are required, the net force points somewhere
between the forward direction (parallel to the track) and the center of the circle.
8.7. (a) The plane is in dynamic equilibrium, so the net force on the plane is zero.
(b) The vertical forces cancel, and so do the horizontal forces, so the net force is zero. The plane is traveling in the
positive x direction.
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8-1
8-2
Chapter 8
(c) As seen from behind, with the velocity and positive x direction into the page.
(d) The net force must be toward the center of the circle during a turn at constant speed and altitude. Note that
the radial component of the lift force provides the centripetal force while the vertical component balances the
gravitational force. The velocity is into the page.
8.8. Yes, the bug is weightless because it, like the projectile it is riding in, is in free fall around the planet.
8.9. When the gravitational force on the ball is greater than the required centripetal force
mv 2
, the ball is no longer
r
in circular motion. As the figure shows, at the top of the circle the net force on the ball is ( Fr ) net = FG + T =
When the string goes slack, T = 0, leaving FG =
mv 2
.
r
mv 2
. If the velocity is not high enough to make this equality true,
r
mv 2
, and the ball begins to fall downward since the net force
r
downward is greater than the centripetal force required for circular motion.
the equation above becomes an inequality, FG >
8.10.
The golfer is swinging the club in circular motion. The club is speeding up as it swings. This motion requires a linear
acceleration in the direction of motion of the club to speed it up and a centripetal acceleration to maintain circular motion.
The vector sum yields a total acceleration pointing approximately toward the golfer’s feet (c), as shown in the figure.
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Dynamics II: Motion in a Plane
8-3
Exercises and Problems
Section 8.1 Dynamics in Two Dimensions
8.1. Model: The model rocket and the target will be treated as particles. The kinematics equations in two
dimensions apply.
Visualize:
Solve: For the rocket, Newton’s second law along the y-direction is
( Fnet ) y = FR − mg = maR
⇒ aR =
1
1
( FR − mg ) =
[15 N − (0.8 kg)(9.8 m/s 2 )] = 8.95 m/s 2
m
0.8 kg
Using the kinematic equation y1R = y0R + (v0R ) y (t1R − t0R ) + 12 aR (t1R − t0R ) 2 ,
30 m = 0 m + 0 m + 12 (8.95 m/s 2 )(t1R − 0 s)2 ⇒ t1R = 2.589 s
For the target (noting t1T = t1R ),
x1T = x0T + (v0T ) x (t1T − t0T ) + 12 aT (t1T − t0T ) 2 = 0 m + (15 m/s)(2.589 s − 0 s) + 0 m = 39 m
You should launch when the target is 39 m away.
Assess: The rocket is to be fired when the target is at x0T . For a net acceleration of approximately 9 m/s 2 in the
vertical direction and a time of 2.6 s to cover a vertical distance of 30 m, a horizontal distance of 39 m is reasonable.
8.2. Model: The model rocket will be treated as a particle. Kinematic equations in two dimensions apply.
Air resistance is neglected.
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8-4
Chapter 8
Visualize:
The horizontal velocity of the rocket is equal to the speed of the car, which is 3.0 m/s.
Solve: For the rocket, Newton’s second law along the y-direction is:
1
[(8.0 N) − (0.5 kg)(9.8 m/s 2 )] = 6.2 m/s 2
( Fnet ) y = FR − mg = maR ⇒ a y =
0.5 kg
Thus using y1 = y0 + (v0 ) y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ,
(20 m) = 0 m + 0 m + 12 (6.2 m/s 2 )(t1R − 0 s) 2 ⇒ (20 m) = (3.1 m/s 2 )t12 ⇒ t1 = 2.54 s
Since t1 is also the time for the rocket to move horizontally up to the hoop,
x1 = x0 + (v0 ) x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (3.0 m/s)(2.54 s − 0 s) + 0 m = 7.6 m
Assess: In view of the rocket’s horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the above-obtained value
for the horizontal distance is reasonable.
8.3. Model: The asteroid and the giant rocket will be treated as particles undergoing motion according to the
constant-acceleration equations of kinematics.
Visualize:
Solve: (a) The time it will take the asteroid to reach the earth is
displacement 4.0 × 106 km
=
= 2.0 × 105 s = 56 h
velocity
20 km/s
(b) The angle of a line that just misses the earth is
⎛ 6400 km ⎞
⎛R⎞
R
⇒ θ = tan −1 ⎜ ⎟ = tan −1 ⎜⎜
tanθ =
⎟⎟ = 0.092°
6
y0
⎝ y0 ⎠
⎝ 4.0 × 10 km ⎠
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Dynamics II: Motion in a Plane
8-5
(c) When the rocket is fired, the horizontal acceleration of the asteroid is
5.0 × 109 N
ax =
= 0.125 m/s 2
10
4.0 × 10 kg
(Note that the mass of the rocket is much smaller than the mass of the asteroid and can therefore be ignored
completely.) The velocity of the asteroid after the rocket has been fired for 300 s is
vx = v0 x + a x (t − t0 ) = 0 m/s + (0.125 m/s 2 )(300 s − 0 s) = 37.5 m/s
After 300 s, the vertical velocity is v y = 2 × 104 m/s and the horizontal velocity is vx = 37.5 m/s. The deflection due
to this horizontal velocity is
tan θ =
vx
⎛ 37.5 m/s ⎞
⇒ θ = tan21 ⎜
⎟ = 0.107°
vy
⎝ 2 × 104 m/s ⎠
That is, the earth is saved.
Section 8.2 Uniform Circular Motion
8.4. Model: We are using the particle model for the car in uniform circular motion on a flat circular track. There
must be friction between the tires and the road for the car to move in a circle.
Visualize:
Solve: The centripetal acceleration is
v 2 (25 m/s) 2
=
= 6.25 m/s 2
r
100 m
The acceleration points to the center of the circle, so the net force is
G
G
Fr = ma = (1500 kg)(6.25 m/s 2, toward center)
ar =
= (9380 N, toward center) ≈ (9400 N, toward center)
This force is provided by static friction
fs = Fr = 9.4 kN
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8-6
Chapter 8
8.5. Model: We will use the particle model for the car which is in uniform circular motion.
Visualize:
Solve: The centripetal acceleration of the car is
ar =
v 2 (15 m/s)2
=
= 4.5 m/s 2
r
50 m
The acceleration is due to the force of static friction. The force of friction is fs = mar = (1500 kg)(4.5 m/s 2 ) =
6750 N = 6.8 kN.
Assess: The model of static friction is ( fs ) max = nμs = mg μs ≈ mg ≈ 15,000 N since μs ≈ 1 for a dry road surface.
We see that fs < ( fs ) max , which is reasonable.
8.6. Model: Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless
table.
Visualize:
Solve: (a) The angular velocity and speed are
rev 2π rad
1min
×
= 471.2 rad/min vt = rω = (0.50 m)(471.2 rad/min) ×
= 3.93 m/s
ω = 75
min 1 rev
60 s
The tangential velocity is 3.9 m/s.
(b) The radial component of Newton’s second law is
mv 2
∑ Fr = T =
r
Thus
(3.93 m/s) 2
T = (0.20 kg)
= 6.2 N
0.50 m
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Dynamics II: Motion in a Plane
8-7
8.7. Solve: Newton’s second law is Fr = mar = mrω 2 . Substituting into this equation yields:
ω=
Fr
8.2 × 1028 N
=
2
mr
(9.1 × 10 31 kg)(5.3 × 10211 m)
rad 1 rev
×
= 6.6 × 1015 rev/s
s 2π rad
Assess: This is a very high number of revolutions per second.
= 4.37 × 1016 rad/s = 4.37 × 1016
8.8. Model: The vehicle is to be treated as a particle in uniform circular motion.
Visualize:
On a banked road, the normal force on a vehicle has a horizontal component that provides the necessary centripetal
acceleration. The vertical component of the normal force balances the gravitational force.
Solve: From the physical representation of the forces in the r-z plane, Newton’s second law can be written
mv 2
∑ Fz = n cosθ − mg = 0 ⇒ n cosθ = mg
∑ Fr = n sin θ =
r
Dividing the two equations and making the conversion 90 km/h = 25 m/s yields:
v2
(25 m/s) 2
=
= 0.128 ⇒ θ = 7.3°
rg (9.8 m/s 2 )500 m
Assess: Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our experience as well.
tan θ =
8.9. Model: The motion of the moon around the earth will be treated through the particle model. The circular
motion is uniform.
Visualize:
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8-8
Chapter 8
Solve: The tension in the cable provides the centripetal acceleration. Newton’s second law is
⎛ 2π ⎞
∑ Fr = T = mrω = mr ⎜
⎟
⎝ Tmoon ⎠
2
2
2
⎡ 2π
1 day
1h ⎤
20
= (7.36 × 1022 kg)(3.84 × 108 m) ⎢
×
×
⎥ = 2.01 × 10 N
⎣ 27.3 days 24 h 3600 s ⎦
Assess: This is a tremendous tension, but clearly understandable in view of the moon’s large mass and the large
radius of circular motion around the earth. This is the same answer we’ll get later with Newton’s law of universal
gravitation.
8.10. Model: Model the person as a particle in uniform circular motion.
Visualize:
Solve: The only force acting on the passengers is the normal force of the wall. Newton’s second law along the
r-axis is:
∑ Fr = n = mrω 2
To create “normal” gravity, the normal force by the inside surface of the space station equals mg. Therefore,
g
r
2π
500 m
mg = mrω 2 ⇒ ω =
=
⇒ T = 2π
= 2π
= 45 s
T
r
g
9.8 m/s 2
Assess: This is a fast rotation. The tangential speed is
2π r 2π (500 m)
=
= 70 m/s ≈ 140 mph
v=
45 s
T
Section 8.3 Circular Orbits
8.11. Model: The satellite is considered to be a particle in uniform circular motion around the moon.
Visualize:
Solve: The radius of the moon is 1.738 × 106 m and the satellite’s distance from the center of the moon is the same
quantity. The angular velocity of the satellite is
2π
2π rad 1min
=
×
= 9.52 × 10−4 rad/s
ω=
T 110 min 60 s
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Dynamics II: Motion in a Plane
8-9
and the centripetal acceleration is
ar = rω 2 = (1.738 × 106 m)(9.52 × 10−4 rad/s)2 = 1.58 m/s 2
The acceleration of a body in orbit is the local “g” experienced by that body.
8.12. Model: The earth is considered to be a particle in uniform circular motion around the sun.
Solve: The earth orbits the sun in 365 days and is 1.5 × 1011 m from the sun. The angular velocity and centripetal
acceleration are
2π rad 1 day
1h
ω=
×
×
= 2.0 × 10−7 rad/s
365 days 24 h 3600 s
ar = g = rω 2 = (1.5 × 1011 m)(2.0 × 10−7 rad/s) 2 = 6.0 × 10−3 m/s 2
Assess: The smallness of this acceleration due to gravity is essentially due to the large earth-sun distance.
Section 8.4 Fictitious Forces
8.13. Model: Use the particle model for the car which is undergoing circular motion.
Visualize:
Solve: The car is in circular motion with the center of the circle below the car. Newton’s second law at the top of the
hill is
mv 2
n⎞
⎛
∑ Fr = ( FG ) r − nr = mg − n = mar =
⇒ v2 = r ⎜ g − ⎟
m⎠
r
⎝
Maximum speed is reached when n = 0 and the car is beginning to lose contact with the road.
vmax = rg = (50 m)(9.8 m/s 2 ) = 22 m/s
Assess: A speed of 22 m/s is equivalent to 49 mph, which seems like a reasonable value.
8.14. Model: The passengers are particles in circular motion.
Visualize:
Solve: The center of the circle of motion of the passengers is directly above them. There must be a net force pointing up
that provides the needed centripetal acceleration. The normal force on the passengers is their weight. Ordinarily their
weight is FG , so if their weight increases by 50%, n = 1.5 FG . Newton’s second law at the bottom of the dip is
∑ Fr = n − FG = (1.5 − 1) FG = 0.5mg =
mv 2
r
⇒ v = 0.5 gr = 0.5(9.8 m/s 2 )(30 m) = 12 m/s
Assess: A speed of 12.1 m/s is 27 mph, which seems very reasonable.
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8-10
Chapter 8
8.15. Model: Model the roller coaster car as a particle at the top of a circular loop-the-loop undergoing uniform
circular motion.
Visualize:
Notice that the r-axis points downward, toward the center of the circle.
G
K
Solve: The critical speed occurs when n goes to zero and FG provides all the centripetal force pulling the car in the
vertical circle. At the critical speed mg = mvc2 /r , therefore vc = rg . Since the car’s speed is twice the critical speed,
vt = 2vc and the centripetal force is
mv 2 m(4vc2 ) m(4rg )
=
=
= 4mg
r
r
r
Thus the normal force is n = 3 mg . Consequently, n/FG = 3.
∑ Fr = n + FG =
8.16. Model: Model the roller coaster car as a particle undergoing uniform circular motion along a loop.
Visualize:
Notice that the r-axis points downward, toward the center of the circle.
Solve: In this problem the normal force is equal to the gravitational force: n = FG = mg . We have
∑ Fr = n + FG =
mv 2
= mg + mg ⇒ v = 2rg = 2(20 m)(9.8 m/s 2 ) = 19.8 m/s ≈ 20 m/s
r
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Dynamics II: Motion in a Plane
8-11
8.17. Model: Model the bucket of water as a particle in uniform circular motion.
Visualize:
Solve: Let us say the distance from the bucket handle to the top of the water in the bucket is 35 cm. This makes the
shoulder to water distance 65 cm + 35 cm = 1.00 m. The minimum angular velocity for swinging a bucket of water in
a vertical circle without spilling any water corresponds to the case when the speed of the bucket is critical. In this
case, n = 0 N when the bucket is in the top position of the circular motion. We get
∑ Fr = n + FG = 0 N + mg =
⇒ ωc = g/r =
mvc 2
= mrωc2
r
9.8 m/s 2
1 rev
60 s
= 3.13 rad/s = 3.13 rad/s ×
×
= 30 rpm
1.00 m
2π rad 1 min
8.18. Model: Use the particle model for yourself while in uniform circular motion.
Visualize:
Solve: (a) The speed and acceleration are
2π r 2π (15 m)
v 2 (3.77 m/s) 2
v=
=
= 3.77 m/s ar =
=
= 0.95 m/s 2
T
25 s
r
15 m
So the speed is 3.8 m/s and the centripetal acceleration is 0.95 m/s 2 .
(b) The weight w = m, the normal force. On the ground, your weight is the same as the gravitational force FG .
Newton’s second law at the top is
∑ Fr = FG − n = mar =
mv 2
r
⎛
⎛
v2 ⎞
(3.77 m/s)2 ⎞
2
⇒ n = w = m ⎜ g − ⎟ = m ⎜ 9.8 m/s 2 −
⎟⎟ = m(8.85 m/s )
⎜
⎜
r ⎟⎠
15 m
⎝
⎝
⎠
⇒
w 8.85 m/s 2
=
= 0.90
FG 9.8 m/s 2
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8-12
Chapter 8
(c) Newton’s second law at the bottom is
∑ Fr = n − FG = mar =
mv 2
r
⎛
⎛
v2 ⎞
(3.77 m/s 2 ) ⎞
2
⇒ n = w = m ⎜ g + ⎟ = m ⎜ 9.8 m/s 2 +
⎟⎟ = (10.75 m/s ) m
⎜
⎟
⎜
r
15
m
⎝
⎠
⎝
⎠
⇒
w 10.75 m/s 2
=
= 1.1
FG
9.8 m/s 2
Section 8.5 Nonuniform Circular Motion
8.19. Model: Use the particle model for the car, which is undergoing nonuniform circular motion.
Visualize:
d
= 100 m. We require
2
Solve: The car is in circular motion with radius r =
ar = ω 2r = 1.5 m/s 2 ⇒ ω =
1.5 m/s 2
1.5 m/s 2
=
= 0.122 s −1
100 m
r
The definition of the angular velocity can be used to determine the time Δt using the angular acceleration
α=
at 1.5 m/s 2
=
= 1.5 × 10−2 s −2 .
100 m
r
ω = ω i + αΔt
⇒ Δt =
ω − ωi
α
=
0.122 s −1 − 0 s −1
0.015 s −2
= 8.2 s
8.20. Model: The train is a particle undergoing nonuniform circular motion.
Visualize:
Solve: (a) Newton’s second law in the vertical direction is
( Fnet ) y = n − FG = 0
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Dynamics II: Motion in a Plane
8-13
from which n = mg . The rolling friction is f R = μ R n = μ R mg . This force provides the tangential acceleration
at = −
fR
= − μR g
m
The angular acceleration is
α=
at − μ R g −(0.10)(9.8 m/s 2 )
=
=
= −1.96 rad/s 2
0.50 m
r
r
The magnitude is 2.0 rad/s 2 .
⎛ rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞
(b) The initial angular velocity is 30 ⎜
⎟⎜
⎟⎜
⎟ = 3.14 rad/s. The time to come to a stop due to the
⎝ min ⎠⎝ 60 sec ⎠⎝ rev ⎠
rolling friction is
ω − ωi 0 − 3.14 rad/s
Δt = f
=
= 1.6 s
α
−1.96 rad/s 2
Assess: The original angular speed of π rad/s means the train goes around the track one time every 2 seconds, so a
stopping time of less than 2 s is reasonable.
Problems
8.21. Model: The object is treated as a particle in the model of kinetic friction with its motion governed by
constant-acceleration kinematics.
Visualize:
Solve: The velocity v1x as the object sails off the edge is related to the initial velocity v0 x by v12x = v02x + 2a x ( x1 − x0 ).
Using Newton’s second law to determine a x while sliding gives
∑ Fx = − f k = ma x ⇒ ∑ Fy = n − mg = 0 N ⇒ n = mg
Using this result and the model of kinetic friction ( f k = μk n), the x-component equation can be written as
− μ k mg = ma x . This implies
a x = − μ k g = −(0.50)(9.8 m/s 2 ) = −4.9 m/s 2
Kinematic equations for the object’s free fall can be used to determine v1x :
g
(t2 − t1) 2 ⇒ (t2 − t1 ) = 0.4518 s
2
x2 = x1 + v1x (t2 − t1 ) = 2.30 m = 2.0 m + v1x (0.4518 s) ⇒ v1x = 0.664 m/s
y2 = y1 + v1 y (t2 − t1 ) + 12 ( − g )(t2 − t1 )2 ⇒ 0 m = 1.0 m + 0 m −
Having determined v1x and a x , we can go back to the velocity equation v12x = v02x + 2a x ( x1 − x0 ):
(0.664 m/s)2 = v02x + 2( −4.9 m/s 2 )(2.0 m) ⇒ v0 x = 4.5 m/s
Assess: v0 x = 4.5 m/s is about 10 mph and is a reasonable speed.
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8-14
Chapter 8
8.22. Model: Treat the motorcycle and rider as a particle.
Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for
the second. The motorcycle’s final velocity at the top of the ramp is its initial velocity as it becomes airborne.
Solve: The motorcycle’s acceleration on the ramp is given by Newton’s second law:
(Fnet ) s = − f r − mgsin20° = − μ r n − mgsin20° = − μ r mgcos20° − mgsin20° = ma0
a0 = − g (μr cos20° + sin20°) = −(9.8 m/s 2 )((0.02)cos20° + sin20°) = −3.536 m/s 2
The length of the ramp is s1 = (2.0 m)/sin 20° = 5.85 m. We can use kinematics to find its speed at the top of the
ramp:
v12 = v02 + 2a0 (s1 − s0 ) = v02 + 2a0 s1
⇒ v1 = (11.0 m/s) 2 + 2( − 3.536 m/s 2 )(5.85 m) = 8.92 m/s
This is the motorcycle’s initial speed into the air, with velocity components v1x = v1cos20° = 8.38 m/s and
v1y = v1sin20° = 3.05 m/s. We can use the y-equation of projectile motion to find the time in the air:
y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 2.0 m + (3.05 m/s)t2 − (4.90 m/s 2 )t22
This quadratic equation has roots t2 = −0.399 s (unphysical) and t2 = 1.021 s. The x-equation of motion is thus
x2 = x1 + v1xt2 = 0 m + (8.38 m/s)t2 = 8.56 m
8.56 m < 10.0 m, so it looks like crocodile food.
8.23. Model: Treat Sam as a particle.
Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for
the second. Sam’s final velocity at the top of the slope is his initial velocity as he becomes airborne.
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Dynamics II: Motion in a Plane
8-15
Solve: Sam’s acceleration up the slope is given by Newton’s second law:
( Fnet ) s = F − mg sin10° = ma0
a0 =
F
200 N
− g sin10° =
− (9.8 m/s 2 )sin10° = 0.965 m/s 2
m
75 kg
The length of the slope is s1 = (50 m)/ sin10° = 288 m. His velocity at the top of the slope is
v12 = v02 + 2a0 ( s1 − s0 ) = 2a0 s1 ⇒ v1 = 2(0.965 m/s 2 )(288 m) = 23.6 m/s
This is Sam’s initial speed into the air, giving him velocity components v1x = v1 cos10° = 23.2 m/s and
v1 y = v1 sin10° = 410 m/s. This is not projectile motion because Sam experiences both the force of gravity and the
thrust of his skis. Newton’s second law for Sam’s acceleration is
(F )
(200 N)cos10°
a1x = net x =
= 2.63 m/s 2
m
75 kg
a1 y =
( Fnet ) y
m
=
(200 N)sin10° − (75 kg)(9.80 m/s 2 )
= −9.34 m/s 2
75 kg
The y-equation of motion allows us to find out how long it takes Sam to reach the ground:
y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 50 m + (4.10 m/s)t2 − (4.67 m/s2 )t22
This quadratic equation has roots t2 = −2.86 s (unphysical) and t2 = 3.74 s. The x-equation of motion—this time
with an acceleration—is
x2 = x1 + v1xt2 + 12 a1xt22 = 0 m + (23.2 m/s)t2 − 12 (2.63 m/s 2 )t22 = 105 m
Sam lands 105 m from the base of the cliff.
8.24. Visualize:
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8-16
Chapter 8
G ⎛1
⎞
Solve: From Chapter 6 the drag on a projectile is D = ⎜ Av 2 , direction opposite to motion ⎟ , where A is the cross4
⎝
⎠
sectional area. Using the free-body diagram above, apply Newton’s second law to each direction.
In the x-direction,
ax =
ay =
Since v =
vx2
+ v 2y ,
1 Av 2 cosθ
( Fnet ) x
D cosθ
=2
=2 4
m
n
m
( Fnet ) y
m
=2
1 Av 2 sin θ
D sin θ − FG
=2 4
−g
m
m
vx = v cosθ , and v y = v sin θ , we can rewrite these as
ax = 2
Avx vx2 + v 2y
A(v cosθ )v
=2
4m
4m
ay = 2
Av y vx2 + v 2y
A(v sin θ )v
− g = −g −
4m
4m
8.25. Model: Use the particle model and the constant-acceleration equations of kinematics for the rocket.
G
G
Solve: (a) The acceleration of the rocket in the launch direction is obtained from Newton’s second law F = ma:
140,700 N = (5000 kg) a ⇒ a = 28.14 m/s 2
Therefore, a x = a cos 44.7° = 20.0 m/s 2 and a y = a sin 44.7° = 19.8 m/s 2 . The net acceleration in the y-direction is thus
(anet ) y = a y − g = (19.8 − 9.8) m/s 2 = 10.0 m/s 2
With this acceleration, we can write the equations for the x- and y-motions of the rocket.
y = y0 + v0 y (t − t0 ) + 12 (anet ) y (t − t0 ) 2 = 0 m + 0 m + 12 (10.0 m/s 2 )t 2 = (5.00 m/s 2 )t 2
x = x0 + v0 x (t − t0 ) + 12 ( anet ) x (t − t0 )2 = 0 m + 0 m + 12 (20.0 m/s 2 )t 2 = (10.0 m/s 2 )t 2
From these two equations,
x (10.0 m/s 2 )t 2
=
=2
y (5.00 m/s 2 )t 2
The equation that describes the rocket’s trajectory is y = 12 x.
(b) It is a straight line with a slope of
1.
2
(c) In general,
v y = v0y + (anet ) y (t1 − t0 ) = 0 + (10.0 m/s 2 )t1
vx = v0x + (anet ) x (t1 − t0 ) = 0 + (20.0 m/s 2 )t1
v = (10.0 m/s 2 ) 2 t12 + (20.0 m/s 2 ) 2 t12 = (22.36 m/s 2 )t1
The time required to reach the speed of sound is calculated as follows:
330 m/s = (22.36 m/s 2 )t1 ⇒ t1 = 14.76 s
We can now obtain the elevation of the rocket. From the y-equation,
y = (5.00 m/s 2 )t12 = (5.00 m/s 2 )(14.76 s)2 = 1090 m
8.26. Model: The hockey puck will be treated as a particle whose motion is determined by constant-acceleration
kinematic equations. We break this problem in two parts, the first pertaining to motion on the table and the second to
free fall.
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Dynamics II: Motion in a Plane
8-17
Visualize:
Solve: Newton’s second law is:
Fx = ma x ⇒ a x =
Fx 2.0 N
=
= 2.0 m/s 2
m 1.0 kg
The kinematic equation v12x = v02x + 2a x ( x1 − x0 ) yields:
v12x = 0 m 2 /s 2 + 2(2.0 m/s 2 )(4.0 m) ⇒ v1x = 4.0 m/s
Let us now find the time of free fall (t2 − t1):
y2 = y1 + v1 y (t2 − t1) + 12 a1 y (t2 − t1 )2
⇒ 0 m = 2.0 m + 0 m + 12 (−9.8 m/s 2 )(t2 − t1 ) 2 ⇒ (t2 − t1 ) = 0.639 s
Having obtained v1x and (t2 − t1 ), we can now find ( x2 − x1 ) as follows:
x2 = x1 + v1x (t2 − t1 ) + 12 ax (t2 − t1 )2
⇒ x2 − x1 = (4.0 m/s)(0.639 s) + 12 (2.0 m/s 2 )(0.639 s) 2 = 3.0 m
Assess: For a modest horizontal thrust of 2.0 N, a landing distance of 3.0 m is reasonable.
8.27. Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration
kinematic equations.
Visualize:
Solve: As the rocket is accidentally bumped v0 x = 0.5 m/s and v0 y = 0 m/s. On the other hand, when the engine is fired
Fx = ma x ⇒ a x =
Fx
20 N
=
= 40 m/s 2
m 0.500 kg
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8-18
Chapter 8
(a) Using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 )2 ,
0 m = 40 m + 0 m + 12 ( −9.8 m/s 2 )t12 ⇒ t1 = 2.857 s
The distance from the base of the wall is
x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (0.5 m/s)(2.857s) + 12 (40 m/s 2 )(2.857 s)2 = 165 m
(b) The x- and y-equations are
y = y0 + v0 y (t − t0 ) + 12 a y (t − t0 ) 2 = 40 − 4.9t 2
x = x0 + v0 x (t − t0 ) + 12 a x (t − t0 ) 2 = 0.5t + 20t 2
Except for a brief interval near t = 0, 20t 2 0.5t . Thus x ≈ 20t 2 , or t 2 = x/20. Substituting this into the y-equation gives
y = 40 − 0.245 x
This is the equation of a straight line, so the rocket follows a linear trajectory to the ground.
8.28. Model: Model the plane as a particle with constant ax and constant a y .
Visualize: The plane is taking off toward the northwest as we can see by plotting the x - y data.
G
But we analyze the data in each direction separately and then apply a = a = a x2 + a 2y .
Solve: In each direction we apply the kinematic equation s (t ) = s0 + (v0 ) s t + 12 ast 2 . With v0 = 0 we can graph s vs.
t 2 and expect to get a straight line whose slope is
1a
2 s
and whose intercept is s0 .
From the x vs. t 2 graph we see that
1a
2 x
= −2.90 m/s 2 ⇒ ax = −5.8 m/s 2 .
From the y vs. t 2 graph we see that
1a
2 y
= 4.00 m/s 2 ⇒ a y = 8.0 m/s 2 .
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Dynamics II: Motion in a Plane
8-19
G
a = a = ax2 + a 2y = (−5.8 m/s 2 ) 2 + (8.0 m/s 2 ) 2 = 9.9 m/s 2
Assess: The intercepts of the best-fit lines are close to what the data table has.
8.29. Model: Assume the particle model for the satellite in circular motion.
Visualize:
To be in a geosynchronous orbit means rotating at the same rate as the earth, which is 24 hours for one complete rotation.
Because the altitude of the satellite is 3.58 × 107 m, r = 3.58 × 107 m, re = 3.58 × 107 m + 6.37 × 106 m = 4.22 × 107 m.
Solve: (a) The period (T) of the satellite is 24.0 hours.
(b) The acceleration due to gravity is
2
2
1 hr ⎞
⎛ 2π ⎞
⎛ 2π
7
2
g = ar = rω 2 = r ⎜
×
⎟ = (4.22 × 10 m) ⎜
⎟ = 0.223 m/s
⎝ T ⎠
⎝ 24.0 hr 3600 s ⎠
(c) There is no normal force on a satellite, so the weight is zero. It is in free fall.
8.30. Model: Treat the man as a particle. The man at the equator undergoes uniform circular motion as the earth
rotates.
Visualize:
Solve: The scale reads the man’s weight FG = n, the force of the scale pushing up against his feet. At the north pole,
where the man is in static equilibrium,
nP = FG = mg = 735 N
At the equator, there must be a net force toward the center of the earth to keep the man moving in a circle. The r-axis
points toward the center, so
∑ Fr = FG − nE = mω 2r ⇒ nE = mg − mω 2r = nP − mω 2 r
The equator scale reads less than the north pole scale by the amount mω 2r . The man’s angular velocity is that of the
equator, or
2π
2π rad
ω=
=
= 7.27 × 10−5 rad/s
24 hours × (3600 s/1 h)
T
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8-20
Chapter 8
Thus the north pole scale reads more than the equator scale by
Δw = (75 kg)(7.27 × 10−5 rad/s) 2 (6.37 × 106 m) = 2.5 N
Assess: The man at the equator appears to have lost Δm = Δw/g ≈ 0.25 kg, or the equivalent of ≈ 12 lb.
8.31. Model: Model the ball as a particle which is in a vertical circular motion.
Visualize:
Solve: At the bottom of the circle,
mv 2
(0.500 kg)v 2
∑ Fr = T − FG =
⇒ (15 N) − (0.500 kg)(9.8 m/s 2 ) =
⇒ v = 5.5 m/s
(1.5 m)
r
8.32. Model: We will use the particle model for the car, which is undergoing uniform circular motion on a banked
highway, and the model of static friction.
Visualize:
Note that we need to use the coefficient of static friction μs , which is 1.0 for rubber on concrete.
Solve: Newton’s second law for the car is
mv 2
∑ Fz = n cosθ − fs sin θ − FG = 0 N
r
Maximum speed is when the static friction force reaches its maximum value ( fs ) max = μs n. Then
∑ Fr = fs cosθ + n sin θ =
n( μs cos15° + sin15°) =
mv 2
n(cos15° − μs sin15°) = mg
r
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Dynamics II: Motion in a Plane
8-21
Dividing these two equations and simplifying, we get
μs + tan15° v 2
μ + tan15°
=
⇒ v = gr s
1 − μs tan15° gr
1 − μs tan15°
= (9.80 m/s 2 )(70 m)
(1.0 + 0.268)
= 34 m/s
(1 − 0.268)
Assess: The above value of 34 m/s ≈ 70 mph is reasonable.
8.33. Model: Use the particle model for the rock, which is undergoing uniform circular motion.
Visualize: L is the hypotenuse of the right triangle. The radius of the circular motion is r = Lcosθ .
Solve:
(a) Apply Newton’s second law in the z - and r -directions.
∑ Fz = T sinθ − mg = 0 ⇒ T =
mg
sinθ
∑ Fr = T cosθ = mω 2 r = mω 2 ( L cosθ ) ⇒ T = mω 2 L
Set the two expressions for T equal to each other and solve for ω.
mg
= mω 2 L ⇒ ω =
sinθ
g
L sinθ
(b) Insert L = 1.0 m and θ = 10°.
ω=
⎛ 1 rev ⎞⎛ 60 s ⎞
g
9.8 m/s 2
=
= 7.51 rad/s ⎜
⎟⎜
⎟ = 72 rpm
L sinθ
(1.0 m)sin10°
⎝ 2π rad ⎠⎝ 1 min ⎠
Assess: Notice that the mass canceled out of the equation so the 500 g was unnecessary information. In other words,
the answer, 72 rpm, would be the same regardless of the mass.
The dependencies of ω on g, L, and θ seem to be in the right directions.
8.34. Model: Use the particle model and static friction model for the coin, which is undergoing circular motion.
Visualize:
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8-22
Chapter 8
Solve: The force of static friction is fs = μs n = μs mg . This force is equivalent to the maximum centripetal force that
can be applied without sliding. That is,
vt 2
μs g
(0.80)(9.8 m/s 2 )
2
=
= 7.23 rad/s
) ⇒ ωmax =
= m(rωmax
r
r
0.15 m
rad 1 rev
60 s
= 7.23
×
×
= 69 rpm
s 2π rad 1 min
So, the coin will stay still on the turntable.
Assess: A rotational speed of approximately 1 rev per second for the coin to stay stationary seems reasonable.
μs mg = m
8.35. Model: Use the particle model for the car, which is in uniform circular motion.
Visualize:
Solve: Newton’s second law is
∑ Fr = T sin 20° = mar =
mv 2
∑ Fz = T cos 20° − FG = 0 N
r
These equations can be written as
T sin 20° =
mv 2
T cos 20° = mg
r
Dividing these two equations gives
tan 20° = v 2 /rg ⇒ v = rg tan 20° = (4.55 m)(9.8 m/s 2 ) tan 20° = 4.03 m/s ≈ 4 m/s
8.36. Use the particle model for the ball, which is undergoing uniform circular motion.
Visualize: We are given L, r, and m, so our answers must be in terms of those variables. L is the hypotenuse of the
right triangle. The ball moves in a horizontal circle of radius r = L cosθ . The acceleration and net force point toward
the center of the circle, not along the string.
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Dynamics II: Motion in a Plane
8-23
Solve:
(a) Apply Newton’s second law in the z -direction.
∑ Fz = T cosθ − mg = 0 ⇒ T =
mg
cosθ
From the right triangle cosθ = L2 − r 2 /L.
T=
mg
=
cosθ
mgL
L2 − r 2
(b) Apply Newton’s second law in the r -direction.
∑ Fr = T sinθ = mω 2r = mω 2 ( L sinθ ) ⇒ T = mω 2 L.
Set the two expressions for T equal to each other, cancel m and one L, and solve for ω.
mgL
L2 − r 2
(c) Insert L = 1.0 m, r = 0.20 m and m = 0.50 kg.
T=
ω=
g
2
L −r
2
=
mgL
L2 − r 2
=
= mω 2 L ⇒ ω =
g
L2 − r 2
(0.50 kg)(9.8 m/s 2 )(1.0 m)
(1.0 m) 2 − (0.20 m) 2
= 5.0 N
9.8 m/s 2
⎛ 1 rev ⎞⎛ 60 s ⎞
= 3.163 rad/s ⎜
⎟⎜
⎟ = 30 rpm
⎝ 2π rad ⎠⎝ 1 min ⎠
(1.0 m) − (0.20 m)
2
2
Assess: Notice that the mass canceled out of the equation for ω , but not for T, so the 500 g was necessary information.
8.37. Model: Assume the particle model for a sphere in circular motion at constant speed.
Visualize:
Solve: (a) Newton’s second law along the r and z axes is:
mv 2
∑ Fr = T1 sin 30° + T2 sin 60° = t ∑ Fz = T1 cos30° + T2 cos 60° − FG = 0 N
r
Since we want T1 = T2 = T , these two equations become
T (sin 30° + sin 60°) =
mvt2
T (cos30° + cos60°) = mg
r
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8-24
Chapter 8
Since sin 30° + sin 60° = cos30° + cos 60°,
mvt2
⇒ vt = rg
r
The triangle with sides L1, L2 , and 1.0 m is isosceles, so L2 = 1.0 m and r = L2 cos30°. Thus
mg =
L2 cos 30° g = (1.0 m)cos 30° g = (0.866 m)(9.8 m/s 2 ) = 2.9 m/s
(b) The tension is
T=
mg
(2.0 kg)(9.8 m/s 2 )
=
= 14.3 N ≈ 14 N
cos30° + cos60°
0.866 + 0.5
8.38. Model: Consider the passenger to be a particle and use the model of static friction.
Visualize:
Solve: The passengers stick to the wall if the static friction force is sufficient to support the gravitational force on
them: fs = FG . The minimum angular velocity occurs when static friction reaches its maximum possible value
( fs ) max = μs n. Although clothing has a range of coefficients of friction, it is the clothing with the smallest coefficient
( μs = 0.60) that will slip first, so this is the case we need to examine. Assuming that the person is stuck to the wall,
Newton’s second law is
∑ Fr = n = mω 2r
∑ Fz = fs − w = 0 ⇒ fs = mg
The minimum frequency occurs when
2
fs = ( fs ) max = μs n = μs mrωmin
Using this expression for fs in the z-equation gives
2
= mg
fs = μs mrωmin
⇒ ωmin =
g
μs r
=
9.80 m/s 2
1 rev
60 s
= 2.56 rad/s = 2.56 rad/s ×
×
= 24 rpm
0.60(2.5 m)
2π rad 1 min
Assess: Note the velocity does not depend on the mass of the individual. Therefore, the minimum mass sign is not
necessary.
8.39. Model: Use the particle model for the marble in uniform circular motion.
Visualize:
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Dynamics II: Motion in a Plane
8-25
Solve: The marble will roll in a horizontal circle if the static friction force is sufficient to support the gravitational on
it: fs = FG . If mg > ( fs ) max then static friction is not sufficient and the marble will slip down the side as it rolls
around the circumference. The r-equation of Newton’s second law is
2
2π rad 1 min ⎞
⎛
∑ Fr = n = mrω 2 = (0.010 kg)(0.060 m) ⎜150 rpm ×
×
⎟ = 0.148 N
1 rev
60 s ⎠
⎝
Thus the maximum possible static friction is ( fs ) max = μs n = (0.80)(0.148 N) = 0.118 N. The friction force needed to
support a 10 g marble is fs = mg = 0.098 N. We see that fs < ( fs ) max , therefore friction is sufficient and the marble
spins in a horizontal circle.
Assess: In reality, rolling friction will cause the marble to gradually slow down until ( fs ) max < mg . At that point, it
will begin to slip down the inside wall.
8.40. Model: Assume uniform circular motion.
Visualize: We expect the centripetal acceleration to be very large because ω is large. This will produce a significant
force even though the mass difference of 10 mg is so small.
A preliminary calculation will convert the mass difference to kg: 10 mg = 1.0 × 10−5 kg. If the two samples are
equally balanced then the shaft doesn’t feel a net force in the horizontal plane. However, the mass difference of
10 mg is what causes the force.
We’ll do another preliminary calculation to convert ω = 70,000 rpm into rad/s.
rev ⎛ 2π rad ⎞⎛ 1 min ⎞
⎜
⎟⎜
⎟ = 7330 rad/s
min ⎝ 1 rev ⎠⎝ 60 s ⎠
Solve: The centripetal acceleration is given by Equation 6.9 and the net force by Newton’s second law.
Fnet = (Δm)(a ) = ( Δm)(ω 2 r ) = (1.0 × 10−5 kg)(7330 rad/s) 2 (0.12 m) = 64 N
ω = 70,000 rpm = 70,000
Assess: As we expected, the centripetal acceleration is large. The force is not huge (because of the small mass
difference) but still enough to worry about. The net force scales with this mass difference, so if the mistake were
bigger it could be enough to shear off the shaft.
8.41. Model: Use the particle model for the car and the model of kinetic friction.
Visualize:
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8-26
Chapter 8
Solve: We will apply Newton’s second law to all three cars.
Car A:
∑ Fx = nx + ( f k ) x + ( FG ) x = 0 N − f k + 0 N = ma x
∑ Fy = n y + ( f k ) y + y y = n + 0 N − mg = 0 N
The y-component equation means n = mg . Since f k = μ k n, we have f k = μ k mg . From the x-component equation,
− f k − μ k mg
=
= − μ k g = −9.8 m/s 2
m
m
Car B: Car B is in circular motion with the center of the circle above the car.
ax =
mv 2
r
∑ Ft = nt + ( f k )t + ( FG )t = 0 N − f k + 0 N = + mat
∑ Fr = nr + ( f k ) r + ( FG )r = n + 0 N − mg = mar =
From the r-equation
n = mg +
⎛
mv 2
v2 ⎞
⇒ f k = μk n = μk m ⎜ g + ⎟
⎜
r
r ⎟⎠
⎝
Substituting back into the t-equation,
at = −
⎛
fk
μ m⎛
v2 ⎞
(25 m/s) 2 ⎞
2
= − k ⎜ g + ⎟ = − μ k ⎜ 9.8 m/s 2 +
⎟⎟ = −12.9 m/s
⎜
m
m ⎜⎝
r ⎟⎠
200
m
⎝
⎠
Car C: Car C is in circular motion with the center of the circle below the car.
∑ Fr = nr + ( f k ) r + ( FG )r = − n + 0 N + mg = mar =
mv 2
r
∑ Ft = nt + ( f k )t + ( FG )t = 0 N − f k + 0 N = mat
2
From the r-equation n = m( g − v /r ). Substituting this into the t-equation yields
at =
− f k − μk n
=
= − μ k ( g − v 2 /r ) = −6.7 m/s 2
m
m
8.42. Model: Model the ball as a particle that is moving in a vertical circle.
Visualize:
Solve: (a) The ball’s gravitational force FG = mg = (0.500 kg)(9.8 m/s 2 ) = 4.9 N.
(b) Newton’s second law at the top is
∑ Fr = T1 + FG = mar = m
v2
r
⎡ ( 4.0 m/s )2
⎤
⎛ v2
⎞
⇒ T1 = m ⎜ − g ⎟ = (0.500 kg) ⎢
− 9.8 m/s 2 ⎥ = 2.9 N
⎜
⎟
⎢⎣ 1.02 m
⎥⎦
⎝ r
⎠
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Dynamics II: Motion in a Plane
8-27
(c) Newton’s second law at the bottom is
∑ Fr = T2 − FG =
mv 2
r
⎛
⎡
v2 ⎞
(7.5 m/s)2 ⎤
⇒ T2 = m ⎜ g + ⎟ = (0.500 kg) ⎢9.8 m/s 2 +
⎥ = 32 N
⎜
r ⎠⎟
1.02 m ⎦⎥
⎝
⎣⎢
8.43. Model: Model a passenger as a particle rotating in a vertical circle.
Visualize:
Solve: (a) Newton’s second law at the top is
∑ Fr = nT + FG = mar =
The speed is
mv 2
mv 2
⇒ nT + mg =
r
r
2π r 2π (8.0 m)
=
= 11.17 m/s
T
4.5 s
⎛ v2
⎞
⎡ (11.17 m/s) 2
⎤
⇒ nT = m ⎜ − g ⎟ = (55 kg) ⎢
− 9.8 m/s2 ⎥ = 319 N
⎜ r
⎟
⎝
⎠
⎣⎢ 8.0 m
⎦⎥
v=
That is, the ring pushes on the passenger with a force of 3.2 × 102 N at the top of the ride. Newton’s second law at
the bottom:
⎛ v2
⎞
mv 2
mv 2
∑ Fr = nB − FG = mar =
⇒ nB =
+ mg = m ⎜ + g ⎟
⎜ r
⎟
r
r
⎝
⎠
2
⎡ (11.17 m/s)
⎤
= (55 kg) ⎢
+ 9.8 m/s 2 ⎥ = 1397 N
⎣⎢ 8.0 m
⎦⎥
Thus the force with which the ring pushes on the rider when she is at the bottom of the ring is 1.4 kN.
(b) To just stay on at the top, nT = 0 N in the r-equation at the top in part (a). Thus,
2
mg =
⎛ 2π ⎞
mv 2
r
8.0 m
= mrω 2 = mr ⎜
= 2π
= 5.7 s
⎟ ⇒ Tmax = 2π
r
T
g
9
8 m/s 2
.
⎝ max ⎠
8.44. Model: Model the chair and the rider as a particle in uniform circular motion.
Visualize:
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8-28
Chapter 8
Solve: Newton’s second law along the r-axis is
∑ Fr = Tr + ( FG ) r = mar ⇒ T sin θ + 0 N = mrω 2
Since r = L sin θ , this equation becomes
2
⎡ 2π rad ⎤
T = mLω 2 = (150 kg)(9.0 m) ⎢
⎥ = 3330 N
⎣ 4.0 s ⎦
Thus, the 3000 N chain is not strong enough for the ride.
8.45. Model: Model the ball as a particle in motion in a vertical circle.
Visualize:
G
Solve: (a) If the ball moves in a complete circle, then there is a tension force T when the ball is at the top of the
circle. The tension force adds to the gravitational force to cause the centripetal acceleration. The forces are along the
r-axis, and the center of the circle is below the ball. Newton’s second law at the top is
mv 2
( Fnet )r = T + FG = T + mg =
L
LT
m
The tension T can’t become negative, so T = 0 N gives the minimum speed vmin at which the ball moves in a circle.
⇒ vtop = Lg +
If the speed is less than vmin , then the string will go slack and the ball will fall out of the circle before it reaches the
top. Thus,
Lg
v
g
vmin = Lg ⇒ ωmin = min =
=
L
L
L
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Dynamics II: Motion in a Plane
8-29
(b) Insert L = 1.0 m.
ωmin =
g
(9.8 m/s 2 )
=
= 3.13 rad/s = 30 rpm
r
(1.0 m)
Assess: Notice that the mass doesn’t appear in the answer, so ωmin is independent of the mass.
8.46. Model: The ball is a particle on a massless rope in circular motion about the point where the rope is attached
to the ceiling.
Visualize:
Solve: Newton’s second law in the radial direction is
(∑ Fr ) = T − FG = T − mg =
mv 2
r
Solving for the tension in the rope and evaluating,
⎛
⎛
v2 ⎞
( 5.5 m/s )2 ⎞⎟ = 168 N
T = m ⎜ g + ⎟ = (10.2 kg) ⎜ 9.8 m/s 2 +
⎜
⎜
r ⎟⎠
4.5 m ⎟
⎝
⎝
⎠
Assess: The tension in the rope is greater than the gravitational force on the ball in order to keep the ball moving
in a circle.
8.47. Model: Model the ball as a particle in uniform circular motion. Rolling friction is ignored.
Visualize:
Solve: The track exerts both an upward normal force and an inward normal force. From Newton’s second law,
n1 = mg = (0.030 kg)(9.8 m/s 2 ) = 0.294 N, up
2
⎡ 60 rev 2π rad 1 min ⎤
n2 = mrω 2 = (0.030 kg)(0.20 m) ⎢
×
×
= 0.2369 N, in
1 rev
60 s ⎥⎦
⎣ min
Fnet = n12 + n22 = (0.294 N)2 + (0.2369 N) 2 = 0.38 N
8.48. Model: Masses m1 and m2 are considered particles. The string is assumed to be massless.
Visualize:
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8-30
Chapter 8
Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth, it
G
G
acts like a pulley. Thus tension forces T1 and T2 act as if they were an action/reaction pair. Mass m1 is in circular
motion of radius r, so Newton’s second law for m1 is
m1v 2
r
Mass m2 is at rest, so the y-equation of Newton’s second law is
∑ Fr = T1 =
∑ Fy = T2 − m2 g = 0 N ⇒ T2 = m2 g
Newton’s third law tells us that T1 = T2 . Equating the two expressions for these quantities:
m1v 2
m2 rg
= m2 g ⇒ v =
r
m1
8.49. Model: Model yourself as a particle in circular motion atop a leg of length L.
Visualize: Set up the coordinate system so that the r -direction is down, toward the center of the circle.
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Dynamics II: Motion in a Plane
8-31
Solve:
(a) Apply Newton’s second law in the downward vertical direction, which is the + r -direction.
mv 2
L
Notice that n < mg. Your body tries to “lift off” as it pivots over your foot, decreasing the normal force exerted on
you by the ground. The normal force becomes smaller as you walk faster, but n cannot be less than zero. Thus the
maximum possible walking speed vmax occurs when n = 0. Setting n = 0,
∑ Fr = mg − n =
mg =
(b) Insert L = 0.70 m.
2
mvmax
⇒ vmax = gL
L
vmax = gL = (9.8 m/s 2 )(0.70 m) = 2.6 m/s = 5.9 mph
Assess: The answer of 5.9 mph is faster than the “normal” walking speed of 3 mph, but we expect vmax to be greater
than the normal speed. This seems reasonable. The units check out.
The maximum walking speed depends on L, so taller people can walk faster than shorter people.
8.50. Model: Model the ball as a particle swinging in a vertical circle, then as a projectile.
Visualize:
Solve: Initially, the ball is moving in a circle. Once the string is cut, it becomes a projectile. The final circularmotion velocity is the initial velocity for the projectile. The free-body diagram for circular motion is shown at the
bottom of the circle. Since T > FG , there is a net force toward the center of the circle that causes the centripetal
acceleration. The r-equation of Newton’s second law is
( Fnet )r = T − FG = T − mg =
mv 2
r
r
0.60 m
(T − mg ) =
[5.0 N − (0.10 kg)(9.8 m/s 2 )] = 4.91 m/s
m
0.100 kg
G
As a projectile the ball starts at y0 = 1.4 m with v0 = 4.91iˆ m/s. The equation for the y-motion is
⇒ vbottom =
y1 = 0 m = y0 + v0 y Δt − 12 g (Δt ) 2 = y0 − 12 gt12
This is easily solved to find that the ball hits the ground at time
2 y0
t1 =
= 0.535 s
g
During this time interval it travels a horizontal distance
x1 = x0 + v0 xt1 = (4.91 m/s)(0.535 s) = 2.63 m
So the ball hits the floor 2.6 m to the right of the point where the string was cut.
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8-32
Chapter 8
8.51. Model: Use the particle model for a ball in motion in a vertical circle and then as a projectile.
Visualize:
Solve: For the circular motion, Newton’s second law along the r-direction is
mv 2
∑ Fr = T + FG = t
r
Since the string goes slack as the particle makes it over the top, T = 0 N. That is,
mvt2
⇒ vt = gr = (9.8 m/s 2 )(0.5 m) = 2.21 m/s
r
The ball begins projectile motion as the string is released. The time it takes for the ball to hit the floor can be found as follows:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ 0 m = 2.0 m + 0 m + 12 ( −9.8 m/s 2 )(t1 − 0 s) 2 ⇒ t1 = 0.639 s
FG = mg =
The place where the ball hits the ground is
x1 = x0 + v0 x (t1 − t0 ) = 0 m + (+2.21 m/s)(0.639 s − 0 s) = + 1.41 m
The ball hits the ground 1.4 m to the right of the point beneath the center of the circle.
8.52. Model: Model yourself as a particle in circular motion.
Visualize: Apply Newton’s second law for circular motion in the vertical direction. The upward normal force is the
scale reading. We are given mg = 588 N. This means m = 60 kg. We seek r.
Solve:
mv 2
r
We can easily get the spreadsheet to subtract mg = 588 N from each of the scale readings on the left side of the
∑ F = n − mg =
equation. Then if we graph n − mg vs. v 2 we should get a straight line whose slope is m/r.
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Dynamics II: Motion in a Plane
8-33
We see from the spreadsheet that the fit is very good and that the slope is m/r = 0.3999 kg/m.
With m = 60 kg,
m
60 kg
= 0.3999 kg/m ⇒ r =
= 150 m
r
0.3999 kg/m
Assess: This radius of curvature does not seem extreme.
8.53. Model: Model the ball as a particle undergoing circular motion in a vertical circle.
Visualize:
Solve: Initially, the ball is moving in circular motion. Once the string breaks, it becomes a projectile. The final
circular-motion velocity is the initial velocity for the projectile, which we can find by using the kinematic equation
v12 = v02 + 2a y ( y1 − y0 ) ⇒ 0 m 2 /s 2 = (v0 ) 2 + 2( −9.8 m/s 2 )(4.0 m − 0 m) ⇒ v0 = 8.85 m/s
This is the speed of the ball as the string broke. The tension in the string at that instant can be found by using the
r-component of the net force on the ball:
⎛ v02 y ⎞
(8.85 m/s) 2
⎟ ⇒ T = (0.100 kg)
= 13 N
∑ Fr = T = m ⎜
⎜ r ⎟
0.60 m
⎝
⎠
8.54. Model: Model the car as a particle on a circular track.
Visualize:
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8-34
Chapter 8
Solve: (a) Newton’s second law along the t-axis is
∑ Ft = Ft = mat ⇒ 1000 N = (1500 kg)at ⇒ at = 2/3 m/s 2
With this tangential acceleration, the car’s tangential velocity after 10 s will be
v1t = v0t + at (t1 − t0 ) = 0 m/s + (2/3 m/s 2 )(10 s − 0 s) = 20/3 m/s
The radial acceleration at this instant is
ar =
v12t (20/3 m/s) 2 16
=
=
m/s 2
r
25 m
9
The car’s acceleration at 10 s has magnitude
a1 = at2 + ar2 = (2/3 m/s 2 ) 2 + (16/9 m/s 2 ) 2 = 1.90 m/s 2
θ = tan −1
at
⎛ 2/3 ⎞
= tan −1 ⎜
⎟ = 21°
ar
⎝ 16/9 ⎠
where the angle is measured from the r-axis.
(b) The car will begin to slide out of the circle when the static friction reaches its maximum possible value
( fs ) max = μs n. That is,
mv22t
⇒ v2t = rg = (25 m)(9.8 m/s 2 ) = 15.7 m/s
r
In the above equation, n = mg follows from Newton’s second law along the z-axis. The time when the car begins to
∑ Fr = ( fs ) max = μs n = μs mg =
slide can now be obtained as follows:
v2t = v0t + at (t2 − t0 ) ⇒ 15.7 m/s = 0 m/s + (2/3 m/s 2 )(t2 − 0) ⇒ t2 = 24 s
8.55. Model: Model the steel block as a particle and use the model of kinetic friction.
Visualize:
G
Solve: (a) The components of thrust ( F ) along the r-, t-, and z-directions are
Fr = Fsin20° = (3.5 N)sin20° = 1.20 N
Ft = Fcos20° = (3.5 N)cos20° = 3.29 N
Fz = 0 N
Newton’s second law is
( Fnet ) r = T + Fr = mrω 2 ( Fnet )t = Ft − f k = mat
( Fnet ) z = n − mg = 0 N
The z-component equation means n = mg . The force of friction is
f k = μ k n = μ k mg = (0.60)(0.500 kg)(9.8 m/s 2 ) = 2.94 N
Substituting into the t-component of Newton’s second law
(3.29 N) − (2.94 N) = (0.500 kg)at ⇒ at = 0.70 m/s 2
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Dynamics II: Motion in a Plane
8-35
Having found at , we can now find the tangential velocity after 10 revolutions = 20π rad as follows:
1⎛ a ⎞
θ1 = ⎜ t ⎟ t12 ⇒ t1 =
2⎝ r ⎠
2rθ1
= 18.95 s
at
⎛ at ⎞
⎟ t1 = 6.63 rad/s
⎝ r ⎠
The block’s angular velocity after 10 rev is 6.6 rad/s.
(b) Substituting ω1 into the r-component of Newton’s second law yields:
ω1 = ω0 + ⎜
T1 + Fr = mrω12 ⇒ T1 + (1.20 N) = (0.500 kg)(2.0 m)(6.63 rad/s) 2 ⇒ T1 = 44 N
8.56. Model: Assume the particle model for a ball in vertical circular motion.
Visualize:
Solve: (a) Newton’s second law in the r- and t-directions is
mv 2
( Fnet )r = T + mg cosθ = mar = t ( Fnet )t = −mg sin θ = mat
r
Substituting into the r-component,
vt2
(20 N) + (2.0 kg)(9.8 m/s 2 )cos30° = (2.0 kg)
⇒ vt = 3.85 m/s
(0.80 m)
The tangential velocity is 3.8 m/s.
(b) Substituting into the t-component,
−(9.8 m/s 2 )sin 30° = at ⇒ at = −4.9 m/s 2
The radial acceleration is
ar =
vt2 (3.85 m/s)2
=
= 18.5 m/s 2
r
0.80 m
Thus, the magnitude of the acceleration is
a = ar2 + at2 = (18.5 m/s 2 ) 2 + (−4.9 m/s 2 )2 = 19.1 m/s 2 ≈ 19 m/s 2
The angle of the acceleration vector from the r-axis is
a
4.9
= 14.8° ≈ 15°
o = tan21 t = tan21
18.5
ar
The angle is below the r-axis.
8.57. Solve: (a) You are spinning a lead fishing weight in a horizontal 1.0 m diameter circle on the ice of a pond
when the string breaks. You know that the test weight (breaking force) of the line is 60 N and that the lead weight has
a mass of 0.30 kg. What was the weight’s angular velocity in rad/s and in rpm?
60 N
rev
60 s
(b)
ω2 =
⇒ ω = 20 rad/s ×
×
= 191 rpm
(0.3 kg)(0.5 m)
2π rad min
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8-36
Chapter 8
8.58. Solve: (a) At what speed does a 1500 kg car going over a hill with a radius of 200 m have a weight of 11,760 N?
(b) The weight is the normal force.
2940 N =
1500 kg v 2
⇒ v = 19.8 m/s
200 m
8.59. Model: Assume the particle model and apply the constant-acceleration kinematic equations.
Visualize:
Solve: (a) Newton’s second law for the projectile is
G
−F
Fnet = − Fwind = ma x ⇒ a x =
x
m
where Fwind is shortened to F. For the y-motion:
(
)
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ 0 m = 0 m + (v0 sin θ )t1 − 12 gt12 ⇒ t1 = 0 s and t1 =
2v0 sin θ
g
Using the above expression for t1 and defining the range as R we get from the x motion:
x1 = x0 + v0 x (t1 − t0 ) + 12 a x (t1 − t0 ) 2
⎛ 2v sin θ ⎞ F ⎛ 2v0 sin θ ⎞
⎛ F⎞
⇒ x1 − x0 = R = v0 xt1 + 12 ⎜ − ⎟ t12 = (v0 cosθ ) ⎜ 0
⎟−
⎜
⎟
g
g
⎝ m⎠
⎝
⎠ 2m ⎝
⎠
2
2v02
2v 2 F
cosθ sin θ − 0 2 sin 2 θ
g
mg
We will now maximize R as a function of θ by setting the derivative equal to 0:
=
dR 2v02
2 Fv02
(cos 2 θ − sin 2 θ ) −
2sin θ cosθ = 0
=
dθ
g
mg 2
⎛ 2 Fv02 ⎞⎛ g ⎞
mg
sin 2θ ⇒ tan 2θ =
⇒ cos 2 θ − sin 2 θ = cos 2θ = ⎜
⎜ mg 2 ⎟⎜
⎟⎜ 2v 2 ⎟⎟
F
⎝
⎠⎝ 0 ⎠
Thus the angle for maximum range is θ = 12 tan −1 (mg/F ).
(b) We have
mg (0.50 kg)(9.8 m/s 2 )
=
= 8.167 ⇒ θ = 12 tan −1 (8.167) = 41.51°
F
0.60 N
The maximum range without air resistance is
2v 2 sin 45° cos 45° v02
R′ = 0
=
g
g
Therefore, we can write the equation for the range R as
2F
R = 2 R′ sin 41.51° cos 41.51° −
R′ sin 2 41.51° = R′(0.9926 − 0.1076) = 0.885 R′
mg
R
R′ − R
= 1 − 0.8850 = 0.115
⇒ = 0.8850 ⇒
R′
R′
Thus R is reduced from R′ by 11.5%.
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Dynamics II: Motion in a Plane
8-37
Assess: The condition for maximum range (tan 2θ = mg/F ) means 2θ → 90° as F → 0. That is, θ = 45° when
F = 0, as is to be expected.
8.60. Model: Use the particle model for the (cart + child) system which is in uniform circular motion.
Visualize:
Solve: Newton’s second law along r and z directions can be written:
∑ Fr = T cos 20° − n sin 20° = mar ∑ Fz = T sin 20° − n cos 20° − mg = 0
The cart’s centripetal acceleration is
2
rev 1 min 2π rad ⎞
⎛
2
ar = rω 2 = (2.0cos20° m) ⎜14
×
×
⎟ = 4.04 m/s
min
60 s
1 rev ⎠
⎝
The above force equations can be rewritten as
0.94T − 0.342n = (25 kg)(4.04 m/s 2 ) = 101 N
0.342T + 0.94n = (25 kg)(9.8 m/s 2 ) = 245 N
Solving these two equations yields T = 179 N ≈ 180 N for the tension in the rope.
Assess: In view of the (child + cart) weight of 245 N, a tension of 179 N is reasonable.
8.61. Model: Use the particle model for the ball, which is in uniform circular motion.
Visualize:
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8-38
Chapter 8
Solve: From Newton’s second law along r and z directions,
mv 2
∑ Fr = n cosθ =
∑ Fz = n sin θ − mg = 0 ⇒ n sin θ = mg
r
Dividing the two force equations gives
gr
tan θ = 2
v
From the geometry of the cone, tan θ = r/y. Thus
r gr
=
⇒ v = gy
y v2
8.62. Model: Model the block as a particle and use the model of kinetic friction.
Visualize:
Solve: The only radial force is tension, so we can use Newton’s second law to find the angular velocity ωmax at
which the tube breaks:
∑ Fr = T = mω 2 r ⇒ ωmax =
Tmax
50 N
=
= 9.12 rad/s
mr
(0.50 kg)(1.2 m)
The compressed air and friction exert tangential forces, and the second law along the tangential direction is
∑ Ft = Ft − f k = Ft − μ k n = Ft − μk mg = mat
4.0 N
Ft
− μk g =
− (0.60)(9.80 m/s 2 ) = 2.12 m/s 2
0.50 kg
m
The time needed to accelerate to 9.12 rad/s is given by
rω
(1.2)(9.12 rad/s)
⎛a ⎞
ω1 = ωmax = 0 + ⎜ t ⎟ t1 ⇒ t1 = max =
= 5.16 s
at
2.12 m/s 2
⎝ r ⎠
During this interval, the block turns through angle
⎛ 2.12 m/s 2 ⎞
1 rev
⎛a ⎞
2
Δθ = θ1 − θ 0 = ω0t1 + 12 ⎜ t ⎟ t12 = 0 + 12 ⎜
= 3.7 rev
⎟⎟ (5.16 s) = 23.52 rad ×
⎜
2π rad
⎝ r ⎠
⎝ 1.2 m ⎠
at =
8.63. Model: Use the particle model for a sphere revolving in a horizontal circle.
Visualize:
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Dynamics II: Motion in a Plane
8-39
Solve: Newton’s second law in the r- and z-directions is
mv 2
∑( F ) r = T1 cos30° + T2 cos30° = t
∑( F ) z = T1 sin 30° − T2 sin 30° − FG = 0 N
r
Using r = (1.0 m)cos30° = 0.886 m, these equations become
T1 + T2 =
mvt2
(0.300 kg)(7.5 m/s) 2
=
= 22.5 N
r cos30°
(0.866 m)(0.866)
T1 − T2 =
mg
(0.300 kg)(9.8 m/s 2 )
=
= 5.88 N
sin 30°
(0.5)
Solving for T1 and T2 yields T1 = 14.2 N ≈ 14 N and T2 = 8.3 N.
8.64. Model: Use the particle model for the ball.
Visualize:
Solve: (a) Newton’s second law along the r- and z-directions is
∑ Fr = n cosθ = mrω 2
∑ Fz = n sin θ − FG = 0 N
Using FG = mg and dividing these equations yields:
tan θ =
g
rω
2
=
R− y
r
where you can see from the figure that tan θ = ( R − y )/r. Thus ω =
g
.
R− y
(b) ω will be minimum when ( R − y ) is maximum or when y = 0 m. Then ωmin = g/R .
(c) Substituting into the above expression,
ω=
g
9.8 m/s 2
rad 60 s
1 rev
=
= 9.9
×
×
= 95 rpm
R− y
0.20 m − 0.10 m
s 1 min 2π rad
8.65. Model: Use the particle model for the airplane.
Visualize:
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8-40
Chapter 8
G
Solve: In level flight, the lift force L balances the gravitational force. When turning, the plane banks so that the
radial component of the lift force can create a centripetal acceleration. Newton’s second law along the r- and
z-directions is
∑ Fr = L sin θ =
mvt2
r
∑ Fz = L cosθ − mg = 0 N
These can be written:
sin θ =
mv 2
mg
cosθ =
rL
L
Dividing the two equations gives:
2
miles 1 hr 1610 m ⎤
⎡
⎢⎣ 400 hour × 3600 s × 1 mile ⎥⎦
v2
v2
tan θ =
⇒r=
=
= 18.5 km
gr
g tan θ
⎡9.8 m/s 2 ⎤ tan10°
⎣
⎦
The diameter of the airplane’s path around the airport is 2 × 18.5 km = 37 km.
8.66. Model: Use the particle model for a small volume of water on the surface.
Visualize:
Solve: Consider a particle of water of mass m at point C on the surface. Newton’s second law along the r- and
z-directions is
mrω 2
( Fnet )r = n cosθ = mrω 2 ⇒ cosθ =
n
mg
( Fnet ) z = n sin θ − mg = 0 N ⇒ sin θ =
n
Dividing both equations gives tanθ = g/rω 2 . For a parabola z = ar 2 . This means
dz
1
1
= 2ar = slope of the curve at C = tan φ = tan(90° − θ ) =
⇒ tan θ =
tan θ
2ar
dr
Equating the two equations for tan θ , we get
1
g
ω2
= 2 ⇒a=
2ar rω
2g
Thus the surface is described by the equation
z=
ω2
2g
r2
which is the equation of a parabola.
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Dynamics II: Motion in a Plane
8-41
8.67. Model: Model the block as a particle in circular motion.
Visualize: Apply Newton’s second law in the r- and t-directions. We seek v(t ) where v is the tangential speed.
Solve:
mv 2
r
∑ Fr = n = mar =
∑ Ft = − f k = mat
Use n = mv 2 /r in f k = μ k n .
− f k = − μk n = − μk
Cancel m and use at =
mv 2
= mat
r
dv
.
dt
μk
−
Separate variables and integrate.
−
−
r
μk
r
v2 =
t
dv
dt
v
dv
0
v2
Ñ0dt = Ñv
v
μk
⎡ −1 ⎤
[t ]t0 = ⎢ ⎥
r
⎣ v ⎦ v0
−
μk
r
t=
1 1
−
v0 v
Solve for v which is a function of t .
−1
⎛1 μ ⎞
1 1 μk
rv0
= +
t ⇒ v =⎜ + k t⎟ =
+
v v0
r
v
r
r
v0 μ kt
⎝ 0
⎠
Assess: The dependencies seem to go in the right direction: as t increases v(t ) decreases; as μk increases v(t )
decreases. The m canceled out, so the result does not depend on the mass.
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9
IMPULSE AND MOMENTUM
Conceptual Questions
9.1. The velocities and masses vary from object to object, so there is no choice but to compute px = mvx for each
one and then compare:
p1x = (20 g)(1 m/s) = 20 g m/s
p2 x = (20 g)(2 m/s) = 40 g m/s
p3 x = (10 g)(2 m/s) = 20 g m/s
p4 x = (10 g)(1 m/s) = 10 g m/s
p5 x = (200 g)(0.1 m/s) = 20 g m/s
So the answer is p2 x > p1x = p3 x = p5 x > p4 x .
9.2. Impulse gives a measure of the effect of a force acting over a period of time. During the time a net force acts on
an object, the object will accelerate and its velocity will change. Impulse gives a measure of the effect of a force
without specifying the detailed time dependence of the force.
9.3. An isolated system is a collection of interacting objects for which all outside forces (i.e., forces from objects
outside the system) are balanced; that is, in an isolated system, all external forces cancel each other out.
9.4. When the question talks about forces, times, and momenta, we immediately think of the impulse-momentum
theorem, which tells us that, to change the momentum of an object, we must exert a net external force on it over a
G G
time interval: Δp = Favg Δt. Because equal forces are exerted over equal times, the impulses are equal and the changes
in momentum are equal. Because both carts start from rest, the change in momentum of each is the same as the final
momentum of each, so their final momenta are equal. Notice that, to answer the question, we do not need to know the
mass of either cart, or even the specific time interval (as long as it is the same for both carts).
9.5. The impulse-momentum theory tells us that the change in momentum of an object is related to the net force on
G G
the object and the length of time the force was applied. Mathematically, Δp = Favg Δt. The same force applied to the
two carts results in a larger acceleration for the less massive plastic cart (Newton’s second law), enabling it to travel
the 1-m distance in a shorter time. Therefore, the plastic cart has a smaller change in momentum than the lead cart.
Because the final momentum of each cart is equal to their change in momentum (zero initial momentum), the final
momentum of the plastic cart is less than that of the lead cart.
9.6. In this story, Carlos is correct. During the short time of the bullet-block collision other forces are negligible
compared to the force between the bullet and block, so in the impulse approximation momentum is conserved. When
the bullet bounces off of the steel block, the bullet’s final momentum is backward. To balance that, the steel block
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9-1
9-2
Chapter 9
must be moving forward faster than the case in which the bullet embeds itself in the wooden block (in which case the
bullet has a final momentum in the forward direction).
9.7. The impulse-momentum theory states that a change in an object’s momentum results when a net force is
G G
applied to the object for some time interval; Δp = Favg Δt. Stopping a hard ball requires changing its momentum, and
this change can be accomplished with a small force over a long time interval or a large force over a short time
interval. The padding in a glove lets the time interval during which the ball is stopped be long, resulting in a smaller
force on the glove and on your hand.
9.8. The impulse-momentum theory states that a change in an object’s momentum results when a net force is
G G
applied to the object for some time interval; Δp = Favg Δt. Stopping an automobile requires changing its momentum
from some to none. This change can be accomplished with a small force over a long time interval or a large force
over a short time interval. The crumple zone that collapses during an automobile collision lengthens the time interval
during which the automobile is stopped, resulting in a smaller force on the passengers as they also come to a stop.
9.9. The impulse is equal to the change in momentum, so
Δpx = mvfx − mvix = 4 Ns
The final velocity is thus
4 Ns
4 Ns
= 1 m/s +
= 3 m/s
2 kg
m
Since the velocity is positive, the object is moving to the right.
vfx = vix +
9.10. The impulse is equal to the change in momentum, so
Δpx = mvfx − mvix = −4 Ns
The final velocity is thus
4 Ns
4 Ns
= 1 m/s −
= −1 m/s
2 kg
m
Since the velocity is negative, the object is now moving to the left with a speed of 1 m/s. Note that the impulse was
negative, which decreases the initially positive velocity.
vfx = vix −
9.11. The club and ball form a system. The interaction force when the club and ball collide is very large compared
to other forces at the time of collision, such as gravity and the force of the golfer on the club. So, in this impulse
approximation, momentum is conserved during the collision. After the club hits the ball, it will give the ball some
of its momentum. The club can continue to move forward as long as the momentum the ball obtains is less than the
initial momentum of the club. Note that the momentum conservation is only valid if we consider the short time
between just before and just after the collision. The wider we make the time window, the more time gravity and
the golfer have to influence the motion of the club and ball, so that momentum conservation would no longer hold for
the club-ball system.
9.12. The impulse one ball receives is equal to the average force on it from the other ball multiplied by the time
during which the force is applied. But by Newton’s third law the force that the rubber ball exerts on the steel ball is
equal to the force the steel ball exerts on the rubber ball. So both balls receive the same amount of impulse, although
the impulses are in opposite directions.
9.13. (a) Both particles cannot be at rest immediately after the collision. If they were both at rest, then the sum of
the momenta after the collision would be zero, and since momentum is conserved in collisions, it would have had to
be zero before as well (and it wasn’t).
(b) If the masses are equal and the collision is elastic, the moving particle will stop and give all of its momentum to
the previously resting particle. A good example of this appears when a billiard ball collides head-on with another
billiard ball that is at rest.
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Impulse and Momentum
9-3
We say momentum is conserved in all collisions because we assume that both colliding objects are part of the system
and we assume the “impulse approximation” prevails, meaning that other forces can be neglected during the short
time interval of the collision. In part (a), if the system contained a third particle that participated in the collision, then
it is possible for the first two particles to end up at rest if the momentum were carried off by the third particle.
9.14. (a) Let Paula and Ricardo be a system. Initially, their total momentum is zero since they are at rest. After they
push off each other, the total momentum must still be zero, so Ricardo and Paula must have equal but opposite
momenta.
(b) Momentum p = mv. Since Paula is less massive than Ricardo, her speed must be higher than Ricardo’s for her to
have the same momentum as Ricardo.
Exercises and Problems
Section 9.1 Momentum and Impulse
9.1. Model: Model the car and the baseball as particles.
Solve: (a) The momentum p = mv = (3000 kg)(15 m/s) = 4.5 × 104 kg m/s.
(b) The momentum p = mv = (0.20 kg)(40 m/s) = 8.0 kg m/s.
9.2. Model: Model the bicycle and its rider as a particle. Also model the car as a particle.
Solve: From the definition of momentum,
⎛ mcar ⎞
⎛ 1500 kg ⎞
(5.0 m/s) = 75 m/s
⇒ vbicycle = ⎜
⎟v =
⎜ mbicycle ⎟ car ⎜⎝ 100 kg ⎟⎠
⎝
⎠
Assess: This is a very high speed (≈ 168 mph). This problem shows the importance of mass in comparing two
pcar = pbicycle
⇒ mcar v car = mbicyclevbicycle
momenta.
9.3. Visualize: Please refer to Figure EX9.3.
tf
Solve: The impulse J x is defined in Equation 9.6 as J x = Ñ Fx (t )dt = area under the Fx (t ) curve between ti and tf .
ti
The area under the force-time curve in Figure EX9.3 is J x = (2 ms)(1000 N) + 12 (6 − 2 ms)(1000 N) = 4 Ns.
9.4. Model: The particle is subjected to an impulsive force.
Visualize: Please refer to Figure EX9.4.
Solve: Using Equation 9.6, the impulse is the area under the force-time curve. From 0 to 2 ms the impulse is
∫ F (t )dt = 12 (−500 N)(2 × 10
−3
s) = −0.5 Ns
From 2 to 8 ms the impulse is
∫ F (t )dt = 12 (+2000 N)(8 ms − 2 ms) = +6.0 Ns
From 8 ms to 10 ms the impulse is
∫ F (t )dt = 12 (−500 N)(10 ms − 8 ms) = −0.5 Ns
Thus, from 0 s to 10 ms the impulse is (−0.5 + 6.0 − 0.5) Ns = 5 Ns.
9.5. Visualize: Please refer to Figure EX9.5.
tf
Solve: The impulse is defined in Equation 9.6 as J x = Ñ Fx (t )dt = area under the Fx (t ) curve between ti and tf .
ti
For the force-time curve shown in Figure EX9.5, the impulse is 6.0 Ns = 12 ( Fmax )(8.0 ms) ⇒ Fmax = 1.5 × 103 N.
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9-4
Chapter 9
9.6. Model: Model the object as a particle and the interaction as a collision.
Visualize: Please refer to Figure EX9.6.
Solve: The momentum bar chart tells us the final momentum ( pfx = 2 kg m/s) and the impulse ( J x = 6 kg m/s).
Using the impulse-momentum theorem pfx = pix + J x , we can find the initial momentum:
pix = pfx − J x = 2 kg m/s − 6 kg m/s = −4 kg m/s
Since pix = mvix we have vix = pix /m = ( −4 kg m/s)/(0.05 kg) = −80 m/s. The speed is thus 80 m/s and the direction
is to the left.
9.7. Model: Model the object as a particle and the interaction as a collision.
Visualize: Please refer to Figure EX9.7.
Solve: The object is initially moving to the right (positive momentum) and ends up moving to the left (negative
momentum). Using the impulse-momentum theorem pfx = pix + J x ,
−2 kg m/s = +6 kg m/s + J x
J x = −8 kg m/s = −8 Ns
⇒
Since J x = Favg Δt , we have
Favg Δt = −8 Ns ⇒
Favg =
−8 Ns
= −8 × 102 N
10 ms
Thus, the force is 800 N to the left.
Section 9.2 Solving Impulse and Momentum Problems
9.8. Model: Model the object as a particle and the interaction with the force as a collision.
Visualize: Please refer to Figure EX9.8.
Solve: Using the equations
tf
pfx = pix + J x and J x = ÑFx (t ) dt = area under force curve
ti
(2.0 kg)vfx = (2.0 kg)(1.0 m/s) + (area under the force curve)
vfx = (1.0 m/s) + 2.01kg (1.0 s)(2.0 N) = 2.0 m/s
Becaue vfx is positive, the object moves to the right at 2.0 m/s.
Assess: For an object with positive velocity, a positive impulse increases the object’s speed. The opposite is true for
an object with negative velocity.
9.9. Model: Model the object as a particle and the interaction with the force as a collision.
Visualize: Please refer to Figure EX9.9.
Solve: Using the equations
tf
pfx = pix + J x and J x = ÑFx (t ) dt = area under force curve
ti
(2.0 kg)vfx = (2.0 kg)(1.0 m/s) + (area under the force curve)
vfx = (1.0 m/s) + 2.01kg ( −8.0 N)(0.50 s) = −1.0 m/s
Because vfx is negative, the object is now moving to the left at 1.0 m/s.
Assess: The direction of the velocity has reversed.
9.10. Model: Use the particle model for the sled, the model of kinetic friction, and the impulse-momentum
theorem.
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Impulse and Momentum
9-5
Visualize:
Note that the force of kinetic friction f k imparts a negative impulse to the sled.
Solve: Using Δp x = J x , we have
tf
tf
ti
ti
pfx − pix = ÑFx (t )dt = − f k Ñdt = − f k Δt ⇒ mvfx − mvix = − μ k nΔt = − μ k mg Δt
We have used the model of kinetic friction f k = μ k n, where μk is the coefficient of kinetic friction and n is the
normal (contact) force by the surface. The force of kinetic friction is independent of time and was therefore taken out
of the impulse integral. Thus,
1
(8.0 m/s − 5.0 m/s)
Δt =
(v − v ) =
= 1.2 s
μk g ix fx
(0.25)(9.8 m/s 2 )
9.11. Model: Model the rocket as a particle, and use the impulse-momentum theorem. The only force acting on the
rocket is due to its own thrust.
Visualize: Please refer to Figure EX9.11.
Solve: (a) The impulse is
J x = ∫ Fx (t )dt = area of the graph of Fx (t ) between t = 0 s and t = 30 s = 12 (1000 N)(30 s) = 1.5 × 104 Ns
(b) From the impulse-momentum theorem, pfx = pix + J x (t ), so the momentum or velocity increases as long as J x
is positive. If J x becomes negative, the speed will stop increasing and start decreasing, so this point will be the
maximum. For the current problem, the impluse is always positive, so the speed increases continuously. The
maximum is therefore at the end of the impluse (t = 30 s). The speed at this point is
mvfx = mvix + 1.5 × 104 Ns ⇒ (425 kg)vfx = (425 kg)(75 m/s) + 1.5 × 104 Ns ⇒ vfx = 110 m/s
9.12. Model: Model the ball as a particle, and its interaction with the wall as a collision in the impulse
approximation.
Visualize: Please refer to Figure EX9.12.
Solve: Using the equations
tf
pfx = pix + J x and J x = ÑFx (t ) dt = area under force curve
ti
(0.250 kg)vfx = (0.250 kg)(−10 m/s) + (500 N)(8.0 ms)
⎛ 4.0 N ⎞
vfx = (−10 m/s) + ⎜
⎟ = 6.0 m/s
⎝ 0.250 kg ⎠
Assess: The ball’s final velocity is positive, indicating it has turned around.
9.13. Model: Model the glider cart as a particle, and its interaction with the spring as a collision. Assume a
frictionless track.
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9-6
Chapter 9
S
Visualize:
Solve: Using the impulse-momentum theorem pfx − pix = ∫ F (t ) dt ,
(0.60 kg)(3 m/s) − (0.60 kg)(−3 m/s) = area under force curve = 12 (36 N)(Δt ) ⇒ Δt = 0.2 s
Section 9.3 Conservation of Momentum
9.14. Model: Choose car + gravel to be the system. Ignore friction in the impulse approximation.
Visualize:
Solve: There are no external horizontal forces on the car + gravel system, so the horizontal momentum is conserved.
This means pfx = pix . Hence,
(10,000 kg + 4000 kg)vfx = (10,000 kg)(2.0 m/s) + (4000 kg)(0.0 m/s) ⇒ vfx = 1.4 m/s
9.15. Model: Choose car + rainwater to be the system.
Visualize:
There are no external horizontal forces on the car + water system, so the horizontal momentum is conserved.
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Impulse and Momentum
9-7
Solve: Conservation of momentum gives pfx = pix . Hence,
(mcar + mwater )(20 m/s) = (mcar )(22 m/s) + (mwater )(0 m/s)
(5000 kg + mwater )(20 m/s) = (5000 kg)(22 m/s) ⇒ mwater = 5.0 × 102 kg
9.16. Model: Choose skydiver + glider to be the system in the impulse approximation. Ignore air resistance.
Visualize:
Note that there are no external forces in the x-direction (ignoring friction in the impulse approximation), implying
conservation of momentum along the x-direction.
Solve: The momentum conservation equation pfx = pix gives
(680 kg − 60 kg)(vG ) x + (60 kg)(vD ) x = (680 kg)(30 m/s)
Immediately after release, the skydiver’s horizontal velocity is still (vD ) x = 30 m/s because he experiences no net
horizontal force. Thus
(620 kg)(vG ) x + (60 kg)(30 m/s) = (680 kg)(30 m/s) ⇒ (vG ) x = 30 m/s
Assess: The skydiver’s motion in the vertical direction has no influence on the glider’s horizontal motion. Notice
that we did not need to invoke conservation of momentum to solve this problem. Because there are no external
horizontal forces acting on either the skydiver or the glider, neither will change their horizontal speed when the
skydiver lets go!
Section 9.4 Inelastic Collisions
9.17. Model: We will define our system to be bird + bug. This is the case of an inelastic collision because the bird
and bug move together after the collision. Horizontal momentum is conserved because there are no external forces
acting on the system during the collision in the impulse approximation.
Visualize:
Solve: The conservation of momentum equation pfx = pix gives
(m1 + m2 )vfx = m1 (vix )1 + m2 (vix )2
⇒ (300 g + 10 g)vfx = (300 g)(6.0 m/s) + (10 g)(−30 m/s) ⇒ vfx = 4.8 m/s
Assess: We left masses in grams, rather than convert to kilograms, because the mass units cancel out from both sides
of the equation. Note that (vix ) 2 is negative because the bug is flying to the left.
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9-8
Chapter 9
9.18. Model: The two cars are not an isolated system because of external frictional forces. But during the collision
friction is not going to be significant. Within the impulse approximation, the momentum of the Cadillac + Volkswagen
system will be conserved in the collision.
Visualize:
Solve: The momentum conservation equation pfx = pix gives
(mC + mVW )vfx = mC (vix )C + mVW (vix ) VW
0 kg mph = (2000 kg)(1.0 mph) + (1000 kg)(vix )VW
⇒ (vix )VW = −2.0 mph
so you need a speed of 2.0 mph.
9.19. Model: Because of external friction and drag forces, the car and the blob of sticky clay are not exactly an
isolated system. But during the collision, friction and drag are not going to be significant. The momentum of the
system will be conserved in the collision, within the impulse approximation.
Visualize:
Solve: The conservation of momentum equation pfx = pix gives
(mC + mB )(vf ) x = mB (vix )B + mC (vix )C
0 kg m/s = (10 kg)(vix ) B + (1500 kg)( −2.0 m/s) ⇒ (vix ) B = 3.0 × 102 m/s
Assess: This speed of the blob is around 600 mph, which is very large. However, a very large speed is expected in
order to stop a car with only 10 kg of clay.
Section 9.5 Explosions
9.20. Model: We will define our system to be archer + arrow. The force of the archer (A) on the arrow (a) is equal
to the force of the arrow on the archer. These are internal forces within the system. The archer is standing on
G
G
frictionless ice, and the normal force by ice on the system balances the weight force. Thus Fext = 0 on the system,
and momentum is conserved.
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Impulse and Momentum
9-9
Visualize:
The initial momentum pix of the system is zero, because the archer and the arrow are at rest. The final moment pfx
must also be zero.
Solve: We have M A vA + ma va = 0 kg m/s. Therefore,
vA =
− ma va −(0.100 kg)(100 m/s)
=
= −0.20 m/s
mA
50 kg
The archer’s recoil speed is 0.20 m/s.
Assess: It is the total final momentum that is zero, although the individual momenta are nonzero. Since the arrow
has forward momentum, the archer will have backward momentum.
9.21. Model: We will define our system to be Dan + skateboard, and their interaction as an explosion. While
friction is present between the skateboard and the ground, it is negligible in the impulse approximation.
Visualize:
The system has nonzero initial momentum pix . As Dan (D) jumps backward off the gliding skateboard (S), the
skateboard will move forward so that the final total momentum of the system pfx is equal to pix .
Solve: We have mS (vfx )S + mD (vfx ) D = ( mS + mD )vix . Thus,
(5.0 kg)(8.0 m/s) + (50 kg)(vfx )D = (5.0 kg + 50 kg)(4.0 m/s) ⇒ (vfx ) D = 3.6 m/s
9.22. Model: We will define our system to be the football player (P) and the football (B). Their interaction is an
explosion because the force involved is internal to the P + B system. There are no external horizontal forces present
on either of the two, so horizontal momentum is conserved.
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9-10
Chapter 9
Visualize:
The system has nonzero initial momentum pix , which must be conserved.
Solve: (a) The final velocity of the ball is (vfx ) B = 15.0 m/s. Equating the initial and final momentum gives
mP (vfx ) P + mB (vfx ) B = ( mB + mP )vix . Solving for (vfx ) P gives
(vfx ) P =
(mB + mP )vix − mB (vfx ) B (0.450 kg + 70.0 kg)(2.00 m/s) − (0.450 kg)(15.0 m/s)
=
= 1.92 m/s
mP
70.0 kg
(b) The final velocity of the ball is (vfx ) B = (vfx ) P + 15.0 m/s. Inserting this into the equation for conservation of
momentum and solving for (vfx ) B gives
mP (vfx ) P + mB [ (vfx ) P + 15.0 m/s ] = (mB + mP )vix .
(vfx ) P =
(mB + mP )vix − mB (15.0 m/s) (0.450 kg + 70.0 kg)(2.00 m/s) − (0.450 kg)(15.0 m/s)
=
= 1.90 m/s
mP + mB
70.0 kg + 0.450 kg
Assess: In part (b), the final velocity of the ball is greater than in part (a), so the player’s final velocity is slightly less
so that momentum is conserved.
Section 9.6 Momentum in Two Dimensions
9.23. Model: Assume that the momentum is conserved in the collision.
Visualize: Please refer to Figure EX9.23.
Solve: Applying conservation of momentum in the x- and y-directions yields
( pfx )1 + ( pfx ) 2 = ( pix )1 + ( pix ) 2
⇒ ( pfx )1 + 0 kg m/s = −2 kg m/s + 4 kg m/s ⇒ ( pfx )1 = 2 kg m/s
( pfy )1 + ( pfy ) 2 = ( piy )1 + ( piy )2
⇒ ( pfy )1 − 1 kg m/s = 2 kg m/s + 1 kg m/s
⇒ ( pfy )1 = 4 kg m/s
G
Thus, the final momentum of particle 1 is ( pf )1 = (2iˆ + 4 ˆj ) kg m/s.
9.24. Since the object is initially at rest, its total momentum is zero. After it explodes the total momentum of the
G
G
G G
G
fragments must also be zero. With p1 = (−2,2) kg m/s and p2 = (3,0) kg m/s, the requirement that p1 + p2 + p3 = 0
G
means that p3 = (−1, − 2) kg m/s.
9.25. Model: This problem deals with the conservation of momentum in two dimensions in an inelastic collision.
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Impulse and Momentum
9-11
Visualize:
G
G
Solve: The conservation of momentum equation pbefore = pafter gives
m1 (vix )1 + m2 (vix ) 2 = (m1 + m2 )vfx , m1 (viy )1 + m2 (viy )2 = (m1 + m2 )vfy
Substituting in the given values, we find
(0.020 kg)(3.0 m/s) + 0.0 kg m/s = (0.020 kg + 0.030 kg)vf cosθ
0.0 kg m/s + (0.030 kg)(2.0 m/s) = (0.020 kg + 0.030 kg)vf sin θ
vf cosθ = 1.2 m/s, vf sinθ = 1.2 m/s
vf = (1.2 m/s) 2 + (1.2 m/s) 2 = 1.7 m/s, θ = tan −1
vy
vx
= tan −1 (1) = 45°
The ball of clay moves 45° north of east at 1.7 m/s.
9.26. Model: Model the tennis ball as a particle, and its interaction with the wall as a collision.
Visualize:
The force increases to Fmax during the first two ms, stays at Fmax for two ms, and then decreases to zero during the
last two ms. The graph shows that Fx is positive, so the force acts to the right.
Solve: Using the impulse-momentum theorem pfx = pix + J x , we find
6 ms
(0.06 kg)(32 m/s) = (0.06 kg)(−32 m/s) +
Ñ
Fx (t )dt
0
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9-12
Chapter 9
The impulse is
6 ms
Jx =
Ñ
Fx (t )dx = area under force curve =
0
Fmax =
1
1
Fmax (0.0020 s) + Fmax (0.0020 s) + Fmax (0.0020 s) = (0.0040 s) Fmax
2
2
(0.060 kg)(32 m/s) + (0.060 kg)(32 m/s)
= 9.6 × 102 N
0.0040 s
9.27. Model: Let the system be ball + racket. During the collision of the ball and racket, momentum is conserved
because all external interactions are insignificantly small.
Visualize:
Solve: (a) The conservation of momentum equation pfx = pix gives
mR (vfx ) R + mB (vfx ) B = mR (vix ) R + mB (vix ) B
(1.000 kg)(vfx ) R + (0.060 kg)(40 m/s) = (1.000 kg)(10 m/s) + (0.060 kg)( −20 m/s) ⇒ (vfx )R = 6.4 m/s
(b) The impulse on the ball is calculated from ( pfx ) B = ( pix ) B + J x as follows:
(0.060 kg)(40 m/s) = (0.060 kg)(−20 m/s) + J x
Favg =
⇒
J x = 3.6 N s = ∫ Fdt = Favg Δt
3.6 Ns
= 3.6 × 102 N
10 ms
Let us now compare this force with the gravitational force on the ball ( FG ) B = mB g = (0.060 kg)(9.8 m/s 2 ) = 0.588 N.
We find Favg = 612( FG ) B.
Assess: This is a significant force and is reasonable because the impulse due to this force not only changes the
direction of the ball also but changes the speed of the ball from approximately 45 mph to 90 mph.
9.28. Model: Model the ball as a particle that is subjected to an impulse when it is in contact with the floor. We
shall also use constant-acceleration kinematic equations. During the collision, ignore any forces other than the
interaction between the floor and the ball in the impulse approximation.
Visualize:
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Impulse and Momentum
9-13
Solve: To find the ball’s velocity just before and after it hits the floor:
v12y = v02y + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2(−9.8 m/s 2 )(0 − 2.0 m) ⇒ v1 y = −6.261 m/s
v32y = v22 y + 2a y ( y3 − y2 ) ⇒ 0 m 2 /s 2 = v22 y + 2(−9.8 m/s 2 )(1.5 m − 0 m) ⇒ v2 y = 5.422 m/s
The force exerted by the floor on the ball can be found from the impulse-momentum theorem:
mv2 y = mv1 y + ∫ Fdt = mv1 y + area under the force curve
(0.200 kg)(5.422 m/s) = −(0.200 kg)(6.261 m/s) + 12 Fmax (5.0 × 10−3 s)
Fmax = 9.3 × 102 N
Assess: A maximum force of 9.3 × 102 N exerted by the floor is reasonable. This force is the same order of magnitude as the force of the racket on the tennis ball in the previous problem.
9.29. Model: Model the cart as a particle sliding down a frictionless ramp. The cart is subjected to an impulsive
force when it comes in contact with a rubber block at the bottom of the ramp. We shall use the impulse-momentum
theorem and the constant-acceleration kinematic equations.
Visualize:
Solve: From the free-body diagram on the cart, Newton’s second law applied to the system before the collision gives
∑( F ) x = FG sinθ = ma x
⇒ ax =
mg sinθ
9.81 m/s 2
= g sin30.0° =
= 4.905 m/s 2
m
2
Using this acceleration, we can find the cart’s speed just before its contact with the rubber block:
v12x = v02x + 2ax ( x1 − x0 ) = 0 m 2 /s 2 + 2(4.905 m/s 2 )(1.00 m − 0 m) ⇒ v1x = 3.132 m/s
Now we can use the impulse-momentum theorem to obtain the velocity just after the collision:
mv2 x = mv1x + ∫ Fx dt = mv1x + area under the force graph
(0.500 kg)v2 x = (0.500 kg)(3.13 m/s) − 12 (200 N)(26.7 × 1023 s) ⇒ v2 x = −2.208 m/s
Note that the given force graph is positive, but in this coordinate system the impulse of the force is to the left (i.e., up
the slope). That is the reason to put a minus sign while evaluating the ∫ Fx dt integral.
We can once again use a kinematic equation to find how far the cart will roll back up the ramp:
v32x = v22x + 2a x ( x3 − x2 ) ⇒ (0 m/s)2 = (−2.208 m/s) 2 + 2( −4.905 m/s 2 )( x3 − x2 ) ⇒ ( x3 − x2 ) = 0.497 m
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9-14
Chapter 9
9.30. Model: Model the balls as particles and ignore air resistance and friction on the table. Apply the impulsemomentum theorem and the constant-acceleration kinematic Equation 4.12.
Visualize:
Solve: (a) Once the ball leaves the table, the time it takes for it to hit the ground is
h = yf − yi = − 12 g Δt 2
⇒ Δt =
2g
h
The spring imparts an impulse J to a stationary ball, so the final momentum the ball is equal to the impulse J. The
horizontal velocity of the ball as it leaves the table is then
mvix = J ⇒ vix = J/m
This velocity remains constant during the free fall, so the range is given by
R = vix Δt =
( mJ )
2h
g
(b) To obtain a linear slope, you should graph the range R as a function of 1/m. The slope of the line will be J
2g
h
.
(c) The data are plotted in the figure below. From the best-fit line, we find a slope of s = 0.252 m kg. This gives an
impluse of J = s
g
2h
= 0.46 kg m/s.
9.31. Solve: Apply Equation 9.7 and Newton’s second law. The latter tells us that the average force used to expel
the grains is
Favg = ma = (1.0 × 10−10 kg)(2.5 × 104 m/s 2 ) = 2.5 × 10−6 N
Inserting this into Equation 9.7 gives
J = Favg Δt = (2.5 × 10−6 N)(3.0 × 104 s) = 7.5 × 10−10 kg m/s
9.32. Solve: Using Newton’s second law for the x-direction, Fx = dpx /dt. Therefore,
d
(6t 2 kg m/s) = 12t N
dt
Assess: The x-component of the net force on an object is equal to the time rate of change of the x-component of the
object’s momentum.
Fx =
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Impulse and Momentum
9-15
9.33. Visualize:
Solve: Using Newton’s second law for the y-direction and the chain rule,
( Fnet ) y =
dp y
dt
=
⎛ dv y ⎞
d
dm
(mv y ) =
(v y ) + m ⎜
⎟
dt
dt
⎝ dt ⎠
= (−0.50 kg/s)(120 m/s) + (48 kg)(18 m/s 2 )
= 8.0 × 102 N
Assess: Since the rocket is losing mass, dm/dt < 0. The time derivative of the velocity is the acceleration.
9.34. Model: Model the train cars as particles. Since the train cars stick together, we are dealing with perfectly
inelastic collisions. Momentum is conserved in the collisions of this problem in the impulse approximation, in which
we ignore external forces during the time of the collision.
Visualize:
Solve: In the collision between the three-car train and the single car:
mv1x + (3m)v2 x = 4mv3 x
⇒ v1x + 3v2 x = 4v3 x
⇒ v3 x = 14 v1x + 34 v2 x
In the collision between the four-car train and the stationary car:
(4m)v3 x + m v4 x = (5m)v5 x
v5 x =
4v
5 3x
=
⇒ 4v3 x + 0 m/s = 5v5 x
1v
5 1x
+ 53 v2 x = v0
9.35. Model: The two blobs of clay will be modeled as particles. Their collision is completely inelastic and
conserves momentum. Newton’s second law and the equations for constant-acceleration kinematics will apply.
Visualize:
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9-16
Chapter 9
Solve: Newton’s second law tells us that the acceleration of the first blob is a = F/m1. Therefore, after covering a
distance d, the speed of blob 1 is
v1′2x = 2ad
⇒ v1′ x = 2ad = 2dF/m1
Momentum is conserved in the collision with blob 2, so
m1v1′ x = vf ( m1 + m2 ) ⇒ vf =
m1v1′ x
m1
2dF
=
(m1 + m2 ) (m1 + m2 ) m1
9.36. Model: Model the gliders as particles and apply conservation of momentum. The 200 g glider will be labeled
1, the 300 g glider will be labeled 2, and the 400 g glider will be labeled 3.
Visualize: See Figure P9.36.
Solve: To apply conservation of momentum, we need to calculate the speed of two gliders from the figure.
Differentiating the position-versus-time curves gives:
dy
dy
v1x =
= 0.53 m/s, v2 x =
= −0.41 m/s
dt
dt
The intial momentum is zero because the three gliders are stationary. Therefore, conservation of momentum gives
m1v1x + m2v2 x + m3v3 x = 0
v3 x =
−1 ( m v
1 1x
m3
−1 [(0.20 kg)(0.53 m/s) + (0.30 kg)( − 0.41 m/s)] = 0.043 m/s
+ m2v2 x ) = 0.40
kg
Thus, the 400 g glider flies of the right (because v3 x > 0) at a speed of 0.043 m/s.
9.37. Model: Model the earth (E) and the asteroid (A) as particles. Earth + asteroid is our system. Since the two
stick together during the collision, this is a case of a perfectly inelastic collision. Momentum is conserved in the
collision since no significant external force acts on the system.
Visualize:
Solve: (a) The conservation of momentum equation pfx = pix gives
mA (vix ) A + mE (vix ) E = (mA + mE )vfx
13
(1.0 × 10
4
kg)(4.0 × 10 m/s) + 0 kg m/s = (1.0 × 1013 kg + 5.98 × 10 24 kg)vfx
⇒ vfx = 6.7 × 10−8 m/s
(b) The speed of the earth going around the sun is
2π r 2π (1.50 × 1011 m)
=
= 3.0 × 104 m/s
vE =
T
3.15 × 107 s
Thus, vfx /vE = 2.2 × 10−12 = 2.2 × 10−10,.
Assess: The earth’s recoil speed is insignificant compared to its orbital speed because of its large mass.
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Impulse and Momentum
9-17
9.38. Model: Model the skaters as particles. The two skaters, one traveling north (N) and the other traveling west
(W), are the system. Since the two skaters hold together after the “collision,” this is a case of a perfectly inelastic
collision in two dimensions. Momentum is conserved since no significant external force in the x-y plane acts on the
system during the “collision.”
Visualize:
Solve: (a) Applying conservation of momentum in the x-direction gives
(mN + mW )vfx = mN (vix ) N + mW (vix ) W
⇒ (75 kg + 60 kg)vfx = 0 kg m/s + (60 kg)(−3.5 m/s)
vfx = −1.556 m/s
Applying conservation of momentum in the y-direction gives
(mN + mW )vfy = mN (viy ) N + mW (viy ) W
⇒ (75 kg + 60 kg)vfy = (75 kg)(2.5 m/s) + 0 kg m/s
vfy = 1.389 m/s
The final speed is therefore
vf = (vfx ) 2 + (vfy ) 2 = 2.085 m/s
The time to glide to the edge of the rink is
radius of the rink
25 m
=
= 12 s
vf
2.085 m/s
(b) The location is θ = tan −1(vfy /vfx ) = 42° north of west.
Assess: A time of 12 s in covering a distance of 25 m at a speed of ≈ 2 m/s is reasonable.
9.39. Model: Model the squid and the water ejected as particles and ignore drag forces during the short time
interval over which the water is expelled (the impulse approximation). Because the external forces are negligible,
momentum will be conserved.
Visualize:
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9-18
Chapter 9
Solve: Applying conservation of momentum gives
mT v1x = mW (vfx ) W + mS (vfx )S
(vfx ) W =
1 [m v
T 1x
mW
− mS (vfx )S ] = 0.101 kg [(1.6 kg)(0.4 m/s) − (1.5 kg )(2.5 m/s)] = −31.1 m/s
This water is ejected in the direction opposite the squid’s initial velocity, so the speed with which the water is ejected
relative to the squid is
vrel = (vfx ) W − (vfx )S = −31.1 m/s − 0.4 m/s = −31.5 m/s
or 32 m/s to two significant figures.
9.40. Model: This problem deals with a case that is the opposite of a collision. The two ice skaters, heavier and
lighter, will be modeled as particles. The skaters (or particles) move apart after pushing off against each other.
During the “explosion,” the total momentum of the system is conserved.
Visualize:
Solve: The initial momentum is zero. Thus the conservation of momentum equation pfx = pix gives
mH (vfx ) H + mL (vfx ) L = 0 kg m/s ⇒ (75 kg)(vfx ) H + (50 kg)(vfx ) L = 0 kg m/s
Using the observation that the heavier skater takes 20 s to cover a distance of 30 m, we find (vfx ) H = (30 m)/(20 s) =
1.5 m/s. Thus,
(75 kg)(1.5 m/s) + (50 kg)(vfx ) L = 0 kg m/s ⇒ (vfx ) L = −2.25 m/s
Thus, the time for the lighter skater to reach the edge is
30 m
30 m
=
= 13 s
(vfx ) L 2.25 m/s
Assess: Conservation of momentum leads to a higher speed for the lighter skater, and hence a shorter time to reach
the edge of the ice rink.
9.41. Model: This problem deals with a case that is the opposite of a collision. Our system is comprised of three
coconut pieces that are modeled as particles. During the explosion, the total momentum of the system is conserved in
the x- and y-directions.
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Impulse and Momentum
9-19
Visualize:
Solve: The initial momentum is zero. From pfx = pix , we get
+ m1 (vf )1 + m3 (vf )3 cosθ = 0 kg m/s ⇒ (vf )3 cosθ =
− m1(vf )1 −m(−v0 ) v0
=
=
m3
2m
2
From pfx = pix , we get
+ m2 (vf )2 + m3 (vf )3 sinθ = 0 kg m/s ⇒ (vf )3 sinθ =
2
− m2 (vf )2 − m(−v0 ) v0
=
=
m3
2m
2
2
v
⎛v ⎞ ⎛v ⎞
(vf )3 = ⎜ 0 ⎟ + ⎜ 0 ⎟ = 0 , θ = tan −1 (1) = 45°
2
⎝ 2⎠ ⎝ 2⎠
The speed to the third piece is 14 m/s at 45° east of north.
9.42. Model: The billiard balls will be modeled as particles. The two balls, m1 (moving east) and m2 (moving
west), together are our system. This is an isolated system because any frictional force during the brief collision period
is going to be insignificant. Within the impulse approximation, the momentum of our system will be conserved in the
collision.
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9-20
Chapter 9
Visualize:
Note that m1 = m2 = m.
Solve: The equation pfx = pix yields:
m1 (vfx )1 + m2 (vfx ) 2 = m1(vix )1 + m2 (vix )2
⇒ m1(vf )1 cosθ + 0 kg m/s = m1 (vix )1 + m2 (vix ) 2
(vf )1 cosθ = (vix )1 + (vix )2 = 2.0 m/s − 1.0 m/s = 1.0 m/s
The equation pfy = piy yields:
+ m1 (vfy )1 sinθ + m2 (vfy ) 2 = 0 kg m/s ⇒ (vf )1 sinθ = −(vfy ) 2 = −1.41 m/s
⎛ 1.41 m/s ⎞
(vf )1 = (1.0 m/s) 2 + ( −1.41 m/s) 2 = 1.7 m/s, θ = tan −1 ⎜
⎟ = 55°
⎝ 1.0 m/s ⎠
The angle is below +x axis, or south of east.
9.43. Model: Model the bullet and block as particles. This is an isolated system because any frictional force during
the brief collision period is going to be insignificant. Within the impulse approximation, the momentum of our
system will be conserved in the collision. After the collision, we will consider the frictional force and apply
Newton’s second law and kinematic equations to find the distance traveled by the block + bullet.
Visualize:
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Impulse and Momentum
9-21
Solve: (a) Applying conservation of momentum to the collision gives
mvbullet + MvW = (m + M )vfx
⇒ vbullet =
m+M
vfx
m
The speed vex can be found from the kinematics equation
v12x = v02x + 2ad = vf2x + 2ad
⇒ vfx = −2ad
The acceleration in the x-direction may be found using Newton’s second law and the friction model. Because the
block does not accelerate in the y-direction, the normal force must be the same magnitude as the force due to gravity
(Newton’s second law). Thus, the frictional force is f k = − μ k n = − μ k ( m + M ) g , where the negative sign indicates
that the force acts in the negative x-direction. Newton’s second law then gives the acceleration of the block as
a = Fnet /( m + M ) = − μk (m + M ) g/(m + M ) = − μ k g
Inserting this into the expression for vex gives
vfx = −2ad = 2 μ k gd
Finally, we insert this expression for vex into the expression for the bullet’s velocity to find
vbullet =
m+M
m
2 μk gd
(b) Inserting the given quantities gives
vbullet =
0.010 kg + 10 kg
2(0.20)(9.8 m/s 2 )(0.050 m) = 4.4 × 102 m/s
0.010 kg
Visualize: If we let the bullet’s mass go to zero, we see that the bullet’s speed goes to infinity, which is reasonable
because a zero-mass bullet would need an infinite speed to make the block move. If the bullet’s mass goes to infinity,
the bullet’s speed would go to
2μ k gd , which is just the result for the initial speed of an object that decelerates to a
stop at a constant rate ( μ k g ) over a distance d. In other words, the block becomes insignificant compared to the
infinite-mass bullet.
9.44. Model: This is a two-part problem. First, we have an inelastic collision between Fred (F) and Brutus (B).
Fred and Brutus are an isolated system. The momentum of the system during collision is conserved since no
significant external force acts on the system. The second part involves the dynamics of the Fred + Brutus system
sliding on the ground.
Visualize:
Note that the collision is head-on and therefore one-dimensional.
Solve: The equation pfx = pix gives
(mF + mB )vfx = mF (vix )F + mB (vix )B
⇒ (60 kg + 120 kg)vfx = (60 kg)( −6.0 m/s) + (120 kg)(4.0 m/s)
vfx = 0.667 m/s
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9-22
Chapter 9
The positive value indicates that the motion is in the direction of Brutus.
The model of kinetic friction yields:
f k = − μ k n = − μk (mF + mB ) g = (mF + mB )a x
⇒ ax = −μk g
Using the kinematic equation v12x = v02x + 2a x ( x1 − x0 ), we get
v12x = v02x − 2μk g x1 ⇒ 0 m 2 /s 2 = vf2x − 2(0.30)(9.8 m/s 2 ) x1
0 m 2 /s 2 = (0.667 m/s) 2 − (5.9 m/s 2 ) x1 ⇒
x1 = 7.6 cm
They slide 7.6 cm in the direction Brutus was running.
Assess: After the collision, Fred and Brutus slide with a small speed but with a good amount of kinetic friction. A
stopping distance of 7.6 cm is reasonable.
9.45. Model: Model the package and the rocket as particles. This is a two-part problem. First we have an inelastic
collision between the rocket (R) and the package (P). During the collision, momentum is conserved since no
significant external force acts on the rocket and the package. However, as soon as the package + rocket system leaves
the cliff they become a projectile motion problem.
Visualize:
Solve: The minimum velocity after collision that the package + rocket must have to reach the explorer is v0 x , which
can be found as follows:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2
⇒ − 200 m = 0 m + 0 m + 12 (−9.8 m/s 2 )t12
⇒ t1 = 6.389 s
With this time, we can now find v0 x using x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 . We obtain
30 m = 0 m + v0 x (6.389 s) + 0 m ⇒ v0 x = 4.696 m/s = vfx
We now use the momentum conservation equation pfx = pix which can be written
(mR + mP )vfx = mR (vix )R + mP (vix )P
(1.0 kg + 5.0 kg)(4.696 m/s) = (1.0 kg)(vix ) R + (5.0 kg)(0 m/s) ⇒ (vix ) R = 28 m/s
9.46. Model: This is a two-part problem. First, we have an explosion that creates two particles. The momentum of
the system, comprised of two fragments, is conserved in the explosion. Second, we will use kinematic equations and
the model of kinetic friction to find the displacement of the lighter fragment.
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Impulse and Momentum
9-23
Visualize:
Solve: The initial momentum is zero. Using momentum conservation pfx = pix during the explosion,
mH (v1x ) H + mL (v1x ) L = mH (v0 x ) H + mL (v0 x ) L
⇒ 7m(v1x ) H + m(v1x ) L = 0 kg m/s ⇒ (v1x ) H = −
( 17 ) (v1x )L
Because mH slides to x2H = −8.2 m before stopping, we have
f k = μk nH = μk wH = μk mH g = mH aH
⇒ aH = μk g
Using kinematics,
(v2 x ) 2H = (v1x ) 2H + 2aH ( x2H − x1H ) ⇒ 0 m 2 /s 2 =
( 17 )
2
(v1x ) L2 + 2 μk g ( −8.2 m − 0 m)
(v1x ) L = −88.74 μk m/s
How far does mL slide? Using the information obtained above in the following kinematic equation,
(v2 x ) 2L = (v1x ) 2L + 2aL ( x2L − x1L ) ⇒ 0 m 2 /s 2 = μk (88.74) 2 − 2μ k gx2L
⇒
x2L = 4.0 × 102 m
Assess: Note that aH is positive, but aL is negative, and both are equal in magnitude to μk g. Also, x2H is
negative but x2L is positive.
9.47. Model: We will model the two fragments of the rocket after the explosion as particles. We assume the
explosion separates the two parts in a vertical manner. This is a three-part problem. In the first part, we will use
kinematic equations to find the vertical position where the rocket breaks into two pieces. In the second part, we will
apply conservation of momentum to the system (that is, the two fragments) in the explosion. In the third part, we
will again use kinematic equations to find the velocity of the heavier fragment just after the explosion.
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9-24
Chapter 9
Visualize:
Solve: The rocket accelerates for 2.0 s from rest, so
v1 y = v0 y + a y (t1 − t0 ) = 0 m/s + (10 m/s 2 )(2.0 s − 0 s) = 20 m/s
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 = 0 m + 0 m + 12 (10 m/s 2 )(2.0 s) 2 = 20 m
At the explosion the equation pfy = piy is
mL (v2 y ) L + mH (v2 y ) H = (mL + mH )v1 y
⇒ (500 kg)(v2 y )L + (1000 kg)(v2 y )H = (1500 kg)(20 m/s)
To find (v2 y ) H we must first find (v2 y ) L , the velocity after the explosion of the upper section. Using kinematics,
(v3 y ) 2L = (v2 y )2L + 2(−9.8 m/s 2 )( y3L − y2L ) ⇒ (v2 y ) L = 2(9.8 m/s 2 )(530 m − 20 m) = 99.98 m/s
Now, going back to the momentum conservation equation we get
(500 kg)(99.98 m/s) + (1000 kg)(v2 y )H = (1500 kg)(20 m/s) ⇒ (v2 y )H = −20 m/s
The negative sign indicates downward motion.
9.48. Model: Let the system be bullet + target. No external horizontal forces act on this system, so the horizontal
momentum is conserved. Model the bullet and the target as particles. Since the target is much more massive than the
bullet, it is reasonable to assume that the target undergoes no significant motion during the brief interval in which the
bullet passes through it.
Visualize:
Solve: Use the conservation of momentum equation p1x = p0 x to find
mT (v1x )T + mB (v1x ) B = mT (v0 x )T + mB (v0 x ) B = 0 + mB (v0 x ) B
(v1x )T =
mB (v0 x ) B − mB (v1x ) B 0.025 kg
=
[(1200 m/s) − (900 m/s )] = 0.021 m/s
mT
350 kg
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Impulse and Momentum
9-25
9.49. Model: Model the two blocks (A and B) and the bullet (L) as particles. This is a two-part problem. First, we
have a collision between the bullet and the first block (A). Momentum is conserved since no external force acts on
the system (bullet + block A). The second part of the problem involves a perfectly inelastic collision between the
bullet and block B. Momentum is again conserved for this system (bullet + block B).
Visualize:
Solve: For the first collision the equation pfx = pix is
mL (v1x ) L + mA (v1x ) A = mL (v0 x ) L + mA (v0 x ) A
(0.010 kg)(v1x ) L + (0.500 kg)(6.0 m/s) = (0.010 kg)(400 m/s) + 0 kg m/s ⇒ (v1x )L = 100 m/s
The bullet emerges from the first block at 100 m/s. For the second collision the equation pfx = pix is
(mL + mB )v2 x = mL (v1x ) L
⇒ (0.010 kg + 0.500 kg)v2 x = (0.010 kg)(100 m/s) ⇒ v2 x = 2.0 m/s
9.50. Model: Model Brian (B) along with his wooden skis as a particle. The “collision” between Brian and Ashley
lasts for a short time, and during this time no significant external forces act on the Brian + Ashley system. Within the
impulse approximation, we can then assume momentum conservation for our system. After finding the velocity of
the system immediately after the collision, we will apply constant-acceleration kinematic equations and the model of
kinetic friction to find the final speed at the bottom of the slope.
Visualize:
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9-26
Chapter 9
Solve: Brian skiing down for 100 m:
2
2
(v1x ) B
= (v0 x ) B
+ 2a x ( x1B − x0B ) = 0 m 2 /s 2 + 2ax (100 m − 0 m) ⇒ (v1x ) B = (200 m)ax
To obtain a x , we apply Newton’s second law to Brian in the x and y directions as follows:
∑ (Fon B ) x = wB sinθ − f k = mBa x ∑( Fon B ) y = n − wB cosθ = 0 N ⇒ n = w cosθ
From the model of kinetic friction, f k = μk n = μk wB cosθ . The x-equation thus becomes
wB sinθ − μk wB cosθ = mBa x
a x = g (sinθ − μ k cosθ ) = (9.8 m/s 2 )[sin 20° − (0.060)cos 20°] = 2.80 m/s 2
Using this value of a x , (v1x ) B = (200 m)(2.80 m/s 2 ) = 23.7 m/s. In the collision with Ashley the conservation of
momentum equation pfx = pix is
(mB + mA )v2 x = mB (v1x ) B
⇒ v2 x =
mB
80 kg
(v1x ) B =
(23.66 m/s) = 14.56 m/s
mB + mA
80 kg + 50 kg
Brian + Ashley skiing down the slope:
v32x = v22x + 2ax ( x3 − x2 ) = (14.56 m/s) 2 + 2(2.80 m/s 2 )(100 m) ⇒ v3 x = 28 m/s
That is, Brian + Ashley arrive at the bottom of the slope with a speed of 28 m/s. Note that we have used the same
value of a x in the first and the last parts of this problem. This is because a x is independent of mass.
Assess: A speed of approximately 60 mph on a ski slope of 200 m length and 20° slope is reasonable.
9.51. Model: This is an inelastic collision. The total momentum of the Volkswagen + Cadillac system is conserved.
Visualize:
Solve: Apply conservation of momentum in the x- and y-directions.
(mC + mVW )v1x = ( mC + mVW )v1 cosθ = mVW (v0 x ) VW
(mC + mVW )v1 y = ( mC + mVW )v1 sinθ = mC (v0 y )C
From the y-equation, we find
v1 =
mC (v0y )C
(mC + mVW )sinθ
=
(2000 kg)(3.0 m/s)
= 3.49 m/s
(3000 kg)sin35°
Inserting this value into the x-equation gives
( m + mVW )v1 cosθ (3000 kg)(3.49 m/s)cos35°
(v0x ) VW = C
=
= 8.6 m/s
mVW
1000 kg
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Impulse and Momentum
9-27
9.52. Model: Model Ann and cart as particles. The initial momentum is pi = 0 kg m/s in a coordinate system
attached to the ground. As Ann begins running to the right, the cart will have to recoil to the left to conserve
momentum.
Visualize:
Solve: The difficulty with this problem is that we are given Ann’s velocity of 5.0 m/s relative to the cart. If the cart
is also moving with velocity vcart then Ann’s velocity relative to the ground is not 5.0 m/s. Using the Galilean
transformation equation for velocity, Ann’s velocity relative to the ground is
(vfx ) Ann = (vfx )cart + 5.0 m/s
Now, the momentum conservation equation pix = pfx is
0 kg m/s = mAnn (vfx ) Ann + mcart (vfx )cart
⇒ 0 kg m/s = (50 kg)[(vfx )cart + 5.0 m/s] + (500 kg)(vfx )cart
(vfx )cart = −0.45 m/s
Using the recoil velocity (vfx )cart relative to the ground, we find Ann’s velocity relative to the ground to be
(vfx ) Ann = 5.00 m/s − 0.45 m/s = 4.55 m/s
The distance Ann runs relative to the ground is Δx = (vfx ) Ann Δt , where Δt is the time it takes to reach the end of the
cart. Relative to the cart, which is 15 m long, Ann’s velocity is 5 m/s. Thus, Δt = (15 m)/(5.0 m/s) = 3.0 s. Her distance
over the ground during this interval is
Δx = (vfx ) Ann Δt = (4.55 m/s)(3.0 s) = 14 m
9.53. Model: Assume that the tube is frictionless and ignore air resistance. Model the two balls as particles.
Visualize:
Solve: The initial momentum of the system is zero because both balls are stationary. Therefore, conservation of
momentum tells us that the final momentum of the system must be zero:
mvm + 3mv0 = 0 ⇒ vm = −3v0
Thus, the speed of the lighter ball is 3v0 .
Assess: The negative sign indicates that the lighter ball moves in the direction opposite the larger ball.
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9-28
Chapter 9
9.54. Model: Change in momentum is given by the impulse-momentum theorem (Equation 9.8).
tf
Solve: Using Δp = mvfx − mvix = J x = ∫ Fx (t ) dx with vix = 0, the velocity after the force has been applied is
ti
2.0 s
vfx =
1 tf
10 N
sin(2π t/4.0 s)dt
Fx (t )dx =
0.25 kg 0.0∫ s
m ∫ti
2.0 s ⎤
⎡ 4.0 s
⎛ 2π t ⎞
⎥
cos ⎜
= (40 N/kg) ⎢ −
⎟
⎢ 2π
⎝ 4.0 s ⎠ 0.0 s ⎥⎦
⎣
= −(25.5 m/s)[cos(π ) − cos(0)]
= −(25.5 m/s)( − 1 − 1)
= 51 m/s
Assess: The force is applied for half the period of 4.0 s. During that time, sin
( ) is positive, so an object initially
2π t
4.0 s
at rest acquires a positive velocity.
9.55. Solve: Apply the impulse-momentum theorem (Equation 9.8) to find the initial velocity:
Δp = ∫ Fdt
vfx = vix +
1
Fdt
m∫
2
= −5.0 m/s +
1
(4 − t 2 )dt
0.500 kg −∫2
= −5.0 m/s +
1
1 3⎞
⎛
⎜ 4t − t ⎟
0.500 kg ⎝
3 ⎠−2
= −5.0 m/s +
1
8 ⎞⎤
⎡ 8 ⎛
8 − − ⎜ −8 + ⎟ ⎥
0.500 kg ⎢⎣ 3 ⎝
3 ⎠⎦
2
= 16 m/s
9.56. Model: The two railcars make up a system. The impulse approximation is used while the spring is expanding,
so friction can be ignored.
Visualize:
Solve: Since the cars are at rest initially, the total momentum of the system is zero. Conservation of momentum
gives
0 = m1 (vfx )1 + m2 (vfx ) 2
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Impulse and Momentum
9-29
We are only told that the relative velocity of the two cars after the spring expands is 4.0 m/s, so
(vfx ) 2 − (vfx )1 = 4.0 m/s
Substitute (vfx ) 2 = (vfx )1 + 4.0 m/s into the conservation of momentum equation, then solve for (vfx )1:
0 = m1(vfx )1 + m2[(vfx )1 + 4.0 m/s]
m2 (4.0 m/s)
(90 tons)(4.0 m/s)
=−
= −3.0 m/s
(m1 + m2 )
(30 tons + 90 tons)
so the speed of the 30 ton car relative to the ground is 3.0 m/s.
Assess: The other more massive railcar has a velocity (vfx ) 2 = (vfx )1 + 4.0 m/s = 1.0 m/s. A slower speed for the more
(vfx )1 = −
massive car makes sense.
9.57. Model: This is a three-part problem. In the first part, the shell, treated as a particle, is launched as a projectile
and reaches its highest point. We will use constant-acceleration kinematic equations for this part. The shell, which is
our system, then explodes at the highest point. During this brief explosion time, momentum is conserved. In the third
part, we will again use the kinematic equations to find the horizontal distance between the landing of the lighter
fragment and the origin.
Visualize:
Solve: The initial velocity is
v0 x = v cosθ = (125 m/s)cos55° = 71.7 m/s
v0 y = v sinθ = (125 m/s)sin55° = 102.4 m/s
At the highest point, v1 y = 0 m/s and v1x = 71.7 m/s. The conservation of momentum equation pfx = pix gives
mL (v1x ) L + mH (v1x ) H = (mL + mH )v1x
The heavier particle falls straight down, so (v1x ) H = 0 m/s. Thus,
(15 kg)(v1x ) L + 0 kg m/s = (15 kg + 60 kg)(71.7 m/s) ⇒ (v1x ) L = 358 m/s
That is, the velocity of the smaller fragment immediately after the explosion is 358 m/s and this velocity is in the
horizontal x-direction. Note that (v1 y )L = 0 m/s. To find x2 , we will first find the displacement x1 − x0 and then
x2 − x1. For x1 − x0 ,
v1 y = v0 y + a y (t1 − t0 ) ⇒ 0 m/s = (102.4 m/s) + (−9.8 m/s2 )(t1 − 0 s) ⇒ t1 = 10.45 s
x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2
⇒
x1 − x0 = (71.7 m/s)(10.45 s) + 0 m = 749 m
For x2 − x1:
x2 = x1 + (v1x ) L (t2 − t1 ) + 12 a x (t2 − t1 ) 2 ⇒ x2 − x1 = (358 m/s)(10.45 s) + 0 m = 3741 m
x2 = ( x2 − x1 ) + ( x1 − x0 ) = 3741 m + 749 m = 4490 m = 4.5 km
Assess: Note that the time of ascent to the highest point is equal to the time of descent to the ground, that is,
t1 − t0 = t2 − t1.
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9-30
Chapter 9
9.58. Model: Model the proton (P) and the gold atom (G) as particles. The two constitute our system, and
momentum is conserved in the collision between the proton and the gold atom.
Visualize:
Solve: The conservation of momentum equation pfx = pix gives
mG (vfx )G + mP (vfx )P = mP (vix ) P + mG (vix )G
(197 u)(vfx )G + (1 u)(−0.90 × 5.0 × 107 m/s) = (1 u)(5.0 × 107 m/s) + 0 u m/s
(vfx )G = 4.8 × 105 m/s
9.59. Model: Model the proton (P) and the target nucleus (T) as particles. The proton and the target nucleus make
our system and in the collision between them momentum is conserved. This is due to the impulse approximation
because the collision lasts a very short time and the external forces acting on the system during this time are not
significant.
Visualize:
Solve: The conservation of momentum equation pfx = pix gives
mT (vfx )T + mP (v fx ) P = mT (vix )T + mP (vix ) P
mT (3.12 × 105 m/s) + (1 u)(−0.750 × 2.50 × 106 m/s) = 0 u m/s + (1 u)(2.50 × 106 m/s)
mT = 14.0 u
Assess: This is the mass of the nucleus of a nitrogen atom.
9.60. Model: This problem deals with an “explosion” in which a
214
Po nucleus (P) decays into an alpha-particle
(A) and a daughter nucleus (N). During the “explosion” or decay, the total momentum of the system is conserved.
Visualize:
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Impulse and Momentum
9-31
Solve: Conservation of mass requires the daughter nucleus to have mass mN = 214 u − 4 u = 210 u. The conservation
of momentum equation pfx = pix gives
mN (vfx ) N + mA (vfx ) A = (mN + mA )vP
⇒ (210 u)(vfx ) N + (4 u)(−1.92 × 107 m/s) = 0 u m/s
(vfx ) N = 3.66 × 105 m/s
9.61. Model: The neutron’s decay is an “explosion” of the neutron into several pieces. The neutron is an isolated
system, so its momentum should be conserved. The observed decay products, the electron and proton, move in
opposite directions.
Visualize:
Solve: (a) The initial momentum is pix = 0 kg m/s. The final momentum pfx = meve + mpvp is
pfx = 2.73 × 10−23 kg m/s − 1.67 × 10−22 kg m/s = −1.4 × 10−22 kg m/s
No, momentum does not seem to be conserved.
(b) and (c) If the neutrino is needed to conserve momentum, then pe + pP + pneutrino = 0 kg m/s. This requires
pneutrino = −( pe + pP ) = +1.4 × 10−22 kg m/s
The neutrino must “carry away” 1.4 × 10−22 kg m/s of momentum in the same direction as the electron.
9.62. Model: Model the two balls of clay as particles. Our system comprises these two balls. Momentum is conserved
in the perfectly inelastic collision.
Visualize:
Solve: Applying conservation of momentum in the x-direction gives
pfx = pix = m1 (vix )1 + m2 (vix ) 2
= (0.020 kg)(2.0 m/s) − (0.030 kg)(1.0 m/s)cos30° = 0.0140 kg m/s
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9-32
Chapter 9
The y-component of the final momentum is
pfy = piy = m1 (viy )1 + m2 (viy ) 2
= (0.02 kg)(0 m/s) − (0.03 kg)(1.0 m/s)sin30° = −0.0150 kg m/s
pf = (0.014 kg m/s) 2 + (−0.015 kg m/s) 2 = 0.0205 kg m/s
Since pf = (m1 + m2 )vf = 0.0205 kg m/s, the final speed is
vf =
0.0205 kg m/s
= 0.41 m/s
(0.02 + 0.03) kg
and the direction is
θ = tan −1
pfy
pfx
= tan −1
0.015
= 47° south of east
0.014
9.63. Model: Model the three balls of clay as particle 1 (moving north), particle 2 (moving west), and particle 3
(moving southeast). The three stick together during their collision, which is perfectly inelastic. The momentum of the
system is conserved.
Visualize:
Solve: The three initial momenta are
G
G
pi1 = m1vi1 = (0.020 kg)(2.0 m/s) ˆj = 0.040 ˆj kg m/s
G
G
pi2 = m2vi2 = (0.030 kg)( −3.0 m/s iˆ) = −0.090iˆ kg m/s
G
G
pi3 = m3vi3 = (0.040 kg)[(4.0 m/s)cos45°iˆ − (4.0 m/s)sin45° ˆj ] = (0.113iˆ − 0.113 ˆj ) kg m/s
G
G
G
G
Since pf = pi = pi1 + pi2 + pi3 , we have
G
G
(m1 + m2 + m3 )vf = (0.023iˆ − 0.073 ˆj ) kg m/s ⇒ vf = (0.256iˆ − 0.811 ˆj ) m/s
vf = (0.256 m/s) 2 + (−0.811 m/s) 2 = 0.85 m/s
θ = tan −1
vfy
vfx
= tan −1
0.811
= 72° below the x-axis.
0.256
9.64. Model: Model the truck (T) and the two cars (C and C′) as particles. The three forming our system stick
together during their collision, which is perfectly inelastic. Since no significant external forces act on the system
during the brief collision time, the momentum of the system is conserved.
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Impulse and Momentum
9-33
Visualize:
Solve: The three momenta are
G
G
piT = mT viT = (2100 kg)(2.0 m/s)iˆ = 4200iˆ kg m/s
G
G
piC = mCviC = (1200 kg)(5.0 m/s) ˆj = 6000 ˆj kg m/s
G
G
piC′ = mC′viC′ = (1500 kg)(10 m/s)iˆ = 15,000iˆ kg m/s
G
G
G
G
G
pf = pi = piT + piC + piC′ = (19,200iˆ + 6000 ˆj ) kg m/s
pf = (mT + mC + mC′ )vf = (19,200 kg m/s)2 + (6000 kg m/s)2
vf = 4.2 m/s, θ = tan −1
py
px
= tan −1
6000
= 17° above the + x -axis
19,200
Assess: A speed of 4.2 m/s for the entangled three vehicles is reasonable since the individual speeds of the cars and
the truck before entanglement were of the same order of magnitude.
9.65. Model: The
14
C atom undergoes an “explosion” and decays into a nucleus, an electron, and a neutrino.
Momentum is conserved in the process of “explosion” or decay.
Visualize:
G
Solve: The conservation of momentum equation pf =
G
G
G
pe + pn + pN = 0 N ⇒
G
pi = 0 kg m/s gives
G
G
G
G
G
pN = −( pe + pn ) = − meve − mn vn
= −(9.11 × 10−31 kg)(5.0 × 107 m/s)iˆ − (8.0 × 10−24 kg m/s) ˆj = −(45.55 × 10−24 iˆ + 8.0 × 10−24 ˆj ) kg m/s
pN = mN vN = (45.55 × 10−24 ) 2 + (8.0 × 10−24 ) 2 kg m/s
(2.34 × 10−26 kg)vN = 4.62 × 10−23 kg m/s ⇒ vN = 2.0 × 103 m/s
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9-34
Chapter 9
9.66. (a) A 100 g ball traveling to the left at 30 m/s is batted back to the right at 40 m/s. The force curve for the
force of the bat on the ball can be modeled as a triangle with a maximum force of 1400 N. How long is the ball in
contact with the bat?
(b)
(c) The solution is Δt = 0.100 s = 10 ms.
9.67. (a) A 200 g ball of clay traveling to the right overtakes and collides with a 400 g ball of clay traveling to the
right at 3.0 m/s. The balls stick and move forward at 4.0 m/s. What was the speed of the 200 g ball of clay?
(b)
(c) The solution is (vix ) 2 = 6.0 m/s.
9.68. (a) A 2000 kg auto traveling east at 5.0 m/s suffers a head-on collision with a small 1000 kg auto traveling
west at 4.0 m/s. They lock bumpers and stick together after the collision. What will be the speed and direction of the
combined wreckage after the collision?
(b)
(c) The solution is vfx = 2.0 m/s along the + x direction.
9.69. (a) A 150 g spring-loaded toy is sliding across a frictionless floor at 1.0 m/s. It suddenly explodes into two
pieces. One piece, which has twice the mass of the second piece, continues to slide in the forward direction at 7.5 m/s.
What is the speed and direction of the second piece?
(b)
(c) The solution is (vfx )1 = −12 m/s. The minus sign tells us that the second piece moves backward at 12 m/s.
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Impulse and Momentum
9-35
9.70. Model: The cart + man (C + M) is our system. It is an isolated system, and momentum is conserved.
Visualize:
Solve: The conservation of momentum equation pfx = pix gives
mM (vfx ) M + mC (vfx )C = mM (vix ) M + mC (vix )C
Note that (vfx ) M and (vfx )C are the final velocities of the man and the cart relative to the ground. What is given in
this problem is the velocity of the man relative to the moving cart. The man’s velocity relative to the ground is
(vfx ) M = (vfx )C − 10 m/s
With this form for (vfx ) M , we rewrite the momentum conservation equation as
mM [(vfx )C − 10 m/s] + mC (vfx )C = mM (5.0 m/s) + mC (5.0 m/s)
(70 kg)[(vfx )C − 10 m/s] + (1000 kg)(vfx )C = (1000 kg + 70 kg)(5.0 m/s)
(vfx )C [1000 kg + 70 kg] = (1070 kg)(5.0 m/s) + (70 kg)(10 m/s) ⇒ (vfx )C = 5.7 m/s
9.71. Model: This is an isolated system, so momentum is conserved in the explosion. Momentum is a vector
G
quantity, so the direction of the initial velocity vector v1 establishes the direction of the momentum vector. The final
momentum vector, after the explosion, must still point in the +x-direction. The two known pieces continue to move
along this line and have no y-components of momentum. The missing third piece cannot have a y-component of
momentum if momentum is to be conserved, so it must move along the x-axis—either straight forward or straight
backward. We can use conservation laws to find out.
Visualize:
Solve: From the conservation of mass, the mass of piece 3 is
m3 = mtotal − m1 − m2 = 7.0 × 105 kg
To conserve momentum along the x-axis, we require
pi = mtotalvi = pf = p1f + p2f + p3f = m1v1f + m2v2f + p3f
p3f = mtotalvi − m1v1f − m2v2f = +1.02 × 1013 kg m/s
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9-36
Chapter 9
Because p3f > 0, the third piece moves in the + x-direction, that is, straight forward. Because we know the mass m3 ,
we can find the velocity of the third piece as follows:
v3f =
p3f 1.02 × 1013 kg m/s
=
= 1.5 × 107 m/s
m3
7.0 × 105 kg
The third piece moves to the right with a speed of 1.5 × 107 m/s.
9.72. Model: The projectile + wood ball are our system. In the collision, momentum is conserved.
Visualize:
Solve: The momentum conservation equation pfx = pix is
(mP + mB )vfx = mP (vix ) P + mB (vix )B
⇒ (1.0 kg + 20 kg)vfx = (1.0 kg)(vix ) P + 0 kg m/s
(vix ) P = 21vfx
We therefore need to determine vfx . Newton’s second law for circular motion is
T − FG = T − (mP + mB ) g =
( mP + mB )vf2x
r
Using Tmax = 400 N, this equation gives
400 N − (1.0 kg + 20 kg)(9.8 m/s) =
(1.0 kg + 20 kg)vf2x
2.0 m
⇒ (vfx )max = 4.3 m/s
Going back to the momentum conservation equation,
(vix ) P = 21vfx = (21)(4.3 m/s) = 90 m/s
That is, the largest speed this projectile can have without causing the cable to break is 90 m/s.
9.73. Model: This is an “explosion” problem and momentum is conserved. The two-stage rocket is our system.
Visualize:
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Impulse and Momentum
9-37
Solve: Relative to the ground, the conservation of momentum equation pfx = pix gives
m1 (vfx )1 + m2 (vfx ) 2 = ( m1 + m2 )v1x
3 m 2 (vfx )1 + m2 (vfx ) 2 = (4 m 2 )(1200 m/s) ⇒ 3(vfx )1 + (vfx ) 2 = 4800 kg m/s
The fact that the first stage is pushed backward at 35 m/s relative to the second can be written
(vfx )1 = −35 m/s + (vfx ) 2
Substituting this form of (vfx )1 in the conservation of momentum equation,
3[−35 m/s + (vfx ) 2 ] + (vfx ) 2 = 4800 kg m/s ⇒ (vfx ) 2 = 1.2 km/s
9.74. Model: Let the system be rocket + bullet. This is an isolated system, so momentum is conserved.
Visualize: The fact that the bullet’s velocity relative to the rocket is 139,000 can be written (vf ) B = (vf ) R + 139,000 m/s.
Solve: Consider the firing of one bullet when the rocket has mass M and velocity vi . The conservation of momentum
equation pf = pi gives
( M − 5kg)vf + (5 kg)(vf + 139,000 m/s) = Mvi
5 kg
(139,000 m/s)
M
The rocket starts with mass M = 2000 kg, which is much larger than 5 kg. If only a few bullets are needed, M will not
Δv = vf − vi = −
change significantly as the rocket slows. If we assume that M remains constant at 2000 kg, the loss of speed per bullet
is Δv = −347.5 m/s = −1250 km/h. Thus exactly 8 bullets will reduce the speed by 10,000 km/h, from 25,000 km/h to
15,000 km/h. If you’re not sure that treating M as a constant is valid, you can calculate Δv for each bullet and reduce
M by 5 kg after each shot. The loss of mass causes Δv to increase slightly for each bullet. An eight-step calculation
then finds that 8 bullets will slow the rocket to 14,900 km/h. Seven bullets wouldn’t be enough, and 9 would slow the
rocket far too much.
9.75. Visualize:
Solve: Ladies and gentlemen of the jury, how far would the chair slide if it was struck with a bullet from my client’s
gun? We know the bullet’s velocity as it leaves the gun is 450 m/s. The bullet travels only a small distance to the
chair, so we will neglect any speed loss due to air resistance. The bullet and chair can be considered an isolated
system during the brief interval of the collision. The bullet embedded itself in the chair, so this was a perfectly
inelastic collision. Momentum conservation allows us to calculate the velocity of the chair immediately after the
collision as follows:
m (v )
(450 m/s)(0.010 kg)
pix = pfx ⇒ mB (vi ) B = (mB + mC )vf ⇒ vf = B i B =
= 0.225 m/s
20.01 kg
mB + mC
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9-38
Chapter 9
This is the velocity immediately after the collision when the chair starts to slide but before it covers any distance. For
the purpose of the problem in dynamics, call this the initial velocity v0 . The free-body diagram of the chair shows
three forces. Newton’s second law applied to the chair (with the embedded bullet) is
ax = a =
( Fnet ) y n − mtot g
( Fnet ) x − f k
μ n
=
= − k , a y = 0 m/s 2 =
=
mtot
mtot
mtot
mtot
mtot
where we’ve used the friction model in the x-equation. The y-equation yields n = mtot g , and the x-equation yields
a = − μk g = −1.96 m/s 2 . We know the coefficient of kinetic friction because it is a wood chair sliding on a wood floor.
Finally, we have to determine the stopping distance of the chair. The motion of the chair ends with v1 = 0 m/s after
sliding a distance Δx, so
v12 = 0 m 2 /s 2 = v02 + 2aΔx ⇒ Δx = −
v02
(0.225 m/s) 2
=−
= 0.013 m = 1.3 cm
2a
2( −1.96 m/s 2 )
If the bullet lost any speed in the air before hitting the chair, the sliding distance would be even less. So you can see
that the most the chair could slide if it had been struck by a bullet from my client’s gun would be 1.3 cm. But in
actuality, the chair slid 3 cm, more than twice as far. The murder weapon, ladies and gentlemen, was a much more
powerful gun than the one possessed by my client. I rest my case.
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10
ENERGY
Conceptual Questions
10.1. Kinetic energy depends on speed. Potential energy depends on position.
10.2. No, kinetic energy can never be negative. Kinetic energy is energy of motion. Motion may stop, but it can’t be
negative. Speed has no direction and cannot be negative. Yes, gravitational potential energy can be negative. Potential energy depends upon position, which can be positive or negative.
1
2
10.3. We must calculate the new kinetic energy and compare it to the original value. Originally, K = mv 2 . With a
1
⎛1
⎞
velocity of 3v, K ′ = m(3v) 2 = 9 ⎜ mv 2 ⎟ = 9 K . The kinetic energy increases by a factor of 9.
2
⎝2
⎠
10.4. We have
K A = 8K B
1
⎛1
⎞
mA vA2 = 8 ⎜ mBvB2 ⎟
2
2
⎝
⎠
1
Since m A = mB ,
2
1⎛1
⎞ 2
⎛1
2⎞
⎜ mB ⎟ vA = 8 ⎜ mBvB ⎟
2⎝ 2
⎠
⎝2
⎠
v
⇒ A =4
vB
10.5. Conservation of energy tells us that U i = K f , since the car starts at rest. Originally, this means that in rolling
down a track of height h,
1
mgh = mv02
2
To go twice as fast at the bottom, we must find the height h′ such that
1
mgh′ = m(2v0 ) 2
2
⎛1
⎞
⇒ mgh′ = 4 ⎜ mv02 ⎟ .
⎝2
⎠
So h′ = 4h. You must increase the track height by a factor of 4.
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10-1
10-2
Chapter 10
10.6. va = vb = vc . They each start with the same kinetic energy and they each have the same change in potential
energy, so they end with the same kinetic energy and, thus, the same speed.
10.7. va = vb = vc . The balls start off with the same kinetic energy and have the same change in potential energy, so
their final kinetic energy is the same.
10.8. (a) We identify the equilibrium position se = 10 cm. At s = 11 cm, Δs = s − se = 11 cm − 10 cm = 1 cm, and
Fsp = F = − k Δs = − k (1 cm). To get Fsp = 3F , we must have 3F = −k ( Δs )′ , which means (Δs )′ = 3 cm. So the
spring must have length 10 cm + 3 cm = 13 cm.
(b) Note that the direction of the force is reversed when the spring is compressed.
To get Fsp = −2 F , we must have −2 F = −k (Δs )′ , which means (Δs )′ = −2 cm. So the spring must have length
10 cm − 2 cm = 8 cm.
10.9. Note that Carlos takes the place of the wall, and that the force on the spring is still 200 N. The spring still
stretches 20 cm.
1
2
energy by a factor of 4. Doubling k doubles the stored energy.
10.10. (U s )d > (U s )c > (U s ) b = (U s )a . U s = k ( Δs ) 2 . Increasing the stretch by a factor of 2 increases the stored
1
2
10.11. The original spring stores energy U = k (1.0 cm)2 . For a spring with spring constant k ′ = 2k ,
If U ′ = U , then
1
1
U ′ = k ′(Δs ) 2 = (2k )(Δs )2
2
2
1
1
k (1.0 cm) 2 = (2k )(Δs ) 2
2
2
1
⇒ Δs =
cm = 0.71 cm
2
10.12. Energy conservation tells us that the initial potential energy stored in the spring is equal to the final kinetic
energy of the ball.
1
1
k (Δs ) 2 = mv02
2
2
When the spring is compressed twice as far,
1
1
k (2Δs )2 = m(2v0 ) 2
2
2
So the ball speed increases by a factor of 2.
10.13. (a) At x = 6 m the particle has the most kinetic energy. The kinetic energy is the difference between the
total energy (TE) and potential energy (PE) curves. At x = 3 m the particle’s speed is locally a maximum, but is not
as fast as at x = 6 m.
(b) The turning points for the particle with total energy (TE) shown are at x = 2 m and x = 8 m.
(c) The particle could remain at rest in stable equilibrium at x = 3 m and x = 6 m. The particle could also remain at
rest in unstable equilibrium at x = 1 m and x = 4 m.
10.14. The problem can be divided into three parts: (1) from when the first ball is released and to just before it hits
the stationary ball, (2) the two balls collide, and (3) the two balls swing up together just after the collision to their
highest point. Energy is conserved in parts (1) and (3) as the balls swing like pendulums, but during the collision in
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Energy
10-3
part (2) momentum is conserved but energy is not. So both energy and momentum conservation are each separately
used as you work through each part of the problem.
Exercises and Problems
Section 10.2 Kinetic Energy and Gravitational Potential Energy
10.1. Model: We will use the particle model for the bullet (B) and the running student (S).
Visualize:
Solve: For the bullet,
1
1
K B = mBvB2 = (0.010 kg)(500 m/s)2 = 1250 J
2
2
For the running student,
1
1
KS = mSvS2 = (75 kg)(5.5 m/s) 2 = 206 J
2
2
Thus, the bullet has the larger kinetic energy.
Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculation shows
this dependence. Although the mass of the bullet is 7500 times smaller than the mass of the student, its speed is more
than 90 times larger.
10.2. Model: Model the hiker as a particle.
Visualize:
The origin of the coordinate system chosen for this problem is at sea level so that the hiker’s position in Death Valley
is y0 = −8.5 m.
Solve: The hiker’s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is
ΔU = U gf − U gi = mgyf − mgyi = mg ( yf − yi )
= (65 kg)(9.8 m/s 2 )[4420 m − (−85 m)] = 2.9 × 106 J
Assess: Note that ΔU is independent of the origin of the coordinate system.
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10-4
Chapter 10
10.3. Model: Model the compact car (C) and the truck (T) as particles.
Visualize:
Solve: For the kinetic energy of the compact car and the kinetic energy of the truck to be equal,
mT
1
1
20,000 kg
K C = K T ⇒ mCvC2 = mT vT2 ⇒ vC =
vT =
(25 km/h) = 112 km/h
mC
2
2
1000 kg
Assess: A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass.
10.4. Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic and
potential energy does not change as the car falls.
Visualize:
Solve: (a) The kinetic energy of the car is
1
1
K C = mCvC2 = (1500 kg)(30 m/s)2 = 6.75 × 105 J
2
2
The car’s kinetic energy is 6.8 × 105 J.
(b) Let us relabel K C as K f and place our coordinate system at yf = 0 m so that the car’s potential energy U gf is
zero, its velocity is vf , and its kinetic energy is K f . At position yi , vi = 0 m/s or Ki = 0 J, and the only energy the
car has is U gi = mgyi . Since the sum K + U g is unchanged by motion, K f + U gf = Ki + U gi . This means
K f + mgyf = Ki + mgyi ⇒ K f + 0 = Ki + mgyi
⇒ yi =
(K f − Ki )
(6.75 × 105 J − 0 J)
=
= 46 m
mg
(1500 kg)(9.8 m/s 2 )
(c) From part (b),
yi =
2
2
(K f − Ki ) 12 mvf − 12 mvi (vf2 − vi2 )
=
=
mg
mg
2g
Free fall does not depend upon the mass.
10.5. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change
as the ball rises and falls.
Visualize:
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Energy
10-5
The figure shows a ball’s before-and-after pictorial representation for the three situations in parts (a), (b), and (c).
Solve: The quantity K + U g is the same during free fall: K f + U gf = Ki + U gi . We have
(a)
1 2
1
mv1 + mgy1 = mv02 + mgy0
2
2
⇒ y1 = (v02 − v12 )/2 g = [(10 m/s)2 − (0 m/s)2 ]/(2 × 9.8 m/s 2 ) = 5.10 m
5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground.
1
1
(b) mv22 + mgy2 = mv02 + mgy0
2
2
Since y2 = y0 = 0, we get for the magnitudes v2 = v0 = 10 m/s.
(c)
1 2
1
mv3 + mgy3 = mv02 + mgy0 ⇒ v32 + 2 gy3 = v02 + 2 gy0 ⇒ v32 = v02 + 2 g ( y0 − y3 )
2
2
⇒ v32 = (10 m/s) 2 + 2(9.8 m/s 2 )[0 m − (−20 m)] = 492 m 2 /s 2
This means the magnitude of v3 is equal to 22 m/s.
Assess: Note that the ball’s speed as it passes the window on its way down is the same as the speed with which it
was tossed up, but in the opposite direction.
10.6. Model: This is a problem of free fall. The sum of the kinetic and gravitational potential energy for the ball,
considered as a particle, does not change during its motion.
Visualize:
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10-6
Chapter 10
The figure shows the ball’s before-and-after pictorial representation for the two situations described in parts (a) and (b).
Solve: The quantity K + U g is the same during free fall. Thus, K f + U gf = Ki + U gi .
(a)
1 2
1
mv1 + mgy1 = mv02 + mgy0 ⇒ v02 = v12 + 2 g ( y1 − y0 )
2
2
⇒ v02 = (0 m/s) 2 + 2(9.8 m/s 2 )(10 m − 1.5 m) = 166.6 m 2 /s 2 ⇒ v0 = 12.9 m/s ≈ 13 m/s
(b)
1 2
1
mv2 + mgy2 = mv02 + mgy0 ⇒ v22 = v02 + 2 g ( y0 − y2 )
2
2
⇒ v22 = 166.6 m 2 /s 2 + 2(9.8 m/s 2 )(1.5 m − 0 m) ⇒ v2 = 14 m/s
Assess: An increase in speed from 12.9 m/s to 14.0 m/s as the ball falls through a distance of 1.5 m is reasonable.
Also, note that mass does not appear in the calculations that involve free fall.
10.7. Model: Model the mother and the son as particles.
Visualize: mmother = 4 mson .
Solve: The energy conservation equation K mother = Kson is
1
1
v
2
2
2
2
mmother vmother
= mson vson
⇒ (4 mson )vmother
= mson vson
⇒ son = 2.0
2
2
vmother
Assess: The result vson = 2vmother , combined with the fact that mson = 14 mmother , is a consequence of the way kinetic
energy is defined: It is directly proportional to the mass and to the square of the speed.
Section 10.3 A Closer Look at Gravitational Potential Energy
10.8. Model: Model the skateboarder as a particle. Assuming that the track offers no rolling friction, the sum of the
skateboarder’s kinetic and gravitational potential energy does not change during his rolling motion.
Visualize:
The vertical displacement of the skateboarder is equal to the radius of the track.
Solve: The quantity K + U g is the same at the upper edge of the quarter-pipe track as it was at the bottom. The energy conservation equation K f + U gf = Ki + U gi is
1 2
1
mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi )
2
2
vi2 = (0 m/s) 2 + 2(9.8 m/s 2 )(3.0 m − 0 m) = 58.8 m 2 /s 2 ⇒ vi = 7.7 m/s
Assess: Note that we did not need to know the skateboarder’s mass, as is the case with free-fall motion.
10.9. Model: Model the puck as a particle. Since the ramp is frictionless, the sum of the puck’s kinetic and gravitational potential energy does not change during its sliding motion.
Visualize:
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Energy
10-7
Solve: The quantity K + U g is the same at the top of the ramp as it was at the bottom. The energy conservation
equation K f + U gf = Ki + U gi is
1 2
1
mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi )
2
2
⇒ vi2 = (0 m/s) 2 + 2(9.8 m/s 2 )(1.03 m − 0 m) = 20.2 m 2 /s 2 ⇒ vi = 4.5 m/s
Assess: An initial push with a speed of 4.5 m/s ≈ 10 mph to cover a distance of 3.0 m up a 20° ramp seems reasonable.
10.10. Model: In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential
energy does not change as the pendulum swings from one side to the other.
Visualize:
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10-8
Chapter 10
The figure shows the pendulum’s before-and-after pictorial representation for the two situations described in parts (a)
and (b).
Solve: (a) The quantity K + U g is the same at the lowest point of the trajectory as it was at the highest point. Thus,
K1 + U g1 = K 0 + U g0 means
1 2
1
mv1 + mgy1 = mv02 + mgy0 ⇒ v12 + 2 gy1 = v02 + 2 gy0
2
2
⇒ v12 + 2 g (0 m) = (0 m/s) 2 + 2 gy0 ⇒ v1 = 2 gy0
From the pictorial representation, we find that y0 = L − Lcos30°. Thus,
v1 = 2 gL(1 − cos30°) = 2(9.8 m/s 2 )(0.75 m)(1 − cos30°) = 1.403 m/s
The speed at the lowest point is 1.4 m/s.
(b) Since the quantity K + U g does not change, K 2 + U g 2 = K1 + U g1. We have
1 2
1
mv2 + mgy2 = mv12 + mgy1 ⇒ y2 = (v12 − v22 )/2 g
2
2
⇒ y2 = [(1.403 m/s) 2 − (0 m/s)2 ]/(2 × 9.8 m/s2 ) = 0.100 m
Since y2 = L − Lcosθ , we obtain
L − y2 (0.75 m) − (0.10 m)
=
= 0.8667 ⇒ θ = cos −1 (0.8667) = 30°
L
(0.75 m)
That is, the pendulum swings to the other side by 30°.
Assess: The swing angle is the same on either side of the rest position. This result is a consequence of the fact that
the sum of the kinetic and gravitational potential energy does not change. This is shown as well in the energy bar
chart in the figure.
cosθ =
10.11. Model: Model the child and swing as a particle, and assume the chain to be massless. In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change during the
swing’s motion.
Visualize:
Solve: The quantity K + U g is the same at the highest point of the swing as it is at the lowest point. That is,
K 0 + U g0 = K1 + U g1. It is clear from this equation that maximum kinetic energy occurs where the gravitational potential energy is the least. This is the case at the lowest position of the swing. At this position, the speed of the swing
and child will also be maximum. The above equation is
1 2
1
mv0 + mgy0 = mv12 + mgy1 ⇒ v12 = v02 + 2 g ( y0 − y1)
2
2
⇒ v12 = (0 m/s) 2 + 2 g ( y0 − 0 m) ⇒ v1 = 2 gy0
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Energy
10-9
We see from the pictorial representation that
y0 = L − Lcos45° = (3.0 m) − (3.0 m)cos45° = 0.879 m
⇒ v1 = 2 gy0 = 2(9.8 m/s 2 )(0.879 m) = 4.2 m/s
Assess: We did not need to know the swing’s or the child’s mass. Also, a maximum speed of 4.2 m/s is reasonable.
10.12. Model: Model the car as a particle with zero rolling friction and no air resistance. The sum of the kinetic
and gravitational potential energy, therefore, does not change during the car’s motion.
Visualize:
Solve: The initial energy of the car is
1
1
K 0 + U g0 = mv02 + mgy0 = (1500 kg)(10.0 m/s) 2 + (1500 kg)(9.8 m/s 2 )(10 m) = 2.22 × 105 J
2
2
The car increases its height to 15 m at the gas station. The conservation of energy equation K 0 + U g0 = K1 + U g1 is
1
1
2.22 × 105 J = mv12 + mgy1 ⇒ 2.22 × 105 J = (1500 kg)v12 + (1500 kg)(9.8 m/s 2 )(15 m)
2
2
⇒ v1 = 1.4 m/s
Assess: A lower speed at the gas station is reasonable because the car has decreased its kinetic energy and increased
its potential energy compared to its starting values.
Section 10.4 Restoring Forces and Hooke’s Law
10.13. Model: Assume that the spring is ideal and obeys Hooke’s law.
Visualize: According to Hooke’s law, the spring force acting on a mass (m) attached to the end of a spring is given
as Fsp = k Δx, where Δx is the change in length of the spring. If the mass m is at rest, then Fsp is also equal to the
gravitational force FG = mg .
Solve: We have Fsp = k Δx = mg . We want a 0.100 kg mass to give Δx = 0.010 m. This means
k = mg/Δx = (0.100 kg)(9.8 N/m)/(0.010 m) = 98 N/m
10.14. Model: Assume an ideal spring that obeys Hooke’s law.
Visualize:
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10-10
Chapter 10
Solve: (a) The spring force on the 2.0 kg mass is Fsp = −k Δy. Notice that Δy is negative, so Fsp is positive. This
force is equal to mg, because the 2.0 kg mass is at rest. We have − k Δy = mg . Solving for k:
k = −( mg/Δy ) = −(2.0 kg)(9.8 m/s 2 )/(−0.15 m − (−0.10 m)) = 392 N/m
The spring constant is 3.9 × 102 N/m.
(b) Again using − k Δy = mg:
Δy = − mg/k = −(3.0 kg)(9.8 m/s 2 )/(392 N/m)
y′ − ye = −0.075 m ⇒ y′ = ye − 0.075 m = −0.10 m − 0.075 m = −0.175 m = −17.5 cm
The length of the spring is 17.5 cm when a mass of 3.0 kg is attached to the spring. The position of the end of the
spring is negative because it is below the origin, but length must be a positive number.
10.15. Model: Model the student (S) as a particle and the spring as obeying Hooke’s law.
Visualize:
Solve: According to Newton’s second law the force on the student is
Σ(Fon S ) y = Fspring on S − FG = ma y
⇒ Fspring on S = FG + ma y = mg + ma y = (60 kg)(9.8 m/s 2 + 3.0 m/s 2 ) = 768 N
Since Fspring on S = FS on spring = k Δy, k Δy = 768 N. This means Δy = (768 N)/(2500 N/m) = 0.31 m.
10.16. Model: Assume the spring is ideal and obeys Hooke’s law.
Visualize: The stretch produced by hanging mass m is L1 − L0 .
Solve: For a hanging mass of m
L1 = L0 + ( L1 − L0 )
If we double the hanging mass to 2m, then
L2 = L0 + 2( L1 − L0 )
and, indeed, for any mass nm,
Ln = L0 + n( L1 − L0 )
In particular for 3m
L3 = L0 + 3( L1 − L0 ) = −2 L0 + 3L1
Assess: Even though one term is negative the answer won’t be because L1 > L0 .
10.17. Model: Assume that the spring is ideal and obeys Hooke’s law. We also model the 5.0 kg mass as a particle.
Visualize: We will use the subscript s for the scale and sp for the spring.
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Energy
10-11
Solve: (a) The scale reads the upward force Fs on m that it applies to the mass. Newton’s second law gives
∑ ( Fon m ) y = Fs on m − FG = 0 ⇒ Fs on m = FG = mg = (5.0 kg)(9.8 m/s2 ) = 49 N
(b) In this case, the force is
∑ ( Fon m ) y = Fs on m + Fsp − FG = 0 ⇒ 20 N + k Δy − mg = 0
⇒ k = (mg − 20 N)/Δy = (49 N − 20 N)/0.02 m = 1450 N/m
The spring constant for the lower spring is 1.45 × 103 N/m.
(c) In this case, the force is
∑ ( Fon m ) y = Fsp − FG = 0 ⇒ k Δy − mg = 0
⇒ Δy = mg/k = (49 N)/(1450 N/m) = 0.0338 m = 3.4 cm
Section 10.5 Elastic Potential Energy
10.18. Model: Assume an ideal spring that obeys Hooke’s law.
Solve: The elastic potential energy of a spring is defined as U s = 12 k ( Δs ) 2 , where Δs is the magnitude of the
stretching or compression relative to the unstretched or uncompressed length. ΔU s = 0 when the spring is at its equilibrium length and Δs = 0. We have U s = 200 J and k = 1000 N/m. Solving for Δs:
Δs = 2U s /k = 2(200 J)/1000 N/m = 0.632 m
10.19. Model: Assume the spring is ideal and obeys Hooke’s law. Then the potential energy of a stretched spring is
1
U sp = k (Δs )2 .
2
Visualize: Use ratios to solve this problem. Use primed variables for the new situation with the spring stretched
three times as far.
Solve:
′
U sp
1 k ( Δs′) 2
2
U sp 1 k ( Δs ) 2
2
′
U sp
= 9U sp
=
=
1 k (3Δs ) 2
2
1 k ( Δs ) 2
2
= 32 = 9
= 9(2.0 J) = 18 J
Assess: The stored energy scales with the square of the spring stretch.
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10-12
Chapter 10
10.20. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energy
K + U s is conserved. Also model the book as a particle.
Visualize:
The figure shows a before-and-after pictorial representation. The compressed spring will push on the book until the
spring has returned to its equilibrium length. We put the origin of our coordinate system at the equilibrium position
of the free end of the spring. The energy bar chart shows that the potential energy of the compressed spring is entirely
transformed into the kinetic energy of the book.
Solve: The conservation of energy equation K 2 + U s2 = K1 + U s1 is
1 2 1
1
1
mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe )2
2
2
2
2
Using x2 = xe = 0 m and v1 = 0 m/s, this simplifies to
kx12
1 2 1
(1250 N/m)(0.040 m) 2
=
= 2.0 m/s
mv2 = k ( x1 − 0 m) 2 ⇒ v2 =
m
2
2
(0.500 kg)
Assess: This problem cannot be solved using constant-acceleration kinematic equations. The acceleration is not a
constant in this problem, since the spring force, given as Fs = − k Δx, is directly proportional to Δx or | x − xe |.
10.21. Model: Assume an ideal spring that obeys Hooke’s law. Since there is no friction, the mechanical energy
K + U s is conserved. Also, model the block as a particle.
Visualize:
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Energy
10-13
The figure shows a before-and-after pictorial representation. We have put the origin of our coordinate system at the
equilibrium position of the free end of the spring. This gives us x1 = xe = 0 cm and x2 = 2.0 cm.
Solve: The conservation of energy equation K 2 + U s2 = K1 + U s1 is
1 2 1
1
1
mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe )2
2
2
2
2
Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 0.020 m, we get
m
1
1
k ( x2 − xe )2 = mv12 ⇒ Δx = ( x2 − xe ) =
v1
k
2
2
That is, the compression is directly proportional to the velocity v1. When the block collides with the spring with
twice the earlier velocity (2v1), the compression will also be doubled to 2( x2 − xe ) = 2(2.0 cm) = 4.0 cm.
Assess: This problem shows the power of using energy conservation over using Newton’s laws in solving problems
involving nonconstant acceleration.
10.22. Model: Model the grocery cart as a particle and the spring as an ideal that obeys Hooke’s law. We will also
assume zero rolling friction during the compression of the spring, so that mechanical energy is conserved.
Visualize:
The figure shows a before-and-after pictorial representation. The “before” situation is when the cart hits the spring in
its equilibrium position. We put the origin of our coordinate system at this equilibrium position of the free end of the
spring. This gives x1 = xe = 0 and (x2 − xe ) = 60 cm.
Solve: The conservation of energy equation K 2 + U s2 = K1 + U s1 is
1 2 1
1
1
mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe )2
2
2
2
2
Using v2 = 0 m/s, (x2 − xe ) = 0.60 m, and x1 = xe = 0 m gives:
k
1
1
250 N/m
k ( x2 − xe )2 = mv12 ⇒ v1 =
( x2 − xe ) =
(0.60 m) = 3.0 m/s
m
2
2
10 kg
10.23. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and thus the mechanical energy K + U s + U g is conserved.
Visualize:
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10-14
Chapter 10
We place the origin of our coordinate system at the spring’s compressed position y1 = 0. The rock leaves the spring
with velocity v2 as the spring reaches its equilibrium position.
Solve: (a) The conservation of mechanical energy equation is
1
1 2 1
1
K 2 + U s2 + U g2 = K1 + U s1 + U g1
mv2 + k (y2 − ye ) 2 + mgy2 = mv12 + k (y1 − ye )2 + mgy1
2
2
2
2
Using y2 = ye , y1 = 0 m, and v1 = 0 m/s, this simplifies to
1 2
1
mv2 + 0 J + mgy2 = 0 J + k (y1 − ye ) 2 + 0
2
2
1
1
2
2
(0.400 kg)v2 + (0.400 kg)(9.8 m/s )(0.30 m) = (1000 N/m)(0.30 m) 2 ⇒ v2 = 14.8 m/s
2
2
(b) Let us use the conservation of mechanical energy equation once again to find the highest position ( y3 ) of the
rock where its speed (v3 ) is zero:
1
1
K3 + U g3 = K 2 + U g2 ⇒ mv32 + mgy3 = mv22 + mgy2
2
2
v 2 (14.8 m/s) 2
1
⇒ 0 + g ( y3 − y2 ) = v22 ⇒ ( y3 − y2 ) = 2 =
= 11.2 m
2
2 g 2(9.8 m/s 2 )
If we assume the spring’s length to be 0.5 m, then the distance between ground and fruit is 11.2 m + 0.5 m = 11.7 m.
This is much smaller than the distance of 15 m between fruit and ground. So, the rock does not reach the fruit, and
the contestants go hungry.
10.24. Model: Model the jet plane as a particle, and the spring as an ideal that obeys Hooke’s law. We will also
assume zero rolling friction during the stretching of the spring, so that mechanical energy is conserved.
Visualize:
The figure shows a before-and-after pictorial representation. The “before” situation occurs just as the jet plane lands on
the aircraft carrier and the spring is in its equilibrium position. We put the origin of our coordinate system at the right
free end of the spring. This gives x1 = xe = 0 m. Since the spring stretches 30 m to stop the plane, x2 − xe = 30 m.
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Energy
10-15
Solve: The conservation of energy equation K 2 + U s2 = K1 + U s1 for the spring-jet plane system is
1 2 1
1
1
mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe )2
2
2
2
2
Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 30 m yields
k
1
1
60,000 N/m
k ( x2 − xe )2 = mv12 ⇒ v1 =
( x2 − x1) =
(30 m) = 60 m/s
m
2
2
15,000 kg
Assess: A landing speed of 60 m/s or ≈ 120 mph is reasonable.
Section 10.6 Energy Diagrams
10.25. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.
Visualize:
The particle is released from rest at x = 1.0 m. That is, K = 0 at x = 1.0 m. Since the total energy is given by
E = K + U , we can draw a horizontal total energy (TE) line through the point of intersection of the potential energy
curve (PE) and the x = 1.0 m line. The distance from the PE curve to the TE line is the particle’s kinetic energy.
These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum
K + U does not change.
Solve: (a) We have E = 4.0 J and this energy is a constant. For x < 1.0, U > 4.0 J and, therefore, K must be negative to keep E the same (note that K = E − U or K = 4.0 J − U ). Since negative kinetic energy is unphysical, the particle cannot move to the left. That is, the particle will move to the right of x = 1.0 m.
(b) The expression for the kinetic energy is E − U . This means the particle has maximum speed or maximum kinetic
energy when U is minimum. This happens at x = 4.0 m. Thus,
K max = E − U min = (4.0 J) − (1.0 J) = 3.0 J
1 2
2(3.0 J)
8.0 J
=
= 17.3 m/s
mvmax = 3.0 J ⇒ vmax =
2
m
0.020 kg
The particle possesses this speed at x = 4.0 m.
(c) The total energy (TE) line intersects the potential energy (PE) curve at x = 1.0 m and x = 6.0 m. These are the
turning points of the motion.
10.26. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.
Visualize:
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10-16
Chapter 10
The particle with a mass of 500 g is released from rest at A. That is, K = 0 at A. Since E = K + U = 0 J + U , we can
draw a horizontal TE line through U = 5.0 J. The distance from the PE curve to the TE line is the particle’s kinetic
energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the
sum K + U does not change.
Solve: The kinetic energy is given by E − U , so we have
1 2
mv = E − U ⇒ v = 2( E − U )/m
2
Using U B = 2.0 J, U C = 3.0 J, and U D = 0 J, we get
vB = 2(5.0 J − 2.0 J)/0.500 kg = 3.5 m/s
vC = 2(5.0 J − 3.0 J)/0.500 kg = 2.8 m/s
vD = 2(5.0 J − 0 J)/0.500 kg = 4.5 m/s
10.27. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.
Visualize:
For the speed of the particle at A that is needed to reach B to be a minimum, the particle’s kinetic energy as it reaches
the top must be zero. Similarly, the minimum speed at B for the particle to reach A obtains when the particle just
makes it to the top with zero kinetic energy.
Solve: (a) The energy equation K A + U A = K top + U top is
1 2
mvA + U A = 0 J + U top
2
⇒ vA = 2(U top − U A )/m = 2(5.0 J − 2.0 J)/0.100 kg = 7.7 m/s
(b) To go from point B to point A, K B + U B = K top + U top is
1 2
mvB + U B = 0 J + U top
2
⇒ vB = 2(U top − U B )/m = 2(5.0 J − 0 J)/0.100 kg = 10.0 m/s
Assess: The particle requires a higher kinetic energy to reach A from B than to reach B from A.
10.28. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.
Visualize:
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Energy
10-17
Since the particle oscillates between x = 2.0 mm and x = 8.0 mm, the speed of the particle is zero at these points. That
is, for these values of x, E = U = 5.0 J, which defines the total energy (TE) line. The distance from the potential
energy (PE) curve to the TE line is the particle’s kinetic energy. These values are transformed as the position
changes, but the sum K + U does not change.
Solve: The equation for total energy E = U + K means K = E − U , so that K is maximum when U is minimum. We
have
1 2
K max = mvmax
= 5.0 J − U min
2
⇒ vmax = 2(5.0 J − U min )/m = 2(5.0 J − 1.0 J)/0.0020 kg = 63 m/s
Section 10.7 Elastic Collisions
10.29. Model: We assume this is a one-dimensional collision that obeys the conservation laws of momentum and
mechanical energy.
Visualize:
Note that momentum conservation alone is not sufficient to solve this problem because the two final velocities (vfx )1
and (vfx ) 2 are unknowns and cannot be determined from one equation.
Solve: Momentum conservation: m1 (vix )1 + m2 (vix ) 2 = m1 (vfx )1 + m2 (vfx ) 2
1
1
1
1
m1 (vix )12 + m2 (vix )22 = m1 (vfx )12 + m2 (vfx ) 22
2
2
2
2
These two equations can be solved for (vfx )1 and (vfx ) 2 , as shown by Equations 10.39 through 10.43, to give
Energy conservation:
(vfx )1 =
m1 − m2
50 g − 20 g
(vix )1 =
(2.0 m/s) = 0.86 m/s
m1 + m2
50 g + 20 g
(vfx ) 2 =
2m1
2(50 g)
(vix )1 =
(2.0 m/s) = 2.9 m/s
m1 + m2
50 g + 20 g
Assess: These velocities are of a reasonable magnitude. Since both these velocities are positive, both balls move
along the +x-direction.
10.30. Model: This is a case of a perfectly elastic collision between a proton and a carbon atom. The collision
obeys the momentum as well as the energy conservation law.
Visualize:
Solve: Momentum conservation: mP (vix ) P + mC (vix )C = mP (vfx ) P + mC (vfx )C
Energy conservation:
1
1
1
1
mP (vix ) 2P + mC (vix )C2 = mP (vfx ) P2 + mC (vfx )C2
2
2
2
2
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10-18
Chapter 10
These two equations can be solved, as described in the text through Equations 10.38 to 10.42:
⎛ m − 12mP ⎞
m − mC
7
7
(vfx ) P = P
mP + mC (vix ) P = ⎜ P
⎟ (2.0 × 10 m/s) = −1.69 × 10 m/s
mP + mC
m
+
12
m
⎝ P
P⎠
(vfx )C =
⎛ 2mP
⎞
2mP
7
6
(vix ) P = ⎜
⎟ (2.0 × 10 m/s) = 3.1 × 10 m/s
+
mP + mC
m
12
m
P⎠
⎝ P
After the elastic collision the proton rebounds at 1.69 × 107 m/s and the carbon atom moves forward at
3.08 × 106 m/s.
10.31. Model: In this case of a one-dimensional collision, the momentum conservation law is obeyed whether the
collision is perfectly elastic or perfectly inelastic. Assume ball 1 is initially moving right, in the positive direction.
Visualize:
Solve: In the case of a perfectly elastic collision, the two velocities (vfx )1 and (vfx ) 2 can be determined by combining the conservation equations of momentum and mechanical energy. By contrast, a perfectly inelastic collision involves only one final velocity vfx and can be determined from just the momentum conservation equation.
(a) Momentum conservation: m1 (vix )1 + m2 (vix )2 = m1 (vfx )1 + m2 (vfx ) 2
1
1
1
1
m1 (vix )12 + m2 (vix )22 = m1 (vfx )12 + m2 (vfx ) 22
2
2
2
2
These two equations can be solved as shown in Equations 10.38 through 10.42:
Energy conservation:
(vfx )1 =
m1 − m2
(100 g) − (300 g)
(vix )1 =
(10 m/s) = −5.0 m/s
m1 + m2
(100 g) + (300 g)
(vfx ) 2 =
2m1
2(100 g)
(vix )1 =
(10 m/s) = +5.0 m/s
m1 + m2
(100 g) + (300 g)
(b) For the inelastic collision, both balls travel with the same final speed vfx . The momentum conservation equation
pfx = pix is
(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2
⎛
⎞
100 g
⇒ vfx = ⎜
⎟ (10 m/s) + 0 m/s = 2.5 m/s
100
g
300
g
+
⎝
⎠
Assess: In the case of the perfectly elastic collision, the two balls bounce off each other with a speed of 5.0 m/s. In
the case of the perfectly inelastic collision, the balls stick together and move together at 2.5 m/s.
10.32. Model: This is the case of a perfectly inelastic collision. Momentum is conserved because no external force
acts on the system (clay + brick). We also represent our system as a particle.
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Energy
10-19
Visualize:
Solve: (a) The conservation of momentum equation pfx = pix is
(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2
Using (vix )1 = v0 and (vix ) 2 = 0, we get
vfx =
m1
0.050 kg
(vix )1 =
(vix )1 = 0.0476(vix )1 = 0.0476 v0
m1 + m2
(1.0 kg + 0.050 kg)
The brick is moving with speed 0.048v0 .
(b) The initial and final kinetic energies are given by
1
1
1
1
Ki = m1 (vix )12 + m2 (vix )22 = (0.050 kg)v02 + (1.0 kg)(0 m/s) 2 = (0.025 kg)v02
2
2
2
2
1
1
K f = ( m1 + m2 )vf2x = (1.0 kg + 0.050 kg)(0.0476) 2 v02 = 0.00119 v02
2
2
⎛ Ki − Kf ⎞
⎛ 0.00119 ⎞
The percent of energy lost = ⎜
⎟ × 100% = ⎜1 −
⎟ × 100% = 95%
0.025 ⎠
⎝
⎝ Ki ⎠
Exercises and Problems
10.33. Model: We will take the system to be the person plus the earth.
Visualize: When a person drops from a certain height, the initial potential energy is transformed to kinetic energy.
When the person hits the ground, if they land rigidly upright, we assume that all of this energy is transformed into
elastic potential energy of the compressed leg bones. The maximum energy that can be absorbed by the leg bones is
200 J; this limits the maximum height.
Solve: (a) The initial potential energy can be at most 200 J, so the height h of the jump is limited by mgh = 200 J
For m = 60 kg, this limits the height to
h = 200 J/mg = 200 J/(60 kg)(9.8 m/s 2 ) = 0.34 m
(b) If some of the energy is transformed to other forms than elastic energy in the bones, the initial height can be
greater. If a person flexes her legs on landing, some energy is transformed to thermal energy. This allows for a
greater initial height.
Assess: There are other tissues in the body with elastic properties that will absorb energy as well, so this limit is
quite conservative.
10.34. Model: Model your vehicle as a particle. Assume zero rolling friction, so that the sum of your kinetic and
gravitational potential energy does not change as the vehicle coasts down the hill.
Visualize:
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10-20
Chapter 10
The figure shows a before-and-after pictorial representation. Note that neither the shape of the hill nor the angle of
the downward slope is given, since these are not needed to solve the problem. All we need is the change in potential
energy as you and your vehicle descend to the bottom of the hill. Also note that
35 km/hr = (35,000 m/3600 s) = 9.722 m/s
Solve: Using yf = 0 and the equation Ki + U gi = K f + U gf we get
1 2
1
mvi + mgyi = mvf2 + mgyf ⇒ vi2 + 2 gyi = vf2
2
2
⇒ vf = vi2 + 2 gyi = (9.722 m/s) 2 + 2(9.8 m/s 2 )(15 m) = 19.7 m/s = 71 km/h
You are driving over the speed limit. Yes, you will get a ticket.
Assess: A speed of 19.7 m/s or 71 km/h at the bottom of the hill, when your speed at the top of the hill was 35 km/s,
is reasonable. From the energy bar chart, we see that the initial potential energy is completely transformed into the
final kinetic energy.
10.35. Model: This is case of free fall, so the sum of the kinetic and gravitational potential energy does not change
as the cannon ball falls.
Visualize:
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Energy
10-21
The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we put the
origin of our coordinate system on the ground below the fortress.
Solve: Using yf = 0 and the equation Ki + U gi = K f + U gf we get
1 2
1
mvi + mgyi = mvf2 + mgyf ⇒ vi2 + 2 gyi = vf2
2
2
vf = vi2 + 2 gyi = (80 m/s) 2 + 2(9.8 m/s 2 )(10 m) = 81 m/s
Assess: Note that we did not need to use the tilt angle of the cannon, because kinetic energy is a scalar. Also note
that using the energy conservation equation, we can find only the magnitude of the final velocity, not the direction of
the velocity vector.
10.36. Model: Assume the spring is ideal and obeys Hooke’s law. Then the potential energy of a stretched spring is
1
U sp = k (Δs )2 .
2
Visualize: The kinetic energy at the beginning is zero, and it is also zero at maximum height, so the spring potential
energy at the beginning equals the gravitational potential energy at maximum height. We are given k = 950N/m.
Solve:
1
(U s )i = (U g )f ⇒ k ( Δs ) 2 = mgy
2
k
(Δs ) 2 leads us to believe that a graph of y vs. (Δs ) 2 would produce a
2mg
straight line whose slope is k/2mg and whose intercept is zero.
Putting this in y = mx + b form y =
The spreadsheet tells us the fit is very good and that the slope is 746.5m −1.
k
k
950 N/m
= 746.5 m −1 ⇒ m =
=
= 0.065 kg = 65 g
−1
2mg
2 g (746.5 m ) 2(9.8 m/s 2 )(746.5 m −1 )
Assess: 65 g seems reasonable, and we were happy to get a very small intercept on our best-fit line.
10.37. Model: For the ice cube sliding around the inside of a smooth pipe, the sum of the kinetic and gravitational
potential energy does not change.
Visualize:
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10-22
Chapter 10
We use a coordinate system with the origin at the bottom of the pipe, that is, y1 = 0. The radius R of the pipe is
10 cm, and therefore ytop = y2 = 2R = 0.20 m. At an arbitrary angle θ , measured counterclockwise from the bottom
of the circle, y = R − Rcosθ .
Solve: (a) The energy conservation equation K 2 + U g2 = K1 + U g1 is
1
1
⇒ mv22 + mgy2 = mv12 + mgy1
2
2
⇒ v2 = v12 + 2 g ( y1 − y2 ) = (3.0 m/s) 2 + 2(9.8 m/s 2 )(0 m − 0.20 m) = 2.25 m/s ≈ 2.3 m/s
(b) Expressing the energy conservation equation as a function of θ :
1
1
K (θ ) + U g (θ ) = K1 + U g1 ⇒ mv 2 (θ ) + mgy (θ ) = mv12 + 0 J
2
2
⇒ v(θ ) = v12 − 2 gy (θ ) = v12 − 2 gR (1 − cosθ )
Using v1 = 3.0 m/s, g = 9.8 m/s 2 , and R = 0.10 m, we get v(θ ) = 9 − 1.96(1 − cosθ ) (m/s)
Assess: Beginning with a speed of 3.0 m/s at the bottom, the marble’s potential energy increases and kinetic energy
decreases as it gets toward the top of the circle. At the top, its speed is 2.25 m/s. This is reasonable since some of the
kinetic energy has been transformed into the marble’s potential energy.
10.38. Model: Assume that the rubber band behaves similar to a spring. Also, model the rock as a particle.
Visualize:
Solve: (a) The rubber band is stretched to the left since a positive spring force on the rock due to the rubber band
results from a negative displacement of the rock. That is, ( Fsp ) x = −kx, where x is the rock’s displacement from the
equilibrium position due to the spring force Fsp .
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Energy
10-23
(b) Since the Fsp versus x graph is linear with a negative slope and can be expressed as Fsp = −kx, the rubber band
obeys Hooke’s law.
(c) From the graph, ΔFsp = 20 N for Δx = 10 cm. Thus,
k=
ΔFsp
Δx
=
20 N
= 200 N/m = 2.0 × 102 N/m
0.10 m
(d) The conservation of energy equation K f + U sf = Ki + U si for the rock is
1 2 1 2 1 2 1 2
1
1
1
1
mvf + kxf = mvi + kxi ⇒ mvf2 + k (0 m) 2 = m(0 m/s) 2 + kxi2
2
2
2
2
2
2
2
2
k
200 N/m
(0.30 m) = 19 m/s
vf =
xi =
m
0.050 kg
Assess: Note that xi is Δx, which is the displacement relative to the equilibrium position, and xf is the equilibrium
position of the rubber band, which is equal to zero.
10.39. Model: We will assume the knee extensor tendon behaves according to Hooke’s Law and stretches in a
straight line.
Visiualize: The elastic energy stored in a spring is given by U s = 12 k ( Δs ) 2 .
Solve: For athletes,
1
U s,athlete = 12 k (Δs ) 2 = (33,000 N/m)(0.041 m) 2 = 27.7 J
2
For non-athletes,
1
U s,non − athlete = 12 k (Δs )2 = (33,000 N/m)(0.033 m)2 = 18.0 J
2
The difference in energy stored between athletes and non-athletes is therefore 9.7 J.
Assess: Notice the energy stored by athletes is over 1.5 times the energy stored by non-athletes.
10.40. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is no friction, hence the mechanical energy K + U s is conserved.
Visualize:
Note that xf = xe and xi − xe = Δx. The before-and-after pictorial representations show that we put the origin of the
coordinate system at the equilibrium position of the free end of the springs.
Solve: The conservation of energy equation K f + U sf = Ki + U si for the single spring is
1 2 1
1
1
mvf + k ( xf − xe ) 2 = mvi2 + k ( xi − xe ) 2
2
2
2
2
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10-24
Chapter 10
Using the value for vf given in the problem, we get
1 2
1
1
1
mv0 + 0 J = 0 J + k (Δx) 2 ⇒ mv02 = k ( Δx) 2
2
2
2
2
Conservation of energy for the two-spring case:
1
1
1
1
mVf2 + 0 J = 0 J + k ( xi − xe ) 2 + k ( xi − xe ) 2
mVf2 = k ( Δx )2
2
2
2
2
Using the result of the single-spring case,
1
mVf2 = mv02 ⇒ Vf = 2v0
2
Assess: The block separates from the spring at the equilibrium position of the spring.
10.41. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is no friction, hence the mechanical energy K + U s is conserved.
Visualize:
The springs in both cases have the same compression Δx. We put the origin of the coordinate system at the equilibrium position of the free end of the spring for the single-spring case (a), and at the free end of the two connected
springs for the two-spring case (b).
Solve: The conservation of energy for the single-spring case:
1
1
1
1
K f + U sf = Ki + U si ⇒ mvf2 + k ( xf − xe ) 2 = mvi2 + k ( xi − xe ) 2
2
2
2
2
Using xf = xe = 0 m, vi = 0 m/s, and vf = v0 , this equation simplifies to
1 2 1
mv0 = k (Δx) 2
2
2
Conservation of energy in the case of the two springs in series, where each spring compresses by Δx /2, is
1
1
1
1
K f + U sf = Ki + U si ⇒ mVf2 + 0 = mvi2 + k (Δx /2) 2 + k ( Δx /2) 2
2
2
2
2
′
Using xf = xe = 0 m and vi = 0 m/s, we get
1
1 ⎡1
⎤
mVf2 = ⎢ k (Δx ) 2 ⎥
2
2 ⎣2
⎦
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Energy
10-25
Comparing the two results we see that Vf = v0 / 2.
Assess: The block pushes on the spring until the spring has returned to its equilibrium length.
10.42. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and therefore the mechanical
energy K + U s + U g is conserved.
Visualize:
The figure shows a before-and-after pictorial representation. We have chosen to place the origin of the coordinate
system at the position where the ice cube has compressed the spring 10 cm. That is, y0 = 0.
Solve: (a) The energy conservation equation K 2 + U s2 + U g2 = K 0 + U s0 + U g0 is
1 2 1
1
1
mv2 + k ( xe − xe ) 2 + mgy2 = mv02 + k ( x − xe ) 2 + mgy0
2
2
2
2
Using v2 = 0 m/s, y0 = 0 m, and v0 = 0 m/s,
1
k ( Δx ) 2
mgy2 = k ( x − xe ) 2 ⇒ y2 =
=h
2
2 mg
(b) Insert the values given
k (Δx) 2
(25 N/m)(0.10 m) 2
=
= 25.5 cm ≈ 26 cm
2 mg
2(0.050 kg)(9.8 m/s 2 )
Assess: The net effect of the launch is to transform the potential energy stored in the spring into gravitational potential energy. The block has kinetic energy as it comes off the spring, but we did not need to know this energy to solve
the problem. The answer is independent of the angle of the slope.
h=
10.43. Model: Model the two packages as particles. Momentum is conserved in both inelastic and elastic collisions. Kinetic energy is conserved only in a perfectly elastic collision.
Visualize:
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10-26
Chapter 10
Solve: For a package with mass m the conservation of energy equation is
1
1
2
K1 + U g1 = K 0 + U g0 ⇒ m(v1) 2m + mgy1 = m(v0 ) m
+ mgy0
2
2
Using (v0 )m = 0 m/s and y1 = 0 m,
1
m(v1 ) 2m = mgy0 ⇒ (v1) m = 2 gy0 = 2(9.8 m/s 2 )(3.0 m) = 7.668 m/s
2
(a) For the perfectly inelastic collision the conservation of momentum equation is
pfx = pix ⇒ (m + 2m)(v2 )3m = m(v1) m + (2m)(v1) 2 m
Using (v1 )2 m = 0 m / s, we get
(v2 )3m = (v1 ) m /3 = 2.56 m/s
The packages move off together at a speed of 2.6 m/s.
(b) For the elastic collision, the mass m package rebounds with velocity
(v3 )m =
m − 2m
1
(v1 ) m = − (7.668 m/s) = −2.56 m/s
m + 2m
3
The negative sign with (v3 )m shows that the package with mass m rebounds and goes to the position y4 . We can
determine y4 by applying the conservation of energy equation as follows. For a package of mass m:
1
1
2
K f + U gf = K i + U gi ⇒ m(v4 ) 2m + mgy4 = m(v3 ) m
+ mgy3
2
2
Using (v3 )m = −2.55 m/s, y3 = 0 m, and (v4 ) m = 0 m/s, we get
1
mgy4 = m(−2.56 m/s) 2 ⇒ y4 = 33 cm
2
10.44. Model: Model the marble and the steel ball as particles. We will assume an elastic collision between the
marble and the ball, and apply the conservation of momentum and the conservation of energy equations. We will also
assume zero rolling friction between the marble and the incline.
Visualize:
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Energy
10-27
This is a two-part problem. In the first part, we will apply the conservation of energy equation to find the marble’s
speed as it exits onto a horizontal surface. We have put the origin of our coordinate system on the horizontal surface
just where the marble exits the incline. In the second part, we will consider the elastic collision between the marble
and the steel ball.
Solve: The conservation of energy equation K1 + U g1 = K 0 + U g0 gives us:
1
1
2
mM (v1) 2M + mM gy1 = mM (v0 ) M
+ mM gy0
2
2
1
Using (v0 ) M = 0 m/s and y1 = 0 m, we get (v1 ) 2M = gy0 ⇒ (v1 ) M = 2 gy0 . When the marble collides with the steel
2
ball, the elastic collision gives the ball velocity
2mM
2mM
(v2 )S =
(v1 ) M =
2gy0
mM + mS
mM + mS
Solving for y0 gives
2
y0 =
⎤
1 ⎡ mM + mS
(v2 )S ⎥ = 0.258 m = 25.8 cm
⎢
2g ⎣ 2mM
⎦
10.45. Model: Assume an ideal spring that obeys Hooke’s law. Since this is a free-fall problem, the mechanical
energy K + U g + U s is conserved. Also, model the safe as a particle.
Visualize:
We have chosen to place the origin of our coordinate system at the free end of the spring, which is neither stretched
nor compressed. The safe gains kinetic energy as it falls. The energy is then converted into elastic potential energy as
the safe compresses the spring. The only two forces are gravity and the spring force, which are both conservative, so
energy is conserved throughout the process. This means that the initial energy—as the safe is released—equals the
final energy—when the safe is at rest and the spring is fully compressed.
Solve: The conservation of energy equation K1 + U g1 + U s1 = K 0 + U g0 + U s0 is
1 2
1
1
1
mv1 + mg ( y1 − ye ) + k ( y1 − ye ) 2 = mv02 + mg ( y0 − ye ) + k ( ye − ye ) 2
2
2
2
2
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10-28
Chapter 10
Using v1 = v0 = 0 m/s and ye = 0 m, the above equation simplifies to
1
mgy1 + ky12 = mgy0
2
2mg ( y0 − y1 )
2(1000 kg)(9.8 m/s 2 )(2.0 m − ( −0.50 m))
= 1.96 × 105 N/m ≈ 2.0 × 105 N/m
y12
(−0.50 m) 2
Assess: By equating energy at these two points, we do not need to find how fast the safe was moving when it hit the spring.
⇒k =
=
10.46. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction and hence the mechanical energy (vfx )1 = −1.2 m/s + 1.0 m/s = −0.2 m/s is conserved.
Visualize:
Solve: (a) When releasing the block suddenly, K 2 + U s2 + U g2 = K1 + U s1 + U g1
1 2 1
1
1
mv2 + k ( y2 − ye ) 2 + mgy2 = mv12 + k ( y1 − ye ) 2 + mgy1
2
2
2
2
Using v2 = 0 m/s, v1 = 0 m/s, and y1 = ye , we get
1
0 J + (490 N/m)( y2 − y1 )2 + mgy2 = 0 J + 0 J + mgy1 ⇒ (245 N/m)( y2 − y1 )2 = mg ( y1 − y2 )
2
⇒ (245 N/m)( y1 − y2 ) 2 = (5.0 kg)(9.8 m/s 2 )( y1 − y2 ) ⇒ ( y1 − y2 ) = 0.20 m
(b) When lowering the block gently until it rests on the spring, the block reaches a point of static equilibrium.
mg (5.0 kg)(9.8 m/s 2 )
Fnet = k Δy − mg = 0 ⇒ Δy =
=
= 0.10 m
k
490 N/m
(c) In part (b), at a point 0.10 m down, the forces balance. But in part (a) the block has kinetic energy as it reaches 0.10 m.
So the block continues on past the equilibrium point until all the gravitational potential energy is stored in the spring.
10.47. Model: Assume an ideal spring that obeys Hooke’s law. Also assume zero rolling friction between the roller
coaster and the track, and a particle model for the roller coaster. Since no friction is involved, the mechanical energy
K + U s + U g is conserved.
Visualize:
We have chosen to place the origin of the coordinate system on the end of the spring that is compressed and touches
the roller coaster car.
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Energy
10-29
Solve: (a) The energy conservation equation for the car going to the top of the hill is
1 2
1
1
1
K 2 + U g2 + U s2 = K 0 + U g0 + U s0
mv2 + mgy2 + k ( xe − xe ) 2 = mv02 + mgy0 + k ( x0 − xe ) 2
2
2
2
2
Noting that y0 = 0 m, v2 = 0 m/s, v1 = 0 m/s, and | x0 − xe| = 2.0 m, we obtain
1
0 J + mgy2 + 0 J = 0 J + 0 J + k (2.0 m) 2
2
⇒k =
2mgy2
(2.0 m)
2
=
2(400 kg)(9.8 m/s 2 )(10 m)
(2.0 m) 2
= 1.96 × 104 N/m
We now increase this value for k by 10% for safety, giving a value of 2.156 × 104 N/m ≈ 2.2 × 104 N/m.
(b) The energy conservation equation K3 + U g3 + U s3 = K 0 + U g0 + U s0 is
1 2
1
1
1
mv3 + mgy3 + k ( xe − xe )2 = mv02 + mgy0 + k ( x0 − xe ) 2
2
2
2
2
Using y3 = −5.0 m, v0 = 0 m/s, y0 = 0 m, and | x0 − xe | = 2.0 m, we get
1 2
1
mv3 + mg (−5.0 m) + 0 J = 0 J + 0 J + k ( x0 − xe ) 2
2
2
1
1
(400 kg)v32 − (400 kg)(9.8 m/s 2 )(5.0 m) = (2.156 × 104 N/m)(2.0 m) 2
2
2
⇒ v3 = 17.7 m/s ≈ 18 m/s
10.48. Model: Since there is no friction, the sum of the kinetic and gravitational potential energy does not change.
Model Julie as a particle.
Visualize:
We place the coordinate system at the bottom of the ramp directly below Julie’s starting position. From geometry,
Julie launches off the end of the ramp at a 30º angle.
1
1
Solve: Energy conservation: K1 + U g1 = K 0 + U g0 ⇒ mv12 + mgy1 = mv02 + mgy0
2
2
Using v0 = 0 m/s, y0 = 25 m, and y1 = 3 m, the above equation simplifies to
1 2
mv1 + mgy1 = mgy0 ⇒ v1 = 2 g ( y0 − y1 ) = 2(9.8 m/s 2 )(25 m − 3 m) = 20.77 m/s
2
We can now use kinematic equations to find the touchdown point from the base of the ramp. First we’ll consider the
vertical motion:
1
1
y2 = y1 + v1 y (t2 − t1) + a y (t2 − t1 )2 0 m = 3 m + (v1sin30°)(t2 − t1 ) + (−9.8 m/s 2 )(t2 − t1 ) 2
2
2
(3 m)
2 ( 20.77 m/s )sin30°
⇒ (t2 − t1 ) −
=0
(t2 − t1) −
(4.9 m/s 2 )
(4.9 m/s 2 )
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10-30
Chapter 10
(t2 − t1) 2 − (2.119 s)(t2 − t1 ) − (0.6122 s 2 ) = 0 ⇒ (t2 − t1 ) = 2.377 s
For the horizontal motion:
1
x2 = x1 + v1x (t2 − t1 ) + ax (t2 − t1) 2
2
x2 − x1 = (v1cos30°)(t2 − t1 ) + 0 m = (20.77 m/s)(cos30°)(2.377 s) = 43 m
Assess: Note that we did not have to make use of the information about the circular arc at the bottom that carries
Julie through a 90° turn.
10.49. Model: We assume the spring to be ideal and to obey Hooke’s law. We also treat the block (B) and the ball
(b) as particles. In the case of an elastic collision, both the momentum and kinetic energy equations apply. On the
other hand, for a perfectly inelastic collision only the equation of momentum conservation is valid.
Visualize:
Place the origin of the coordinate system on the block that is attached to one end of the spring. The before-and-after
pictorial representations of the elastic and perfectly inelastic collision are shown in figures (a) and (b), respectively.
Solve: (a) For an elastic collision, the ball’s rebound velocity is
m − mB
−80 g
(vf ) b = b
(vi ) b =
(5.0 m/s) = −3.33 m/s
120 g
mb + mB
The ball’s speed is 3.3 m/s.
(b) An elastic collision gives the block speed
2mB
40 g
(vf ) B =
(vi ) b =
(5.0 m/s) = 1.667 m/s
120 g
mb + mB
To find the maximum compression of the spring, we use the conservation equation of mechanical energy for the
block + spring system. That is K1 + U s1 = K 0 + U s0 :
1
1
1
1
2
+ k ( x0 − x0 ) 2
mB (vf′ ) 2B + k ( x1 − x0 ) 2 = mB (vf ) B
2
2
2
2
2
+0
0 + k ( x1 − x0 ) 2 = mB (vf ) B
( x1 − x0 ) = (0.100 kg)(1.667 m/s)2 /(20 N/m) = 11.8 cm
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Energy
10-31
(c) Momentum conservation pf = pi for the perfectly inelastic collision means
(mB + mB )vf = mb (vi ) b + mB (vi ) B
(0.100 kg + 0.020 kg)vf = (0.020 kg)(5.0 m/s) + 0 m/v ⇒ vf = 0.833 m/s
The maximum compression in this case can now be obtained using the conservation of energy equation
K1 + U s1 = K 0 + U s0:
0 J + (1/2)k (Δx) 2 = (1/2)(mB + mb )vf 2 + 0 J
⇒ Δx =
mB + mb
0.120 kg
vf =
(0.833 m/s) = 0.0645 m = 6.5 cm
k
20 N/m
10.50. Model: Assume an ideal spring that obeys Hooke’s law. We treat the bullet and the block in the particle
model. For a perfectly inelastic collision, the momentum is conserved. Furthermore, since there is no friction, the
mechanical energy of the system (bullet + block + spring) is conserved.
Visualize:
We place the origin of our coordinate system at the end of the spring that is not anchored to the wall.
Solve: (a) Momentum conservation for perfectly inelastic collision states pf = pi . This means
⎛ m ⎞
(m + M )vf = m(vi ) m + M (vi )M ⇒ (m + M )vf = mvB + 0 kg m/s ⇒ vf = ⎜
⎟ vB
⎝m+M ⎠
where we have used vB for the initial speed of the bullet. The mechanical energy conservation equation
K1 + U s1 = K e + U se as the bullet embedded block compresses the spring is:
1
1
1
1
m(v′f ) 2 + k ( x1 − xe )2 = (m + M )(vf ) 2 + k ( xe − xe ) 2
2
2
2
2
2
1
1
( m + M ) kd 2
⎛ m ⎞ 2
0 J + kd 2 = (m + M ) ⎜
⎟ vB + 0 J ⇒ vB =
2
2
m2
⎝m+M ⎠
(b) Using the above formula with m = 5.0 g, M = 2.0 kg, k = 50 N/m, and d = 10 cm,
vB = (0.0050 kg + 2.0 kg)(50 N/m)(0.10 m)2 /(0.0050)2 = 2.0 × 102 m/s
(c) The fraction of energy lost is
1 2 1
2
2
mvB − (m + M )vf2
m + M ⎛ vf ⎞
m+M ⎛ m ⎞
2
2
1
=1−
=
−
⎜ ⎟
⎜
⎟
1 2
m ⎝ vB ⎠
m ⎝m+M ⎠
mvB
2
m
0.0050 kg
=1−
=1−
= 0.9975
m+M
(0.0050 kg + 2.0 kg)
That is, during the perfectly inelastic collision 99.75% of the bullet’s energy is lost. The energy is dissipated inside
the block. Although it is common to say, “The energy is lost to heat,” in the next chapter we’ll see that it is more
accurate to say, “The energy is transformed to thermal energy.”
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10-32
Chapter 10
10.51. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energy
K + U g + U s is conserved.
Visualize:
We have chosen to place the origin of the coordinate system on the free end of the spring that is neither stretched nor
compressed, that is, at the equilibrium position of the end of the unstretched spring. The bullet’s mass is m and the
block’s mass is M.
Solve: (a) The energy conservation equation K 2 + U s2 + U g2 = K1 + U s1 + U g1 becomes
1
1
1
1
(m + M )v22 + k ( y2 − ye )2 + (m + M ) g ( y2 − ye ) = ( m + M )v12 + k ( y1 − ye )2 − (m + M ) g ( y1 − ye )
2
2
2
2
Noting v2 = 0 m/s, we can rewrite the above equation as
k (Δy2 )2 + 2( m + M ) g ( Δy2 + Δy1) = (m + M )v12 + k (Δy1) 2
Let us express v1 in terms of the bullet’s initial speed vB by using the momentum conservation equation pf = pi
which is (m + M )v1 = mvB + Mvblock . Since vblock = 0 m/s, we have
⎛ m ⎞
v1 = ⎜
⎟ vB
⎝m+M ⎠
We can also find the magnitude of y1 from the equilibrium condition k ( y1 − ye ) = Mg .
Mg
k
With these substitutions for v1 and Δy1, the energy conservation equation simplifies to
Δy1 =
k ( Δy2 ) 2 + 2(m + M ) g (Δy1 + Δy2 ) =
m 2vB2
⎛ Mg ⎞
+ k⎜
⎟
(m + M )
⎝ k ⎠
2
2
(m + M ) M 2 g 2
⎛m+M ⎞
⎛m+M
⇒ vB2 = 2 ⎜
+ k⎜
⎟ g (Δy1 + Δy2 ) −
k
m2
⎝ m ⎠
⎝ m2
⎞
2
⎟ ( Δy2 )
⎠
We still need to include the spring’s maximum compression (d) into this equation. We assume that d = Δy1 + Δy2 ,
that is, maximum compression is measured from the initial position ( y1) of the block. Thus, using Δy2 = d − Δy1 =
(d − Mg/k ), we have
2 2
⎡ ⎛ m + M ⎞2
⎛m+M ⎞M g
⎛m+M
+ k⎜
vB = ⎢ 2 ⎜
⎟ gd − ⎜
⎟
2
m
k
m
⎝
⎠
⎝
⎠
⎝ m2
⎣⎢
1/2
⎤
⎞
2
⎟ (d − Mg/k ) ⎥
⎠
⎦⎥
(b) Using m = 0.010 kg, M = 2.0 kg, k = 50 N/m, and d = 0.45 m,
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Energy
2
10-33
2
⎛ 2.010 kg ⎞
⎛ 2.0 kg ⎞
2
2 2
vB2 = 2 ⎜
⎟ (9.8 m/s )(0.45 m) − (2.010 kg) ⎜
⎟ (9.8 m/s ) /(50 N/m)
0.010
kg
0.010
kg
⎝
⎠
⎝
⎠
1
[0.45 m − (2.0 kg) × (9.8 m/s 2 ) / 50 N/m]2
+ (50 N/m)(2.010 kg)
(0.010 kg) 2
⇒ vB = 453 m/s
The bullet has a speed of 4.5 × 102 m/s.
Assess: This is a reasonable speed for the bullet.
10.52. Model: The track is frictionless.
Visualize: vc = Rg
Solve: (a) First use conservation of momentum during the collision, then conservation of energy as the combined
block goes to the top of the loop.
G
G
∑ pi = ∑ pf
mvm + 0 = (m + M )vtot
m+M
vtot
m
is the speed of the combined block just after the collision.
vm =
Now use the conservation of energy. vtot
U i + Ki = U f + K f
1
1
2
0 + (m + M )vtot
= ( m + M ) g (2 R) + ( m + M )(vc ) 2
2
2
Cancel (m + M ) and replace vc with
Rg .
2
vtot
= 4 Rg + Rg = 5Rg ⇒ vtot = 5Rg
m+M
m+M
vtot =
5Rg
m
m
(b) First use conservation of momentum and kinetic energy during the collision, then conservation of mechanical
energy as the big block goes to the top of the loop. Call the speed of M just after the elastic collision V and the speed
′.
of m just after the collision vm
G
G
∑ pi = ∑ pf
We drop the vectors because this is one-dimensional motion, but v′ may be negative.
mvm + 0 = mv′ + MV
vm =
M
V
m
Now use the conservation of energy as the block goes to the top of the loop.
U i + Ki = U f + K f
vm = v′ +
1
1
1
1
′2 + MV 2 = Mg (2 R ) + mv′m2 + M
0 + mvm
2
2
2
2
Subtract
1 mv′2
m
2
(
Rg
)
2
from both sides, cancel M, and solve for V.
V = 4 Rg + Rg = 5Rg
Now go back to the conservation of kinetic energy in the elastic collision.
∑ Ki = ∑ Kf
1 2 1
1
′2 + MV 2
mvm = mvm
2
2
2
Cancel
1
2
and divide by m.
′2 +
vm2 = vm
M
5Rg
m
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10-34
Chapter 10
′2 from the momentum equation.
Find vm
2
M ⎞
M
⎛
vm2 = ⎜ vm − V ⎟ + 5Rg
m ⎠
m
⎝
vm2 = vm2 − 2
2
M
M
⎛M ⎞
vmV + ⎜ V ⎟ + 5Rg
m
m
⎝m ⎠
M
, and solve for vm .
m
1M
1 5 Rg
1M
1
1 m+M
=
vm =
V+
5 Rg +
5 Rg =
5 Rg
2m
2 5 Rg 2 m
2
2 m
Subtract vm2 from both sides, cancel
Assess: We expected the initial speed needed to be greater for the inelastic case because kinetic energy isn’t conserved in the collision.
10.53. Model: This is a two-part problem. In the first part, we will find the critical velocity for the block to go over
the top of the loop without falling off. Since there is no friction, the sum of the kinetic and gravitational potential
energy is conserved during the block’s motion. We will use this conservation equation in the second part to find the
minimum height the block must start from to make it around the loop.
Visualize:
We place the origin of our coordinate system directly below the block’s starting position on the frictionless track.
Solve: The free-body diagram on the block implies
mv 2
FG + n = c
R
For the block to just stay on track, n = 0. Thus the critical velocity vc is
FG = mg =
The block needs kinetic energy
1 mv 2
c
2
mvc2
⇒ vc2 = gR
R
= 12 mgR to go over the top of the loop. We can now use the conservation of
mechanical energy equation to find the minimum height h.
1
1
K f + U gf = Ki + U gi ⇒ mvf2 + mgyf = mvi2 + mgyi
2
2
Using vf = vc = gR , yf = 2 R, vi = 0 m/s, and yi = h, we obtain
1
gR + g (2 R ) = 0 + gh ⇒ h = 2.5R
2
10.54. Model: Model Lisa (L) and the bobsled (B) as particles. We will assume the ramp to be frictionless, so that
the mechanical energy of the system (Lisa + bobsled + spring) is conserved. Furthermore, during the collision, as
Lisa leaps onto the bobsled, the momentum of the Lisa + bobsled system is conserved. We will also assume the
spring to be an ideal one that obeys Hooke’s law.
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Energy
10-35
Visualize:
We place the origin of our coordinate system directly below the bobsled’s initial position.
Solve: (a) Momentum conservation in Lisa’s collision with bobsled states p1 = p0 , or
(mL + mB )v1 = mL (v0 ) L + mB (v0 ) B ⇒ (mL + mB )v1 = mL (v0 ) L + 0
⎛ mL ⎞
⎛
⎞
40 kg
⇒ v1 = ⎜
⎟ (v0 )L = ⎜
⎟ (12 m/s) = 8.0 m/s
+
+
m
m
40
kg
20
kg
⎝
⎠
⎝ L
B⎠
The energy conservation equation: K 2 + U s2 + U g2 = K1 + U s1 + U g1 is
1
1
1
1
(mL + mB )v22 + k ( x2 − xe ) 2 + ( mL + mB ) gy2 = (mL + mB )v12 + k ( xe − xe )2 + (mL + mB ) gy1
2
2
2
2
Using v2 = 0 m/s, k = 2000 N/m, y2 = 0 m, y1 = (50 m)sin20° = 17.1 m, v1 = 8.0 m/s, and (mL mB ) = 60 kg, we get
1
1
0 J + (2000 N/m)( x2 − xe ) 2 + 0 J = (60 kg)(8.0 m/s) 2 + 0 J + (60 kg)(9.8 m/s 2 )(17.1 m)
2
2
Solving this equation yields (x2 − xe ) = 3.5 m.
(b) As long as the ice is slippery enough to be considered frictionless, we know from conservation of mechanical
energy that the speed at the bottom depends only on the vertical descent Δy. Only the ramp’s height h is important,
not its shape or angle.
10.55. Model: We can divide this problem into two parts. First, we have an elastic collision between the 20 g ball
(m) and the 100 g ball (M). Second, the 100 g ball swings up as a pendulum.
Visualize:
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10-36
Chapter 10
The figure shows three distinct moments of time: the time before the collision, the time after the collision but before
the two balls move, and the time the 100 g ball reaches its highest point. We place the origin of our coordinate system on the 100 g ball when it is hanging motionless.
Solve: For a perfectly elastic collision, the ball moves forward with speed
(v1 ) M =
2mm
1
(v0 ) m = (v0 ) m
mm + mM
3
In the second part, the sum of the kinetic and gravitational potential energy is conserved as the 100 g ball swings up
after the collision. That is, K 2 + U g2 = K1 + U g1. We have
1
1
M (v2 ) 2M + Mgy2 = M (v1 ) 2M + Mgy1
2
2
Using (v2 ) M = 0 J, (v1) M =
(v0 ) m
, y1 = 0 m, and y2 = L − Lcosθ , the energy equation simplifies to
3
g ( L − Lcosθ ) =
1 (v0 ) 2m
2 9
⇒ (v0 ) m = 18 g L(1 − cosθ ) = 18(9.8 m/s 2 )(1.0 m)(1 − cos50°) = 7.9 m/s
10.56. Model: Model the balls as particles. We will use the Galilean transformation of velocities to analyze the
problem of elastic collisions. We will transform velocities from the lab frame L to a frame M in which one ball is at
rest. This allows us to apply equations to the case of a perfectly elastic collision in M, find the final velocities of the
balls in M, and then transform these velocities back to the lab frame L.
Visualize: Let M be the frame of the 200 g ball. Denoting masses as m1 = 100 g and m2 = 200 g, the initial velocities in the L frame are (vix )1L = 4 m/s and (vix ) 2L = −3 m/s.
Figure (a) shows the before-collision situation as seen in frame L, and figure (b) shows the before-collision situation
as seen in frame M. The after-collision velocities in M are shown in figure (c), and figure (d) indicates velocities in L
after they have been transformed to frame L from M.
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Energy
10-37
Solve: (a) In the L frame, (vix )1L = 4 m/s and (vix ) 2L = −3 m/s. M is the reference frame of the 200 g ball, so
(vx ) ML = −3 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of
velocities (vx )OM = (vx )OL + (vx ) LM . So,
(vix )1M = (vix )1L − (vix ) ML = 4 m/s − (−3 m/s) = 7 m/s
(vix ) 2M = (vix ) 2L − (vix )ML = −3 m/s − (−3 m/s) = 0 m/s
Figure (b) shows the “before” situation, where ball 2 is at rest.
Now we can use Equations 10.42 to find the after-collision velocities in frame M:
m − m2
100 g − 200 g
7
(vfx )1M = 1
(vix )1M =
(7 m/s) = − m/s
m1 + m2
100 g + 200 g
3
(vfx ) 2M =
2m1
2(100) g
14
(vix )1M =
(7 m/s) =
m/s
m1 + m2
100 g + 200 g
3
Finally, we need to apply the reverse Galilean transformation (vx )OM = (vx )OL + (vx )LM with the same (vx )LM , to
transform the after-collision velocities back to the lab frame S:
7
(vfx )1L = (vfx )1M + (vx ) ML = − m/s − 3 m/s = −5.33 m/s
3
14
(vfx ) 2L = (vfx ) 2M + (vx ) ML = m/s − 3 m/s = 1.67 m/s
3
Figure (d) shows the “after” situation in the lab frame. The 100 g ball is moving left at 5.3 m/s; the 200 g ball is moving right at 1.7 m/s.
(b) The momentum conservation equation pfx = pix for a perfectly inelastic collision is
(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2
(0.100 kg + 0.200 kg)vfx = (0.100 kg)(4.0 m/s) + (0.200 kg)(−3.0 m/s) ⇒ vfx = −0.667 m/s
Both balls are moving left at 0.67 m/s.
10.57. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation 10.43)
to analyze the problem of elastic collisions. We will transform velocities from the lab frame L to a frame M in which
one ball is at rest. This allows us to apply Equations 10.43 to a perfectly elastic collision in M. After finding the final
velocities of the balls in M, we can then transform these velocities back to the lab frame L.
Visualize: Let M be the frame of the 400 g ball. Denoting masses as m1 = 100 g and m2 = 400 g, the initial velocities in the S frame are (vix )1L = +4.0 m/s and (vix ) 2L = +1.0 m/s.
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10-38
Chapter 10
Figures (a) and (b) show the before-collision situations in frames L and M, respectively. The after-collision velocities
in M are shown in figure (c). Figure (d) indicates velocities in L after they have been transformed to L from M.
Solve: In frame L, (vix )1L = 4.0 m/s and (vix ) 2L = 1.0 m/s. Because M is the reference frame of the 400 g ball,
(vx )ML = 1.0 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of
velocities (vx )OM = (vx )OL − (vx ) ML . So,
(vix )1M = (vix )1L − (vx ) ML = 4.0 m/s − 1.0 m/s = 3.0 m/s
(vix ) 2M = (vix ) 2L − (vx ) ML = 1.0 m/s − 1.0 m/s = 0 m/s
Figure (b) shows the “before” situation in frame M where the ball 2 is at rest.
Now we can use Equations 10.43 to find the after-collision velocities in frame M.
(vfx )1M =
m1 − m2
100 g − 400 g
(vix )1M =
(3.0 m/s) = −1.80 m/s
m1 + m2
100 g + 400 g
(vfx ) 2M =
2m1
2(100 g)
(vix )1M =
(3.0 m/s) = 1.20 m/s
m1 + m2
100 g + 400 g
Finally, we need to apply the reverse Galilean transformation (vx )OM = (vx )OL + (vx ) LM with the same (vx )LM , to
transform the after-collision velocities back to the lab frame L.
(vfx )1L = (vfx )1M + (vx ) ML = −1.80 m/s + 1.0 m/s = −0.80 m/s
(vfx ) 2L = (vfx ) 2M + (vx ) ML = 1.20 m/s + 1.0 m/s = 2.20 m/s
Figure (d) shows the “after” situation in frame L. The 100 g ball moves left at 0.80 m/s, the 400 g ball right at 2.2 m/s.
Assess: The magnitudes of the after-collision velocities are similar to the magnitudes of the before-collision velocities.
10.58. Model: Use the model of the conservation of mechanical energy.
Visualize:
Solve: (a) The turning points occur where the total energy line crosses the potential energy curve. For E = 12 J, this
occurs at the points x = 1 m and x = 8 m.
(b) The equation for kinetic energy K = E − U gives the distance between the potential energy curve and total energy line. U = 4 J at x = 2 m, so K = 12 J − 4 J = 8 J. The speed corresponding to this kinetic energy is
v=
2K
2(8 J)
=
= 5.7 m/s
m
0.5 kg
(c) Maximum speed occurs for minimum U. This occurs at x = 6 m where U = 0 J and K = 12 J. The speed at this
point is
v=
2K
2(12 J)
=
= 6.9 m/s
m
0.500 kg
(d) The particle leaves x = 1 m with v = 6.3 m/s. It gradually slows down, reaching x = 4 m with a speed of 4.0 m/s. After
x = 4 m, it speeds up again, achieving a speed of 6.9 m/s as it crosses x = 6 m. Then it slows again, coming instantaneously
to a halt (v = 0 m/s) at the x = 8 m turning point. Now it will reverse direction and move back to the left.
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Energy
10-39
(e) If the particle has E = 4 J it cannot cross the 8 J potential energy “mountain” in the center. It can either oscillate
back and forth over the range 1.0 m ≤ x ≤ 2 m or over the range 5 m ≤ x ≤ 6.7 m.
dU
= 0.
dx
dU
1
= 0 = 1 + 2cos(2 x ) ⇒ cos(2 x) = −
dx
2
1
1
⎛
⎞
⇒ x = cos −1 ⎜ − ⎟
2
⎝ 2⎠
10.59. Solve: (a) The equilibrium positions are located at points where
1
2π
4π
⎛ 1⎞
is in radians and x is in meters. The function cos −1 ⎜ − ⎟ may have values
and
. Thus there
2
3
3
⎝ 2⎠
are two values of x,
π
2π
x1 = and x2 =
3
3
within the interval 0 m ≤ x ≤ π m.
(b) A point of stable equilibrium corresponds to a local minimum, while a point of unstable equilibrium corresponds
to a local maximum. Compute the concavity of U(x) at the equilibrium positions to determine their stability.
d 2U
= −4sin(2 x )
dx 2
Note that −
⎛ 3⎞
d 2U
π
π
( x1 ) < 0, x1 =
=
−
=
−
.
(
x
)
4
2
3
Since
is a local maximum, so x1 =
is a point of
⎜
⎟
1
2
2
⎜
⎟
3
3
3 dx
dx
⎝ 2 ⎠
unstable equilibrium.
⎛
2π d 2U
3⎞
d 2U
2π
2π
,
(− x2 ) = −4 ⎜⎜ −
> 0, x2 =
is a local minimum, so x2 =
is a point
At x2 =
⎟⎟ = +2 3. Since
2
2
3
3
2
3
dx
dx
⎝
⎠
of stable equilibrium.
At x1 =
π
,
d 2U
10.60. Model: The potential energy of two nucleons interacting via the strong force is
U = U 0 [1 − e − x/x0 ]
where x is the distance between the centers of the two nucleons, x0 = 2.0 × 10−15 m, and U 0 = 6.0 × 10−11 J.
Visualize: Nucleons are protons and neutrons, and they are held together in the nucleus by a force called the strong
force. This force exists between nucleons at very small separations.
Solve: (a)
(b) For x = 5.0 × 10−15 m, U = 55.1 × 10−12 J. This energy is represented by a total energy line.
(c) Due to conservation of total energy, the potential energy when x = 5.0 × 10−15 J is transformed into kinetic energy
until x = twice the radius = 1.0 × 10−15 m. At this separation, u = 23.6 × 10−12 J. Thus,
1 2 1 2
mv + mv + 23.6 × 10−12 J = 55.1× 10−12 J ⇒ v = 1.94 × 108 m/s
2
2
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10-40
Chapter 10
Assess: A speed of 1.94 × 108 m/s is approximately 0.65 c where c is the speed of light. This speed is understandable
for the present model.
10.61. Model: Assume U = c/x. Use conservation of energy.
Solve:
(a)
U i + Ki = U f + K f
Ki is zero.
Kf = U i − U f
1 2 c c
mv = −
xi xf
2
xf = 1
v2 =
This suggests that a graph of v 2 vs.
⎞ 2c 1 2c
2c ⎛ 1
−
⎜ − 1⎟ =
m ⎝ xi
⎠ m xi m
1
would be a straight line with a slope of 2c/m and intercept of −2c/m.
xi
1
is extraordinary and that the slope and the intercept have the
xi
same magnitude, as our theoretical considerations predicted.
(b) We can find the value of c from either the slope or the intercept. The spreadsheet reports the slope as
The spreadsheet graph shows the linear fit of v 2 vs.
0.040m3/s 2 .
c=−
0.050 kg
m
(slope) = −
(0.040 m3/s 2 ) = 0.0010 J/m = 1.0 mJ/m
2
2
Assess: The units of the rise/run of the graph are m3/s 2 and from the original statement of U the units of c should
be J/m, and those are consistent with each other in part (b).
10.62. A 2.5 kg ball is thrown upward at a speed of 4.0 m/s from a height of 82 cm above a vertical spring. When
the ball comes down it lands on and compresses the spring. If the spring has a spring constant of k = 600 N/m, by
how much is it compressed?
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Energy
10-41
10.63. (a) A 1500 kg auto coasts up a 10.0 m high hill and reaches the top with a speed of 5.0 m/s. What initial
speed must the auto have had at the bottom of the hill?
(b)
We place the origin of our coordinate system at the bottom of the hill.
(c) The solution of the equation is vi = 14.9 m/s ≈ 15 m/s. This is approximately 30 mph and is a reasonable value
for the speed at the bottom of the hill.
10.64. (a) A spring gun is compressed 15 cm to launch a 200 g ball on a horizontal, frictionless surface. The ball
has a speed of 2.0 m/s as it loses contact with the spring. Find the spring constant of the gun.
(b)
We place the origin of our coordinate system on the free end of the spring in the equilibrium position. Because the
surface is frictionless, the mechanical energy for the system (ball + spring) is conserved.
(c) The conservation of energy equation is
K f + U sf = Ki + U si
1 2 1
1
1
mv1 + k (0 m) 2 = m(0 m/s) 2 + k (−0.15 m) 2
2
2
2
2
(0.200 kg)(2.0 m/s) 2 = k (−0.15 m)2
k = 36 N/m
10.65. (a) A 100 g lump of clay traveling at 3.0 m/s strikes and sticks to a 200 g lump of clay at rest on a frictionless surface. The combined lumps smash into a horizontal spring with k = 3.0 N/m. The other end of the spring is
firmly anchored to a fixed post on the surface. How far will the spring compress?
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10-42
Chapter 10
(b)
(c) Solving the conservation of momentum equation we get v1x = 1.0 m/s. Substituting this value into the conserva-
tion of energy equation yields Δx2 = 32 cm.
10.66. (a) A spring with spring constant 400 N/m is anchored at the bottom of a frictionless 30° incline. A 500 g
block is pressed against the spring, compressing the spring by 10 cm, then released. What is the speed with which the
block is launched up the incline?
(b) The origin is placed at the end of the uncompressed spring. This is the point from which the block is launched as
the spring expands.
(c) Solving the energy conservation equation, we get vf = 2.6 m/s.
10.67. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energy
K + U g + U s is conserved.
Visualize:
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Energy
10-43
Place the origin of the coordinate system at the end of the unstretched spring, making ye = 0 m.
Solve: The clay is in static equilibrium while resting in the pan. The net force on it is zero. We can start by using this
to find the spring constant.
mg
(0.10 kg)(9.8 m/s 2 )
Fsp = FG ⇒ − k ( y1 − ye ) = − ky1 = mg ⇒ k = −
=−
= 9.8 N/m
y1
−0.10 m
Now apply conservation of energy. Initially, the spring is unstretched and the clay ball is at the ceiling. At the end,
the spring has maximum stretch and the clay is instantaneously at rest. Thus
K 2 + (U g ) 2 + (U s ) 2 = K 0 + (U g )0 + (U s )0 ⇒ 12 mv22 + mgy2 + 12 ky22 = 12 mv02 + mgy0 + 0 J
Since v0 = 0 m/s and v2 = 0 m/s, this equation becomes
mgy2 + 12 ky22 = mgy0 ⇒ y22 +
2mg
2mgy0
y2 −
=0
k
k
y22 + 0.20 y2 − 0.10 = 0
The numerical values were found using known values of m, g, k, and y0 . The two solutions to this quadratic equation
are y2 = 0.231 m and y2 = −0.432 m. The point we’re looking for is below the origin, so we need the negative root.
The distance of the pan from the ceiling is
L = | y2 | + 50 cm = 93 cm
10.68. Model: This is a two-part problem. In the first part, we will find the critical velocity for the ball to go over
the top of the peg without the string going slack. Using the energy conservation equation, we will then obtain the
gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus determining the
angle θ .
Visualize:
We place the origin of our coordinate system on the peg. This choice will provide a reference to measure gravitational potential energy. For θ to be minimum, the ball will just go over the top of the peg.
Solve: The two forces in the free-body force diagram provide the centripetal acceleration at the top of the circle.
Newton’s second law at this point is
mv 2
FG + T =
r
where T is the tension in the string. The critical speed vc at which the string goes slack is found when T → 0. In this case,
mvC2
⇒ vC2 = gr = gL/3
r
The ball should have kinetic energy at least equal to
1 2 1
⎛L⎞
mvC = mg ⎜ ⎟
2
2
⎝3⎠
for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to get the
minimum angle θ . The equation for the conservation of energy is
1
1
K f + U gf = K i + U gi ⇒ mvf2 + mgyf = mvi2 + mgyi
2
2
mg =
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10-44
Chapter 10
Using vf = vc , yf = 13 L, vi = 0, and the above value for vC2 , we get
1
L
L
L
mg + mg = mgyi ⇒ yi =
2
3
3
2
That is, the ball is a vertical distance 12 L above the peg’s location or a distance of
⎛ 2L L ⎞ L
− ⎟=
⎜
⎝ 3 2⎠ 6
below the point of suspension of the pendulum, as shown in the figure on the right. Thus,
L /6 1
= ⇒ θ = 80.4°
cosθ =
L
6
10.69. Model: Assume an ideal spring that obeys Hooke’s law. The mechanical energy K + U s + U g is conserved
during the launch of the ball.
Visualize:
This is a two-part problem. In the first part, we use projectile equations to find the ball’s velocity v2 as it leaves the
spring. This will yield the ball’s kinetic energy as it leaves the spring.
Solve: Using the equations of kinematics,
1
x3 = x2 + v2 x (t3 − t2 ) + ax (t3 − t2 ) 2 ⇒ 5.0 m = 0 m + (v2cos30°)(t3 − 0 s) + 0 m
2
(v2cos30°)t3 = 5.0 m ⇒ t3 = (5.0 m/v2cos30°)
1
y3 = y2 + v2 y (t3 − t2 ) + a y (t3 − t2 ) 2
2
1
−1.5 m = 0 + (v2sin30°)(t3 − 0 s) + (9.8 m/s 2 )(t3 − 0 s) 2
2
⎛ 5.0 m ⎞
2 ⎛ 5.0 m ⎞
Substituting the value for t3 , (−1.5 m) = (v2sin30°) ⎜
⎟ − (4.9 m/s ) ⎜
⎟
v
cos30
°
⎝ 2
⎠
⎝ v2cos30° ⎠
163.33
⇒ (−1.5 m) = + (2.887 m) −
⇒ v2 = 6.102 m/s
v22
2
The conservation of energy equation K 2 + U s2 + U g2 = K1 + U s1 + U g1 is
1 2 1
1
1
mv2 + k (0 m)2 + mgy2 = mv12 + k (Δs ) 2 + mgy1
2
2
2
2
Using y2 = 0 m, v1 = 0 m/s, Δs = 0.20 m, and y1 = −(Δs )sin g 30°, we get
1 2
1
mv2 + 0 J + 0 J = 0 J + k (Δs ) 2 − mg ( Δs )sin30°
2
2
( Δs ) 2 k = mv22 + 2mg (Δs )sin30°
(0.20 m) 2 k = (0.020 kg)(6.102 m/s)2 + 2(0.020 kg)(9.8 m/s 2 )(0.20)(0.5) ⇒ k = 19.6 N/m
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Energy
10-45
The final answer rounds to 20 N/m.
Assess: Note that y1 = −(Δs )sin30° is with a minus sign and hence the gravitational potential energy mgy1 is
− mg (Δs )sin30°.
10.70. Model: Choose yourself + spring + earth as the system. There are no forces from outside this system, so it
is an isolated system. The interaction forces within the system are the spring force of the bungee cord and the gravitational force. These are both conservative forces, so mechanical energy is conserved.
Visualize:
We can equate the system’s initial energy, as you step off the bridge, to its final energy when you reach the lowest
point. We do not need to compute your speed at the point where the cord starts to stretch. We do, however, need to
note that the end of the unstretched cord is at y0 = y1 − 30 m = 70 m, so U 2s = 12 k ( y2 − y0 ) 2 . Also note that U1s = 0,
since the cord is not stretched. The energy conservation equation is
1
K 2 + U 2g + U 2s = K1 + U1g + U1s ⇒ 0 J + mgy2 + k ( y2 − y0 ) 2 = 0 J + mgy1 + 0 J
2
Multiply out the square of the binomial and rearrange:
1
1
mgy2 + ky22 − ky0 y2 + ky02 = mgy1
2
2
⎞
⎛ 2 2mgy1 ⎞
2 ⎛ 2mg
2
⇒ y2 + ⎜
− 2 y0 ⎟ y2 + ⎜ y0 −
⎟ = y2 − 100.8 y2 + 980 = 0
k
k
⎝
⎠
⎝
⎠
This is a quadratic equation with roots 89.9 m and 10.9 m. The first is not physically meaningful because it is a
height above the point where the cord started to stretch. So we find that your distance from the water when the
bungee cord stops stretching is 10.9 m which is 11 m to two sig figs.
10.71. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energy
K + U g + U s is conserved.
Visualize:
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10-46
Chapter 10
We have chosen to place the origin of the coordinate system at the point of maximum compression. We will use
lengths along the ramp with the variable s rather than x.
Solve: (a) The conservation of energy equation K 2 + U g2 + U s2 = K1 + U g1 + U s1 is
1 2
1
1
mv2 + mgy2 + k ( Δs ) 2 = mv12 + mgy1 + k (0 m) 2
2
2
2
1
1
1
2
2
m(0 m/s) + mg (0 m) + k (Δs ) = m(0 m/s) 2 + mg (4.0 m + Δs )sin30° + 0 J
2
2
2
1
⎛1⎞
(250 N/m)(Δs ) 2 = (10 kg)(9.8 m/s 2 )(4.0 m + Δs ) ⎜ ⎟
2
⎝2⎠
This gives the quadratic equation:
(125 N/m)(Δs ) 2 − (49 kg ⋅ m/s 2 )Δs − 196 kg ⋅ m 2 /s 2 = 0
⇒ Δs = 1.46 m and − 1.07 m (unphysical)
The maximum compression is 1.46 m which rounds to 1.5 m.
(b) We will now apply the conservation of mechanical energy to a point where the vertical position is y and the
block’s velocity is v. We place the origin of our coordinate system on the free end of the spring when the spring is
neither compressed nor stretched.
1 2
1
1
1
mv + mgy + k ( Δs ) 2 = mv12 + mgy1 + k (0 m)2
2
2
2
2
1 2
1
2
mv + mg (−Δs sin30°) + k (Δs ) = 0 J + mg (4.0 m sin30°) + 0 J
2
2
1
1
2
k (Δs ) − ( mg sin30°)Δs + mv 2 − mgsin30°(4.0 m) = 0
2
2
To find the compression where v is maximum, take the derivative of this equation with respect to Δs:
1
1
dv
k 2(Δs ) − (mg sin30°) + m 2v
−0=0
2
2
d Δs
dv
= 0 at the maximum, we have
Since
d Δs
Δs = (mg sin30°)/k = (10 kg)(9.8 m/s 2 )(0.5)/(250 N/m) = 19.6 cm
This is 20 cm to two sig figs.
10.72. Model: Assume an ideal, massless spring that obeys Hooke’s law. Let us also assume that the cannon (C)
fires balls (B) horizontally and that the spring is directly behind the cannon to absorb all motion.
Visualize:
The before-and-after pictorial representation is shown, with the origin of the coordinate system located at the spring’s
free end when the spring is neither compressed nor stretched. This free end of the spring is just behind the cannon.
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Energy
10-47
Solve: The momentum conservation equation pfx = pix is
mB (vfx ) B + mC (vfx )C = mB (vix )B + mC (vix )C
Since the initial momentum is zero,
⎛ 200 kg ⎞
mC
(vfx )C = − ⎜
⎟ (vfx )C = −20(vfx )C
mB
⎝ 10 kg ⎠
The mechanical energy conservation equation for the cannon + spring K f + U sf = Ki + U si is
1
1
1
1
1
m(vf )C2 + k (Δx) 2 = m(vi )C2 + 0 J ⇒ 0 J + k (Δx) 2 = m(vfx )C2
2
2
2
2
2
(20,000 N/m)
k
(0.50 m) = ±5.0 m/s
⇒ (vfx )C = ±
Δx = ±
200 kg
m
(vfx ) B = −
To make this velocity physically correct, we retain the minus sign with (vfx )C . Substituting into the momentum conservation equation yields:
(vfx ) B = −20(−5.0 m/s) = 100 m/s
10.73. Model: This is a collision between two objects, and momentum is conserved in the collision. In addition,
because the interaction force is a spring force and the surface is frictionless, energy is also conserved.
Visualize:
Let part 1 refer to the time before the collision starts, part 2 refer to the instant when the spring is at maximum compression, and part 3 refer to the time after the collision. Notice that just for an instant, when the spring is at maximum
compression, the two blocks are moving side by side and have equal velocities: vA2 = vB2 = v2 . This is an important
observation.
Solve: First relate part 1 to part 2. Conservation of energy is
1
1
1
1
1
1
2
2
2
mA vA1
= mA vA2
+ mBvB2
+ k (Δxmax ) 2 = ( mA + mB )v22 + k ( Δxmax ) 2
2
2
2
2
2
2
where Δxmax is the spring’s compression. U g is not in the equation because there are no elevation changes. Also note
that K 2 is the sum of the kinetic energies of all moving objects. Both v2 and Δxmax are unknowns. Now add the
conservation of momentum:
mA vA1 = mA vA2 + mBvB2 = (mA + mB )v2 ⇒ v2 =
mA vA1
= 2.667 m/s
mA + mB
Substitute this result for v2 into the energy equation to find:
1
1
1
2
k (Δxmax ) 2 = mA vA1
− (mA + mB )v22
2
2
2
2
mA vA1
− (mA + mB )v22
= 0.046 m = 4.6 cm
k
Notice how both conservation laws were needed to solve this problem.
⇒ Δxmax =
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10-48
Chapter 10
(b) Again, both energy and momentum are conserved between “before” and “after.” Energy is
1
1
1
1 2
2
2
2
2
mA vA1
= mA vA3
+ mBvB3
⇒ 16 = vA3
+ vB3
2
2
2
2
The spring is no longer compressed, so the energies are purely kinetic. Momentum is
mA vA1 = mA vA3 + mBvB3 ⇒ 8 = 2vA3 + vB3
We have two equations in two unknowns. From the momentum equation, we can write vB3 = 2(4 − vA3 ) and use this
in the energy equation to obtain:
1
2
2 2
2
2
16 = vA3
+ ⋅ 4(4 − vA3
) = 3vA3
− 16vA3 + 32 ⇒ 3vA3
− 16vA3 + 16 = 0
2
This is a quadratic equation for vA3 with roots vA3 = (4 m/s, 1.33 m/s). Using vB3 = 2(4 − vA3 ), these two roots give
vB3 = (0 m/s, 5.333 m/s). The first pair of roots corresponds to a “collision” in which A misses B, so each keeps its
initial speed. That’s not the situation here. We want the second pair of roots, from which we learn that the blocks’
speeds after the collision are vA3 = 1.33 m/s and vB3 = 5.3 m/s.
10.74. Model: Mechanical energy and momentum are conserved during the expansion of the spring.
Visualize: Please refer to Figure CP10.74.
Solve: Example 10.8 is a very similar problem, except that the objects are initially at rest. We can use the solution
from Example 10.8 for this problem in a reference frame S′ in which the two carts are initially at rest, then transform
the answer to the frame S in which the carts are initially moving.
Thus in the S′ frame,
(vfx )′2 =
k (Δxi ) 2
m2 (1 + m2 m )
1
(vfx )1′ = −
m2
(vfx )′2
m1
Let the 100 g cart be Cart 1 and the 300 g cart be Cart 2. With k = 120 N/m and Δxi = 4.0 cm,
(vfx )′2 = 0.40 m/s, (vfx )1′ = −1.2 m/s
An object at rest in the S′ frame is traveling to the right at 1.0 m/s in the S frame. The equation of transformation is
therefore
vx = v′x + 1.0 m/s
In the S frame, the velocities of the carts are
(vfx )1 = −1.2 m/s + 1.0 m/s = −0.2 m/s
(vfx ) 2 = 0.40 m/s + 1.0 m/s = 1.4 m/s
Assess: Cart 1 is moving slowly to the left while the heavier Cart 2 is moving quickly to the right.
10.75. Model: Model the balls as particles, and assume a perfectly elastic collision. After the collision is over, the
balls swing out as pendulums. The sum of the kinetic energy and gravitational potential energy does not change as
the balls swing out.
Visualize:
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Energy
10-49
In the pictorial representation we have identified before-and-after quantities for both the collision and the pendulum
swing. We have chosen to place the origin of the coordinate system at a point where the two balls at rest barely touch
each other.
Solve: As the ball with mass m1, whose string makes an angle θil with the vertical, swings through its equilibrium
position, it lowers its gravitational energy from m1gy0 = m1g ( L − Lcosθi1) to zero. This change in potential energy
transforms into a change in kinetic energy. That is,
1
m1g ( L − Lcosθi1 ) = m1(v1 )12 ⇒ (v1 )1 = 2 gL(1 − cosθi1 )
2
Similarly, (v1 )2 = 2 gL(1 − cosθi2 ) . Using θi1 = θi2 = 45°, we get (v1 )1 = 2.396 m/s = (v1) 2 . Both balls are moving at
the point where they have an elastic collision. Since our analysis of elastic collisions was for a situation in which
ball 2 is initially at rest, we need to use the Galilean transformation to change to a frame S′ in which ball 2 is at rest. Ball 2
is at rest in a frame that moves with ball 2, so choose S′ to have V = −2.396 m/s, with the minus sign because this frame
(like ball 2) is moving to the left. In this frame, ball 1 has velocity (v1′ )1 = (v1 )1 − V = 2.396 m/s + 2.396 m/s = 4.792 m/s
and ball 2 is at rest. The elastic collision causes the balls to move with velocities
(v2′ )1 =
m1 − m2 ′
1
(v1 )1 = − (4.792 m/s) = −1.597 m/s
m1 + m2
3
(v2′ ) 2 =
2m1
2
(v1′ )1 = (2.396 m/s) = 3.194 m/s
m1 + m2
3
We can now use v = v′ + V to transform these back into the laboratory frame:
(v2 )1 = −1.597 m/s − 2.396 m/s = −3.99 m/s
(v2 ) 2 = 3.195 m/s − 2.396 m/s = 0.799 m/s
Having determined the velocities of the two balls after collision, we will once again use the conservation equation
K f + U gf = Ki + U gi for each ball to solve for the θf1 and θ f 2 .
1
1
m1 (v3 )12 + m1gL(1 − cosθf1 ) = m1 (v2 )12 + 0 J
2
2
Using (v3 )1 = 0, this equation simplifies to
1
1
gL(1 − cosθ f1 ) = (−3.99 m/s) 2 ⇒ cosθf1 = 1 −
(−3.99 m/s)2 ⇒ θ f1 = 79.3°
2
2 gL
The 100 g ball rebounds to 79º. Similarly, for the other ball:
1
1
m2 (v3 )22 + m2 gL(1 − cosθ f 2 ) = m2 (v2 ) 22 + 0 J
2
2
Using (v3 ) 2 = 0, this equation becomes
⎛ 1 ⎞
2
cosθf 2 = 1 − ⎜
⎟ (0.799) ⇒ θ f 2 = 14.7°
2gL
⎝
⎠
The 200 g ball rebounds to 14.7°.
10.76. Model: Model the sled as a particle. Because there is no friction, the sum of the kinetic and gravitational
potential energy is conserved during motion.
Visualize:
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10-50
Chapter 10
Place the origin of the coordinate system at the center of the hemisphere. Then y0 = R and, from geometry,
y1 = Rcosφ .
Solve: The energy conservation equation K1 + U1 = K 0 + U 0 is
1 2
1
1
mv1 + mgy1 = mv02 + mgy0 ⇒ mv12 + mgRcosφ = mgR ⇒ v1 = 2 gR (1 − cosφ )
2
2
2
G
(b) If the sled is on the hill, it is moving in a circle and the r-component of Fnet has to point to the center with magnitude Fnet = mv 2 /R. Eventually the speed gets so large that there is not enough force to keep it in a circular trajectory, and that is the point where it flies off the hill. Consider the sled at angle φ . Establish an r-axis pointing toward
the center of the circle, as we usually do for circular motion problems. Newton’s second law along this axis requires:
mv 2
R
⎛
mv 2
v2 ⎞
⇒ n = mg cosφ −
= m ⎜ g cosφ − ⎟
⎜
R
R ⎟⎠
⎝
The normal force decreases as v increases. But n can’t be negative, so the fastest speed at which the sled stays on the
hill is the speed vmax that makes n → 0. We can see that vmax = gR cosφ .
(c) We now know the sled’s speed at angle φ , and we know the maximum speed it can have while remaining on the
hill. The angle at which v reaches vmax is the angle φmax at which the sled will fly off the hill. Combining the two
( Fnet )r = FG cosφ − n = mg cosφ − n = mar =
expressions for v1 and vmax gives:
2 gR (1 − cosφ ) = gR cosφ ⇒ 2 R(1 − cosφmax ) = R cosφmax
⇒ cosφmax =
2
⎛2⎞
⇒ φmax = cos −1 ⎜ ⎟ = 48°
3
⎝3⎠
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WORK
11
Conceptual Questions
11.1. There is not enough information to tell. The lost potential energy and the work done by the environment could
increase the kinetic energy or it is possible that all the work and energy are converted to thermal energy.
11.2. There is not enough information to tell. The work done could cause some or all of the potential energy change
or some of the work could be converted to thermal energy. Without more information, it is impossible to say whether
a kinetic energy change is present.
11.3. The system is doing work on the environment. The total mechanical energy of the system is lower.
11.4. The ball’s kinetic energy is equal to the work done on it by gravity. Since work is force × distance, the kinetic
energy of the ball increases by equal amounts in equal distance intervals.
11.5. No work was done by gravity. Wg = −mg Δy. Here, Δy = 0. Any work done during a downward part of the
motion was undone during the upward parts.
11.6. The kinetic energies are equal. Equal forces are applied over equal displacements so that the same work is done
on each. Thus, the change in kinetic energy is the same. Because Ki = 0, ΔK = K f . (The plastic will be moving 10
times faster, however.)
11.7. The work is the same in both cases, since the work done against gravity is −mg Δy, and Δy, the change in
height, is the same in both cases.
11.8. (a) No, the rate of change of potential energy with respect to position will be zero at that point, but the value
of the potential energy is not known without specifying it at some reference point.
(b) No, the zero point for the potential energy is arbitrary. There will be a force present if the rate of change of the
potential energy with position is nonzero.
11.9. The kinetic energy was dissipated as thermal energy by friction between the tires and the road and in the
brakes.
11.10. Gravitational potential energy is transformed into thermal energy. There is no change in the kinetic energy.
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11-1
11-2
Chapter 11
11.11. (a) Push a puck with force F across a frictionless level surface. With the puck as the system,
Wext = F Δx = ΔK . The gravitational potential energy does not change because Δy = 0. Since the surface is
frictionless, ΔEth = 0.
(b) Push a box across a rough level surface at constant speed. The system is the box. Again, Δy = 0, but now
ΔK = 0, and friction dissipates the external work done by the push as thermal energy.
11.12. Power is energy per time. The energy required to lift a beam a height Δy is the same as the change in
gravitational potential energy of the beam. Power P =
different power P′ =
W mg Δy
=
. So doubling Δy and halving Δt requires a
Δt
Δt
mg (2Δy )
mg Δy
=4
= 4 P. The power must be increased by a factor of 4.
Δt
(Δt/2)
Exercises and Problems
Section 11.2 Work and Kinetic Energy
Section 11.3 Calculating and Using Work
G G
11.1. Solve: (a) A ⋅ B = Ax Bx + Ay B y = (3)(2) + ( 4)(−6) = −18.
G G
(b) A ⋅ B = Ax Bx + Ay By = (3)(6) + (−2 )( 4 ) = 10.
G G
11.2. Solve: (a) A ⋅ B = Ax Bx + Ay B y = (4)(−2) + (−2)(−3) = −2.
G G
(b) A ⋅ B = Ax Bx + Ay By = ( −4)(2) + (2)(4) = 0.
G
G
G
G G
11.3. Solve: (a) The length of A is A = A = A ⋅ A = (3)2 + (4)2 = 25 = 5. The length of B is B = (2)2 + (−6)2 =
G G
40 = 2 10 . Using the answer A ⋅ B = −18 from Ex 11.1(a),
G G
A ⋅ B = AB cos α
−18 = (5)(2 10)cos α
α = cos −1 (−18/ 40) = 125°
G
G
G G
G
(b) The length of A is A = A = A ⋅ A = (3) 2 + (−2 )2 = 13. The length of B is B = (6) 2 + ( 4)2 = 52 = 2 13.
G G
Using the answer A ⋅ B = 10 from Ex 11.1(b),
G G
A ⋅ B = AB cos α
10 = ( 13)(2 13)cos α
α = cos −1 (10/26) = 67°
G
G
G
G G
11.4. Solve: (a) The length of A is A = A = A ⋅ A = (4)2 + (−2)2 = 20. The length of B is B = (−2)2 + (−3)2 =
G G
13. Using the answer A ⋅ B = −2 from EX11.2(a),
G G
A ⋅ B = AB cos α
−2 = ( 20)( 13)cos α
G G
(b) From EX11.2(b), A ⋅ B = 0. Thus
α = cos −1 (−2/ 260) = 97°
cos α = 0
α = 90°
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Work
11-3
11.5. Visualize:
Please refer to Figure EX11.5.
G G
Solve: (a) A ⋅ B = AB cos α = (5)(3)cos 40° = 11.
G G
(b) C ⋅ D = CD cos α = (2)(3)cos140° = −4.6.
G G
(c) E ⋅ F = EF cos α = (3)(4)cos90° = 0.
11.6. Visualize:
Please refer to Figure EX11.6.
G G
Solve: (a) A ⋅ B = AB cos α = (2)(4)cos110° = −2.7.
G G
(b) C ⋅ D = CD cos α = (5)(4)cos180° = −20.
G G
(c) E ⋅ F = EF cos α = (4)(3)cos30° = 10.
=0 ⎞
P
⎛
G G
⎜
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
11.7. Solve: (a) W = F ⋅ Δr = (−3.0i + 6.0 j ) ⋅ (2.0i ) N m = −6.0i ⋅ i + 12 j ⋅ iˆ ⎟ J = −6.0 J.
⎜
⎟
⎝
⎠
=0
P
⎛
⎞
G G
⎜
ˆ
ˆ
ˆ
ˆ
(b) W = F ⋅ Δr = (−3.0i + 6.0 j ) ⋅ (2.0 j ) N m = −6.0 i ⋅ ˆj + 12 ˆj ⋅ ˆj ⎟ J = 12 J.
⎜
⎟
⎝
⎠
=0 ⎞
P
⎛
G G G
⎜
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
11.8. Solve: (a) W = F ⋅ Δr = (−4.0i − 6.0 j ) ⋅ (−3.0i ) N m = 12i ⋅ i + 18.0 j ⋅ iˆ ⎟ J = 12 J.
⎜
⎟
⎝
⎠
G G G
ˆ
ˆ
ˆ
ˆ
(b) W = F ⋅ Δr = (−4.0i − 6.0 j ) ⋅ (3.0i − 2.0 j ) N m = (−12 + 12) J = 0 J.
11.9. Model: Use the work-kinetic energy theorem to find the net work done on the particle.
Visualize:
Solve: From the work-kinetic energy theorem,
2
1
1
1
1
W = ΔK = mv12 − mv02 = m(v12 − v02 ) = (0.020 kg) ⎡(30 m/s) 2 − (−30 m/s )⎤ = 0J
⎣
⎦
2
2
2
2
Assess: Negative work is done in slowing down the particle to rest, and an equal amount of positive work is done in
bringing the particle to the original speed but in the opposite direction.
G
G
G
G
G
11.10. Model: Work done by a force F on a particle is defined as W = F ⋅ Δr , where Δr is the particle’s
displacement.
Visualize:
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11-4
Chapter 11
Solve: (a) The work done by gravity is
G
G
Wg = FG ⋅ Δr = ( −mgjˆ) ⋅ (2.25 − 0.75) ˆj N m = −(2.0 kg)(9.8 m/s 2 )(1.50 m) J = −29 J
G
G
(b) The work done by hand is WH = Fhand on book ⋅ Δr . As long as the book does not accelerate,
G
G
Fhand on book = − Fearth on book = −(−mgjˆ) = mgjˆ
WH = (mgjˆ) ⋅ ( 2.25 − 0.75) ˆj N m = (2.0 kg)(9.8 m/s 2 )(1.50 m) = 29 J
G
G G
11.11. Model: Model the piano as a particle and use W = F ⋅ Δr , where W is the work done by the force F
G
through the displacement Δr .
Visualize:
G
Solve: For the force FG :
G G G
G
W = F ⋅ Δr = FG ⋅ Δr = ( Fg ) ⋅ (Δr )cos(0°) = (255 kg)(9.81 m/s 2 )(5.00 m)(1.00) = 1.25 × 104 J
G
For the tension T1 :
G G
W = T1 ⋅ Δr = (T1 )( Δr )cos(150°) = (1830 N )(5.00 m )(−0.8660 ) = −7.92 × 103 J
G
For the tension T2 :
G
G
W = T2 ⋅ Δr = (T2 )(Δr )cos(135°) = (1295 N )(5.00 m )(−0.7071) = −4.58 × 103 J
G
Assess: Note that the displacement Δr in all the above cases is directed downwards along − ĵ.
G
G
G
11.12. Model: Model the crate as a particle and use W = F ⋅ Δr , where W is the work done by a force F on a
G
particle and Δr is the particle’s displacement.
Visualize:
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Work
G
Solve: For the force f k :
G
For the tension T1 :
G
For the tension T2 :
11-5
G
G
W = f k ⋅ Δr = f k (Δr )cos(180°) = ( 660 N )(3.0 m )(−1.0) = −2.0 kJ
G G
W = T1 ⋅ Δr = (T1 )( Δr )cos(20°) = ( 600 N )(3.0 m )( 0.9397 ) = 1.7 kJ
G
G
W = T2 ⋅ Δr = (T2 )(Δr )cos(30°) = ( 410 N )(3.0 m )( 0.866 ) = 1.1 kJ
G
Assess: Negative work done by the force of kinetic friction f k means that 1.95 kJ of energy has been transferred out
of the crate.
11.13. Model: Model the 2.0 kg object as a particle, and use the work–kinetic-energy theorem.
Visualize: Please refer to Figure EX11.13. For each of the five intervals the velocity-versus-time graph gives the
initial and final velocities. The mass of the object is 2.0 kg.
Solve: According to the work–kinetic-energy theorem:
W = ΔK = 12 mvf2 = 12 mvi2 = 12 m(vf2 − vi2 )
Interval AB: vi = 2 m/s, vf = −2 m/s ⇒ W = 12 (2.0 kg) ⎡( −2 m/s) 2 − ( 2 m/s )2 ⎤ = 0 J
⎣
⎦
2
Interval BC: vi = −2 m/s, vf = −2 m/s ⇒ W = 12 (2.0 kg) ⎡(−2 m/s)2 − (−2 m/s )⎤ = 0 J
⎣
⎦
2
2⎤
1
⎡
Interval CD: vi = −2 m/s, vf = 0 m/s ⇒ W = 2 (2.0 kg) (0 m/s) − (−2 m/s ) = −4 J
⎣
⎦
2
2⎤
1
⎡
Interval DE: vi = 0 m/s, vf = 2 m/s ⇒ W = 2 (2.0 kg) (2 m/s) − ( 0 m/s ) = +4 J
⎣
⎦
Assess: The work done is zero in intervals AB and BC. In the interval CD + DE the total work done is zero. It is not
whether v is positive or negative that counts because K ∝ v 2 . What is important is the magnitude of v and how v
changes.
Section 11.4 The Work Done by a Variable Force
11.14. Model: Use the definition of work.
Visualize: Please refer to Figure EX11.14.
Solve: Work is defined as the area under the force-versus-position graph:
sf
W = Ñ Fs ds = area under the force curve
si
Interval 0 − 1 m: W = ( 4 N )(1 m − 0 m ) = 4 J
Interval 1 − 2 m: W = (4 N)(0.5 m) + (−4 N )( 0.5 m ) = 0 J
Interval 2 − 3 m: W = 12 (−4.0 N)(1 m) = −2 J
11.15. Model: Use the work–kinetic-energy theorem to find velocities.
Visualize: Please refer to Figure EX11.15.
Solve: The work–kinetic-energy theorem is
ΔK = 12 mvf2 − 12 mvi2 = W =
1 mv 2
f
2
xf
∫ Fx dx =
area under the force curve from xi to xf
xi
− 12 (0.500 kg)(2.0 m/s)2 = 12 mvf2 − 1.0 J =
vf =
xf
∫
0m
Fx dx = 52 x 2 N m
2
5x N m
+ 4.0 m 2 /s 2
0.500 kg
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11-6
Chapter 11
At x = 1 m:
At x = 2 m:
At x = 3 m:
⇒ vf = 3.7 m/s
⇒ vf = 6.6 m/s
⇒ vf = 9.7 m/s
11.16. Model: Use the work–kinetic-energy theorem.
Visualize: Please refer to Figure EX11.16.
Solve: The work–kinetc-energy theorem is
ΔK = 12 mvf2 − 12 mvi2 =
vf =
xf
∫
0m
Fx dx = 10 x − 52 x 2
20 x − 5 x 2
+ 4.0 m/s
2.0 kg
At x = 2 m: ⇒ vf = 5 m/s
At x = 4 m: ⇒ vf = 4 m/s
11.17. Model: Use the work–kinetic-energy theorem.
Visualize: Please refer to Figure EX11.17.
Solve: The work–kinetic-energy theorem is
xf
ΔK = W = Ñ Fx dx = area of the Fx -versus- x graph between xi and xf
xi
1 mv 2
f
2
− 12 mvi2 = 12 ( Fmax )(2 m)
Using m = 0.500 kg, vf = 6.0 m/s, and vi = 2.0 m/s, the above equation yields Fmax = 8 N.
Assess: Problems in which the force is not a constant cannot be solved using constant-acceleration kinematic
equations.
Section 11.5 Work and Potential Energy
Section 11.6 Finding Force from Potential Energy
11.18. Model: Use the definition Fs = −dU/ds.
Visualize: Please refer to Figure EX11.18.
Solve: Fx is the negative of the slope of the potential energy graph at position x. Between x = 0 cm and x = 10 cm
the slope is
slope = (U f − U i )/( xf − xi ) = ( 0 J − 10 J )/(0.10 m − 0.0 m) = −100 N
Thus, Fx = 100 N at x = 5 cm. The slope between x = 10 cm and x = 20 cm is zero, so Fx = 0 N at x = 15 cm.
Between 20 cm and 40 cm,
slope = (10 J − 0 J)/(0.40 m − 0.20 m) = 50 N
At x = 25 cm and x = 35 cm, therefore, Fx = −50 N.
11.19. Model: Use the definition Fs = −dU/ds.
Visualize: Please refer to Figure EX11.19.
Solve: Fx is the negative of the slope of the potential energy graph at position x.
⎛ dU ⎞
Fx = − ⎜
⎟
⎝ dx ⎠
Between y = 0 m and x = 1 m, the slope is
slope = (U f − U i )/( xf − xi ) = (60 J − 0 J)/(1 m − 0 m) = 60 N
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11-7
Thus, Fx = 260 N at x = 1 m. Between x = 1 m and x = 5 m, the slope is
slope = (U f − U i )/( xf − xi ) = (0 J − 60 J)/(5 m − 1 m) = −15 N
Thus, Fx = 15 N at x = 4 m.
11.20. Model: Use the negative derivative of the potential energy to determine the force acting on a particle.
Solve: The y-component of the force is
dU
d
Fy = −
= − (4 y 3 ) = −12 y 2
dy
dy
At y = 0 m, Fy = 0 N; at y = 1 m, Fy = −12 N; and at y = 2 m, Fy = −48 N.
11.21. Model: Use the negative derivative of the potential energy to determine the force acting on a particle.
Solve: (b) The x-component of the force is
dU
d ⎛ 10 ⎞ 10
Fx = −
=− ⎜
⎟=
dx
dx ⎝ x ⎠ x 2
F ( x = 2 m) =
10
x2
= 2.5 N, F ( x = 5 m) =
x=2 m
10
x2
= 0.40 N, F ( = 8 m) =
x =5 m
10
x2
= 0.16 N
x =8 m
Section 11.7 Thermal Energy
11.22. Model: Assume the carbon-carbon bond acts like an ideal spring that obeys Hooke’s law.
Visualize:
The quantity ( x − xe ) is the stretching relative to the spring’s equilibrium length. In the present case, bond stretching
is analogous to spring stretching.
Solve: (a) The kinetic energy of the carbon atom is
1
1
K = mv 2 = (2.0 × 10−26 kg)(500 m/s) 2 = 2.5 × 10−21 J
2
2
(b) The energy of the spring is given by
1
U s = k ( x − xe ) 2 = K
2
k=
2K
( x − xe ) 2
=
2(2.5 × 10−21 J)
(0.050 × 10−9 m) 2
= 2.0 N/m
11.23. Visualize: One mole of helium atoms in the gas phase contains N A = 6.02 × 1023 atoms.
Solve: If each atom moves with the same speed v, the microscopic total kinetic energy will be
2 K micro
2(3700 J)
⎛1
⎞
K micro = N A ⎜ mv 2 ⎟ = 3700 J ⇒ v =
=
= 1360 m/s
−
mN A
(6.68 × 10 27 kg)(6.02 × 1023 )
⎝2
⎠
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11-8
Chapter 11
11.24. Visualize:
Solve: (a) Ki = K 0 = 12 mv02 = 0 J , U i = U g0 = mgy0 = (20 kg)(9.8 m/s 2 )(3.0 m) = 5.9 × 102 J
Wext = 0 J, K f = K1 = 12 mv12 = 12 (20 kg)(2.0 m/s)2 = 40 J, U f = U g1 = mgy1 = 0 J
At the top of the slide, the child has gravitational potential energy of 5.9 × 102 J. This energy is transformed into the
thermal energy of the child’s pants and the slide and the kinetic energy of the child. This energy transfer and
transformation is shown on the energy bar chart.
(b)
The change in the thermal energy of the slide and of the child’s pants is 5.9 × 102 J − 40 J = 5.5 × 102 J.
Section 11.8 Conservation of Energy
11.25. Visualize: The system loses 400 J of potential energy. In the process of losing this energy, it does 400 J of
work on the environment, which means Wext = −400 J. Since the thermal energy increases 100 J, we have
ΔEth = 100 J, which must have been 100 J of kinetic energy originally. This is shown in the energy bar chart.
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11-9
11.26. Visualize:
Note that the conservation of energy equation
Ki + U i + Wext = K f + U f + ΔEth
requires that Wext be equal to −400 J.
11.27. Solve: Please refer to Figure EX11.27. The energy conservation equation yields
Ki + U i + Wext = K f + U f + ΔEth ⇒ 4 J + 1 J + Wext = 1 J + 2 J + 1 J ⇒ Wext = −1 J
Thus, the work done to the environment is −1 J. In other words, 1 J of energy is transferred from the system into the
environment. This is shown in the energy bar chart.
11.28. Visualize: The tension of 20.0 N in the cable is an external force that does work on the block Wext =
(20.0 N)(2.00 m) = 40.0 J, increasing the gravitational potential energy of the block. We placed the origin of our
coordinate system on the initial resting position of the block, so we have U i = 0 J and U f = mgyf =
(1.02 kg)(9.8 m/s2 )(2.00 m) = 20.0 J. Also, Ki = 0 J, and ΔEth = 0 J. The energy bar chart shows the energy
transfers and transformations.
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11-10
Chapter 11
Solve: The conservation of energy equation is
Ki + U i + Wext = K f + U f + ΔEth
⇒ 0 J + 0 J + 40.0 J = 12 mvf2 + 20.0 J + 0 J
vf = (20.0 J)(2)/(1.02 kg) = 6.26 m/s
Section 11.9 Power
11.29. Model: Model the elevator as a particle, and apply the conservation of energy.
Solve: The tension in the cable does work on the elevator to lift it. Because the cable is pulled by the motor, we say
that the motor does the work of lifting the elevator.
(a) The energy conservation equation is Ki + U i + Wext = K f + U f + ΔEth . Using K i = 0 J, K f = 0 J, and ΔEth = 0 J
gives
Wext = (U f − U i ) = mg ( yf − yi ) = (1000 kg )(9.8 m/s 2 )(100 m ) = 9.80 × 105 J
(b) The power required to give the elevator this much energy in a time of 50 s is
W
9.80 × 105 J
P = ext =
= 1.96 × 104 W
Δt
50 s
Assess: Since 1 horsepower (hp) is 746 W, the power of the motor is 26 hp. This is a reasonable amount of power to
lift a mass of 1000 kg to a height of 100 m in 50 s.
11.30. Model: Model the steel block as a particle subject to the force of kinetic friction and use energy conservation.
Visualize:
G
G
G
Solve: (a) The work done on the block is Wnet = Fnet ⋅ Δr where Δr is the displacement. We will find the
displacement using kinematic equations and the force using Newton’s second law of motion. The displacement in the
x-direction is
Δx = x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 )2 = 0 m + (1.0 m/s)(3.0 s − 0 s) + 0 m = 3.0 m
G
Thus Δr = 3.0iˆ m.
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11-11
The equations for Newton’s second law along the x and y components are
( Fnet ) y = n − FG = 0 N ⇒ n = FG = mg = (10 kg)(9.8 m/s 2 ) = 98.0 N
G G
( Fnet ) x = F − f k = 0 N ⇒ F = f k = μ k n = (0.6)(98.0 N) = 58.8 N
G
G
Wnet = Fnet ⋅ Δr = F Δx cos (0°) = (58.8 N)(3.0 m)(1) = 176 J
(b) The power required to do this much work in 3.0 s is
W 176 J
P= =
= 59 W
t
3.0 s
11.31. Solve: The power of the solar collector is the solar energy collected divided by time. The intensity of the
solar energy striking the earth is the power divided by area. We have
ΔE 150 × 106 J
=
= 41,667 W and intensity = 1000 W/m 2
Δt
3600 s
41,667 W
Area of solar collector =
= 42 m2
1000 W/m 2
P=
11.32. Solve: The night light consumes more energy than the hair dryer. The calculations are
1.2 kW × 10 min = 1.2 × 103 × 10 × 60 J = 7.2 × 105 J
10 W × 24 hours = 10 × 24 × 60 × 60 J = 8.6 × 105 J
11.33. Solve: Using the conversion 746 W = 1 hp,
we
have
a
power
of
1492
J/s.
This
means
6
W = Pt = (1492 J/s)(1 h) = 5.3712 × 10 J is the total work done by the electric motor in one hour. Furthermore,
Wmotor = −Wg = U gf − U gi = mg ( yf − yi ) = mg (10 m)
m=
Wmotor
5.3712 × 106 J
1 liter
=
= 5.481 × 104 kg = 5.481 × 104 kg ×
= 5.5 × 104 liters
g (10 m) (9.8 m/s 2 )(10 m)
1 kg
11.34. Model: Model the sprinter as a particle, and use the constant-acceleration kinematic equations and the
definition of power in terms of velocity.
Visualize:
Solve: (a) We can find the acceleration from the kinematic equations and the horizontal force from Newton’s second
law. We have
x = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 ⇒ 50 m = 0 m + 0 m + 12 ax (7.0 s − 0 s) 2 ⇒ ax = 2.04 m/s 2
Fx = max = (50 kg)(2.04 m/s 2 ) = 10 × 101 N
G G
G
(b) We obtain the sprinter’s power output by using P = F ⋅ v , where v is the sprinter’s velocity. At t = 2.0 s the
power is
P = ( Fx )[v0 x + a x (t − t0 )] = (102 N)[0 m/s + ( 2.04 m/s 2 )( 2.0 s − 0 s )] = 0.42 kW
The power at t = 4.0 s is 0.83 kW, and at t = 6.0 s the power is 1.3 kW.
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11-12
Chapter 11
11.35. Visualize: We place the origin of the coordinate system at the base of the stairs on the first floor.
Solve: (a) We might estimate y2 − y1 ≈ 4.0 m ≈ 12 ft ≈ y3 − y2 , thus, y3 − y1 ≈ 8.0 m.
(b) We might estimate the time to run up these two flights of stairs to be 20 s.
(c) Estimate your mass as m ≈ 70 kg ≈ 150 lb. Your power output while running up the stairs is
work done by you change in potential energy mg ( y3 − y1)
=
=
time
time
time
(70 kg)(9.8 m/s 2 )(8.0 m)
⎛ 1 hp ⎞
≈ 270 W = ( 270 W ) ⎜
⎟ ≈ 0.35 hp
20 s
⎝ 746 W ⎠
Assess: Your estimate may vary, depending on your mass and how fast you run.
=
11.36. Visualize: See figure below.
Solve: Average power output is the change in energy of the system divided by the time interval. For the runner, the
change in energy is just the change Δ K in the kinetic energy because the potential energy remains unchanged. Thus,
ΔE = ΔK = K f − K 0 = 12 mvf2 − 12 mv02 = 12 (70 kg)(10 m/s) 2 = 3500 J. For the greyhound, the change in energy
is ΔE = ΔK = 12 (30 kg)(20 m/s) 2 = 6000 J. Thus, the average power output of the runner is P = ΔE/Δt =
(3500 J ) / (3.0 s ) = 1.2 kW and the average power output of the greyhound is P = (6000 J)/(3.0 s) = 2.0 kW.
G
G G
G
G
Visualize: Please refer to Figure P11.37. The force F = (6iˆ + 8 ˆj ) N on the particle is constant.
G
G
G
G
Solve: (a) WABD = WAB + WBD = F ⋅ ( Δs ) AB + F ⋅ (Δs ) BD
11.37. Model: Use the definition of work for a constant force F , W = F ⋅ Δs , where Δs is the displacement.
= (6iˆ + 8ˆ j ) N ⋅ (3iˆ) m + (6iˆ + 8 ˆj ) N ⋅ (4 ˆj ) m = 18 J + 32 J = 50 J
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Work
11-13
G
G
G
G
(b) WACD = WAC + WCD = F ⋅ (Δs ) AC + F ⋅ ( Δs )CD
(c) WAD
= (6iˆ + 8ˆ j ) N ⋅ (4 ˆj ) m + (6iˆ + 8 ˆj ) N ⋅ (3 ˆj ) m = 32 J + 18 J = 50 J
G
G
= F ⋅ (Δs ) AD = (6iˆ + 8ˆ j ) N ⋅ (3iˆ + 4 ˆj ) m = 18 J + 32 J = 50 J
The force is conservative because the work done is independent of the path.
11.38. Model: The force is conservative, so it has a potential energy.
Visualize: Please refer to Figure P11.38 for the graph of the force.
Solve: (a) The definition of potential energy is ΔU = −W (i → f ). In addition, work is the area under the force-
versus-displacement graph. Thus ΔU = U f − U i = −(area under the force curve). Since U i = 0 at x = 0 m, the
potential energy at position x is U ( x) = −(area under the force curve from 0 to x). From 0 m to 3 m, the area increases
linearly from 0 Nm to −60 Nm, so U increases from 0 J to 60 J. At x = 4 m, the area is −70 J. Thus U = 70 J at
x = 4 m, and U doesn’t change after that since the force is then zero. Between 3 m and 4 m, where F changes
linearly, U must have a quadratic dependence on x (i.e., the potential energy curve is a parabola). This information is
shown on the potential energy graph below.
(b) Mechanical energy is E = K + U . From the graph, U = 20 J at x = 1.0 m.
The kinetic energy is K = 12 mv 2 = 12 (0.100 kg)(25 m/s) 2 = 31.25 J. Thus E = 51 J.
(c) The total energy line at 51 J is shown on the graph above.
(d) The turning point occurs where the total energy line crosses the potential energy curve. We can see from the
graph that this is at approximately 2.5 m. For a more accurate value, the potential energy function is U = 20x J. The
TE line crosses at the point where 20 x = 51.25, which is x = 2.6 m.
11.39. Model: Use the relationship between force and potential energy and the work–kinetic-energy theorem.
Visualize: Please refer to Figure P11.39. We will find the slope in the following x regions: 0 cm < x < 1 cm,
1 < x < 3 cm, 3 < x < 5 cm, 5 < x < 7 cm, and 7 < x < 8 cm.
Solve: (a) Fx is the negative slope of the U-versus-x graph, for example, for 0 m < x < 2 m
dU
−4 J
=
= −400 N ⇒ Fx = +400 N
dx 0.01 m
Calculating the values of Fx in this way, we can draw the force-versus-position graph as shown below.
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11-14
Chapter 11
(b) Since W = ∫ xxf Fx dx = area of the Fx -versus-x graph between xi and xf , the work done by the force as the
i
particle moves from xi = 2 cm to xf = 6 cm is −2 J.
(c) The conservation of energy equation is K f + U f = Ki + U i . We can see from the graph that U i = 0 J and
U f = 2 J in moving from x = 2 cm to x = 6 cm. The final speed is vf = 10 m/s, so
2 J + 12 (0.010 kg)(10.0 m/s)2 = 0 J +
1 (0.010
2
kg)vi2
⇒ vi = 22 m/s
11.40. Model: Use the relationship between a conservative force and potential energy.
Visualize: Please refer to Figure P11.40. We will obtain U as a function of x and Fx as a function of x by using the
calculus techniques of integration and differentiation.
Solve: (a) For the interval 0 m < x < 0.5 m, Fx = ( 4 x ) N, where x is in meters. This means
dU
= − Fx = −4 x ⇒ U = −2 x 2 + C1 = −2 x 2
dx
where we have used U = 0 J at x = 0 m to obtain C1 = 0. For the interval 0.5 m < x < 1 m, Fx = (−4 x + 4) N.
Likewise,
dU
= 4 x − 4 ⇒ U = 2 x 2 − 4 x + C2
dx
Since U should be continuous at the junction, we have the continuity condition
(−2 x 2 ) x = 0.5 m = ( 2 x 2 − 4 x + C2 ) x = 0.5 m ⇒ − 0.5 = 0.5 − 2 + C2
⇒ C2 = 1
U remains constant for x ≥ 1 m.
(b) For the interval 0 m < x < 0.5 m, U = +4 x, and for the interval 0.5 m < x < 1.0 m, U = −4 x + 4, where x is in
meters. The derivatives give Fx = −4 N and Fx = +4 N, respectively. The slope is zero for x ≥ 1 m.
11.41. Model: Use ax = dvx /dt , x = ∫ vx dt , K = 12 mvx2 , and F = max .
Visualize: Please refer to Figure P11.41. We know a x = slope of the vx -versus-t graph and x = area under the
vx -versus-x graph between 0 and x.
Solve: Using the above definitions and methodology, we can generate the following table:
t(s)
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
a x (m/s2)
10
10
10
10
+10 or −10
10
−10 or 0
0
0
x(m)
0
1.25
5
11.25
20
28.75
35
40
45
K(J)
0
6.25
25
56.25
100
56.25
25
25
25
F(N)
5
5
5
5
−5 or +5
5
−5 or 0
0
0
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11-15
(e) Let J1 be the impulse from t = 0 s to t = 2 s and J 2 be the impulse from t = 2 s to t = 4 s. We have
2s
4s
0s
0s
J1 = Ñ Fx dt = (5 N)(2 s) = 10 N ⋅ s and J 2 = Ñ Fx dt = (−5 N)(1 s) = −5 N ⋅ s
(f) J = Δp = mvf − mvi
⇒ vf = vi + J/m
At t = 2 s, vx = 0 m/s + (10 N ⋅ s )/(0.5 kg) = 0 m/s + 20 m/s = 20 m/s
At t = 4 s, vx = 20 m/s + (−5 N ⋅ s)/(0.5 kg) = 20 m/s − 10 m/s = 10 m/s
The vx -versus-t graph also gives vx = 20 m/s at t = 2 s and vx = 10 m/s at t = 4 s.
(g)
(h) From t = 0 s to t = 2 s, W = ∫ Fx dx = (5 N)(20 m) = 100 J
From t = 2 s to t = 4 s, W = ∫ Fx dx = ( −5 N)(15 m) = −75 J
(i) At t = 0 s, vx = 0 m/s so the work–kinetic-energy theorem for calculating vx at t = 2 s is
1
1
1
1
W = ΔK = mvf2 − mvi2 ⇒ 100 J = (0.5 kg)vx2 − (0.5 kg)(0 m/s) 2
2
2
2
2
To calculate vx at t = 4 s, we use vx at t = 2 s as the initial velocity:
1
1
−75 J = (0.5 kg)vx2 − (0.5 kg)(20 m/s) 2
2
2
Both of these values agree with the values on the velocity graph.
⇒ vx = 20 m/s
⇒ vx = 10 m/s
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11-16
Chapter 11
11.42. Model: Model the elevator as a particle.
Visualize:
+
Solve: (a) The work done by gravity on the elevator is
Wg = −ΔU g = mgy0 − mgy1 = − mg ( y1 − y0 ) = −(1000 kg)(9.8 m/s 2 )(10 m) = −9.8 × 104 J
(b) The work done by the tension in the cable on the elevator is
WT = T (Δy )cos(0°) = T ( y1 − y0 ) = T (10 m)
To find T we write Newton’s second law for the elevator:
∑ Fy = T − FG = ma y ⇒ T = FG + ma y = m( g + a y ) = (1000 kg)(9.8 m/s2 + 1.0 m/s2 )
= 1.08 × 104 N ⇒ WT = (1.08 × 104 N)(10 m) = 1.1 × 105 J
(c) The work–kinetic-energy theorem is
Wnet = Wg + WT = ΔK = K f − Ki = K f − 12 mv02 ⇒ K f = Wg + WT + 12 mv02
K f = (−9.8 × 104 J) + (1.08 × 105 J ) + 12 (1000 kg)(0 m/s) 2 = 1.0 × 104 J
(d) K f = 12 mvf2 ⇒ 1.0 × 104 J = 12 (1000 kg)vf2 ⇒ vf = 4.5 m/s
11.43. Model: Model the rock as a particle, and apply the work–kinetic-energy theorem.
Visualize:
Solve: (a) The work done by Bob on the rock is
WBob = ΔK = 12 mv12 − 12 mv02 = 12 mv12 = 12 (0.500 kg)(30 m/s) 2 = 225 J = 2.3 × 102 J
(b) For a constant force, WBob = FBob Δx ⇒
FBob = WBob /Δx = 2.3 × 102 N.
(c) Bob’s power output is PBob = FBobvrock and will be a maximum when the rock has maximum speed. This is just
as he releases the rock with vrock = v1 = 30 m/s. Thus, Pmax = FBobv1 = (225 J)(30 m/s) = 6750 W = 6.8 kW.
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Work
11-17
11.44. Model: Model the crate as a particle, and use the work–kinetic-energy theorem.
Visualize:
Solve: (a) The work–kinetic-energy theorem is ΔK = 12 mv12 − 12 mv02 = 12 mv12 = Wtotal . Three forces act on the box, so
Wtotal = Wgrav + Wn + Wpush . The normal force is perpendicular to the motion, so Wn = 0 J. The other two forces do
the following amount of work:
G
⎛ h ⎞
G
Wpush = Fpush ⋅ Δr = Fpush x1 cosθ = Fpush ⎜
⎟ cosθ = Fpush h cot θ
⎝ sin θ ⎠
G
⎛ h ⎞
G
Wgrav = FG ⋅ Δr = − mgx1 sin θ = − mg ⎜
⎟ sin θ = − mgh
⎝ sin θ ⎠
Thus, the speed at the top of the ramp is
2( Fpush h cot θ − mgh)
2Wtotal
=
m
m
(b) Insert the given quantities into the expression for the speed to find
v1 =
v1 =
2[(25 N)(2.0 m)cot (20°) − ( 5.0 kg )( 9.8 m/s 2 )( 2.0 m )]
= 4.0 m/s
5.0 kg
Assess: Note that Fpush cosθ > mg sin θ for the radical to remain positive. This means that the component of the
pushing force up the slope must be greater than the component of gravity down the slope for the crate to move
upwards, which is the assumption with which we started. Furthermore, if we take the limit h → 0, we get
v1
lim h →0
2 ⎡ Fh
⎣
=
( 1h ) − mgh ⎤⎦
=
m
2F
m
h =0
which is the expected result for pushing the crate along a horizontal frictionless surface.
11.45. Model: Model Sam strapped with skis as a particle, and apply the law of conservation of energy.
Visualize:
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11-18
Chapter 11
Solve: (a) The conservation of energy equation is
K1 + U g1 + Δ Eth = K 0 + U g0 + Wext
The snow is frictionless, so ΔEth = 0 J. However, the wind is an external force doing work on Sam as he moves
down the hill. Thus,
Wext = Wwind = ( K1 + U g1 ) − ( K 0 + U g0 )
=
(
1 mv 2
1
2
) (
+ mgy1 −
1 mv 2
0
2
) (
+ mgy0 =
1 mv 2
1
2
)
+ 0 J − (0 J + mgy0 ) = 12 mv12 − mgy0
2Wwind
m
We compute the work done by the wind as follows:
G
G
Wwind = Fwind ⋅ Δr = Fwind Δr cos(160°) = (200 N)(146 m)cos (160°) = −27,400 J
where we have used Δr = h/ sin(20°) = 146 m. Now we can compute
v1 = 2 gy0 +
v1 = 2(9.8 m/s 2 )(50 m) +
2(−27,400 J)
= 16 m/s
75 kg
Assess: We used a vertical y-axis for energy analysis, rather than a tilted coordinate system, because U g is determined
by its vertical position.
11.46. Model: Model Paul and the mat as a particle, assume the mat to be massless, use the model of kinetic friction,
and apply the work–kinetic-energy theorem.
Visualize:
We define the x-axis along the floor and the y-axis perpendicular to the floor.
Solve: We first need to determine f k . Newton’s second law in the y-direction gives
n + T sin (30°) = FG = mg
⇒ n = mg − T sin (30°) = (10 kg)(9.8 m/s 2 ) − (30 N)sin (30°) = 83.0 N.
Using n and the model of kinetic friction gives f k = μ k n = (0.2)(83.0 N) = 16.60 N. The net force on Paul and the
mat is therefore Fnet = T cos(30°) − f k = (30 N)cos(30°) − 16.6 N = 9.4 N . Thus,
Wnet = Fnet Δr = (9.4 N)(3.0 m) = 28 J
G
G
G
The other forces n and FG make an angle of 90° with Δr and do zero work. We can now use the work–kineticenergy theorem to find the final velocity as follows:
1
Wnet = K f − Ki = K f − 0 J = mvf2 ⇒ vf = 2Wnet /m = 2(28 J)/(10 kg) = 2.4 m/s
2
Assess: A speed of 2.4 m/s or 5.4 mph is reasonable for the present problem.
11.47. Model: Assume an ideal spring that obeys Hooke’s law. Model the box as a particle and use the model of
kinetic friction.
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Work
11-19
Visualize:
Solve: When the horizontal surface is frictionless, conservation of energy means
2
1 k ( x − x ) 2 = 1 mv 2 = K
1
0
e
1x
1 ⇒ K1 = 2 (100 N/m)(0.20 m − 0 m) = 2.0 J
2
2
That is, the box is launched with 2.0 J of kinetic energy. It will lose 2.0 J of kinetic energy on the rough surface. The
G
G
net force on the box is Fnet = − f k = − μk mgiˆ. The work–kinetic-energy theorem is
G
G
Wnet = Fnet ⋅ Δr = K 2 − K1 = 0 J − 2.0 J = −2.0 J
(− μ k mg )( x2 − x1) = −2.0 J
( x2 − x1) =
2.0 J
μk mg
=
2. 0 J
(0.15)(2.5 kg)(9.8 m/s 2 )
= 0.54 m
Assess: Because the force of friction transforms kinetic energy into thermal energy, energy is transferred out of the
box into the environment. In response, the box slows down and comes to rest.
11.48. Model: Model the suitcase as a particle, use the model of kinetic friction, and use the work–kinetic-energy
theorem.
Visualize:
G
G
The net force on the suitcase is Fnet = f k .
Solve: (a) The work–kinetic-energy theorem gives
Wnet = ΔK = 12 mv12 − 12 mv02
G
G G
G
⇒ Fnet ⋅ Δr = f k ⋅ Δr = 0 J − 12 mv02
v02
2 gd
(b) Inserting the given quantities into the expression for the coefficient of kinetic friction gives
v2
(1.2 m/s) 2
= 0.037
μk = 0 =
2 gd 2(9.8 m/s 2 )(2.0 m)
Assess: Friction transforms kinetic energy of the suitcase into thermal energy. In response, the suitcase slows down and
comes to rest. Notice that the coefficient of friction does not depend on the mass of the object, which is reasonable.
( f k )d cos(180°) = − 12 mv02 − μ k mgd = − 12 mv02
⇒ μk =
11.49. Model: Identify the truck and the loose gravel as the system. We need the gravel inside the system because
friction increases the temperature of the truck and the gravel. We will also use the model of kinetic friction and the
conservation of energy equation.
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11-20
Chapter 11
Visualize:
We place the origin of our coordinate system at the base of the ramp in such a way that the x-axis is along the ramp
and the y-axis is vertical so that we can calculate potential energy. The free-body diagram of forces on the truck is
shown.
Solve: The conservation of energy equation is K1 + U g1 + Δ Eth = K 0 + U g0 + Wext . In the present case, Wext = 0 J,
v1x = 0 m/s, U g 0 = 0 J, v0 x = 35 m/s. The thermal energy created by friction is
ΔEth = f k ( x1 − x0 ) = ( μ k n)( x1 − x0 ) = μk mg cos (6.0°)( x1 − x0 )
= (0.40)(15,000 kg)(9.8 m/s 2 )cos(6.0°)( x1 − x0 ) = (58,478 J/m)( x1 − x0 )
Thus, the energy conservation equation simplifies to
0 J + mgy1 + (58,478 J/m )( x1 − x0 ) = 12 mv02x + 0 J + 0 J
(15,000 kg)(9.8 m/s 2 )( x1 − x0 )sin (6.0°) + (58,478 J/m)( x1 − x0 ) = 12 (15,000 kg)(35 m/s) 2
( x1 − x0 ) = 124 m = 0.12 km
Assess: A length of 124 m at a slope of 6° seems reasonable.
11.50. Model: We will use the spring, the package, and the ramp as the system. We will model the package as a
particle.
Visualize:
We place the origin of our coordinate system on the end of the spring when it is compressed and is in contact with the
package to be shot.
Model: (a) The energy conservation equation is
K1 + U g1 + U s1 + Δ Eth = K 0 + U g0 + U s0 + Wext
1 mv 2
1
2
+ mgy1 + 12 k ( xe − xe ) 2 + Δ Eth = 12 mv02 + mgy0 + 12 k (Δx) 2 + Wext
Using y1 = 1.0 m, ΔEth = 0 J (the frictionless ramp), v0 = 0 m/s, y0 = 0 m, Δx = 30 cm, and Wext = 0 J, we get
1 (2.0
2
1 mv 2 + mg (1.0 m) + 0 J + 0 J = 0 J + 0 J + 1 k (0.30 m) 2
1
2
2
2
2
1
kg)v1 + (2.0 kg)(9.8 m/s )(1.0 m) = 2 (500 N/m)(0.30 m)2
+0 J
v1 = 1.7 m/s
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Work
11-21
(b) How high can the package go after crossing the sticky spot? If the package can reach y1 ≥ 1.0 m before stopping
(v1 = 0), then it makes it. But if y1 < 1.0 m when v1 = 0, the package does not make it. The friction of the sticky spot
generates thermal energy
ΔEth = ( μ k mg )Δx = (0.30)(2.0 kg)(9.8 m/s 2 )(0.50 m) = 2.94 J
The energy conservation equation is now
1 mv 2
1
2
+ mgy1 + ΔEth = 12 k (Δx) 2
If we set v1 = 0 m/s to find the highest point the package can reach, we get
y1 =
( 12 k Δx2 − ΔEth )
( mg ) = [ 12 (500 N/m)(0.30 m) 2 − 2.94 J]/[(2.0 kg)(9.8 m/s 2 )] = 0.998 m
The package does not make it. It just barely misses.
11.51. Model: Model the two blocks as particles. The two blocks make our system.
Visualize:
We place the origin of our coordinate system at the location of the 3.0 kg block.
Solve: (a) The conservation of energy equation gives K f + U gf + ΔEth = Ki + U gi + Wext . The thermal energy is
ΔEth = μ k Mg Δx and the external work done is Wext = 0. The initial potential energy is U gi = myi , and the final
potential energy is U gf = myf , where we have ignored the gravitational potential energy of block M because its
height does not change. The initial and final kinetic energy are Ki = 0, and K f = 12 ( M + m)vf2 , respectively. The
energy conservation equation thus takes the form
1 ( m + m )v 2 + m y
3
2 f
3 table + m2 gyf + μ k m3 g Δx = m3 ytable + m2 gyi
2
Note that Δx = −Δy = h because the blocks are constrained by the cable to move the same distance. Solving for vf
gives
2g
2 gh
(− mΔy − μk M Δx ) =
(m − μk M )
M +m
M +m
(b) If the table is frictionless, this expression takes the form
2gmh
vf =
M +m
Assess: It is reasonable that the speed is reduced when friction is present compared with when there is no friction.
vf =
G
G
11.52. Model: Use the particle model, the definition of work W = F ⋅ Δs , and the model of kinetic friction.
Visualize: We place the coordinate frame on the incline so that its x-axis is along the incline.
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11-22
Chapter 11
G G
Solve: (a) WT = T ⋅ Δr = T Δx cos(18°) = (120 N)(5.0 m)cos(18°) = 0.57 kJ
G
G
Wg = FG ⋅ Δr = mg Δx cos(120°) = (8.0 kg)(9.8 m/s2 )(5.0 m)cos(120°) = −0.20 kJ
G G
Wn = n ⋅ Δr = nΔx cos(90°) = 0.0 J
(b) The amount of energy transformed into thermal energy is ΔEth = f k Δx = μk nΔx.
To find n, we write Newton’s second law as follows:
∑ Fy = n − FG cos(30°) + T sin (18°) = 0 N ⇒
n = FG cos(30°) − T sin (18°)
n = mg cos(30°) − T sin (18°) = (8.0 kg)(9.8 m/s 2 )cos(30°) − (120 N ) sin (18°) = 30.814 N
Thus, ΔEth = (0.25)(30.814 N)(5.0 m) = 39 J.
Assess: Any force that acts perpendicular to the displacement does no work.
11.53. Model: Model the water skier as a particle, apply the law of conservation of mechanical energy, and use the
constant-acceleration kinematic equations.
Visualize:
We placed the origin of the coordinate system at the base of the frictionless ramp.
Solve: We’ll start by finding the smallest speed v1 at the top of the ramp that allows her to clear the shark tank. From
the vertical motion for jumping the shark tank,
y2 = y1 + v1 y Δt + 12 a y Δt 2
0 m = 2.0 m + 0 m + 12 (9.8 m/s 2 )Δt 2
⇒ Δt = 0.639 s
From the horizontal motion,
x2 = x1 + v1x Δt + 12 ax Δt 2
5.0 m
= 7.825 m/s
0.639 s
Having found the v1 that will take the skier to the other side of the tank, we now use the energy equation to find the
( x1 + 5.0 m) = x1 + v1Δt + 0 m ⇒ v1 =
minimum speed v0 . We have
K1 + U g1 = K 0 + U g0
⇒
1 mv 2
1
2
+ mgy1 = 12 mv02 + mgy0
v0 = v12 + 2 g ( y1 − y0 ) = (7.825 m/s) 2 + 2(9.8 m/s2 )(2.0 m) = 10 m/s
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Work
11-23
11.54. Model: Use the particle model for the ice skater, the friction model, and the work–kinetic-energy theorem.
Visualize:
Solve: (a) The work–kinetic-energy theorem gives
ΔK = 12 mv12 − 12 mv02 = Wnet = Wwind
There is no kinetic friction along her direction of motion. Static friction acts to prevent her skates from slipping
sideways on the ice, but this force is perpendicular to the motion and does not contribute to a change in thermal
G
G
energy. The angle between Fwind and Δr is θ = 135°, so
G
G
Wwind = Fwind ⋅ Δr = Fwind Δy cos(135°) = (4.0 N)(100 m)cos(135°) = −282.8 J
Thus, her final speed is
v1 = v02 +
2Wwind
= 2.2 m/s
m
(b) If the skates don’t slip, she has no acceleration in the x-direction and so ( Fnet ) x = 0 N. That is:
fs − Fwind cos (45°) = 0 N ⇒
fs = Fwind cos(45°) = 2.83 N
Now there is an upper limit to the static friction: fs ≤ ( fs )max = μs mg . To not slip requires
μs ≥
fs
2.83 N
=
= 0.0058
mg (50 kg)(9.8 m/s 2 )
Thus, the minimum value of μs is 0.0058.
Assess: The work done by the wind on the ice skater is negative, because the wind slows the skater down.
11.55. Model: Model the ice cube as a particle, the spring as an ideal that obeys Hooke’s law, and the law of
conservation of energy.
Visualize:
Solve: (a) The normal force does no work and the slope is frictionless, so mechanical energy is conserved. We’ve
drawn two separate axes: a vertical y-axis to measure potential energy and a tilted s-axis to measure distance along
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11-24
Chapter 11
the slope. Both have the same origin which is at the point where the spring is not compressed. Thus, the two axes are
related by y = s sin θ . Also, this choice of origin makes the elastic potential energy simply U s = 12 k ( s − s0 ) 2 = 12 ks 2 .
Because energy is conserved, we can relate the initial point—with the spring compressed—to the final point where
the ice cube is at maximum height. We do not need to find the speed with which it leaves the spring. We have
K 2 + U g2 + U s2 = K1 + U g1 + U s1
1 mv 2
2
2
+ mgy2 + 12 ks02 = 12 mv12 + mgy1 + 12 ks12
It is important to note that at the final point, when the ice cube is at y2 , the end of the spring is only at s0 . The spring
does not stretch to s2 , so U s2 is not
mgy2 = + mgy1 + 12 ks12
1 ks 2 .
2 2
⇒
Three of the terms are zero, leaving
y2 − y1 = Δy = height gained =
ks12
= 0.255 m = 25.5 cm
2mg
The distance traveled is Δs = Δy/ sin (30°) = 0.51 m.
(b) Using the energy equation and the expression for thermal energy:
K 2 + U g2 + U s2 + ΔEth = K1 + U g1 + U s1 + Wext , ΔEth = f k Δs = μk nΔs
From the free-body diagram,
G
( Fnet ) y = 0 N = n − mg cos(30°) ⇒ n = mg cos(30°)
Now, having found ΔEth = μ k mg cos(30°)Δs, the energy equation can be written
0 J + mgy2 + 0 J + μ k mg cos(30°)Δs = 0 J + mgy1 + 12 ks12 + 0 J
mg ( y2 − y1 ) − 12 ks12 + μk mg cos(30°)Δs = 0
Using Δy = Δs sin (30°), the above equation simplifies to
mg Δs sin (30°) + μk mg cos(30°)Δs = 12 ks12
⇒ Δs =
ks12
= 0.38 m
2mg[sin (30°) + μ k cos(30°)]
11.56. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the box as a particle and apply the energy
conservation law. Box, spring, and the ground make our system, and we also use the model of kinetic friction.
Visualize: We place the origin of the coordinate system on the ground directly below the box’s starting position.
Solve: (a) The energy conservation equation gives
K1 + U g1 + U s1 + ΔEth = K 0 + U g0 + U s0 + Wext
1 mv 2
1
2
+ mgy1 + 0 J + 0 J = 12 mv02 + mgy0 + 0 J + 0 J ⇒
1 mv 2
1
2
+ 0 J = 0 J + mgy0
v1 = 2 gy0 = 2(9.8 m/s 2 )(5.0 m) = 9.9 m/s
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Work
11-25
(b) The friction creates thermal energy. The energy conservation equation for this part of the problem is
K 2 + U g2 + U s2 + ΔEth = K1 + U g1 + U s1 + Wext , 12 mv22 + 0 J + 0 J + μ k mg ( x2 − x1) = 12 mv12 + 0 J + 0 J + 0 J
1 mv 2
2
2
+ μ k n( x2 − x1 ) = 12 mv12
⇒
1 mv 2
2
2
+ μ k mg ( x2 − x1 ) = 12 mv12
v2 = v12 − 2 μk g ( x2 − x1 ) = (9.9 m/s) 2 − 2(0.25)(9.8 m/s 2 )(2.0 m) = 9.4 m/s
(c) To find how much the spring is compressed, we apply the energy conservation once again:
1
1
K3 + U g3 + U s3 + ΔEth = K 2 + U g2 + U s2 + Wext , 0 J + 0 J + k ( x3 − x2 ) 2 + 0 J = mv22 + 0 J + 0 J + 0 J
2
2
Using v2 = 9.4 m/s, k = 500 N/m, and m = 5.0 kg, the above equation yields ( x3 − x2 ) = Δx = 0.94 m.
(d) The initial energy = mgy0 = (5.0 kg)(9.8 m/s 2 )(5.0 m) = 254 J. The energy transformed to thermal energy during
each passage is
μ k mg ( x2 − x1) = (0.25)(5.0 kg)(9.8 m/s 2 )(2.0 m) = 24.5 J
The number of passages is equal to 245 J/24.5 J or 10.
11.57. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the physics student as a particle and apply
the law of conservation of energy. Our system is comprised of the spring, the student, and the ground. We also use
the model of kinetic friction.
Visualize: We place the origin of the coordinate system on the ground directly below the end of the compressed
spring that is in contact with the student.
Solve: (a) The energy conservation equation gives
K1 + U g1 + U s1 + ΔEth = K 0 + U g0 + U s0 + Wext
1 mv 2
1
2
+ mgy1 + 12 k ( x1 − xe )2 + 0 J = 12 mv02 + mgy0 + 12 k ( x1 − x0 ) 2 + 0 J
Since y1 = y0 = 10 m, x1 = xe , v0 = 0 m/s, k = 80,000 N/m, m = 100 kg, and ( x1 − x0 ) = 0.5 m,
1 mv 2
1
2
= 12 k ( x1 − x0 ) 2
⇒ v1 =
k
80,000 N/m
( x1 − x0 ) =
(0.50 m) = 14 m/s
100 kg
m
(b) Friction creates thermal energy. Applying the conservation of energy equation once again:
K 2 + U g2 + U s2 + ΔEth = K 0 + U g0 + U s0 + Wext
1 mv 2
2
2
+ mgy2 + 0 J + f k Δs = 0 J + mgy0 + 12 k ( x1 − x0 ) 2 + 0 J
With v2 = 0 m/s and y2 = Δs sin (30°), the above equation is simplified to
mg Δs sin (30°) + μk nΔs = mgy0 + 12 k ( x1 − x0 ) 2
From the free-body diagram for the physics student, we see that n = FG cos(30°) = mg cos(30°). Thus, the conservation
of energy equation gives
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11-26
Chapter 11
Δs[mg sin (30°) + μk mg cos(30°)] = mgy0 + 12 k ( x1 − x0 ) 2
Using m = 100 kg, k = 80,000 N/m, ( x1 − x0 ) = 0.50 m, y0 = 10 m, and μ k = 0.15, we get
Δs =
mgy0 + 12 k ( x1 − x0 )2
mg[sin (30°) + μ k cos(30°)]
= 32 m
Assess: y2 = Δs sin (30°) = 16 m, which is greater than y0 = 10 m. The higher value is due to the transformation of
the spring energy into gravitational potential energy.
11.58. Model: Treat the block as a particle, use the model of kinetic friction, and apply the energy conservation
law. The block and the incline comprise our system.
Visualize: We place the origin of the coordinate system directly below the block’s starting position at the same
level as the horizontal surface. On the horizontal surface the model of kinetic friction applies.
Solve: (a) For the first incline, the conservation of energy equation gives
K1 + U g1 + ΔEth = K 0 + U g0 + Wext , 12 mv12 + 0 J + 0 J = 0 J + mgy0 + 0 J ⇒ v1 = 2 gy0 = 2 gh
(b) The friction creates thermal energy. Applying once again the conservation of energy equation, we have
K3 + U g3 + ΔEth = K1 + U g1 + Wext , 12 mv32 + mgy3 + μk mg ( x2 − x1 ) = 12 mv12 + mgy1 + Wext
Using v3 = 0 m/s, y1 = 0 m, Wext = 0 J, v1 = 2 gh , and ( x2 − x1) = L, we get
mgy3 + μ k mgL = 12 m(2 gh) ⇒
y3 = h − μ k L
Assess: For μ k = 0, y3 = h which is predicted by the law of the conservation of energy.
11.59. Model: Assume an ideal spring, so Hooke’s law is obeyed.
Visualize:
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Work
11-27
Solve: For a conservative force the work done on a particle as it moves from an initial to a final position is
independent of the path. We will show that WA →C→B = WA →B for the spring force. Work done by a spring force
F = −kx is given by
xf
W = ∫ Fdx = − Ñ kx dx
xi
This means
x
x
x
C
B
B
k
k
k
WA → B = − Ñ kx dx = − ( xB2 − xA2 ), WA →C = − Ñ kx dx = − ( xC2 − xA2 ), and WC→ B = − Ñ kx dx = − ( xB2 − xC2 )
2
2
2
x
x
x
A
A
C
Adding the last two:
k
WA →C→ B = WA →C + WC→ B = − ( xC2 − xA2 + xB2 − xC2 ) = WA → B
2
Assess: Because the paths are arbitrary, we have shown that the work done on a particle is independent of the path,
so the spring force is conservative.
11.60. Model: A “sprong” obeys the force law Fx = −q ( x − xe )3 , where q is the sprong constant and xe is the
equilibrium position.
Visualize: We place the origin of the coordinate system on the free end of the sprong, that is, xe = xf = 0 m.
Solve: (a) The units of q are N/m3 .
x
(b) Since Fx = − dU/dx, we have U ( x ) = − ∫ Fx dx = − Ñ ( −qx3 )dx = qx 4 /4.
0
(c) Applying the energy conservation equation to the ball and sprong system gives
K f + U f = Ki + U i
1 mv 2
f
2
+ 0 J = 0 J + 14 qxi4
vf =
qx 4
(40,000 N/m3 )( −0.10 m) 4
=
= 10 m/s
2m
2(0.020 kg)
11.61. Solve: (a) Because sin (cx) is dimensionless, F0 must have units of force in newtons.
(b) The product cx is an angle because we are taking the sine of it. An angle is dimensionless. If x has units of m and
the product cx is dimensionless, then c has to have units of m −1.
(c) The force is a maximum when sin (cx) = 1. This occurs when cx = π /2, or for xmax = π /(2c).
(d) The graph is the first quarter of a sine curve.
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11-28
Chapter 11
(e) We can find the velocity vf at xf = xmax from the work–kinetic-energy theorem:
ΔK = 12 mvf2 − 12 mvi2 = 12 mvf2 − 12 mv02 = W
This is a variable force. As the particle moves from xi = 0 m to xf = xmax
xf
xf
xi
xi
W = Ñ F ( x)dx = F0 Ñ sin(cx) dx = −
2W
m
= π /(2c) the work done on it is
⇒ vf = v02 +
π /(2c )
F0
F
F
cos(cx)
= − 0 [cos(π /2) − cos(0) ] = 0
c
c
c
0
Thus, the particle’s speed at xf = xmax = π /(2c) is vf = v02 + 2 F0 /( mc) .
11.62. Solve: The average power output during the push-off period is equal to the work done by the cat divided by the
time the cat applied the force. Since the force on the floor by the cat is equal in magnitude to the force on the cat by the
floor, work done by the cat can be found using the work–kinetic-energy theorem during the push-off period;
Wnet = Wfloor = ΔK . We do not need to explicitly calculate Wcat , since we know that the cat’s kinetic energy is transformed
into its potential energy during the leap. That is,
ΔU g = mg ( y2 − y1) = (5.0 kg)(9.8 m/s 2 )(0.95 m) = 46.55 J
Thus, the average power output during the push-off period is
W
46.55 J
= 0.23 kW
P = net =
t
0.20 s
11.63. Model: The heart provides the pressure to move blood through the body and therefore does work on the
blood. We assume all the work goes into pushing the blood through the body.
Solve: (a) Using the hint, W = PAd = PV = (1.3 × 104 N/m 2 )(6.0 × 10−3 m3 ) = 78 J (in this equation, P represents
pressure, not power).
(b) Using P to represent power now, we can calculate the average power output of the heart as follows:
W 78 J
P=
=
= 1.3 W
Δt 60 s
11.64. Model: We will ignore rolling friction because it is much less than the drag force (and becaue we are not
given the mass of the bicyclist + bicycle). Therefore, model the system as a particle with the given cross-sectional
area and that is moving through the air at the given speed.
Solve: (a) From Eq. 6.16, we know that the drag force D has the magnitude
D = 12 C ρ Av 2
where C = 0.90, A = 0.45 m 2 , and ρ = 1.2 kg/m3 (the density of air). To overcome this force, the cyclist must
G
G
G G
generate the force F = − D, or a power P = F ⋅ v = Dv, where the last equality follows because the drag force acts in
parallel to the velocity. Thus, the power is
P = 12 C ρ Av3 = 12 (0.90)(1.2 kg/m3 )(0.45 m 2 )(7.3 m/s)3 = 95 W
(b) The metabolic power output PM is PM = P/0.25 = 3.8 × 102 W.
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11-29
(c) The number of food calories c burned riding for one hour is
⎛ 1 cal ⎞⎛ 3600 s ⎞
2
c = (378 W) ⎜
⎟⎜
⎟ (1 h ) = 3.2 × 10 cal
4190
J
1
h
⎝
⎠⎝
⎠
11.65. Solve: (a) The change in the potential energy of 1.0 kg of water in falling 25 m is
ΔU g = − mgh = −(1.0 kg)(9.8 m/s2 )(25 m) = −245 J ≈ −0.25 kJ
(b) The power required of the dam is
P=
W W
=
= 50 × 106 Watts ⇒ W = 50 × 106 J
t 1s
That is, 50 × 106 J of energy is required per second for the dam. Out of the 245 J of lost potential energy,
(245 J)(0.80) = 196 J is converted to electrical energy. Thus, the amount of water needed per second is
(50 × 106 J)(1.0 kg/196 J) = 255,000 kg ≈ 2.6 × 105 kg.
11.66. Solve: The force required to tow a water skier at a speed v is Ftow = Av. Since power P = Fv, the power
required to tow the water skier is Ptow = Ftow v = Av 2 . We can find the constant A by noting that a speed of v = 2.5 mph
requires a power of 2 hp. Thus,
2 hp = A(2.5 mph) 2
⇒
A = 0.32
hp
mph 2
Now, the power required to tow a water skier at 7.5 mph is
hp
Ptow = Av 2 = 0.32
(7.5 mph) 2 = 18 hp
2
mph
Assess: Since P ∝ v 2 , a three-fold increase in velocity leads to a nine-fold increase in power.
11.67. Model: Use the model of static friction, kinematic equations, and the definition of power.
Solve: (a) The rated power of the Porsche is 217 hp = 161,882 W and the gravitational force on the car is
(1480 kg)(9.8 m/s 2 ) = 14,504 N. The amount of that force on the drive wheels is (14,504)(2/3) = 9670 N. Because
the static friction of the tires on road pushes the car forward,
Fmax = fs,max = μs n = μs mg = (1.00)(9670 N) = mamax
9670 N
= 6.53 m/s 2
1480 kg
(b) Only 70% of the power generated by the motor is applied at the wheels.
P (0.70)(161,882 W)
P = Fvmax ⇒ vmax = =
= 11.7 m/s
F
9670 N
(c) Using the kinematic equation, vmax = v0 + amax (tmin − t0 ) with v0 = 0 m/s and t0 = 0 s, we obtain
amax =
vmax 11.7 m/s
=
= 1.79 s
amax 6.53 m/s 2
Assess: An acceleration time of 1.79 s for the Porsche to reach a speed of ≈ 26 mph from rest is reasonable.
tmin =
11.68. Solve: (a) A student uses a string to pull her 2.0 kg physics book, starting from rest, across a 2.0-m-long lab
bench. The coefficient of kinetic friction between the book and the lab bench is 0.15. If the book’s final speed is 4.0 m/s,
what is the tension in the string?
(b)
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11-30
Chapter 11
(c) The tension does external work Wext . This work increases the book’s kinetic energy and also causes an increase
ΔEth in the thermal energy of the book and the lab bench. Solving the equation gives T = 11 N.
11.69. (a) A 20 kg chicken crate slides down a 2.5-m-high, 40° ramp from the back of a truck to the ground. The
coefficient of kinetic friction between the crate and the ramp bench is 0.15. How fast are the chickens going at the
bottom of the ramp?
(b)
(c) v1 = 6.3 m/s.
11.70. (a) If you expend 75 W of power to push a 30 kg sled on a surface where the coefficient of kinetic friction
between the sled and the surface is μ k = 0.20, what speed will you be able to maintain?
(b)
(c) Fpush = (0.20)(30 kg)(9.8 m/s 2 ) = 58.8 N ⇒ 75 W = (58.8 N)v ⇒ v =
75 W
= 1.3 m/s
58.8 N
11.71. (a) A 1500 kg object is being accelerated upward at 1.0 m/s 2 by a rope. How much power must the motor
supply at the instant when the velocity is 2.0 m/s?
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11-31
(b)
(c) T = (1500 kg)(9.8 m/s 2 ) + 1500 kg(1.0 m/s 2 ) = 16,200 N = 16.2 kN
P = T (2 m/s) = (16,200 N)(2.0 m/s) = 32,400 W = 32 kW
11.72. Model: Assume the spring is ideal so that Hooke’s law is obeyed, and model the weather rocket as a particle.
Visualize:
The origin of the coordinate system is placed on the free end of the spring. Note that the bottom of the spring is
anchored to the ground.
Solve: (a) The rocket is initially at rest. The free-body diagram on the rocket helps us write Newton’s second law as
∑ Fy = 0 N ⇒ Fsp = FG = mg ⇒ k Δy = mg
(
)
mg (10.2 kg)(9.81 m/s 2 )
=
= 20.0 cm
k
(500 N/m)
(b) The thrust does work. Using the energy conservation equation when y2 − ye = 40 cm = 0.40 m:
Δy =
K 2 + U g2 + U sp2 = K1 + U g1 + U sp1 + Wext
Wext =
1 mv 2
2
2
+ mgy2 + 12 k ( y2 − ye ) 2 = 12 mv12 + mgy1 + 12 k ( y1 − ye ) 2 + ( 200 N )( 0.60 m )
(5.10 kg)v22 + 40.0 J + 40.0 J = 0 − 20.0 J + 10.0 J + 120 J ⇒ v2 = 2.43 m/s
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11-32
Chapter 11
If the rocket were not attached to the spring, the energy conservation equation would not involve the spring energy
term U sp2 . That is,
K 2 + U g2 = K1 + U g1 + U sp1 + Wext
1 (10.2
2
kg)v22 + (10.2 kg)(9.81 m/s 2 )(0.40 m) = 0 J − (10.2 kg)(9.81 m/s 2 )(0.20 m)
+ 12 (500 N/m)(0.20 m)2 + ( 200 N )( 0.60 m )
(5.10 kg)v22 = 70.0 J ⇒ v2 = 3.70 m/s
Assess: (a) The rocket has greater speed at y2 when it is not attached to the spring because, as the spring extends, it
contributes a downward force to the rocket.
11.73. Model: Assume the spring to be ideal that obeys Hooke’s law, and model the block as a particle.
Visualize: We place the origin of the coordinate system on the free end of the compressed spring which is in contact
with the block. Because the horizontal surface at the bottom of the ramp is frictionless, the spring energy appears as
kinetic energy of the block until the block begins to climb up the incline.
Solve: Although we could find the speed v1 of the block as it leaves the spring, we don’t need to. We can use energy
conservation to relate the initial potential energy of the spring to the energy of the block as it begins projectile motion
at point 2. However, friction requires us to calculate the increase in thermal energy. The energy equation is
K 2 + U g2 + ΔEth = K0 + U g0 + Wext ⇒ 12 mv22 + mgy2 + f k Δs = 12 k ( x0 − xe ) 2
The distance along the slope is Δs = y2 / sin(45°). The friction force is f k = μ k n, and we can see from the free-body
diagram that n = mg cos(45°). Thus
v2 =
k
( x0 − xe ) 2 − 2 gy2 − 2μk gy2 cot(45°)
m
1/ 2
⎡1000 N/m
⎤
=⎢
(0.15 m)2 − 2(9.8 m/s 2 )(2.0 m) − 2(0.20)(9.8 m/s 2 )(2.0 m)cot(45°) ⎥ = 8.091 m/s
⎣ 0.20 kg
⎦
Having found the velocity v2 , we can now find ( x3 − x2 ) = d using the kinematic equations of projectile motion:
1
y3 = y2 + v2 y (t3 − t2 ) + a2 y (t3 − t2 ) 2
2
2.0 m = 2.0 m + v2 sin (45°)(t3 − t2 ) + 12 ( −9.8 m/s 2 )(t3 − t2 ) 2
t3 − t2 = 0 s and 1.168 s
Finally,
1
x3 = x2 + v2 x (t3 − t2 ) + a2 x (t3 − t2 ) 2
2
d = ( x3 − x2 ) = v2 cos(45°)(1.168 s) + 0 m = 6.7 m
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Work
11-33
11.74. Solve: (a)
The graph is a hyperbola.
(b) The separation for zero potential energy is r = ∞, since
Gm m
U = − 1 2 → 0 J as r → ∞
r
This makes sense because two masses don’t interact at all if they are infinitely far apart.
(c) Due to the absence of nonconservative forces in our system of two particles, the mechanical energy is conserved.
The equations of energy and momentum conservation are
K f + U gf = Ki + U gi
pf = pi
⎛ Gm m ⎞
⎛ Gm m ⎞
2
2
+ 12 m2v2f
+ ⎜ − 1 2 ⎟ = 12 m1v1i2 + 12 m2v2i
+⎜− 1 2 ⎟
rf ⎠
ri ⎠
⎝
⎝
1 m v 2 + 1 m v 2 = Gm m ⎛ 1 − 1 ⎞
⎟
1 2⎜
2 1 1f
2 2 2f
⎝ rf ri ⎠
1 m v2
2 1 1f
⇒ m1v1f + m2v2f = 0 kg m/s ⇒ v2f = −
m1
v1f
m2
Substituting this expression for v2f into the energy equation, we get
2
1 m v2
2 1 1f
With
⎛m
⎞
⎛ 1 1⎞
2Gm2 ⎛ 1 1 ⎞
+ 12 m2 ⎜ 1 v1f ⎟ = Gm1m2 ⎜ − ⎟ ⇒ v1f2 =
⎜ − ⎟
m
r
r
(1
m1/m2 ) ⎝ rf ri ⎠
+
i⎠
⎝ 2 ⎠
⎝ f
G = 6.67 × 10−11 N ⋅ m(kg) −2 , rf = R1 + R2 = 18 × 108 m, ri = 1.0 × 1014 m, m1 = 8.0 × 1030 kg, and m2 = 2.0 × 1030 kg,
the above equation can be simplified to yield
v1f = 1.72 × 105 m/s, and v2f = −
⎛ 8.0 × 1030 kg ⎞
m1
v1f = ⎜
(1.72 × 105 m/s) = 6.89 × 105 m/s
⎜ 2.0 × 1030 kg ⎟⎟
m2
⎝
⎠
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11-34
Chapter 11
The speed of the heavier star is 1.7 × 105 m/s. That of the lighter star is 6.9 × 105 m/s.
11.75. Model: Model the lawnmower as a particle and use the model of kinetic friction.
Visualize:
We placed the origin of our coordinate system on the lawnmower and drew the free-body diagram of forces.
G
G
G
Solve: The normal force n, which is related to the frictional force, is not equal to FG . This is due to the presence of F.
G
G
The rolling friction is f r = μ r n, or n = f r /μ r . The lawnmower moves at constant velocity, so Fnet = 0. The two
components of Newton’s second law are
(∑ Fy ) = n − FG − F sin (37°) = ma y = 0 N ⇒ f r /μ r − mg − F sin (37°) = 0 N ⇒
( ∑ Fx ) = F cos(37°) − f r = 0 N ⇒
F=
f r = μ r mg + μr F sin 37°
F cos(37°) − μr mg − μ r F sin (37°) = 0 N
μr mg
(0.15)(12 kg)(9.8 m/s 2 )
=
= 24.9 N
cos(37°) − μ r sin (37°) 0.7986 − ( 0.15)( 0.6018)
G G
Thus, the power supplied by the gardener in pushing the lawnmower at a constant speed of 1.2 m/s is P = F ⋅ v =
Fv cosθ = (24.9 N)(1.2 m/s)cos (37°) = 24 W.
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12
ROTATION OF A RIGID BODY
Conceptual Questions
12.1. As suggested by the figure, we will assume that the larger sphere is more massive. Then the center of gravity
would be at point a because if we suspend the dumbbell from point a then the counterclockwise torque due to the
large sphere (large weight times small lever arm) will be equal to the clockwise torque due to the small sphere (small
weight times large lever arm).
Look at the figure and mentally balance the dumbbell on your finger; your finger would have to be at point a. The
sun-earth system is similar to this except that the sun’s mass is so much greater than the earth’s that the center of
mass (called the barycenter for astronomical objects orbiting each other) is only 450 km from the center of the sun.
12.2. To double the rotational energy without changing ω requires doubling the moment of inertia. The moment of
inertia is proportional to R 2 so R must increase by 2.
1
2
1
2
12.3. The rotational kinetic energy is K rot = I ω 2 . For a disk, I = MR 2 . Since the mass is the same for all three
disks, the quantity R 2ω 2 determines the ranking. Thus K a = K b > K c .
12.4. No. The moment of inertia does not have any dependence on a quantity that indicates an object is rotating,
such as ω or α , so an object does not have to be rotating to have a moment of inertia.
12.5. Mass that is farther away from the axis of rotation contributes more to the moment of inertia I = ∫ r 2dm.
Here, r is the distance from the axis of rotation to the mass element dm. Note r 2 is always positive. For a rod, there is
more mass farther away from an axis through the rod’s end than one through its middle.
4
3
12.6. Because sphere 2 has twice the radius, its mass is greater by a factor of 23 = 8, since m = π r 3 ρsteel . The
added mass is also distributed farther from the center, so, I ∝ mr 2 leads to I 2 ∝ (8m1 )(2r1 ) 2 = 32 I1.
12.7. It will be easier to rotate the solid sphere because the hollow sphere’s mass is generally distributed farther
from its center. If you roll both simultaneously down an incline, the solid sphere will win.
12.8. τ e > τ a = τ b > τ c = τ d > τ f The torque τ = rF sin θ . We must calculate each torque:
⎛L⎞
⎝ ⎠
L
2
F sin 45° =
LF
2
4
τ e = L(2 F )
τa = ⎜ ⎟ F
2
τd =
⎛L⎞
⎝ ⎠
τb = ⎜ ⎟F
4
⎛L⎞
2
LF
τ c = ⎜ ⎟ F sin 45° =
4
⎝2⎠
τ f = LF sin 0° = 0
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12-1
12-2
Chapter 12
12.9. (a) Negative, since it causes the ball to rotate clockwise.
(b) The angular velocity holds steady (iii) after the push has ended, since the torque has ended, and there is no more
angular acceleration.
(c) The torque is zero after the push has ended because there is no longer an applied force.
τ
F
12.10. Since τ = I α, α = . Also, τ = Fr and I = mr 2 ⇒ α = . Calculate α for each case:
I
mr
αa =
F0
m0 r0
αc =
F0
1
= αa
m0 (2r0 ) 2
αb =
2 F0
= αa
2m0 r0
αd =
2 F0
1
= αa
(2m0 )(2r0 ) 2
So α a = α b > α c = α d .
12.11. The block attached to the solid cylinder hits first. The solid cylinder has a smaller moment of inertia since
more of its mass is closer to the rotation axis, so it has less resistance to a change in its rotational motion. The torque
applied by the string attached to the block makes the solid cylinder change its rotation and unwind the string faster.
12.12. The moment of inertia for the tuck position is smaller than that of the pike position. Since the angular
momentum of the diver is conserved, any initial angular velocity is increased more when the diver moves to the tuck
position relative to the pike position.
12.13. The angular momentum L of disk b is larger than the angular momentum of disk a. Calculate L for each:
1
La = I aωa = mra2ωa
2
1
Lb = I bωb = m 2ra2
2
( ) ⎛⎜⎝ 12 ω ⎞⎟⎠ = 2 ⎛⎜⎝ 12 mr ω ⎞⎟⎠ = 2L
2
a
2
a a
a
Exercises and Problems
Section 12.1 Rotational Motion
12.1. Model: A spinning skater, whose arms are outstretched, is a rigid rotating body.
Visualize:
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Rotation of a Rigid Body
12-3
Solve: The speed v = rω , where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2π /60 rad/s = 6π rad/s. Thus,
v = (0.70 m)(6π rad/s) = 13.2 m/s.
Assess: A speed of 13.2 m/s ≈ 26 mph for the hands is a little high, but reasonable.
12.2. Model: Assume constant angular acceleration.
⎛ 2π rad ⎞⎛ min ⎞
Solve: (a) The final angular velocity is ωf = (2000 rpm) ⎜
⎟⎜
⎟ = 209.4 rad/s. The definition of angular
⎝ rev ⎠⎝ 60 s ⎠
acceleration gives us
Δω ωf − ωi 209.4 rad/s − 0 rad/s
α=
=
=
= 419 rad/s 2
Δt
Δt
0.50 s
The angular acceleration of the drill is 4.2 × 102 rad/s 2 .
1
1
(b) θ f = θi + ωi Δt + α ( Δt ) 2 = 0 rad + 0 rad + (419 rad/s 2 )(0.50 s) 2 = 52.4 rad
2
2
⎛ rev ⎞
The drill makes (52.4 rad) ⎜
⎟ = 8.3 revolutions.
⎝ 2π rad ⎠
12.3. Model: Assume constant angular acceleration.
Visualize:
⎛ 2π rad ⎞⎛ min ⎞
Solve: The initial angular velocity is ωi = (60 rpm) ⎜
⎟⎜
⎟ = 2π rad/s.
⎝ rev ⎠⎝ 60 s ⎠
The angular acceleration is
ω − ωi 0 rad/s − 2π rad/s
α= f
=
= 20.251 rad/s 2
Δt
25 s
The angular velocity of the fan blade after 10 s is
ωf = ωi + α (t − t0 ) = 2π rad/s + (−0.251 rad/s 2 )(10 s − 0 s) = 3.77 rad/s
The tangential speed of the tip of the fan blade is
vt = rω = (0.40 m)(3.77 rad/s) = 1.5 m/s
1
1
(b) θ f = θi + ωi Δt + α (Δt ) 2 = 0 rad + (2π rad/s)(25 s) + ( −0.251 rad/s 2 )(25 s) 2 = 78.64 rad
2
2
The fan turns 78.64 rad = 12.5 rev ≈ 13 rev while coming to a stop.
12.4. Model: Assume constant angular acceleration.
Visualize:
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12-4
Chapter 12
Solve: (a) Since at = rα , find α first. With 90 rpm = 9.43 rad/s and 60 rpm = 6.28 rad/s,
Δω 9.43 rad − 6.28 rad/s
=
= 0.314 rad/s 2
Δt
10 s
The angular acceleration of the sprocket and pedal are the same. So
at = rα = (0.18 m)(0.314 rad/s 2 ) = 0.057 m/s 2
(b) The length of chain that passes over the sprocket during this time is L = r Δθ . Find Δθ :
α=
1
2
θ f = θi + ωi Δt + α (Δt ) 2
1
2
The length of chain which has passed over the top of the sprocket is
θ f − θi = Δθ = (6.28 rad/s)(10 s) + (0.314 rad/s 2 )(10 s) 2 = 78.5 rad
L = (0.10 m)(78.5 rad) = 7.9 m
Section 12.2 Rotation About the Center of Mass
12.5. Model: The earth and moon are particles.
Visualize:
Choosing xE = 0 m sets the coordinate origin at the center of the earth so that the center of mass location is the
distance from the center of the earth.
Solve:
m x + mM xM (5.98 × 1024 kg)(0 m) + (7.36 × 1022 kg)(3.84 × 108 m)
xcm = E E
=
mE + mM
5.98 × 1024 kg + 7.36 × 1022 kg
= 4.67 × 106 m ≈ 4.7 × 106 m
Assess: The center of mass of the earth-moon system is called the barycenter and is located beneath the surface of
the earth. Even though xE = 0 m the earth influences the center of mass location because mE is in the denominator of
the expression for xcm .
12.6. Visualize: The coordinates of the three masses mA , mB , and mC are (0 cm, 0 cm), (0 cm, 10 cm), and (10 cm, 0 cm),
respectively.
Solve: The coordinates of the center of mass are
m x + mB xB + mC xC (100 g)(0 cm) + (200 g)(0 cm) + (300 g)(10 cm)
xcm = A A
=
= 5.0 cm
mA + mB + mC
(100 g + 200 g + 300 g)
ycm =
mA yA + mB yB + mC yC (100 g)(0 cm) + (200 g)(10 cm) + (300 g)(0 cm)
=
= 3.3 cm
mA + mB + mC
(100 g + 200 g + 300 g)
12.7. The coordinates of the three masses mA , mB , and mC are (0 cm, 10 cm), (10 cm, 10 cm), and (10 cm, 0 cm),
respectively.
Solve: The coordinates of the center of mass are
m x + mB xB + mC xC (200 g)(0 cm) + (300 g)(10 cm) + (100 g)(10 cm)
xcm = A A
=
= 6.7 cm
mA + mB + mC
(200 g + 300 g + 100 g)
ycm =
mA yA + mB yB + mC yC (200 g)(0 cm) + (300 g)(10 cm) + (100 g)(0 cm)
=
= 5.0 cm
mA + mB + mC
(200 g + 300 g + 100 g)
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Rotation of a Rigid Body
12-5
12.8. Model: The balls are particles located at the ball’s respective centers.
Visualize:
Solve: The center of mass of the two balls measured from the left hand ball is
(100 g)(0 cm) + (200 g)(30 cm)
xcm =
= 20 cm
100 g + 200 g
The linear speed of the 100 g ball is
⎛ 2π rad ⎞⎛ min ⎞
v1 = rω = xcmω = (0.20 m)(120 rev/min) ⎜
⎟⎜
⎟ = 2.5 m/s
⎝ rev ⎠⎝ 60 s ⎠
Section 12.3 Rotational Energy
12.9. Model: The earth is a rigid, spherical rotating body.
Solve: The rotational kinetic energy of the earth is K rot = 12 I ω 2 . The moment of inertia of a sphere about its diameter
(see Table 12.2) is I = 25 M earth R 2 and the angular velocity of the earth is
ω=
2π rad
= 7.27 × 10−5 rad/s
24 × 3600 s
Thus, the rotational kinetic energy is
1⎛ 2
⎞
K rot = ⎜ M earth R 2 ⎟ ω 2
2⎝ 5
⎠
1
= (5.98 × 1024 kg)(6.37 × 106 m) 2 (7.27 × 1025 rad/s)2 = 2.57 × 1029 J
5
12.10. Model: The disk is a rigid body rotating about an axis through its center.
Visualize:
Solve: The speed of the point on the rim is given by vrim = Rω . The angular velocity ω of the disk can be
determined from its rotational kinetic energy which is K = 12 I ω 2 = 0.15 J. The moment of inertia I of the disk about
its center and perpendicular to the plane of the disk is given by
1
1
I = MR 2 = (0.10 kg)(0.040 m) 2 = 8.0 × 1025 kg m 2
2
2
2(0.15 J)
0.30 J
2
⇒ω =
=
⇒ ω = 61.237 rad/s
I
8.0 × 10−5 kg m 2
Now, we can go back to the first equation to find vrim . We get vrim = Rω = (0.040 m)(61.237 rad/s) = 2.4 m/s.
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12-6
Chapter 12
12.11. Model: The triangle is a rigid body rotating about an axis through the center.
Visualize: Please refer to Figure EX12.11. Each 200 g mass is a distance r away from the axis of rotation, where r is
given by
0.20 m
0.20 m
= cos30° ⇒ r =
= 0.2309 m
r
cos30°
Solve: (a) The moment of inertia of the triangle is I = 3 × mr 2 = 3(0.200 kg)(0.2309 m) 2 = 0.032 kg m 2.
(b) The frequency of rotation is given as 5.0 revolutions per s or 10π rad/s. The rotational kinetic energy is
1
1
K rot . = I ω 2 = (0.0320 kg m 2 )(10.0π rad/s) 2 = 15.8 J = 16 J
2
2
12.12. Model: The baton is a thin rod rotating about a perpendicular axis through its center of mass.
Solve: The moment of inertia of a thin rod rotating about its center is I =
1
ML2 . For the baton,
12
1
(0.400 kg)(0.96 m) 2 = 0.031 kg m 2
12
The rotational kinetic energy of the baton is
I=
2
K rot =
1 2 1
⎛
⎛ 2π rad ⎞⎛ min ⎞ ⎞
I ω = (0.031 kg m 2 ) ⎜ (100 rev/min) ⎜
⎟⎜
⎟ ⎟ = 1.68 J ≈ 1.7 J
2
2
⎝ rev ⎠⎝ 60 s ⎠ ⎠
⎝
Section 12.4 Calculating Moment of Inertia
12.13. Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the rotation
axis passes through mass A and is perpendicular to the page.
∑ mi xi mA xA + mB xB + mC xC + mD xD
=
Solve: (a) xcm =
∑ mi
mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)
= 0.057 m
100 g + 200 g + 200 g + 200 g
m y + mB yB + mC yC + mD yD
= A A
mA + mB + mC + mD
=
ycm
(100 g)(0 m) + (200 g)(0.08 m) + (200 g)(0.08 cm) + (200 g )(0 m)
= 0.046 m
700 g
(b) The distance from the axis to mass C is 12.81 cm. The moment of inertia through A and perpendicular to
the page is
=
I A = ∑ mi ri2 = mA rA2 + mBrB2 + mCrC2 + mD rD2
i
= (0.100 kg)(0 m)2 + (0.200 kg)(0.08 m)2 + (0.200 kg)(0.1281 m)2 + (0.200 kg)(0.10 m)2 = 0.0066 kg m 2
12.14. Model: The moment of inertia of any object depends on the axis of rotation.
Visualize:
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Rotation of a Rigid Body
Solve: (a) xcm =
12-7
∑ mi xi mA xA + mB xB + mC xC + mD xD
=
∑ mi
mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)
= 0.057 m
100 g + 200 g + 200 g + 200 g
m y + mB yB + mC yC + mD yD
= A A
mA + mB + mC + mD
=
ycm
(100 g)(0 m) + (200 g)(0.08 m) + (200 g)(0.08 cm) + (200 g )(0 m)
= 0.046 m
700 g
(b) The moment of inertia about a diagonal that passes through B and D is
I BD = mA rA2 + mC rC2
=
where we must compute rA = rC which are the distances from the diagonal. From triangle ABD we see that
⎛8⎞
⎟ = 38.66°. Now rA = (0.10 m)sin 38.66° = 0.06247 m Thus,
⎝ 10 ⎠
θ = tan −1 ⎜
I BD = (0.100 kg)(0.06247) 2 + (0.200 kg)(0.06247) 2 = 0.0012 kg m 2
Assess: Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.
12.15. Model: The three masses connected by massless rigid rods are a rigid body.
Solve: (a) xcm =
ycm
∑ mi xi (0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m)
=
= 0.060 m
∑ mi
0.100 kg + 0.200 kg + 0.100 kg
∑ mi yi
=
=
∑ mi
(0.100 kg)(0 m) + (0.200 kg)
(
)
(0.10 m) 2 − (0.06 m) 2 + (0.100 kg)(0 m)
0.100 kg + 0.200 kg + 0.100 kg
= 0.040 m
(b) The moment of inertia about an axis through A and perpendicular to the page is
I A = ∑ mi ri2 = mB (0.10 m)2 + mC (0.10 m) 2 = (0.100 kg)[(0.10 m)2 + (0.10 m) 2 ] = 0.0020 kg m 2
(c) The moment of inertia about an axis that passes through B and C is
I BC = mA
(
(0.10 m) 2 − (0.06 m) 2
) = 0.00128 kg m
2
2
≈ 0.0013 kg m 2
Assess: Note that mass mA does not contribute to I A , and the masses mB and mC do not contribute to I BC .
12.16. Model: The door is a slab of uniform density.
Solve: (a) The hinges are at the edge of the door, so from Table 12.2,
1
I = (25 kg)(0.91 m) 2 = 6.9 kg m 2
3
(b) The distance from the axis through the center of mass along the height of the door is
⎛ 0.91 m
⎞
− 0.15 m ⎟ = 0.305 m. Using the parallel–axis theorem,
d =⎜
2
⎝
⎠
1
I = I cm + Md 2 = (25 kg)(0.91 m)2 + (25 kg)(0.305 cm) 2 = 4.1 kg m 2
12
Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass.
12.17. Model: The CD is a disk of uniform density.
Solve: (a) The center of the CD is its center of mass. Using Table 12.2,
1
1
I cm = MR 2 = (0.021 kg)(0.060 m) 2 = 3.8 × 10−5 kg m 2
2
2
(b) Using the parallel–axis theorem with d = 0.060 m,
I = I cm + Md 2 = 3.8 × 10−5 kg m 2 + (0.021 kg)(0.060 m) 2 = 1.14 × 10−4 kg m 2
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12-8
Chapter 12
Section 12.5 Torque
12.18. Visualize:
G
G
Solve: Torque by a force is defined as τ = Fr sin o where φ is measured counterclockwise from the r vector to the F
vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force:
(30 N)r1 sinφ1 + (20 N)r2 sinφ2 = (30 N)(0.02 m) sin ( − 90°) + (20 N)(0.02 m) sin (90°)
= ( − 0.60 N m) + (0.40 N m) = −0.20 N m
Assess: A negative torque causes a clockwise acceleration of the pulley.
12.19. Visualize:
The two equal but opposite 50 N forces, one acting at point P and the other at point Q, make a couple that causes a
net torque.
Solve: The distance between the lines of action is l = d cos30°. The net torque is given by
τ = lF = (d cos30°) F = (0.10 m)(0.866)(50 N) = 4.3 N m
12.20. Model: The disk is a rotating rigid body.
Visualize:
The radius of the disk is 10 cm and the disk rotates on an axle through its center.
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Rotation of a Rigid Body
12-9
Solve: The net torque on the axle is
τ = FA rA sin φA + FBrB sin φB + FC rC sin φC + FD rD sin φD
= (30 N)(0.10 m)sin(−90°) + (20 N)(0.050 m)sin 90° + (30 N)(0.050 m)sin135° + (20 N)(0.10 m)sin 0°
= −3 N m + 1 N m + 1.0607 N m = −0.94 N m
Assess: A negative torque means a clockwise rotation of the disk.
12.21. Model: The beam is a solid rigid body.
Visualize:
G
The steel beam experiences a torque due to the gravitational force on the construction worker ( FG )C and the
G
gravitational force on the beam ( FG ) B . The normal force exerts no torque since the net torque is calculated about the
point where the beam is bolted into place.
G
Solve: The net torque on the steel beam about point O is the sum of the torque due to ( FG )C and the torque due to
G
( FG ) B . The gravitational force on the beam acts at the center of mass.
τ = (( FG )C )(4.0 m)sin(−90°) + (( FG ) B )(2.0 m)sin( −90°)
= −(70 kg)(9.80 m/s2 )(4.0 m) − (500 kg)(9.80 m/s2 )(2.0 m) = −12.5 kN m
The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is
12.5 kN m.
12.22. Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass.
Visualize:
Solve: (a) The torque is due both to the gravitational force on the ball and the gravitational force on the arm:
τ = τ ball + τ arm = (mb g )rb sin 90° + ( ma g )ra sin 90°
= (3.0 kg)(9.8 m/s 2)(0.70 m)+(4.0 kg)(9.8 m/s 2)(0.35 m) = 34 N m
(b) The torque is reduced because the moment arms are reduced. Both forces act at φ = 45° from the radial line, so
τ = τ ball + τ arm = (mb g )rb sin 45° + ( ma g )ra sin 45°
= (3.0 kg)(9.8 m/s 2) (0.70 m)(0.707) + (4.0 kg)(9.8 m/s 2) (0.35 m)(0.707) = 24 N m
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12-10
Chapter 12
Section 12.6 Rotational Dynamics
Section 12.7 Rotation About a Fixed Axis
12.23. Solve: τ = Iα is the rotational analog of Newton’s second law F = ma. We have
τ = (2.0 kg m 2 )(4.0 rad/s2 ) = 8.0 kg m 2 /s2 = 8.0 N m.
12.24. Visualize: Since α = τ /I , a graph of the angular acceleration looks just like the torque graph with the numerical
values divided by I = 4.0 kg m 2 .
Solve: From the discussion in Chapter 4,
ωf = ωi + area under the angular acceleration α curve between ti and tf
The area under the curve between t = 0 s and t = 3 s is 0.50 rad/s. With ω1 = 0 rad/s, we have
ωf = 0 rad/s + 0.50 rad/s = 0.50 rad/s
12.25. Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the
center of mass. Assume that the size of the balls is small compared to 1 m.
Visualize:
We placed the origin of the coordinate system on the 1.0 kg ball.
Solve: The center of mass and the moment of inertia are
(1.0 kg)(0 m) + (2.0 kg)(1.0 m)
= 0.667 m and ycm = 0 m
xcm =
(1.0 kg + 2.0 kg)
I about cm = ∑ mi ri2 = (1.0 kg)(0.667 m)2 + (2.0 kg)(0.333 m)2 = 0.667 kg m 2
We have ωf = 0 rad/s, tf − ti = 5.0 s, and ωi = −20 rpm = −20(2π rad/60 s) = − 23 π rad/s, so ωf = ωi + α (tf − ti ) becomes
2π
⎛ 2π
⎞
0 rad/s = ⎜ −
rad/s ⎟ + α (5.0 s) ⇒ α =
rad/s 2
15
⎝ 3
⎠
Having found I and α , we can now find the torque τ that will bring the balls to a halt in 5.0 s:
⎛2
⎞⎛ 2π
⎞ 4π
rad/s 2 ⎟ =
N m = 0.28 N m
τ = I about cmα = ⎜ kg m 2 ⎟⎜
⎝3
⎠⎝ 15
⎠ 45
The magnitude of the torque is 0.28 N m, applied in the counterclockwise direction.
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Rotation of a Rigid Body
12-11
12.26. Model: The compact disk is a rigid body rotating about its center.
Visualize:
Solve: (a) The rotational kinematic equation ω1 = ω0 + α (t1 − t0 ) gives
200π
⎛ 2π ⎞
(2000 rpm) ⎜
rad/s 2
⎟ rad/s = 0 rad + α (3.0 s − 0 s) ⇒ α =
9
⎝ 60 ⎠
The torque needed to obtain this operating angular velocity is
⎛ 200π
⎞
rad/s 2 ⎟ = 1.75 × 10−3 N m
τ = Iα = (2.5 × 10−5 kg m 2 ) ⎜
⎝ 9
⎠
(b) From the rotational kinematic equation,
1
1 ⎛ 200π
⎞
2
rad/s 2 ⎟ ( 3.0 s − 0 s )
θ1 = θ0 + ω0 (t1 − t0 ) + α (t1 − t0 ) 2 = 0 rad + 0 rad + ⎜
2
2⎝ 9
⎠
100π
revolutions = 50 rev
= 100π rad =
2π
Assess: Fifty revolutions in 3 seconds is a reasonable value.
12.27. Model: Model the rod as thin enough to use I = 13 ML2 .
Visualize:
Solve: From Newton’s second law we have ΔL = τΔt. Combine with ΔL = I Δω and then solve for ω1.
τΔt = I Δω
With ω0 = 0 we have Δω = ω1 − ω0 = ω1. Also τ = rF where r = 0.25 m because the rod is hit perpendicular to it.
ω1 = Δω =
τΔt
I
=
rF Δt
1 ML2
3
=
(0.25 m)(1000 N)(0.0020 s)
1 (0.75 kg)(0.50 m) 2
3
= 8.0 rad/s
Assess: The units check out, and 8.0 rad/s seems like a reasonable answer.
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12-12
Chapter 12
Section 12.8 Static Equilibrium
12.28. Model: The rod is in rotational equilibrium, which means that τ net = 0.
Visualize:
As the gravitational force on the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the
figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right
end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with
your thumb on top (pushing down) and your forefinger underneath (pushing up).
Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this
G
point, so it exerts no torque. To prevent rotation, the pin’s normal force npin exerts a positive torque (ccw about the
left end) to balance the negative torques (cw) of the gravitational force on the mass and rod. The gravitational force
on the rod acts at the center of mass, so
τ net = 0 Nm = τ pin − (0.40 m)(2.0 kg)(9.8 m/s 2) − (0.80 m)(0.50 kg)(9.8 m/s 2)
⇒ τ pin = 11.8 Nm ≈ 12 N m
12.29. Model: The massless rod is a rigid body.
Visualize:
G
Solve: To be in equilibrium, the object must be in both translational equilibrium ( Fnet = 0 N) and rotational
equilibrium (τ net = 0 N m). We have ( Fnet ) y = (40 N) − (100 N) + (60 N) = 0 N, so the object is in translational
equilibrium. Measuring τ net about the left end,
τ net = (60 N)(3.0 m)sin(+90°) + (100 N)(2.0 m)sin(−90°) = −20 N m
The object is not in equilibrium.
12.30. Model: The object balanced on the pivot is a rigid body.
Visualize:
Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium.
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Rotation of a Rigid Body
12-13
G
Solve: There are three forces acting on the object: the gravitational force FG acting through the center of mass of
1
G
the long rod, the gravitational force FG acting through the center of mass of the short rod, and the normal force
2
G
P on the object applied by the pivot. The translational equilibrium equation ( Fnet ) y = 0 N is
( )
( )
−( FG )1 − ( FG ) 2 + P = 0 N ⇒ P = ( FG )1 + ( FG ) 2 = (1.0 kg)(9.8 m/s 2 ) + (4.0 kg)(9.8 m/s 2 ) = 49 N
Measuring torques about the left end, the equation for rotational equilibrium τ net = 0 N m is
Pd − w1 (1.0 m) − w2 (1.5 m) = 0 N m
⇒ (49 N)d − (1.0 kg)(9.8 m/s 2 )(1.0 m) − (4.0 kg)(9.8 m/s2 )(1.5 m) = 0 N ⇒ d = 1.40 m
Thus, the pivot is 1.4 m from the left end.
12.31. Model: The see-saw is a rigid body. The cats and bowl are particles.
Visualize:
Solve: The see-saw is in rotational equilibrium. Calculate the net torque about the pivot point.
τ net = 0 = ( FG )1(2.0 m) − ( FG ) 2 (d ) − ( FG ) B (2.0 m)
m2 gd = m1g (2.0 m) − mB g (2.0 m)
d=
(m1 − mB )(2.0 m) (5.0 kg − 2.0 kg)(2.0 m)
=
= 1.5 m
m2
4.0 kg
Assess: The smaller cat is close but not all the way to the end by the bowl, which makes sense since the combined
mass of the smaller cat and bowl of tuna is greater than the mass of the larger cat.
Section 12.9 Rolling Motion
12.32. Solve: (a) According to Equation 12.36, the speed of the center of mass of the tire is
vcm = Rω = 20 m/s ⇒ ω =
vcm 20 m/s
⎛ 60 ⎞
2
=
= 66.67 rad/s = (66.7) ⎜
⎟ rpm = 6.4 × 10 rpm
R
0.30 m
⎝ 2π ⎠
(b) The speed at the top edge of the tire relative to the ground is vtop = 2vcm = 2(20 m/s) = 40 m/s.
(c) The speed at the bottom edge of the tire relative to ground is vbottom = 0 m/s.
12.33. Model: The can is a rigid body rolling across the floor.
Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass.
The moment of inertia of the can about the center of mass is 12 MR 2 , where R is the radius of the can. Also vcm = Rω ,
where ω is the angular velocity of the can. The total kinetic energy of the can is
1
1
1
1⎛1
⎞⎛ v ⎞
2
2
K = K cm + K rot = Mvcm
+ I cmω 2 = Mvcm
+ ⎜ MR 2 ⎟⎜ cm ⎟
2
2
2
2⎝ 2
⎠⎝ R ⎠
3
3
2
2
= Mvcm = (0.50 kg)(1.0 m/s) = 0.38 J
4
4
2
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12-14
Chapter 12
12.34. Model: The sphere is a rigid body rolling down the incline without slipping.
Visualize:
The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down.
Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation
will be K f = U i . The kinetic energy consists of both translational and rotational energy. This means
Kf =
1
1
1⎛ 2
1
⎞
2
= Mgh ⇒ ⎜ MR 2 ⎟ ω 2 + M ( Rω ) 2 = Mgh
I cmω 2 + Mvcm
2
2
2⎝ 5
2
⎠
7
⇒ MR 2ω 2 = Mg (2.1 m)sin 25°
10
⇒ω =
10
7
g (2.1 m)(sin 25°)
R
2
=
10
7
g (2.1 m)(sin 25°)
(0.04 m) 2
= 88 rad/s
(b) From part (a)
K total =
1
1
7
1
1⎛ 2
1
⎞
2
= MR 2ω 2 and K rot = I cmω 2 = ⎜ MR 2 ⎟ ω 2 = MR 2ω 2
I cmω 2 + Mvcm
2
2
10
2
2⎝ 5
5
⎠
⇒
K rot
=
K total
1 MR 2ω 2
5
7 MR 2ω 2
10
1 10 2
= × =
5 7 7
12.35. Model: The mechanical energy of both the hoop (h) and the sphere (s) is conserved. The initial gravitational
potential energy is transformed into kinetic energy as the objects roll down the slope. The kinetic energy is a
combination of translational and rotational kinetic energy. We also assume no slipping of the hoop or of the sphere.
Visualize:
The zero of gravitational potential energy is chosen at the bottom of the slope.
Solve: The energy conservation equation for the sphere or hoop K f + U gf = Ki + U gi is
1
1
1
1
I (ω1) 2 + m(v1 ) 2 + mgy1 = I (ω0 )2 + m(v0 ) 2 + mgy0
2
2
2
2
For the sphere, this becomes
2
1⎛ 2
1
2 ⎞ (v1 )s
2
mR
⎜
⎟ 2 + m(v1 )s + 0 J = 0 J + 0 J + mghs
2⎝ 5
2
⎠ R
⇒
7
(v1)s2 = gh ⇒ (v1 )s = 10 gh/7 = 10(9.8 m/s 2 )(0.30 m)/7 = 2.05 m/s
10
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Rotation of a Rigid Body
12-15
For the hoop, this becomes
1
(v ) 2 1
(mR 2 ) 12h + m(v1 ) 2h + 0 J = 0 J + 0 J + mghhoop
2
2
R
⇒ hhoop =
(v1 ) 2h
g
For the hoop to have the same velocity as that of the sphere,
(v ) 2 (2.05 m/s) 2
hhoop = 1 s =
= 42.9 cm
g
9.8 m/s 2
The hoop should be released from a height of 43 cm.
Section 12.10 The Vector Description of Rotational Motion
12.36. Visualize: Please refer to Figure EX12.36. To determine angle α , put the tails of the vectors together.
G G
G G
Solve: (a) The magnitude of A × B is AB sin α = (6)(4)sin 45° = 17. The direction of A × B, using the right-hand
G G
rule, is out of the page. Thus, A × B = (17, out of the page).
G G
G G G
(b) The magnitude of C × D is CD sin α = (6)(4)sin180° = 0. Thus C × D = 0.
G
G
G
G
12.37. Solve: (a) The magnitude of A × B is AB sin α = (6)(4)sin 45° = 21.21. The direction of A × B is given by
G
G G
G
the right-hand rule. To curl our fingers from A to B, we have to point our thumb into the page. Thus, A × B =
(21, into the page).
G G
(b) C × D = ((6)(4)sin 90°, out of the page) = (24, out of the page).
12.38. Solve: (a) (iˆ × ˆj ) × iˆ = kˆ × iˆ = ˆj
(b) iˆ × ( ˆj × iˆ) = iˆ × ( −kˆ) = −iˆ × kˆ = −( − ˆj ) = ˆj
12.39. Solve: (a) iˆ × (iˆ × ˆj ) = iˆ × kˆ = − ˆj
G
(b) (iˆ × ˆj ) × kˆ = kˆ × kˆ = 0
G
G
12.40. Solve: A × B = (3iˆ + ˆj ) × (3iˆ − 2 ˆj + 2kˆ)
= 9iˆ × iˆ − 6iˆ × ˆj + 6iˆ × kˆ + 3 ˆj × iˆ − 2 ˆj × ˆj + 2 ˆj × kˆ
= 0 − 6kˆ + 6(− ˆj ) + 3(− kˆ) − 0 + 2iˆ = 2iˆ − 6 ˆj − 9kˆ
G
G
G
G
because iˆ × iˆ = 0. Thus D = niˆ, where n could be any real number.
G G
G
G
(b) C × E = 6kˆ implies that E must be along the ĵ vector, because iˆ × ˆj = kˆ. Thus E = 2 ˆj.
G G
G
G
(c) C × F = −3 ˆj implies that F must be along the k̂ vector, because iˆ × kˆ = − ˆj. Thus F = 1kˆ.
G
12.41. Solve: (a) C × D = 0 implies that D must also be in the same or opposite direction as the C vector or zero,
G
G
G
12.42. Solve: τ = r × F = (5iˆ + 5 ˆj ) × (−10 ˆj ) N m
G
= [−50(iˆ × ˆj ) − 50( ˆj × ˆj )] N m = [−50(+ kˆ) − 0] N m = −50kˆ N m
G
G
G
12.43. Solve: L = r × mv = (3.0iˆ + 2.0 ˆj ) m × (0.1 kg)(4.0ˆj ) m/s
= 1.20(iˆ × ˆj ) kg m 2 /s + 0.8( ˆj × ˆj ) kg m 2 /s = 1.20kˆ kg m 2 /s + 0 kg m 2 /s
= 1.20kˆ kg m 2 /s or (1.20 kg m 2 /s, out of page)
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12-16
Chapter 12
12.44.
Solve:
G
G
G
L = r × mv = (1.0iˆ + 2.0 ˆj ) m × (0.200 kg)(3.0 m/s)(cos 45°iˆ − sin 45° ˆj )
= (0.42iˆ × iˆ − 0.42iˆ × ˆj + 0.85 ˆj × iˆ − 0.85 ˆj × ˆj ) kg m 2 /s = −(1.27 kˆ) kg m 2 /s or (1.27 kg m 2 /s, into page)
Section 12.11 Angular Momentum
12.45. Model: The bar is a rotating rigid body. Assume that the bar is thin.
Solve: The angular velocity ω = 120 rpm = (120)(2π )/60 rad/s = 4π rad/s. From Table 12.2, the moment of inertia
1 ML2 . The angular momentum is
of a rod about its center is I = 12
⎛1⎞
L = I ω = ⎜ ⎟ (0.50 kg)(2.0 m) 2 (4π rad/s) = 2.1 kg m 2 /s
⎝ 12 ⎠
If we wrap our fingers in the direction of the rod’s rotation, our thumb will point in the z direction or out of the page.
Consequently,
G
L = (2.1 kg m 2 /s, out of the page)
12.46. Model: The disk is a rotating rigid body.
Solve: From Table 12.2, the moment of inertia of the disk about its center is
1
1
I = MR 2 = (2.0 kg)(0.020 m)2 = 4.0 × 10−4 kg m 2
2
2
The angular velocity ω is 600 rpm = 600 × 2π /60 rad/s = 20π rad/s. Thus, L = I ω = (4.0 × 10−4 kg m 2 )(20π rad/s) =
0.025 kg m 2 /s. If we wrap our right fingers in the direction of the disk’s rotation, our thumb will point in the
− x direction. Consequently,
G
L = −0.025 iˆ kg m 2 /s = (0.025 kg m 2 /s, into page)
12.47. Model: The bowling ball is a solid sphere.
Solve: From Table 12.2, the moment of inertia about a diameter of a solid sphere is
2
2
I = MR 2 = (5.0 kg)(0.11 m)2 = 0.0242 kg m 2
5
5
Require
L = 0.23 kg m 2 /s = I ω = (0.0243 kg m 2 )ω
⇒ ω = (9.5 rad/s)
⎛ rev ⎞ ⎛ 60 s ⎞
In rpm, this is (9.5 rad/s) ⎜
⎟⎜
⎟ = 91 rpm.
⎝ 2π rad ⎠ ⎝ min ⎠
12.48. Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from above
and stick on the turntable, the turntable slows down due to increased rotational inertia of the (turntable + blocks)
system. Any torques between the turntable and the blocks are internal to the system, so angular momentum of the
system is conserved.
Visualize: The initial moment of inertia is I1 and the final moment of inertia is I 2 .
Solve: The initial moment of inertia is I1 = I disk = 12 mR 2 = 12 (2.0 kg)(0.10 m) 2 = 0.010 kg m 2 and the final moment
of inertia is
I 2 = I1 + 2mR 2 = 0.010 kg m 2 + 2(0.500 kg) × (0.10 m) 2 = 0.010 kg m 2 + 0.010 kg m 2 = 0.020 kg m 2
Let ω1 and ω2 be the initial and final angular velocities. Then
Lf = Li ⇒ ω2 I 2 = ω1I1 ⇒ ω2 =
I1ω1 (0.010 kg m 2 )(100 rpm)
=
= 50 rpm
I2
0.020 kg m 2
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Rotation of a Rigid Body
12-17
12.49. Model: Model all the rest of the body other than the arms as one object (call it the trunk, even though it
includes head and legs), then we can write
ytrunk mtrunk + 2 yarm marm
M
+ marm = 70 kg (the mass of the whole body).
ycm =
where M = mtrunk
Visualize: The language “by how much does he raise his center of mass” makes us think of writing Δycg.
Since we have modeled the arm as a uniform cylinder 0.75 m long, its own center of gravity is at its geometric center,
0.375 m from the pivot point at the shoulder. So raising the arm from hanging down to straight up would change the
height of the center of gravity of the arm by twice the distance from the pivot to the center of gravity: (Δycg)arm =
2(0.375 m) = 0.75 m.
Solve:
( ycm ) trunk mtrunk + 2( ycm )arm, up marm
Δ ( ycm ) body = ( ycm ) with arms up − ( ycm ) with arms down =
M
( ycm ) trunk mtrunk + 2( ycm )arm, down marm
−
M
2marm
2marm
2(3.5 kg)
=
(( ycm )arm, up − ( ycm )arm, down ) =
(Δycm )arm =
(0.75 m) = 0.075 m = 7.5 cm
M
M
70 kg
Assess: 7.5 cm seems like a reasonable amount, not a lot, but not too little. The trunk term subtracted out, which is
both expected and good because we didn’t know ( ycg ) trunk.
12.50. Model: The structure is a rigid body rotating about its center of mass.
Visualize:
We placed the origin of the coordinate system on the 300 g ball.
Solve: First, we calculate the center of mass:
(300 g)(0 cm) + (600 g)(40 cm)
xcm =
= 26.67 cm
300 g + 600 g
Next, we will calculate the moment of inertia about the structure’s center of mass:
I = (300 g)( xcm ) 2 + (600 g)(40 cm − xcm ) 2
= (0.300 kg)(0.2667 m) 2 + (0.600 kg)(0.1333 m) 2 = 0.032 kg m 2
Finally, we calculate the rotational kinetic energy:
2
K rot =
1 2 1
⎛ 100 × 2π
⎞
I ω = (0.032 kg m 2 ) ⎜
rad/s ⎟ = 1.75 J ≈ 1.8 J
2
2
⎝ 60
⎠
12.51. Model: The wheel is a rigid rolling body.
Visualize:
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12-18
Chapter 12
Solve: The front of the disk is moving forward at velocity vcm . Also, because of rotation the point is moving
downward at velocity vrel = Rω = vcm . So, this point has a speed
2
2
v = vcm
+ vcm
= 2vcm = 2(20 m/s) = 28 m/s
Assess: The speed v is independent of the radius of the wheel.
12.52. Visualize:
Solve: We will consider a vertical strip of width dx and of mass dm at a position x from the origin. The formula for
the x component of the center of mass is
1
xcm =
x dm
M∫
The area of the steel plate is A = 12 (0.2 m)(0.3 m) = 0.030 m 2 . Mass dm in the strip is the same fraction of M as dA is
of A. Thus
dm dA
M
⎛ 0.800 kg ⎞
2
=
⇒ dm = dA = ⎜
⎟ dA = (26.67 kg/m )l dx
M
A
A
⎝ 0.030 m 2 ⎠
The relationship between l and x is
l
x
2
=
⇒l = x
0.20 m 0.30 m
3
Therefore,
xcm =
1
(17.78 kg/m 2 ) x3
2 ⎛2⎞ 2
(26.67
kg/m
)
x
dx
=
⎜ ⎟
M∫
M
3
⎝3⎠
0.3 m
=
0m
(17.78 kg/m 2 ) (0.3 m) 2
= 20 cm
0.8 kg
3
Due to symmetry ycm = 0 cm.
12.53. Model: The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk.
Visualize:
Solve: (a) From Table 12.2, the moment of inertia of a disk about its center is
1
1
I = MR 2 = (2.0 kg)(0.10 m) 2 = 0.010 kg m 2
2
2
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Rotation of a Rigid Body
12-19
(b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem:
I = I center + Mh 2 = (0.010 kg m 2 ) + (2.0 kg)(0.10 m) 2 = 0.030 kg m 2
Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the edge
than about the center.
12.54. Model: The object is a rigid rotating body. Assume the masses m1 and m2 are small and the rod is thin.
Visualize: Please refer to Figure P12.54.
Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1, and
mass m2 . Using Table 12.2 for the moment of inertia of the rod, we get
2
I object = I rod about center + I m1 + I m 2 =
=
1
⎛L⎞
⎛L⎞
ML2 + m1 ⎜ ⎟ + m2 ⎜ ⎟
12
⎝2⎠
⎝4⎠
2
1
1
1
L2 ⎛ M
m ⎞
ML2 + m1L2 + m2 L2 = ⎜ + m1 + 2 ⎟
12
4
16
4⎝ 3
4 ⎠
1 ML2 , as expected.
Assess: With m1 = m2 = 0 kg, I rod 12
12.55. Visualize:
We chose the origin of the coordinate system to be on the axis of rotation, that is, at a distance d from one end of the rod.
Solve: The moment of inertia can be calculated as follows:
I=
x2
∫x
2
dm
and
x1
⇒I =
For d = 0 m, I =
1 ML2 ,
3
M
L
L−d
∫
−d
⎛M
x 2dx = ⎜
⎝ L
and for d =
3 L−d
⎞x
⎟
⎠ 3
−d
dm dx
M
=
⇒ dm = dx
M
L
L
1⎛ M
= ⎜
3⎝ L
M
⎞
3
3
3
3
⎟[( L − d ) − (− d ) ] = [( L − d ) + d ]
3L
⎠
1 L,
2
3
3
M ⎡⎛ L ⎞ ⎛ L ⎞ ⎤ 1
⎢⎜ ⎟ + ⎜ ⎟ ⎥ = ML2
3L ⎢⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ 12
Assess: The special cases d = 0 m and d = L/2 of the general formula give the same results that are found in Table 12.2.
I=
12.56. Visualize:
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12-20
Chapter 12
Solve: We solve this problem by dividing the disk between radii r1 and r2 into narrow rings of mass dm. Let
dA = 2π rdr be the area of a ring of radius r. The mass dm in this ring is the same fraction of the total mass M as dA
is of the total area A.
(a) The moment of inertia can be calculated as follows:
M
M
I disk = ∫ r 2dm and dm = dA =
(2π r )dr
2
A
π ( r2 − r12 )
⇒ I disk =
M
r2
2
∫ r (2π r ) dr =
π ( r22 − r12 ) r
1
2M
r2
3
∫ r dr =
(r22 − r12 ) r
1
=
2M
( r24
2
2
4( r2 − r1 )
r4
2
2 4
(r2 − r1 )
− r14 ) =
2M
r2
r1
M 2 2
( r2 − r1 )
2
Replacing r1 with r and r2 with R, the moment of inertia of the disk through its center is I disk = 12 M ( R 2 + r 2 ).
(b) For r = 0 m, I disk = 12 MR 2 . This is the moment of inertia for a solid disk or cylinder about the center.
Additionally, for r ≅ R, we have I = MR 2 . This is the expression for the moment of inertia of a cylindrical hoop or
ring about the center.
(c) The initial gravitational potential energy of the disk is transformed into kinetic energy as it rolls down. If we
choose the bottom of the incline as the zero of potential energy, and use vcm = ω R, the energy conservation equation
K f = U i is
1 2 1
1⎛ M
2
I ω + Mvcm
= Mgh ⇒ ⎜
2
2
2⎝ 2
2
⎞ 2 2 vcm 1
2
⎟ ( R + r ) 2 + Mvcm = Mgyi = Mg (0.50 m)sin 20°
2
R
⎠
2
2
r2 ⎞
2 ⎛R +r ⎞ 1 2
2 ⎛1 1
v
v
⇒ vcm
+
=
+
+
= 1.6759 m 2 /s 2
⎜⎜
⎟
⎜
cm
cm
2 ⎟ 2
⎜ 2 4 4 R 2 ⎟⎟
R
4
⎝
⎠
⎝
⎠
2 ⎞
⎛
3
(0.015
m)
2
vcm
= 1.6759 m 2 /s 2 ⇒ vcm = 1.37 m/s ≈ 1.4 m/s
⎜⎜ +
2⎟
⎟
4
4(0.020
m)
⎝
⎠
For a sliding particle on a frictionless surface K f = U i , so
1 2
v
mvf = mgyi ⇒ vf = 2 gyi = 2 g (0.50 m)sin 20° = 1.83 m/s ⇒ cm = 0.75
2
vf
That is, vcm is 75% of the speed of a particle sliding down a frictionless ramp.
12.57. Model: The plate has uniform density.
Visualize:
Solve: The moment of inertia is
I = ∫ r 2dm.
M
M
dA = 2 dx dy.
A
L
L
L
L
L
The distance from the axis of rotation to the point (x, y) is r = x 2 + y 2 . With − ≤ x ≤ and − ≤ y ≤ ,
2
2
2
2
Let the mass of the plate be M. Its area is L2 . A region of area dA located at (x, y) has mass dm =
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Rotation of a Rigid Body
I=
L
2
L
2
∫ ∫ (x
2
L L
− −
2 2
M
⎛M ⎞
+ y 2 ) ⎜ 2 ⎟ dx dy = 2
L
⎝L ⎠
L
2
⎛ x3
∫ ⎜⎜ 3
L
−
2
⎝
⎞
+ y2x ⎟
⎟
⎠
L
2
dy
−
L
2
L
⎛
⎛ L3 y 2 L ⎛ − L3 y 2 L ⎞ ⎞
M 2 ⎛ L3
M ⎜ L4
y3
2⎞
= 2 ∫⎜ +
−⎜
−
⎟dy = 2 ∫ ⎜ + Ly ⎟dy = 2 ⎜ + L
⎟
⎜
⎟
2
2 ⎠⎟ ⎠⎟
3
L L ⎝⎜ 24
L L ⎝⎜ 12
L ⎜ 12
⎝ 24
⎠
−
−
⎝
2
2
M
=
12-21
L
2
⎞
⎟
⎟
L
− ⎟
2⎠
L
2
⎛ L3 L3 ⎞ ⎞ 1 2
M ⎛ L4
⎜
+
+
⎟ = ML
L
⎜
⎜ 24 24 ⎟⎟ ⎟ 6
L2 ⎝⎜ 12
⎝
⎠⎠
12.58. Solve: From Equation 12.16,
I = ∫ ( x 2 + y 2 ) dm
A small region of area dA has mass dm, and
dm =
The area of the plate is
1 (0.20
2
M
M
dA = dx dy
A
A
m)(0.30 m) = 0.030 m 2 . So
M ⎛ 0.800 kg ⎞
2
=⎜
⎟ = 26.67 kg m
A ⎝ 0.030 m 2 ⎠
1
1
1
The limits for x are 0 ≤ x ≤ 30 cm. For a particular value of x, − x ≤ y ≤ x. Note that ± is the slope of the top
3
3
3
and bottom edges of the triangle. Therefore,
I=
1
x
30 cm 3
∫
0
∫
( x 2 + y 2 )(26.67 kg/m 2 )dx dy
1
− x
3
1
2
= (26.67 kg/m )
= (26.67 kg/m 2 )
30 cm ⎛
x
0
− x
3
∫
y3 ⎞ 3
2
dx
⎜⎜ x y + ⎟⎟
3 ⎠ 1
⎝
30 cm ⎛
∫
0
2 3 2 x3 ⎞
⎜⎜ x +
⎟dx
81 ⎟⎠
⎝3
30 cm
⎛ 56 ⎞⎛ 1 ⎞
= (26.67 kg/m 2 ) ⎜ ⎟⎜ x 4 ⎟
⎝ 81 ⎠⎝ 4 ⎠ 0
= 0.037 kg m 2
12.59. Model: Assume the woman is in equilibrium, so ∑ F = 0 and ∑τ = 0.
Visualize: Choose the axis to be the left end of the board.
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12-22
Chapter 12
Solve: Use ∑τ = 0.
L
(mb ) g = 0 N ⋅ m
2
L
1
L(scalereading) g − ( mb ) g L[(scalereading) − mb ]
2
2
d=
=
=
mw g
mw
∑τ = L(scalereading) g − d (mw ) g −
1
(2.5 m)[(25 kg) − (6.1 kg)]
2
= 0.91 m
60 kg
Assess: This is a little more than halfway up the body of a woman of average height.
12.60. GModel: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium
G
( Fnet = 0 N) and rotational equilibrium (τ net = 0 N m). We also apply the model of static friction.
Visualize:
G
Since the wall is frictionless, the only force from the wall on the ladder is the normal force n2 . On the other hand, the
G
G
G
floor exerts both the normal force n1 and the static frictional force fs . The gravitational force FG on the ladder acts
through the center of mass of the ladder.
G
G
Solve: The x- and y-components of Fnet = 0 N are
∑ Fx = n2 − fs = 0 N ⇒ fs = n2
∑ Fy = n1 − FG = 0 N ⇒ n1 = FG
The minimum angle occurs when the static friction is at its maximum value fs max = μs n1. Thus we have
n2 = fs = μs n1 = μs mg . We choose the bottom corner of the ladder as a pivot point to obtain τ net , because two forces
pass through this point and have no torque about it. The net torque about the bottom corner is
τ net = d1mg − d 2n2 = (0.5L cosθ min )mg − ( L sin θ min ) μs mg = 0 N m
⇒ 0.5cosθ min = μs sin θ min ⇒ tan θ min =
0.5
μs
=
0.5
= 1.25 ⇒ θ min = 51°
0.4
12.61. Model: The beam is a rigid body of length 3.0 m and the student is a particle.
Visualize:
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Rotation of a Rigid Body
12-23
G
G
Solve: To stay in place, the beam must be in both translational equilibrium ( Fnet = 0 N) and rotational equilibrium
(τ net = 0 N m). The first condition is
∑ Fy = −( FG )beam − ( FG )student + F1 + F2 = 0 N
⇒ F1 + F2 = ( FG ) beam + ( FG )student = (100 kg + 80 kg)(9.80 m/s 2 ) = 1764 N
Taking the torques about the left end of the beam, the second condition is
−( FG ) beam (1.5 m) − ( FG )student (2.0 m) + F2 (3.0 m) = 0 N m
−(100 kg)(9.8 m/s 2 )(1.5 m) − (80 kg)(9.8 m/s 2 )(2.0 m) + F2 (3.0 m) = 0 N m
⇒ F2 = 1013 N ≈ 1000 N
From F1 + F2 = 1764 N, we get F1 = 1764 N − 1013 N = 0.75 kN.
Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point
on the body of interest.
12.62. Model: The structure is a rigid body.
Visualize:
Solve: We pick the left end of the beam as our pivot point. We don’t need to know the forces Fh and Fv because the
pivot point passes through the line of application of Fh and Fv and therefore these forces do not exert a torque. For the
beam to stay in equilibrium, the net torque about this point is zero. We can write
τ about left end = −( FG ) B (3.0 m) − ( FG ) W (4.0 m) + (T sin150°)(6.0 m) = 0 N m
Using ( FG ) B = (1450 kg)(9.8 m/s 2 ) and ( FG ) W = (80 kg)(9.8 m/s 2 ), the torque equation can be solved to yield
T = 15,300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried.
12.63. Model: Model
the beam as a rigid body. For the beam not to fall over, it must be both in translational
G
G
equilibrium ( Fnet = 0 N) and rotational equilibrium (τ net = 0 Nm).
Visualize:
The boy walks along the beam a distance x, measured from the left end of the beam. There are four forces acting on
G
G
the beam. F1 and F2 are from the two supports, ( FG ) b is the gravitational force on the beam, and ( FG ) B is the
gravitational force on the boy.
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12-24
Chapter 12
Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium is
−( FG ) b (2.5 m) + F2 (3.0 m) − ( FG ) B x = 0 N m
−(40 kg)(9.80 m/s 2 )(2.5 m) + F2 (3.0 m) − (20 kg)(9.80 m/s 2 ) x = 0 N m
The equation for translation equilibrium is
∑ Fy = 0 N = F1 + F2 − ( FG ) b − ( FG ) B
⇒ F1 + F2 = ( FG ) b + ( FG ) B = (40 kg + 20 kg)(9.8 m/s 2 ) = 588 N
Just when the boy is at the point where the beam tips, F1 = 0 N. Thus F2 = 588 N. With this value of F2 , we can
simplify the torque equation to:
−(40 kg)(9.80 m/s 2 )(2.5 m) + (588 N)(3.0 m) − (20 kg)(9.80 m/s 2 ) x = 0 N m
⇒ x = 4.0 m
Thus, the distance from the right end is 5.0 m − 4.0 m = 1.0 m.
12.64. Solve: The bricks are stable when the net gravitational torque on each individual brick or combination of
bricks is zero. This is true as long as the center of gravity of each individual brick and any combination is over a base
of support. To determine the relative positions of the bricks, work from the top down. The top brick can extend past
the second brick by L/2. For maximum extension, their combined center of gravity will be at the edge of the third
brick, and the combined center of gravity of the three upper bricks will be at the edge of the fourth brick. The
combined center of gravity of all four bricks will be over the edge of the table.
Measuring from the left edge of brick 2, the center of gravity of the top two bricks is
⎛L⎞
m ⎜ ⎟ + mL
m1x1 + m2 x2
3
2
( x12 )com =
= ⎝ ⎠
= L
2m
4
m1 + m2
Thus the top two bricks can extend L/4 past the edge of the third brick. The top three bricks have a center of mass
⎛L⎞
⎛ 3L ⎞
⎛ 5L ⎞
m⎜ ⎟ + m⎜ ⎟ + m⎜ ⎟
m1x1 + m2 x2 + m3 x3
2
4
⎝ ⎠
⎝ 4 ⎠ = 5L
( x123 )com =
= ⎝ ⎠
3m
6
m1 + m2 + m3
Thus the top three bricks can extend past the edge of the fourth brick by L/6. Finally, the four bricks have a combined
center of mass at
⎛L⎞
⎛ 4L ⎞
⎛ 11L ⎞
⎛ 17L ⎞
m⎜ ⎟ + m⎜ ⎟ + m⎜
⎟ + m⎜
⎟
2⎠
6 ⎠
12 ⎠
⎝
⎝
⎝
⎝ 12 ⎠ = 7 L
(x1234 )com =
4m
8
The center of gravity of all four bricks combined is 7 L/8 from the left edge of the bottom brick, so brick 4 can
extend L/8 past the table edge. Thus the maximum distance to the right edge of the top brick from the table edge is
L L L L 25
d max = + + + =
L
8 6 4 2 24
Thus, yes, it is possible that no part of the top brick is directly over the table because d max > L.
Assess: As crazy as this seems, the center of gravity of all four bricks is stably supported, so the net gravitational
torque is zero, and the bricks do not fall over.
12.65. Model: The pole is a uniform rod. The sign is also uniform.
Visualize:
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Rotation of a Rigid Body
12-25
Solve: The geometry of the rod and cable give the angle that the cable makes with the rod.
⎛ 250 ⎞
⎟ = 51.3°
⎝ 200 ⎠
θ = tan21 ⎜
The rod is in rotational equilibrium about its left-hand end.
⎛1⎞
⎝ ⎠
⎛1⎞
⎝ ⎠
τ net = 0 = −(100 cm)(FG ) P − (80 cm) ⎜ ⎟ (FG )S − (200 cm) ⎜ ⎟ (FG )S + (200 cm)T sin 51.3°
2
2
= −(100 cm)(5.0 kg)(9.8 m/s 2 ) − mS (9.8 m/s 2 )(140 cm) + (156 cm)T
With T = 300 N , mS = 30.6 kg ≈ 31 kg.
Assess: A mass of 30.6 kg is reasonable for a sign.
12.66. Model: The sacs are constrained by the stamens. The gravitational torque is 2000 times less than the
straightening torque, so ignore it in part b.
Visualize: Refer to the diagram below.
Solve: (a) The moment of inertia of the stamen and sac can be calculated using kinematic equations.
2
⎛R⎞
I = mstamen ⎜ ⎟ + msac R 2 = (1.0 × 10−6 kg)(5.0 × 10−4 m)2 + (1.0 × 10−6 kg)(1.0 × 10−3 m) 2 = 1.25 × 10−12 kg ⋅ m 2
⎝2⎠
An additional significant figure has been kept in this intermediate result.
The “straightening torque” is
τ = Iα = (1.25 × 10−12 kg ⋅ m 2 )(2.32 × 107 rad/s 2 ) = 2.91 × 10−5 N ⋅ m ≈ 2.9 × 10−5 N ⋅ m
(b) We can use the kinematic equations to find the angular acceleration of a sac, and then the corresponding
tangential acceleration.
α=
2Δθ
Δt
2
=
2(60°)
(
) = 2.32 ×10 rad/s
2π rad
360°
−3 2
(0.30 × 10
7
2
s)
The tangential acceleration is
at = α R = (2.32 × 107 rad/s 2 )(1.0 × 10−3 m) = 2.32 × 104 m/s 2
Using the kinematic equations with the result above gives
vf = at Δt = (2.32 × 104 m/s 2 )(0.30 × 10−3 s) = 7.0 m/s
Assess: These results seem reasonable. The accelerations are huge, while the torque is relatively small. This is due to
the relatively low moment of inertia of the structure.
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12-26
Chapter 12
12.67. Model: The bar is a solid body rotating through its center.
Visualize:
Solve: (a) The two forces form a couple. The net torque on the bar about its center is
Iα
τ net = LF = Iα ⇒ F =
L
where F is the force produced by one of the air jets. We can find I and α as follows:
1
1
I = ML2 = (0.50 kg)(0.60 m)2 = 0.015 kg m 2
12
12
ω1 = ω0 + α (t1 − t0 ) ⇒ 150 rpm = 5.0 π rad/s = 0 rad + α (10 s − 0 s) ⇒ α = 0.50 π rad/s2
⇒F=
(0.015 kg m 2 )(0.5π rad/s 2 )
= 0.0393 N
(0.60 m)
The force F = 39 mN.
(b) The torque of a couple is the same about any point. It is still τ net = LF . However, the moment of inertia has
changed.
LF
1
1
where I = ML2 = (0.500 kg)(0.6 m) 2 = 0.060 kg m 2
I
3
3
(0.0393 N) × (0.60 m)
⇒α =
= 0.393 rad/s 2
0.060 kg m 2
τ net = LF = Iα ⇒ α =
Finally,
ω1 = ω0 + α (t1 − t0 ) = 0 rad/s + (0.393 rad/s 2 )(10 s − 0 s)
= 3.93 rad/s =
(3.93)(60)
rpm = 37.5 rpm
2π
The angular speed is 38 rpm.
Assess: Note that ω ∝ α and α ∝ 1/I . Thus, ω ∝ 1/I . I about the center of the rod is 4 times smaller than I about one
end of the rod. Consequently, ω is 4 times larger.
12.68. Model: The flywheel is a rigid body rotating about its central axis.
Visualize:
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Rotation of a Rigid Body
12-27
Solve: (a) The radius of the flywheel is R = 0.75 m and its mass is M = 250 kg. The moment of inertia about the
axis of rotation is that of a disk:
1
1
I = MR 2 = (250 kg)(0.75 m) 2 = 70.31 kg m 2
2
2
The angular acceleration is calculated as follows:
τ net = Iα ⇒ α = τ net /I = (50 N m)/(70.31 kg m 2 ) = 0.711 rad/s 2
Using the kinematic equation for angular velocity gives
ω1 = ω0 + α (t1 − t0 ) = 1200 rpm = 40 π rad/s = 0 rad/s + 0.711 rad/s 2 (t1 − 0 s)
⇒ t1 = 177 s
(b) The energy stored in the flywheel is rotational kinetic energy:
1
1
K rot = I ω12 = (70.31 kg m 2 )(40π rad/s) 2 = 5.55 × 105 J
2
2
The energy stored is 5.6 × 105 J.
energy delivered (5.55 × 105 J)/2
=
= 1.39 × 105 W ≈ 140 kW
time interval
2.0 s
⎛ ω full energy − ω half energy ⎞
Δω
(d) Because τ = Iα , ⇒ τ = I
= I⎜
⎟ . ωfull energy = ω1 (from part (a)) = 40π rad/s. ωhalf energy
Δt
Δt
⎝
⎠
can be obtained as:
(c) Average power delivered =
1 2
1
I ω half energy = K rot ⇒ ωhalf energy =
2
2
K rot
5.55 × 105 J
=
= 88.85 rad/s
I
70.31 kg m 2
Thus
⎛ 40 π rad/s − 88.85 rad/s ⎞
⎟ = 1.30 kN m
2.0 s
⎝
⎠
τ = (70.31 kg m 2 ) ⎜
12.69. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a
disk and that the string does not slip.
Visualize:
Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same.
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12-28
Chapter 12
Solve: Applying Newton’s second law to m1, m2 , and the pulley yields the three equations:
T1 − ( FG )1 = m1a1
− ( FG ) 2 + T2 = m2 a2
T2 R − T1R − 0.50 Nm = Iα
Noting that − a2 = a1 = a, I = 12 mp R 2 , and α = a/R, the above equations simplify to
T1 − m1g = m1a
0.50 Nm
⎛1
⎞⎛ a ⎞ 1 0.50 Nm 1
T2 − T1 = ⎜ mp R 2 ⎟⎜ ⎟ +
= mp a +
R
2
0.060 m
⎝2
⎠⎝ R ⎠ R
m2 g − T2 = m2a
Adding these three equations,
1 ⎞
⎛
( m2 − m1) g = a ⎜ m1 + m2 + mp ⎟ + 8.333 N
2 ⎠
⎝
⇒a=
( m2 − m1) g − 8.333 N (4.0 kg − 2.0 kg)(9.8 m/s 2 ) − 8.333 N
=
= 1.610 m/s 2
m1 + m2 + 12 mp
2.0 kg + 4.0 kg + (2.0 kg/2)
We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor:
1
1
y1 = y0 + v0 (t1 − t0 ) + a2 (t1 − t0 ) 2 ⇒ 0 = 1.0 m + 0 + (−1.610 m/s 2 )(t1 − 0 s) 2
2
2
2(1.0 m)
⇒ t1 =
= 1.1 s
(1.610 m/s 2 )
12.70. Model: Assume the string does not slip on the pulley.
Visualize:
The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force
on the block m2 , but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block
m1 to the right, but a leftward force on the outer edge of the pulley.
Solve: (a) Newton’s second law for m1 and m2 is T = m1a1 and T − m2 g = m2a2 . Using the constraint − a2 = + a1 = a,
we have T = m1a and −T + m2 g = m2a. Adding these equations, we get m2 g = (m1 + m2 )a, or
a=
m2 g
mm g
⇒ T = m1a = 1 2
m1 + m2
m1 + m2
(b) When the pulley has mass m, the tensions (T1 and T2 ) in the upper and lower portions of the string are different.
Newton’s second law for m1 and the pulley are:
T1 = m1a
and
T1R − T2 R = − Iα
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Rotation of a Rigid Body
12-29
We are using the minus sign with α because the pulley accelerates clockwise. Also, a = Rα . Thus, T1 = m1a and
T2 − T1 =
I a aI
=
R R R2
Adding these two equations gives
I ⎞
⎛
T2 = a ⎜ m1 + 2 ⎟
R ⎠
⎝
Newton’s second law for m2 is T2 − m2 g = m2a2 = − m2 a. Using the above expression for T2 ,
I ⎞
m2 g
⎛
a ⎜ m1 + 2 ⎟ + m2a = m2 g ⇒ a =
R ⎠
m1 + m2 + I/R 2
⎝
Since I = 12 mp R 2 for a disk about its center,
a=
m2 g
m1 + m2 + 12 mp
With this value for a we can now find T1 and T2:
T1 = m1a =
m1m2 g
m1 + m2 + 12 mp
T2 = a (m1 + I/R 2 ) =
1
1 ⎞ m2 (m1 + 2 mp ) g
m2 g
⎛
+
=
m
m
1
p
⎜
⎟
(m1 + m2 + 12 mp ) ⎝
2 ⎠ m1 + m2 + 12 mp
Assess: For m = 0 kg, the equations for a, T1 and T2 of part (b) simplify to
a=
m2 g
m1 + m2
and T1 =
m1m2 g
m1 + m2
and T2 =
m1m2 g
m1 + m2
These agree with the results of part (a).
12.71. Model: The disk is a rigid spinning body.
Visualize: Please refer to Figure P12.71. The initial angular velocity is 300 rpm or (300)(2π )/60 = 10π rad/s. After
3.0 s the disk stops.
Solve: Using the kinematic equation for angular velocity,
ω1 = ω0 + α (t1 − t0 ) ⇒ α =
ω1 − ω0
t1 − t0
=
(0 rad/s − 10π rad/s) −10π
rad/s 2
=
(3.0 s − 0 s)
3
Thus, the torque due to the force of friction that brings the disk to rest is
τ = Iα = − fR ⇒ f = 2
( 1 mR 2 )α
Iα
1
1
π
⎛
⎞
=2 2
= − (mR )α = − (2.0 kg)(0.15 m) ⎜ −10 rad/s 2 ⎟ = 1.57 N ≈ 1.6 N
R
R
2
2
3
⎝
⎠
The minus sign with τ = − fR indicates that the torque due to friction acts clockwise.
12.72. Model: Assume the turbine is a rigid rotating body.
Visualize: Start with Newton’s second law: ∑τ = Iα . The net torque is just the frictional torque we seek.
Solve: Apply the definition of α .
τ = Iα = I
Δω
Δt
When the turbine has reduced its rotation speed by 50% then Δω = 12 ωi .
Δt =
I
ωi
2τ
This suggests that a graph of Δt vs. ωi should be a straight line whose slope is I/2τ .
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12-30
Chapter 12
We see that the fit is quite good and that the slope is 0.1188 s 2 , so the frictional torque is
I
2.6 kg ⋅ m 2
=
= 10.94 N ⋅ m ≈ 11 N ⋅ m
2 ⋅ slope 2(0.1188 s 2 )
Assess: Your boss wants the frictional torque to be as small as possible, but 11 N ⋅ m seems reasonable.
τ=
12.73. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without
slipping. We also assume that the coefficient of rolling friction is zero.
Visualize:
The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in
gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy.
Solve: The conservation of energy equation K f + U gf = Ki + U gi is
1
1
1
1
2
2
M (v1 )cm
+ I cm (ω1 ) 2 + Mgy1 = M (v0 )cm
+ I cm (ω0 )2 + Mgy0
2
2
2
2
1
1⎛ 2
1
1
(v ) 2
⎞
2
2
2
+ ⎜ MR 2 ⎟ (ω0 )cm
+ 0 J ⇒ Mgy1 = M (v0 )cm
+ MR 2 0 2cm
0 J + 0 J + Mgy1 = M (v0 )cm
2
2⎝ 3
2
3
R
⎠
5 (v ) 2
5
5 (5.0 m/s) 2
0 cm
2
⇒ y1 = 6
=
= 2.126 m
⇒ gy1 = (v0 )cm
g
6
6 9.8 m/s 2
The distance traveled along the incline is
y1
2.126 m
s=
=
= 4.3 m
sin 30°
0.5
Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the
incline is approximately 10 mph.
12.74. Model: The disk is a rigid body rotating on an axle passing through one edge. The gravitational potential
energy is transformed into rotational kinetic energy as the disk is released.
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Rotation of a Rigid Body
12-31
Visualize:
We placed the origin of the coordinate system at a distance R just below the axle. In the initial position, the center of
mass of the disk is at the same level as the axle. The center of mass of the disk in the final position is coincident with
the origin of the coordinate system.
Solve: (a) The torque is due to the gravitational force on the disk acting at the center of mass. Thus
τ = (mg ) R = (5.0 kg)(9.8 m/s 2 )(0.30 m) = 14.7 Nm
The moment of inertia about the disk’s edge is obtained using the parallel-axis theorem:
1
3
⎛3⎞
I = I cm + mR 2 = mR 2 + mR 2 = mR 2 = ⎜ ⎟ (5.0 kg)(0.30 m)2 = 0.675 kg m 2
2
2
⎝2⎠
τ
14.7 Nm
⇒α = =
= 22 rad/s 2
I 0.675 kg m 2
(b) The energy conservation equation K f + U gf = Ki = U gi is
1 2
1
1
I ω1 + mgy1 = I ω02 + mgy0 ⇒ I ω12 + 0 J = 0 J + mgR
2
2
2
ω1 =
2mgR
2(5.0 kg)(9.8 m/s 2 )(0.30 m)
=
= 6.6 rad/s
I
0.675 kg m 2
Assess: An angular velocity of 6.6 rad/s (or 1.05 revolutions/s) as the center of mass of the disk reaches below the
axle is reasonable.
12.75. Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational torque on
the hoop causes it to rotate, transforming the gravitational potential energy of the hoop’s center of mass into
rotational kinetic energy.
Visualize:
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12-32
Chapter 12
We placed the origin of the coordinate system at the hoop’s edge on the axle. In the initial position, the center of
mass is a distance R above the origin, but it is a distance R below the origin in the final position.
Solve: (a) Applying the parallel-axis theorem, I edge = I cm + mR 2 = mR 2 + mR 2 = 2mR 2 . Using this expression in the
energy conservation equation K f + U gf = Ki + U gi yields:
1
1
I edgeω12 + mgy1 = I edgeω02 + mgy0
2
2
(b) The speed of the lowest point on the hoop is
1
2g
(2mR 2 )ω12 − mgR = 0 J + mgR ⇒ ω1 =
R
2
2g
(2 R ) = 8 gR
R
Assess: Note that the speed of the lowest point on the loop involves a distance of 2R instead of R.
v = (ω1)(2 R ) =
12.76. Model: The long, thin rod is a rigid body rotating about a frictionless pivot on the end of the rod. The
gravitational torque on the rod causes it to rotate, transforming the gravitational potential energy of the rod’s center
of mass into rotational kinetic energy.
Visualize:
We placed the origin of the coordinate system at the pivot point. In the initial position, the center of mass is a
distance 12 L above the origin. In the final position, the center of mass is at y = 0 m and thus has zero gravitational
potential energy.
Solve: (a) The energy conservation equation for the rod K f + U gf = Ki + U gi is
1⎛1 2⎞ 2
⎜ mL ⎟ ω1 + 0 J = 0 J + mg ( L/2) ⇒ ω1 = 3g/L
2⎝ 3
⎠
1 2
1
I ω1 + mgy1 = I ω02 + mgy0
2
2
(b) The speed at the tip of the rod is vtip = (ω1) L = 3 gL .
12.77. Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is vertical
and the sphere solid.
Visualize: Please refer to Figure P12.77. The sphere rotates because the string wrapped around the rod exerts a torque τ .
Solve: The torque exerted by the string on the rod is τ = Tr.
From the parallel-axis theorem, the moment of inertia of the sphere about the rod’s axis is
2
2
MR 2 13
⎛R⎞
= MR 2
I off center = I cm + M ⎜ ⎟ = MR 2 +
5
4
20
⎝2⎠
From Newton’s second law,
α=
τ
I
=
Tr
(13MR 2 /20)
=
20Tr
13MR 2
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Rotation of a Rigid Body
12-33
12.78. Model: The angular momentum of the satellite in the elliptical orbit is a constant.
Visualize:
Solve: (a) Because the gravitational force is always along the same direction as the direction of the moment arm
G G G
vector, the torque τ = r × Fg is zero at all points on the orbit.
(b) The angular momentum of the satellite at any point on the elliptical trajectory is conserved. The velocity is
G
perpendicular to r at points a and b, so β = 90° and L = mvr. Thus
⎛r ⎞
Lb = La ⇒ mvb rb = mva ra ⇒ vb = ⎜ a ⎟ va
⎝ rb ⎠
30,000 km
30,000 km
− 9000 km = 6000 km and rb =
+ 9000 km = 24,000 km
ra =
2
2
⎛ 6000 km ⎞
⇒ vb = ⎜
⎟ (8000 m/s) = 2000 m/s
⎝ 24,000 km ⎠
(c) Using the conservation of angular momentum Lc = La , we get
⎛r ⎞
mvc rc sin β c = mva ra ⇒ vc = ⎜ a ⎟ va /sin β c
rc = (9000 km)2 + (12,000 km)2 = 1.5 × 107 m
r
⎝ c⎠
From the figure, we see that sin β c = 12,000/15,000 = 0.80. Thus
⎛ 6000 km ⎞ (8000 m/s)
vc = ⎜
= 4000 m/s
⎟
0.80
⎝ 15,000 km ⎠
12.79. Model: For the (bullet + door) system, the angular momentum is conserved in the collision.
Visualize:
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12-34
Chapter 12
G
G
Solve: As the bullet hits the door, its velocity v is perpendicular to r. Thus the initial angular momentum about the
rotation axis, with r = L, is
Li = mBvB L = (0.010 kg)(400 m/s)(1.0 m) = 4.0 kg m 2 /s
After the collision, with the bullet in the door, the moment of inertia about the hinges is
1
1
I = I door + I bullet = mD L2 + mB L2 = (10.0 kg)(1.0 m) 2 + (0.010 kg)(1.0 m) 2 = 3.343 kg m 2
3
3
Therefore, Lf = I ω = (3.343 kg m 2 )ω. Using the angular momentum conservation equation Lf = Li (3.343 kg m 2 )ω =
4.0 kg m 2 /s and thus ω = 1.2 rad/s.
12.80. Model: Model the turntable as a rigid disk rotating on frictionless bearings. For the (turntable + block)
system, no external torques act as the block moves outward towards the outer edge. Angular momentum is thus
conserved.
Visualize: The initial moment of inertia of the turntable is I1 and the final moment of inertia is I 2 .
Solve: The initial moment of inertia is I1 = I disk = 12 mR 2 = 12 (0.2 kg)(0.2 m) 2 = 0.0040 kg m 2 . As the block reaches
the outer edge, the final moment of inertia is
I 2 = I1 + mB R 2 = 0.0040 kg m 2 + (0.020 kg)(0.20 m) 2
= 0.0040 kg m 2 + 0.0008 kg m 2 = 0.0048 kg m 2
Let ω1 and ω2 be the initial and final angular velocities, then the conservation of angular momentum equation is
Lf = Li ⇒ ω2 I 2 = ω1I1 ⇒ ω2 =
I1ω1 (0.0040 kg m 2 )(60 rpm)
=
= 50 rpm
I2
(0.0048 kg m 2 )
Assess: A change of angular velocity from 60 rpm to 50 rpm with an increase in the value of the moment of inertia
is reasonable.
12.81. Model: Model the merry-go-round as a rigid disk rotating on frictionless bearings about an axle in the center
and John as a particle. For the (merry-go-round + John) system, no external torques act as John jumps on the merrygo-round. Angular momentum is thus conserved.
Visualize: The initial angular momentum is the sum of the angular momentum of the merry-go-round and the
angular momentum of John. The final angular momentum as John jumps on the merry-go-round is equal to
I finalωfinal .
Solve: John’s initial angular momentum is that of a particle: LJ = mJ vJ R sin β = mJ vJ R. The angle β = 90° since
John runs tangent to the disk. The conservation of angular momentum equation Lf = Li is
⎛1
⎞
I finalωfinal = Ldisk + LJ = ⎜ MR 2 ⎟ ωi + mJ vJ R
⎝2
⎠
2π ⎛ rad ⎞
⎛1⎞
2
= ⎜ ⎟ (250 kg)(1.5 m) 2 (20 rpm) ⎜
⎟ + (30 kg)(5.0 m/s)(1.5 m) = 814 kg m /s
2
60
rpm
⎝ ⎠
⎝
⎠
⇒ ωfinal =
814 kg m 2 /s
I final
1
1
I final = I disk + I J = MR 2 + mJ R 2 = (250 kg)(1.5 m) 2 + (30 kg)(1.5 m)2 = 349 kg m 2
2
2
ωfinal =
814 kg m 2 /s
349 kg m 2
= 2.33 rad/s = 22 rpm
12.82. Model: Model the skater as a cylindrical torso with two rod-like arms that are perpendicular to the axis of
the torso in the initial position and collapse into the torso in the final position.
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Rotation of a Rigid Body
12-35
Visualize:
Solve: For the initial position, the moment of inertia is I1 = I Torso + 2 I Arm . The moment of inertia of each arm is that
of a 66-cm-long rod rotating about a point 10 cm from its end, and can be found using the parallel-axis theorem. In
the final position, the moment of inertia is I 2 = 12 MR 2 . The equation for the conservation of angular momentum
Lf = Li can be written I 2ω2 = I1ω1 ⇒ ω2 = ( I1/I 2 )ω1. Calculating I1 and I 2 ,
1
⎡1
⎤
I1 = M T R 2 + 2 ⎢ M A L2A + M A d 2 ⎥
2
⎣12
⎦
1
⎡1
⎤
2
= (40 kg)(0.10 m) + 2 ⎢ (2.5 kg)(0.66 m) 2 + (2.5 kg)(0.33 m + 0.10 m) 2 ⎥ = 1.306 kg m 2
2
12
⎣
⎦
1
1
(1.306 kg m 2 )
I 2 = MR 2 = (45 kg)(0.10 m) 2 = 0.225 kg m 2 ⇒ ω2 =
(1.0 rev/s) = 5.8 rev/s
2
2
(0.225 kg m 2 )
12.83. Model: The toy car is a particle located at the rim of the track. The track is a cylindrical hoop rotating about
its center, which is an axis of symmetry. No net torques are present on the track, so the angular momentum of the car
and track is conserved.
Visualize:
Solve: The toy car’s steady speed of 0.75 m/s relative to the track means that
vc − vt = 0.75 m/s ⇒ vc = vt + 0.75 m/s,
where vt is the velocity of a point on the track at the same radius as the car. Conservation of angular momentum
implies that
Li = Lf
0 = I cωc + I tωt = (mr 2 )ωc + ( Mr 2 )ωt = mωc + M ωt
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12-36
Chapter 12
The initial and final states refer to before and after the toy car was turned on. Table 12.2 was used for the track.
v
v
Since ωc = c , ωt = t , we have
r
r
0 = mvc + Mvt
⇒ m(vt + 0.75 m/s) + Mvt = 0
M
(0.200 kg)
(0.75 m/s) = −
(0.75 m/s) = −0.125 m/s
m+M
(0.200 kg + 1.0 kg)
The minus sign indicates that the track is moving in the opposite direction of the car. The angular velocity of the track is
v (0.125 m/s)
= 0.417 rad/s clockwise.
ωt = t =
r
0.30 m
In rpm,
⎛ rev ⎞ ⎛ 60 s ⎞
ωt = (0.417 rad/s) ⎜
⎟⎜
⎟
⎝ 2π rad ⎠ ⎝ min ⎠
⇒ vt = −
= 4.0 rpm
Assess: The speed of the track is less than that of the car because it is more massive.
12.84. Model: Assume that the marble does not slip as it rolls down the track and around a loop-the-loop. The
mechanical energy of the marble is conserved.
Visualize:
Solve: The marble’s center of mass moves in a circle of radius R − r. The free-body diagram on the marble at its
highest position shows that Newton’s second law for the marble is
mv12
mg + n =
R−r
The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when the
normal force of the track on the marble tends to zero. Then the weight will provide the centripetal acceleration
needed for the circular motion. For n → 0 N,
mg =
mv 2
⇒ v12 = g ( R − r )
(R − r)
Since rolling motion requires v12 = r 2ω12 , we have
ω12r 2 = g ( R − r ) ⇒ ω12 =
g (R − r)
r2
The conservation of energy equation is
1
1
( K f + U gf ) top of loop = ( Ki + U gi )initial ⇒ mv12 + I ω12 + mgy1 = mgy0 = mgh
2
2
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Rotation of a Rigid Body
12-37
Using the above expressions and I = 25 mr 2 the energy equation simplifies to
1
1⎛ 2⎞
⎛ g (R − r) ⎞
mg ( R − r ) + ⎜ ⎟ mr 2 ⎜
⎟ + mg 2( R − r ) = mgh ⇒ h = 2.7( R − r )
2
2⎝ 5⎠
⎝ r2 ⎠
12.85. Model: The Swiss cheese wedge is of uniform density—or at least uniform enough that its center of mass is
at the same location as that of a solid piece. To find the angle at which the cheese starts sliding, the cheese will be
treated as a particle, and the model of static friction will be used.
Visualize:
Solve: The angle at which the cheese starts sliding, θS , will be compared to the critical angle θ c for stability. Use
Newton’s second law with the free body diagram.
( Fnet ) x = 0 = fs − FG sin θs
( Fnet ) y = 0 = n − FG cosθs
With FG = mg , the y-direction equation gives n = mg cosθ . The cheese starts sliding when μs is at its maximum
value. Combining that with the x-direction equation and fs = μs n,
0 = μs ( mg cosθs ) − mg sin θs
⇒ θs = tan −1 ( μs ) = tan −1 (0.90) = 42°
The cheese will start sliding at an angle of 42°.
The center of mass of the cheese wedge can be found using the result of Problem 12.52. There, the center of mass of
a triangle with the same proportions as the cheese wedge was found. So xcm is at the center of the cheese wedge
(by symmetry). The ycm can be found by proportional reasoning.
ycm
(30 cm − 20 cm)
=
⇒ ycm = 4.0 cm
12 cm
30 cm
Note that here we have measured ycm from the base of the wedge.
Stability considerations require that the center of mass be no farther than the left corner of the wedge. At the critical
angle geometry shown in the figure above, the right triangle formed by the wedge’s center of mass, lower-left corner,
and center point of the base is a 45°-45°-90° triangle. So θ c = 45°.
The cheese will slide first as the incline reaches 42°. It would not topple until the angle reaches 45°. So Emily is
correct.
Assess: Both Emily’s and Fred’s suppositions are plausible. The calculation must be done to find out which is right.
12.86. Model: Define the system as the rod and cube. Energy and angular momentum are conserved in a perfectly
elastic collision in the absence of a net external torque. The rod is uniform.
Visualize: Please refer to Figure CP12.86.
Solve: Let the final speed of the cube be vf , and the final angular velocity of the rod be ω . Energy is conserved,
and angular momentum around the rod’s pivot point is conserved.
1
1
1
Ei = Ef ⇒ mv02 = mvf2 + I rodω 2
2
2
2
⎛d ⎞
⎛d⎞
Li = Lf ⇒ mv0 ⎜ ⎟ = mvf ⎜ ⎟ + I rodω
⎝2⎠
⎝2⎠
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12-38
Chapter 12
This is two equations in the two unknowns vf and ω. From Table 12.2,
I rod =
1
1
1
Md 2 = (2m)d 2 = md 2
12
12
6
From the angular momentum equation,
v0 = vf +
d
ω
3
⇒ω =
3
(v0 − vf )
d
Substituting into the energy equation,
1 2 1 2 1⎛1
⎞⎛ 9 ⎞
mv0 = mvf + ⎜ md 2 ⎟⎜ 2 ⎟ (v0 − vf ) 2
2
2
2⎝ 6
⎠⎝ d ⎠
3
v02 = vf2 + (v0 − vf ) 2
2
6
1
0 = vf2 − v0vf + v02
5
5
This is a quadratic equation in vf . The roots are
2
⎛ v2 ⎞
6
⎛6 ⎞
v0 ± ⎜ v0 ⎟ − 4 ⎜ 0 ⎟
⎜ 5 ⎟
5
⎝5 ⎠
2
⎝ ⎠ 3
vf =
= v0 ± v0
2
5
5
⎧ 1 v0
= ⎨5
⎩ v0
1
The answer vf = v0 means the ice cube missed the rod. So vf = v0 to the right.
5
12.87. Model: The clay ball is a particle. The rod is a uniform thin rod rotating about its center. Angular momentum
is conserved in the collision.
Visualize:
Solve: This is a two-part problem. Angular momentum is conserved in the collision, and energy is conserved as the
ball rises like a pendulum. The angular momentum conservation equation about the rod’s pivot point is
Li = Lf ⇒ mv0r = ( I ball + rod )ω
Note r =
L
1
= 15 cm. The rod and ball are a composite object. From Table 12.2, I rod = ML2 , so
12
2
1
L2 1
L2 ⎛
M⎞
ML2 = m + ML2 = ⎜ m + ⎟
12
4 12
4⎝
3 ⎠
v
2v
If vf is the final velocity of the clay ball, ω = f = f since the ball sticks to the rod. Thus
r
L
I ball + rod = I ball + I rod = mr 2 +
mv0 L L2 ⎛
M ⎞⎛ 2v ⎞
= ⎜ m + ⎟⎜ f ⎟
2
4⎝
3 ⎠⎝ L ⎠
mv0
(0.010 kg)(2.5 m/s)
⇒ vf =
=
= 0.714 m/s
M
(0.075 kg)
m+
(0.010 kg) +
3
3
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Rotation of a Rigid Body
12-39
Energy is conserved as the clay ball rises. Compare the energy of the ball-rod system just after the collision to when
the ball reaches the maximum height. Note that the center of mass of the rod does not change position.
1
Ei = Ef ⇒ ( I rod + ball )ω 2 = mgh
2
Thus
2
M ⎞⎛ 2vf ⎞
M
1 L2 ⎛
2⎛
⎜ m + ⎟⎜
⎟ = mgh ⇒ vf ⎜ m +
2 4⎝
3 ⎠⎝ L ⎠
3
⎝
⎞
⎟ = mgL(1 − cosθ )
⎠
vf2 ⎛
M⎞
⎜ m + ⎟ = cosθ
mgL ⎝
3 ⎠
Using the various values, cosθ = 0.393 ⇒ θ = 67°.
L
Assess: The clay ball rises h = (1 − cosθ ) = 9.1 cm. This is about 2/3 of the height of the pivot point, and is
2
reasonable.
⇒1−
12.88. Model: Because no external torque acts on the star during gravitational collapse, its angular momentum is
conserved. Model the star as a solid rotating sphere.
Solve: (a) The equation for the conservation of angular momentum is
⎛2
⎞
⎛2
⎞
Li = Lf ⇒ I iωi = I f ωf ⇒ ⎜ mRi2 ⎟ ωi = ⎜ mRf2 ⎟ ωf
⎝5
⎠
⎝5
⎠
⇒ Rf = Ri
ωi
ωf
The angular velocity is inversely proportional to the period T. We can write
T
0.10 s
= 1.3749 × 105 m = 137 km
Rf = Ri f = (7.0 × 108 m)
Ti
2.592 × 106 s
(b) A point on the equator rotates with r = Rf . Its speed is
v=
2π Rf 2π (137,490 m)
=
= 8.6 × 106 m/s
T
0.10 s
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13
NEWTON’S THEORY OF GRAVITY
Conceptual Questions
13.1. Newton’s third law tells us that the forces are equal. They are also clearly equal when Newton’s law of gravity
is examined: F12 = Gm1m2 /r 2 has the same value whether m1 = Earth and m2 = sun or vice versa.
13.2. The force of the star on planet 1 is F1 = G
M s m1
r12
. The force of the star on planet 2 is F2 = G
M s m2
r2 2
. Since
m2 = 2m1, and r2 = 2r1,
M (2m )
G s 21
2 1
F2
(2r1 )
=
= =
M
m
4 2
F1
G s2 1
r1
13.3. (a) The force of the Earth on the first satellite is F1 = G
M e m1
r12
, while F2 = G
M e m2
r2 2
. Since r1 = r2 ,
F1 m1 1000 kg 1
=
=
= .
F2 m2 2000 kg 2
(b) With Fnet = ma, F1 = m1a1 and F2 = m2a2 , so
a1 ⎛ F1 ⎞⎛ m2 ⎞ ⎛ 1 ⎞ ⎛ 2000 kg ⎞
= ⎜ ⎟⎜
⎟ = ⎜ ⎟⎜
⎟ =1
a2 ⎝ F2 ⎠⎝ m1 ⎠ ⎝ 2 ⎠ ⎝ 1000 kg ⎠
The free-fall acceleration is the same for objects of different mass.
13.4. The astronauts can be “weightless” at any distance because an object is said to be weightless if it is in free fall
(as in orbit). For the gravitational force to become zero, the spacecraft would have to be an infinite distance away.
13.5. The hammer will not fall to the Earth. The astronaut, shuttle, and hammer are all in free fall around the Earth
(in an orbit), so the hammer has the same acceleration as the astronaut and does not move away from him.
13.6. The acceleration due to gravity is g = G
g2 = G
M2
R22
M
R2
=G
. With g1 = 20 m/s 2 , M 2 = 2 M1, and R2 = 2 R1,
(2M1 )
(2 R1 )
2
=G
M1 ⎛ 1 ⎞ 1
2
⎜ ⎟ = g1 = 10 m/s
R12 ⎝ 2 ⎠ 2
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13-1
13-2
Chapter 13
13.7. The gravitational potential energy is negative because we choose to place the zero point of potential energy at
infinity (U at ∞ = 0). With this choice, the gravitational potential energy is negative because the conservative force of
gravity is attractive. The two masses will gain kinetic energy as they approach each other.
13.8. The escape speed from Planet X is
vescape X =
2GM X
= 10,000 m/s.
RX
Planet Y is twice as dense as Planet X but the same size so has twice the mass. Therefore
2G (2 M X )
= 2vescape X = 14,142 m/s
RX
vescape Y =
13.9. The Earth’s new orbital period would be about 11.9 years. For circular orbits, Kepler’s third law relates orbital
2
2 ⎛ 4π ⎞ 3
period to radius and can be written T = ⎜
r . Here M is the mass of the body being orbited (the Sun in this
⎜ GM ⎟⎟
⎝
⎠
problem.) The orbital period is independent of the mass of the orbiting body.
13.10. For a circular orbit, v = GM/r (Equation 13.22). As r decreases, v increases, so the satellite speeds up as
the satellite spirals inward.
Exercises and Problems
Section 13.3 Newton’s Law of Gravity
13.1. Model: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size
and mass and that of either the sun or the earth, a human body can be treated as a particle.
GM s M y
GM e M y
Solve: Fs on you =
and Fe on you =
2
rs-e
re2
Dividing these two equations gives
2
2
Fs on y
Fe on y
⎛ M ⎞⎛ r ⎞ ⎛ 1.99 × 1030 kg ⎞⎛ 6.37 × 106 m ⎞
−4
= ⎜ s ⎟⎜ e ⎟ = ⎜
⎟⎜
⎟⎟ = 6.00 × 10
24
11
⎜
⎟⎜
⎝ M e ⎠⎝ rs-e ⎠ ⎝ 5.98 × 10 kg ⎠⎝ 1.50 × 10 m ⎠
13.2. Model: Assume the two lead balls are spherical masses.
Gm1m2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(10 kg)(0.100 kg)
= 6.7 × 10−9 N
r2
(0.10 m) 2
(b) The ratio of the above gravitational force to the gravitational force on the 100 g ball is
Solve: (a) F1 on 2 = F2 on 1 =
=
6.67 × 10−9 N
(0.100 kg)(9.8 m/s 2 )
= 6.8 × 10−9
Assess: The answer in part (b) shows how small is the gravitational force between two lead balls separated by 10 cm
compared to the gravitational force on the 100 g ball.
13.3. Model: Model the sun (s), the moon (m), and the earth (e) as spherical masses.
Solve: Fs on m =
GM s M m
2
rs-m
and Fe on m =
GM e M m
2
re-m
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Newton’s Theory of Gravity
13-3
Dividing the two equations and using the astronomical data from Table 13.2,
2
2
Fs on m ⎛ M s ⎞⎛ re-m ⎞ ⎛ 1.99 × 1030 kg ⎞⎛ 3.84 × 108 m ⎞
=⎜
⎟⎜
⎟⎟ = 2.18
⎟⎜
⎟ =⎜
11
Fe on m ⎝ M e ⎠⎝ rs-m ⎠ ⎜⎝ 5.98 × 1024 kg ⎟⎜
⎠⎝ 1.50 × 10 m ⎠
Note that the sun-moon distance is not noticeably different from the tabulated sun-earth distance.
13.4. Solve: Fsphere on particle =
GM s M p
2
rs-p
and Fearth on particle =
GM e M p
re2
Dividing the two equations,
Fsphere on particle
Fearth on particle
2
2
⎛ M ⎞ ⎛ r ⎞ ⎛ 5900 kg ⎞ ⎛ 6.37 × 106 m ⎞
−7
= ⎜ s ⎟ ⎜ e ⎟ = ⎜⎜
⎟⎟ = 1.6 × 10
⎟⎟⎜⎜
24
⎜
⎟
.
M
r
0
50
m
⎝ e ⎠ ⎝ s-p ⎠ ⎝ 5.98 × 10 kg ⎠⎝
⎠
13.5. Model: Model the woman (w) and the man (m) as spherical masses or particles.
Solve: Fw on m = Fm on w =
GM w M m
2
rm-w
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(50 kg)(70 kg)
(1.0 m) 2
= 2.3 × 10−7 N
13.6. Model: Model the earth (e) as a sphere.
Visualize:
The space shuttle or a 1.0 kg sphere (s) in the space shuttle is Re + rs = 6.37 × 106 m + 0.30 × 106 m = 6.67 × 106 m away
from the center of the earth.
GM e M s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.0 kg)
Solve: (a) Fe on s =
=
= 9.0 N
( Re + rs ) 2
(6.67 × 106 m) 2
Section 13.4 Little g and Big G
13.7. Model: Model the sun (s) as a spherical mass.
Solve: (a) gsun surface =
(b) gsun at earth =
GM s
2
rs-e
=
GM s
Rs2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
(6.96 × 108 m) 2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
(1.50 × 1011 m) 2
= 274 m/s 2
= 5.90 × 10−3 m/s 2
13.8. Model: Model the moon (m) and Jupiter (J) as spherical masses.
Solve: (a) g moon surface =
GM m
2
Rm
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)
6
(1.74 × 10 m)
2
= 1.62 m/s 2
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13-4
Chapter 13
(b) g Jupiter surface =
GM J
RJ2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg)
(6.99 × 107 m) 2
= 25.9 m/s 2
13.9. Model: Model the earth (e) as a spherical mass.
Visualize: The acceleration due to gravity at sea level is 9.83 m/s 2 (see Table 13.1) and Re = 6.37 × 106 m (see
Table 13.2).
Solve: gobservatory =
GM e
( Re + h)
2
=
GM e
2
Re (1 + h/Re )2
=
g earth
(1 + h/Re ) 2
= (9.83 − 0.0075) m/s 2
Here g earth = GM e /Re2 is the acceleration due to gravity on a non-rotating earth, which is why we’ve used the value
9.83 m/s 2 . Solving for h,
⎛ 9.83
⎞
h = ⎜⎜
− 1⎟⎟ Re = 2.4 km
9
8225
.
⎝
⎠
13.10. Model: Model the earth (e) as a spherical mass.
Solve: Let the acceleration due to gravity be 3.0gsurface when the earth is shrunk to a radius of x. Then,
gsurface =
3
GM e
Re2
=
GM e
Re2
GM e
x
2
and 3.0 gsurface =
x=
⇒
GM e
x2
Re
= 0.58 Re
3. 0
The earth’s radius would need to be 0.58 times its present value.
13.11. Model: Model Planet Z as a spherical mass.
Solve: (a) g Z surface =
GM Z
⇒ 8.0 m/s 2 =
(6.67 × 10−11 N ⋅ m 2 /kg 2 ) M Z
RZ2
(b) Let h be the height above the north pole. Thus,
g above N pole =
GM Z
( RZ + h)
2
=
GM Z
RZ2 (1 + h/RZ ) 2
(5.0 × 106 m) 2
=
g Z surface
(1 + h/RZ )
2
=
⇒ M Z = 3.0 × 1024 kg
8.0 m/s 2
⎛ 10.0 × 10 m ⎞
⎜⎜1 +
⎟
5.0 × 106 m ⎟⎠
⎝
6
2
= 0.89 m/s 2
Section 13.5 Gravitational Potential Energy
13.12. Model: Model Mars (M) as a spherical mass. Ignore air resistance. Also consider Mars and the ball as an
isolated system, so mechanical energy is conserved.
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Newton’s Theory of Gravity
13-5
Visualize:
Solve: A height of 15 m is very small in comparison with the radius of earth or Mars. We can use flat-earth gravitational potential energy to find the speed with which the astronaut can throw the ball. On earth, with yi = 0 m and
vf = 0 m/s, energy conservation gives
1 mv 2
f
2
+ mgyf = 12 mvi2 + mgyi
⇒ vi = 2 g e yf = 2(9.81 m/s 2 )(15 m) = 17.2 m/s
Energy is also conserved on Mars, but the acceleration due to gravity is different.
g Mar’s surface =
GM M
2
RM
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(6.42 × 1023 kg)
(3.37 × 106 m) 2
= 3.77 m/s 2
On Mars, with yi = 0 m and vf = 0 m/s, energy conservation is
1 mv 2
f
2
+ mgyf = 12 mvi2 + mgyi
⇒
yf =
vi2
(17.2 m/s) 2
=
= 39 m
2 g m 2(3.77 m/s 2 )
13.13. Model: Model Jupiter as a spherical mass and the object as a point particle. The object and Jupiter form an
isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, allows an object to escape to an infinite distance from Jupiter (or, in general, from its partner object).
Visualize:
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13-6
Chapter 13
Solve: The energy conservation equation K 2 + U 2 = K1 + U1 is
1 m v2
2 o 2
−
GM J mo 1
GM J mo
= 2 mov12 −
r2
RJ
where RJ and M J are the radius and mass of Jupiter. Using the asymptotic condition v2 = 0 m/s as r2 → ∞,
0 J = 12 mov12 −
GM J mo
RJ
⇒ v1 =
2GM J
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg)
=
= 6.02 × 104 m/s
RJ
6.99 × 107 m
Thus, the escape velocity from Jupiter is 60.2 km/s.
13.14. Model: Model the earth (e) as a spherical mass. Compared to the earth’s size and mass, the rocket (r) is
modeled as a particle. This is an isolated system, so mechanical energy is conserved.
Visualize:
Solve: The energy conservation equation K 2 + U 2 = K1 + U1 is
1 m v2
2 r 2
−
GM e mr 1
GM e mr
= 2 mr v12 −
r2
Re
In the present case, r2 → ∞, so
1 m v2
2 r 2
= 12 mr v12 −
v2 = (1.5 × 104 m/s)2 −
GM e mr
Re
⇒ v22 = v12 −
2GM e
Re
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
6.37 × 106 m
= 10 km/s
13.15. Model: The probe and the sun form an isolated system, so mechanical energy is conserved. The minimum
launch speed for escape, which is called the escape speed, allows an object to escape to an infinite distance from the
sun, where the object will have slowed to zero speed with respect to the sun.
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Newton’s Theory of Gravity
13-7
Visualize:
We denote by mp the mass of the probe. M s is the sun’s mass, and Rs-p is the separation between the centers of the
sun and the probe.
Solve: The conservation of energy equation K 2 + U 2 = K1 + U1 is
GM s mp 1
GM s mp
1
= mpv12 −
mpv22 −
r2
Rs-p
2
2
Using the condition v2 = 0 m/s asymptotically as r2 → ∞,
GM s mp
1
2
mpvescape
=
2
Rs-p
⇒ vescape =
2GM s
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
=
= 4.21 × 104 m/s
Rs-p
1.50 × 1011 m
13.16. Model: Model the distant planet (p) and the earth (e) as spherical masses. Because both are isolated, the
mechanical energy of the object on both the planet and the earth is conserved.
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13-8
Chapter 13
Visualize:
Let us denote the mass of the planet by M p and that of the earth by M e . Your mass is m0 . Also, acceleration due to
gravity on the surface of the planet is g p and on the surface of the earth is g e . Rp and Re are the radii of the planet
and the earth, respectively.
Solve: (a) We are given that M P = 2 M e and g p = 14 g e .
Since g p =
GM p
Rp2
and g e =
GM p
Rp2
=
GM e
Re2
, we have
1 GM e 1 G ( M p / 2)
=
4 Re2
4
Re2
⇒
Rp = 8 Re = 8(6.37 × 106 m) = 1.80 × 107 m
(b) The conservation of energy equation K 2 + U 2 = K1 + U1 is
GM p m0 1
GM p m0
1
m0v22 −
= m0v12 −
2
r2
2
Rp
Using v2 = 0 m/s as r2 → ∞, we have
GM p m0
1
2
m0vescape
=
2
Rp
vescape =
2GM p
Rp
=
2G (2M e )
4(6.67 × 10211 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
= 9.41 km/s
Rp
1.80 × 107 m
Section 13.6 Satellite Orbits and Energies
13.17. Model: Model the sun (s) as a spherical mass and the asteroid (a) as a point particle.
Visualize: The asteroid, having mass ma and velocity va , orbits the sun in a circle of radius ra . The asteroid’s time
period is Ta = 5.0 earth years = 1.58 × 108 s.
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Newton’s Theory of Gravity
13-9
Solve: The gravitational force between the sun (mass = M s ) and the asteroid provides the centripetal acceleration
required for circular motion.
GM s ma
=
ra2
1/3
2
ma va2
ra
⎛ GM sTa2 ⎞
GM s ⎛ 2π ra ⎞
=⎜
⎟ ⇒ ra = ⎜⎜
2 ⎟
⎟
ra
⎝ Ta ⎠
⎝ 4π
⎠
⇒
Substituting G = 6.67 × 10−11 N ⋅ m 2 /kg 2 , M s = 1.99 × 1030 kg, and the time period of the asteroid, we obtain
ra = 4.4 × 1011 m. The velocity of the asteroid in its orbit will therefore be
va =
2π ra 2π (4.4 × 1011 m)
=
= 1.7 × 104 m/s
Ta
1.58 × 108 s
Solve: We give the answer to two significant figures because the asteroid period is given to two significant figures.
13.18. Model: Model the sun (s) and the earth (e) as spherical masses.
Visualize: The earth orbits the sun with velocity ve in a circular path with a radius denoted by rs-e . The sun’s and
the earth’s masses are denoted by M s and me , respectively.
Solve: The gravitational force provides the centripetal acceleration required for circular motion.
GM s me
2
rs-e
Ms =
3
4π 2rs-e
GTe2
=
=
meve2 me (2π rs-e ) 2
=
rs-e
rs-eTe2
4π 2 (1.50 × 1011 m)3
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(365 × 24 × 3600 s) 2
= 2.01 × 1030 kg
Assess: The tabulated value is 1.99 × 1030 kg. The slight difference can be ascribed to the fact that the earth’s orbit
isn’t exactly circular.
13.19. From Kepler’s third law, the orbital period squared is proportional to the orbital radius cubed: T 2 ∝ r 3 .
Thus, at ry = 4rx , Ty2 ∝ (4rx )3 = 64rx3 ∝ (8Tx ) 2 . Therefore Ty = 8Tx . A year on Planet Y is 8 × 200 = 1600 earth days
long.
13.20. Model: Model the star (s) and the planet (p) as spherical masses.
Solve: From the planet’s acceleration due to gravity, we find its mass to be
gp =
Mp =
g p Rp2
G
GM p
Rp2
(12.2 m/s 2 )(9.0 × 106 m) 2
=
6.67 × 10−11 N ⋅ m 2 /kg 2
= 1.5 × 1025 kg
(b) From the planet’s orbital period, we find the mass of the star Omega to be
⎛ 4π 2 ⎞ 3
T2 =⎜
r
⎜ GM ⎟⎟
s⎠
⎝
Ms =
4π 2r 3
GT
2
=
4π 2 (2.2 × 1011 m)3
211
(6.67 × 10
N ⋅ m 2 /kg 2 )(402 × 24 × 3600 s) 2
= 5.2 × 1030 kg
13.21. Model: Model the planet and satellites as spherical masses.
Visualize: Please refer to Figure EX13.21.
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13-10
Chapter 13
Solve: (a) The period of a satellite in a circular orbit is T = [4π 2r 3/(GM )]1/2 This is independent of the satellite’s
mass, so we can find the ratio of the periods of two satellites a and b:
⎛r ⎞
Ta
= ⎜ a⎟
Tb
⎝ rb ⎠
3
Satellite 2 has r2 = r1, so T2 = T1 = 250 min. Satellite 3 has r3 = (3/2) r1, so T3 = (3/2)3/2 T1 = 459 min.
(b) The force on a satellite is F = GMm/r 2 . Thus the ratio of the forces on the two satellites a and b is
2
Fa ⎛ rb ⎞ ⎛ ma ⎞
=⎜ ⎟ ⎜
⎟
Fb ⎝ ra ⎠ ⎝ mb ⎠
Satellite 2 has r2 = r1 and m2 = 2m1, so F2 = (1) 2 (2) F1 = 20,000 N. Similarly, satellite 3 has r3 = (3/2) r1 and
m3 = m1, so F3 = (2/3) 2 (1) F1 = 4440 N.
(c) The speed of a satellite in a circular orbit is v = (GM/r ) 2 , so its kinetic energy is K = 12 mv 2 = GMm/2r. Thus the
ratio of the kinetic energy of two satellites a and b is
K a ⎛ rb ⎞⎛ ma ⎞
= ⎜ ⎟⎜
⎟
K b ⎝ ra ⎠⎝ mb ⎠
Satellite 3 has r3 = (3/2)r1 and m3 = m1, so K1/K3 = (3/2)(1/1) = 3/2 = 1.50.
13.22. Model: Model the sun (S) as a spherical mass and the satellite (s) as a point particle.
Visualize: The satellite, having mass ms and velocity vs , orbits the sun with a mass M S in a circle of radius rs .
Solve: The gravitational force between the sun and the satellite provides the necessary centripetal acceleration for
circular motion. Newton’s second law gives
GM Sms
rs2
=
msvs2
rs
Because vs = 2π rs /Ts where Ts is the period of the satellite, this equation simplifies to
GM S
rs2
=
(2π rs ) 2
⇒ rs3 =
Ts2rs
GM STs2
4π 2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(24 × 3600 s) 2
4π 2
⇒ rs = 2.9 × 109 m
13.23. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle.
Visualize: The satellite has a mass is ms and orbits the earth with a velocity vs . The radius of the circular orbit is
denoted by rs and the mass of the earth by M e .
Solve: The satellite experiences a gravitational force that provides the centripetal acceleration required for circular
motion:
GM e ms
rs2
=
msvs2
rs
⇒ rs =
Ts =
GM e
vs2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
(5500 m/s)2
= 1.32 × 107 m
2π Rs 2π (1.32 × 107 m)
=
= 1.51 × 104 s = 4.2 h
vs
5500 m/s
13.24. Model: Model Mars (m) as a spherical mass and the satellite (s) as a point particle.
Visualize: The geosynchronous satellite whose mass is ms and velocity is vs orbits in a circle of radius rs around
Mars. Let us denote mass of Mars by M m .
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Newton’s Theory of Gravity
13-11
Solve: The gravitational force between the satellite and Mars provides the centripetal acceleration needed for circular motion:
GM m ms
rs2
=
msvs2 ms (2π rs ) 2
=
rs
rsTs2
1/3
⎛ GM mTs2 ⎞
⇒ rs = ⎜
⎜ 4π 2 ⎟⎟
⎝
⎠
Using G = 6.67 × 10−11 N ⋅ m 2 /kg 2 , M m = 6.42 × 1023 kg, and Ts = (24.8 hrs) = (24.8)(3600) s = 89,280 s, we obtain
rs = 2.05 × 107 m. Thus, altitude = rs − Rm = 1.72 × 107 m. The speed is the orbital circumference divided by the
period, or
vs =
2π rs 2π (2.05 × 107 m/s)
=
= 1.44 km/s
Ts
89,280 s
13.25. Solve: We are given M1 + M 2 = 150 kg which means M1 = 150 kg − M 2 . We also have
GM1M 2
(0.20 m) 2
= 8.00 × 10−6 N ⇒ M1M 2 =
(8.00 × 10−6 N)(0.20 m)2
6.67 × 10−11 N ⋅ m 2 /kg 2
= 4798 kg 2
Thus, (150 kg − M 2 ) M 2 = 4798 kg 2 or M 22 − (150 kg) M 2 + (4798 kg 2 ) = 0. Solving this equation gives M 2 = 103.75 kg
and 46.25 kg. The two masses are 104 and 46 kg.
13.26. Visualize: We placed the origin of the coordinate system on the 20.0 kg mass (m1) so that the 5.0 kg mass
(m3 ) is on the y-axis and the 10.0 kg mass (m2 ) is on the x-axis.
Solve: (a) The forces acting on the 20.0 kg mass (m1) are
G
Fm2
on m1
=
gm1m2 ˆ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(20.0 kg)(10.0 kg) ˆ
i=
i = 1.33 × 10−6 iˆ N
(0.10 m) 2
r122
gm1m3 ˆ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(20.0 kg)(5.0 kg) ˆ
j=
j = 1.67 × 10−7 ˆj N
(0.20 m)2
r132
G
G
G
Fon m1 = Fm2 on m1 + Fm3 on m1 = 1.33 × 10−6 iˆ N +1.67 ×10−7 ˆj N ⇒ Fon m1 = 1.34 × 10−6 N
G
⎛ ( Fon m ) x ⎞
⎛ 1.33 × 10−6 N ⎞
−1
θ = tan ⎜ G 1 ⎟ = tan −1 ⎜
= 83°
⎜ 1.67 × 10−7 N ⎟⎟
⎜ ( Fon m ) y ⎟
⎝
⎠
1
⎝
⎠
G
Thus the force is Fon m1 = 1.3 × 10−6 N, 83° cw from the +y-axis
G
Fm3
on m1
=
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13-12
Chapter 13
(b) The forces acting on the 5.0 kg mass (m3 ) are
G
G
Fm1 on m3 = − Fm3 on m1 = −1.67 × 10−7 ˆj N
gm2m3
on m3
=
G
Fm2
on m3
= (6.67 × 10−8 N)sinφ iˆ − (6.67 × 10−8 N)cosφ ˆj
G
Fon m3
2
r23
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(10.0 kg)(5.0 kg)
Fm2
(0.20 m) 2 + (0.10 m) 2
= 6.67 × 10−8 N
⎛ 10 cm ⎞ ˆ
⎛ 20 cm ⎞ ˆ
−8
= (6.67 × 10−8 N) ⎜
⎟ i − (6.67 × 10 N) ⎜
⎟j
⎝ 22.36 cm ⎠
⎝ 22.36 cm ⎠
= (2.983 × 10−8 N)iˆ − (5.966 × 10−8 N) ˆj
G
G
= Fm1 on m3 + Fm2 on m3 = 2.983 × 10−8 iˆ N − 2.264 × 10−7 ˆj N
Fon m3 = (2.983 × 10−8 N) 2 + (−2.264 × 10−7 N) 2 = 2.284 × 10−7 N
G
⎛ ( Fon m ) x ⎞
⎛ 2.983 × 10−8 N ⎞
−1
θ ′ = tan ⎜ G 3 ⎟ = tan −1 ⎜
⎟⎟ = 7.51°
−7
⎜
⎜ ( Fon m ) y ⎟
⎝ 2.264 × 10 N ⎠
3
⎝
⎠
G
Thus Fon m3 = 2.3 × 1027 N, 7.5° ccw from the −y-axis.
13.27. Visualize:
⎛ 5 ⎞
Solve: The angle θ = tan −1 ⎜ ⎟ = 14.04°. The distance r1 = r2 = (0.050 m)2 + (0.200 m)2 = 0.206 m. The forces
⎝ 20 ⎠
on the 20.0 kg mass are
G
Mm
F1 = G 2 1 ( − sinθ iˆ + cosθ ˆj )
r1
G
Mm2
F2 = G 2 (sinθ iˆ + cosθ ˆj )
r2
Note m1 = m2 and r1 = r2 . Thus, the net force on the 20.0 kg mass is
G
G G
Mm
Fnet = F1 + F2 = 2G 2 1 cosθ ˆj
r1
=
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(20.0 kg)(5.0 kg)cos(14.04°) ˆ
j
(0.206 m)2
= 3.0 × 10−7 ˆj N
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Newton’s Theory of Gravity
13-13
13.28. Visualize:
Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs
of masses.
U = U12 + U13 + U 23
= −G
m1m2
mm
mm
−G 1 3 −G 2 3
r12
r13
r23
With m1 = 20.0 kg, m2 = 10.0 kg, m3 = 5.0 kg, r12 = 0.20 m, r13 = 0.10 m, and r23 = 0.2236 m,
U = −1.48 × 10−7 J
Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there
are no vector calculations required.
13.29. Visualize:
Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs
of masses.
U = U12 + U13 + U 23
= −G
m1m2
mm
mm
−G 1 3 −G 2 3
r12
r13
r23
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13-14
Chapter 13
With m1 = 20.0 kg, m2 = m3 = 10.0 kg, r12 = r13 = (0.050 m) 2 + (0.200 m)2 = 0.206 m, and r23 = 0.100 m,
U = −1.96 × 10−7 J
Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there
are no vector calculations required.
13.30. Visualize: Because of the gravitational force of attraction between the lead spheres, the cables will make an
angle of θ with the vertical. The distance between the sphere centers is therefore going to be less than 1 m. The freebody diagram shows the forces acting on the lead sphere.
Solve: We can see from the diagram that the distance between the centers is d = 1.000 m − 2 L sin θ . Each sphere is
in static equilibrium, so Newton’s second law is
∑ Fx = Fgrav − T sin θ = 0 ⇒ T sin θ = Fgrav
∑ Fy = T cosθ − mg = 0 ⇒ T cosθ = mg
Dividing these two equations to eliminate the tension T yields
Fgrav Gmm/d 2 Gm
sin θ
= tan θ =
=
= 2
cosθ
mg
mg
d g
We know that d is going to be only very, very slightly less than 1.00 m. The very slight difference is not going to be
enough to affect the value of Fgrav , the gravitational attraction between the two masses, so we’ll evaluate the right
side of this equation by using 1.00 m for d. This gives
tan θ =
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(100 kg)
(1.00 m) 2 (9.81 m/s 2 )
= 6.80 × 10−10
⇒ θ = (3.90 × 10−8 )°
This small angle causes the two spheres to move closer by 2 L sin θ = 1.4 × 10−7 m = 0.00000014 m. Consequently,
the distance between their centers is d = 0.99999986 m.
13.31. Visualize:
We placed the origin of the coordinate system on the 20 kg sphere (m1 ). The sphere (m2 ) with a mass of 10 kg is
20 cm away on the x-axis. The point at which the net gravitational force is zero must lie between the masses m1
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Newton’s Theory of Gravity
13-15
and m2 . This is because on such a point, the gravitational forces due to m1 and m2 are in opposite directions. As the
gravitational force is directly proportional to the two masses and inversely proportional to the square of distance between them, the mass m must be closer to the 10-kg mass. The small mass m, if placed either to the left of m1 or to
the right of m2 , will experience gravitational forces from m1 and m2 pointing in the same direction, thus always
leading to a nonzero force.
Solve:
Fm1 on m = Fm2
on m
⇒ G
m1m
x
2
=G
10 x 2 − 8 x + 0.8 = 0 ⇒
m2m
(0.20 − x)
2
⇒
20
x
2
=
10
(0.20 − x) 2
x = 0.683 m and 0.117 m
The value x = 68.3 cm is unphysical in the current situation, since this point is not between m1 and m2 . Thus, the
mass should be placed 11.7 cm to the right of the larger mass. To two significant figures, this is 12 cm.
13.32. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle.
Visualize: Let h be the height from the surface of the earth where the acceleration due to gravity ( g altitude ) is 10%
of the surface value ( gsurface ).
Solve: (a) Since g altitude = (0.10) gsurface , we have
GM e
( Re + h) 2
h = 2.162 Re
= (0.10)
GM e
Re2
⇒ ( Re + h) 2 = 10 Re2
⇒ h = (2.162)(6.37 × 106 m) = 1.377 × 107 m = 1.4 × 107 m
(b) For a satellite orbiting the earth at a height h above the surface of the earth, the gravitational force between the
earth and the satellite provides the centripetal acceleration necessary for circular motion. For a satellite orbiting with
velocity vs ,
GM e ms
( Re + h)2
=
msvs2
( Re + h)
⇒ vs =
GM e
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
= 4.5 km/s
Re + h
(6.37 × 106 m + 1.377 × 107 m)
13.33. Model: Model the earth as a spherical mass and the object (o) as a point particle. Ignore air resistance. This
is an isolated system, so mechanical energy is conserved.
Visualize:
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13-16
Chapter 13
Solve: (a) The conservation of energy equation K 2 + U g2 = K1 + U g1 is
1
GM e mo 1
GM e mo
= mov12 −
mov22 −
2
2
Re
Re + y1
⎛ 1
1 ⎞
v2 = 2GM e ⎜ −
⎟
⎝ Re Re + y1 ⎠
1
1
⎛
⎞
= 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) ⎜
−
⎟ = 3.02 km/s
⎝ 6.37 × 106 m 6.87 × 106 m ⎠
(b) In the flat-earth approximation, U g = mgy. The energy conservation equation thus becomes
1
1
mov22 + mo gy2 = mov12 + mo gy1
2
2
v2 = v12 + 2 g ( y1 − y2 ) = 2(9.81 m/s 2 )(5.00 × 105 m − 0 m) = 3.13 km/s
(c) The percent error in the flat-earth calculation is
3130 m/s − 3020 m/s
≈ 3.6%
3020 m/s
13.34. Model: Consider the object as a particle and take the planet to be a spherical mass.
Solve: Conservation of mechanical energy of the object gives
−G
Mm 1 2
Mm
= mv − G
R+h 2
R
The object’s mass drops out. Solving for the speed as it hits the ground,
⎛R+h−R⎞
1 ⎞
⎛1
v = 2GM ⎜ −
⎟=
⎟ = 2GM ⎜
⎝R R+h⎠
⎝ R ( R + h) ⎠
Assess: Compare this to v = 2 gh =
2GMh
R2
2GMh
R ( R + h)
, which is the result if the potential U = mgh is used.
13.35. Model: Model the earth and the projectile as spherical masses. Ignore air resistance. This is an isolated system, so mechanical energy is conserved.
Visualize:
A pictorial representation of the before-and-after events is shown.
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Newton’s Theory of Gravity
13-17
Solve: After using v2 = 0 m/s, the energy conservation equation K 2 + U 2 = K1 + U1 is
0 J−
GM e mp
GM e mp
1
= mpv12 −
2
Re + h
Re
The projectile mass cancels. Solving for h, we find
⎡1
v2 ⎤
h=⎢ − 1 ⎥
⎢⎣ Re 2GM e ⎥⎦
−1
− Re = 4.2 × 105 m
13.36. Model: Model the earth as a spherical mass and the meteoroids as point masses.
Visualize:
Solve: (a) The energy conservation equation K 2 + U 2 = K1 + U1 gives
1 mv 2
2
2
−
GM e m 1 2 GM e m
= 2 mv1 −
Re
rm
1/ 2
⎡
⎛ 1
1 ⎞⎤
⇒ v2 = ⎢v12 + 2GM e ⎜ − ⎟ ⎥
⎢⎣
⎝ Re rm ⎠ ⎥⎦
1/2
⎡
1
1
⎛
⎞⎤
−11
2
2
2
24
−
⎟⎥
⎢(2000 m/s) + 2(6.67 × 10 N ⋅ m /kg )(5.97 × 10 kg) ⎜
6
8
⎝ 6.37 × 10 m 3.85 × 10 m ⎠ ⎦
⎣
= 1.1 × 104 m/s = 11 km/s
The speed does not depend on the meteoroid’s mass.
(b) This part differs in that r2 = Re + 5000 km = 1.137 × 107 m. The shape of the meteoroid’s trajectory is not important for using energy conservation. Thus
1 mv 2
2
2
−
GM e m
GM e m
= 1 mv12 −
Re + 5000 km 2
rm
1/ 2
⎡
⎛
1
1 ⎞⎤
⇒ v2 = ⎢v12 + 2GM e ⎜
− ⎟⎥
⎢⎣
⎝ Re + 5000 km rm ⎠ ⎥⎦
= 8.9 × 103 m/s = 8.9 km/s
13.37. Model: Model the two stars as spherical masses, and the comet as a point mass. This is an isolated system,
so mechanical energy is conserved.
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13-18
Chapter 13
Visualize:
In the initial state, the comet is far away from the two stars and thus it has neither kinetic energy nor potential energy.
In the final state, as the comet passes through the midpoint connecting the two stars, it possesses both kinetic energy
and potential energy.
Solve: The conservation of energy equation K f + U f = Ki + U i gives
1 2 GMm GMm
mvf −
−
=0 J+0 J
rf1
rf 2
2
vf =
4GM
4(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
=
= 32,600 m/s = 33 km/s
rf
0.50 × 1012 m
Assess: Note that the final velocity of 33 km/s does not depend on the mass of the comet.
13.38. Model: Model the asteroid as a spherical mass and yourself as a point mass. This is an isolated system, so
mechanical energy is conserved.
Visualize: The radius of the asteroid is M a and its mass is Ra .
Solve: The conservation of energy equation K f + U f = Ki + U i for the asteroid gives
1 2 GM a m 1 2 GM a m
mvf −
= mvi −
2
Ra + r 2
Ra
The minimum speed for escape is the one that will cause you to stop only when the separation between you and the
asteroid becomes very large. Noting that vf → 0 m/s as r → ∞, we have
vi2 =
2GM a 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1014 kg)
=
Ra
2.0 × 103 m
⇒ vi = 2.58 m/s
That is, you need a speed of 2.58 m/s to escape from the asteroid. We can now calculate your jumping speed on the
earth. The conservation of energy equation gives
1
0 J − mvi2 = −mg (0.50 m) ⇒ vi = 2(9.8 m/s 2 )(0.50 m) = 3.13 m/s
2
This means you can escape from the asteroid.
13.39. Model: The projectile is a particle. The earth and moon are spherical masses.
Solve: The projectile is attracted to both the moon and earth. Its final velocity and potential energy are zero. Since
the projectile is fired from the far side of the moon, its initial distance from the center of the earth is the earth-moon
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Newton’s Theory of Gravity
13-19
distance Re-m plus the radius of the moon Rm . Let the projectile have mass m. The conservation of mechanical energy equation is
1 2
M em
M m
mvescape − G
−G m = 0 J +0 J
2
( Re-m + Rm )
Rm
⎛
Me
M ⎞
2
vescape
= 2G ⎜
+ m⎟
⎝ Re-m + Rm Rm ⎠
From Table 13.2, Re-m = 3.84 × 108 m, Rm = 1.74 × 106 m, M e = 5.98 × 1024 kg, and M m = 7.36 × 1022 kg, which
gives vescape = 2.78 km/s.
Assess: The escape velocity does not depend on the mass of the object which is trying to escape.
13.40. Model: The two asteroids make an isolated system, so mechanical energy is conserved. We will also use the
law of conservation of momentum for our system.
Visualize:
Solve: The conservation of momentum equation pfx = pix is
M (vfx )1 + 2 M (vfx ) 2 = 0 kg m/s ⇒ (vfx )1 = −2(vfx ) 2
The equation for mechanical energy conservation K f + U f = Ki + U i is
1
1
G ( M )(2 M )
G ( M )(2M )
1
0.8GM
⇒
M (vfx )12 + (2 M )(vfx ) 22 −
[2(vfx )2 ]2 + (vfx ) 22 =
=−
2
2
2R
10 R
2
R
GM
GM
⇒ (vfx )1 = −2(vfx )2 = 1.032
(vfx ) 2 = −0.516
R
R
The heavier asteroid has a speed of 0.516(GM/R)1/2 and the lighter one a speed of 1.032(GM/R )1/2 .
13.41. Model: Gravity is a conservative force, so we can use conservation of energy.
Visualize:
The planets will be pulled together by gravity and each will have speed v2 as they crash and the separation between
their centers will be 2R.
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13-20
Chapter 13
Solve: The planets begin with only gravitational potential energy. When they crash, they have both potential and
kinetic energy. Thus,
1
1
GMM
GMM
K 2 + U 2 = Mv22 + Mv22 −
= K1 + U1 = 0 J −
2
2
r2
r1
⎛1 1⎞
v2 = GM ⎜ − ⎟
⎝ r2 r1 ⎠
Because the planet is “Jupiter-size,” we’ll use M = M Jupiter = 1.9 × 1027 kg and r2 = 2 RJupiter = 1.4 × 108 m. Inserting
these values into the expression above gives the crash speed of each planet as v2 = 3.0 × 104 m/s.
Assess: Note that the force is not constant, because it varies with distance, so the motion is not constant acceleration
motion. The formulas from constant-acceleration kinematics do not work for problems such as this.
13.42. Model: Model the distant planet (P) as a spherical mass.
Solve: The acceleration at the surface of the planet and at the altitude h are
gsurface =
2
GM P
R
2
( R + h) = 2 R 2
and g altitude =
⇒
1
MP
1 GM P
=
gsurface ⇒ G
2
2
2 R2
( R + h)
R + h = 2 R ⇒ h = ( 2 − 1) R = 0.414 R
That is, the starship is orbiting at an altitude of 0.414R.
13.43. Model: The stars are spherical masses.
Visualize:
Solve: The starts are identical, so their final speeds vf are the same. They collide when their centers are 2R apart.
From energy conservation,
⎛
⎞
⎛ M2 ⎞
M2
⎛1
2⎞
3 ⎜ −G
⎟⎟ = 3 ⎜ Mvf ⎟ − 3 ⎜⎜ G
⎟
9
⎜
(5.0 × 10 m) ⎠
⎝2
⎠ ⎝ 2 R ⎟⎠
⎝
1
⎛ 1
⎞
vf2 = 2GM ⎜
−
⎟
9
⎝ 2 R 5.0 × 10 m ⎠
vf = 3.7 × 105 m/s
13.44. Model: The stars are spherical masses. They each rotate about the system’s center of mass.
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Newton’s Theory of Gravity
13-21
Visualize:
Solve: (a) The stars rotate about the system’s center of mass with the same period: T1 = T2 = T . We can locate the
center of mass by letting the origin be at the smaller-mass star. Then
r1 = rcm =
(2.0 × 1030 kg)(0 m) + (6.0 × 1030 kg)(2.0 × 1012 m)
2.0 × 1030 kg + 6.0 × 1030 kg
= 1.5 × 1012 m
Mass m2 undergoes uniform circular motion with radius r2 = 0.5 × 1012 m due to the gravitational force of mass m1
at distance R = 2.0 × 1012 m. The gravitational force is responsible for the centripetal acceleration, so
Fgrav =
1/2
⎡ 4π 2
⎤
T =⎢
r2 R 2 ⎥
⎣⎢ Gm1
⎦⎥
Gm1m2
R2
2
= m2acentripetal =
m2v22 m2 ⎛ 2π r2 ⎞
4π 2m2r2
=
⎜
⎟ =
r2
r2 ⎝ T ⎠
T2
1/2
⎡ 4π 2 (0.5 × 1012 m)(0.5 × 1012 m)2
⎤
=⎢
⎥
2
2
30
−11
⎣⎢ (6.67 × 10 N ⋅ m /kg )(2.0 × 10 kg) ⎦⎥
= 7.693 × 108 s = 24 years
(b) The speed of each star is v = (2π r )/T . Thus
v1 =
2π r1 2π (1.5 × 1012 m)
=
= 12.3 km/s
T
7.693 × 108 s
v2 =
2π r2 2π (0.5 × 1012 m)
=
= 4.1 km/s
T
7.693 × 108 s
13.45. Model: Model the moon (m) as a spherical mass and the lander (l) as a particle. This is an isolated system,
so mechanical energy is conserved.
Visualize: The initial position of the lunar lander (mass = m1 ) is at a distance r1 = Rm + 50 km from the center of
the moon. The final position of the lunar lander is the orbit whose distance from the center of the moon is
r2 = Rm + 300 km.
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13-22
Chapter 13
Solve: The external work done by the thrusters is
Wext = ΔEmech = 12 ΔU g
where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at
ri = Rm + 50 km to the final orbit rf = Rm + 300 km. Thus
1 ⎛ −GM m m −GM m m ⎞ GM m m ⎛ 1 1 ⎞
Wext = ⎜
−
⎟=
⎜ − ⎟
2⎝
rf
ri
2 ⎝ ri rf ⎠
⎠
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)(4000 kg) ⎛
1
1
⎞
−
⎜
⎟
2
⎝ 1.79 × 106 m 2.04 × 106 m ⎠
= 6.7 × 108 J
13.46. Model: Model the earth (e) as a spherical mass and the space shuttle (s) as a point particle. This is an isolated system, so the mechanical energy is conserved.
Visualize: The space shuttle (mass = ms ) is at a distance of r1 = Re + 250 km.
Solve: The external work done by the thrusters is
Wext = ΔEmech = 12 ΔU g
where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at
r1 = Re + 250 km to the final orbit rf = Re + 610 km. Thus
1 ⎛ −GM e m −GM e m ⎞ GM e m ⎛ 1 1 ⎞
Wext = ⎜
−
⎟=
⎜ − ⎟
2 ⎝ rf
ri
2 ⎝ ri rf ⎠
⎠
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(75,000 kg) ⎛
1
1
⎞
−
⎜
⎟
6
6
2
⎝ 6.62 × 10 m 6.98 × 10 m ⎠
= 1.2 × 1011 J
This much energy must be supplied by burning the on-board fuel.
13.47. Model: Assume a spherical asteroid and a point mass model for the satellite. This is an isolated system, so
mechanical energy is conserved.
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Newton’s Theory of Gravity
13-23
Visualize:
The orbital radius of the satellite is
r = Ra + h = 8,800 m + 5,000 m = 13,800 m
Solve: (a) The speed of a satellite in a circular orbit is
1/2
GM ⎡ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg) ⎤
v=
=⎢
⎥
r
13,800 m
⎢⎣
⎦⎥
= 7.0 m/s
(b) The minimum launch speed for escape (vi ) will cause the satellite to stop asymptotically (vf = 0 m/s) as
rf → ∞. Using the energy conservation equation K 2 + U 2 = K1 + U1, we get
1
GM a ms 1
GM a ms
= msvi2 −
msvf2 −
2
2
rf
Ra
1 2
GM a
⇒ 0 J − 0 J = vescape
−
2
Ra
2GM a
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg)
=
= 12 m/s
8800 m
Ra
vescape =
13.48. Model: Model the moon as a spherical mass and the satellite as a point mass.
Visualize: The rotational period of the satellite is the same as the rotational period of the moon around its own axis.
This time happens to be 27.3 days.
Solve: The gravitational force between the moon and the satellite provides the centripetal acceleration necessary for
circular motion around the moon. Therefore,
GM m m
r2
r3 =
GM mT
4π 2
2
=
(6.67 × 10
−11
⎛ 2π ⎞
= mrω 2 = mr ⎜
⎟
⎝ T ⎠
2
N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)(27.3 × 24 × 3600 s)2
4π 2
r = 8.84 × 107 m
Since r = Rm + h, then h = r − Rm = 8.84 × 107 m − 1.74 × 106 m = 8.67 × 107 m.
13.49. Model: Model the earth as a spherical mass and the satellite as a point mass.
Visualize: The satellite is directly over a point on the equator once every two days. Thus, T = 2Te = 2 × 24 ×
3600 s = 1.728 × 105 s.
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13-24
Chapter 13
Solve: From Kepler’s laws applied to a circular orbit:
⎛ 4π 2 ⎞ 3
T2 =⎜
r
⎜ GM ⎟⎟
e⎠
⎝
r3 =
GM eT 2
4π 2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.728 × 105 )2
4π 2
7
r = 6.71 × 10 m
Assess: The radius of the orbit is larger than the geosynchronous orbit.
13.50. Visualize: Please refer to Figure P13.50.
Solve: The gravitational force on one of the masses is due to the star and the other planet. Thus
G
Mm
r2
+
Gmm
(2r )2
=
mv 2 m ⎛ 2π r ⎞
= ⎜
⎟
r
r⎝ T ⎠
G⎛
m ⎞ 4π 2r 2
⎜M + ⎟ =
r⎝
4⎠
T2
2
⇒
GM Gm 4π 2r 2
+
=
r
4r
T2
1/2
⎡ 4π 2r 3
⎤
1
⇒ T =⎢
⎥
⎢⎣ G ( M + m/4) ⎥⎦
13.51. Solve: (a) Taking the logarithm of both sides of v p = Cu q gives
[log(v p ) = p log v] = [log(Cu q ) = log C + q log u ] ⇒ log v =
q
log C
log u +
p
p
But x = log u and y = log v, so x and y are related by
⎛q⎞
log C
y = ⎜ ⎟x +
p
⎝ p⎠
(b) The previous result shows there is a linear relationship between x and y, so there is a linear relationship between
log u and log v. The graph of a linear relationship is a straight line, so the graph of log v-versus-log u will be a
straight line.
(c) The slope of the straight line represented by the equation y = (q/p ) x + log C/p is q/p. Thus, the slope of the log vversus-log u graph will be q/p.
(d) The predicted y-intercept of the graph is log C/p, and the experimentally determined value is 9.264. Equating
these, we can solve for M. Because the planets all orbit the sun, the mass we are finding is M = M sun .
⎛ 4π 2
1
1
log C = log ⎜
⎜ GM
2
2
sun
⎝
M sun
⎞
4π 2
1
= 10−18.528 = 18.528
⎟⎟ = −9.264 ⇒
GM
10
sun
⎠
2
⎛ 4π ⎞ 18.528
=⎜
(10
) = 1.996 × 1030 kg
⎜ G ⎟⎟
⎝
⎠
The tabulated value, to three significant figures, is M sum = 1.99 × 1030 kg. We have used the orbits of the planets to
“weigh the sun!”
13.52. Solve: (a) Dividing the circumference of the orbit by the period gives
v=
2π Rs 2π (1.0 × 104 m)
=
= 6.3 × 104 m/s
T
1.0 s
(b) Using the formula for the acceleration at the surface,
gsurface =
GM s
Rs2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
(1.0 × 104 m) 2
= 1.3 × 1012 m/s 2
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Newton’s Theory of Gravity
13-25
(c) The mass of an object on the earth will be the same as its mass on the star. The gravitational force is
( FG )star = mg surface = 1.3 × 1012 N
(d) The radius of the orbit of the satellite is r = 1 × 104 m + 1.0 × 103 m = 1.1 × 104 m. The period is
T2 =
4π 2r 3
4π 2 (1.1 × 104 m)3
=
GM s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
⇒ T = 6.29 × 10−4 s
This means there are 1589 revolutions per second or 9.5 × 104 orbits per minute.
(e) Applying Equation 13.25 for a geosynchronous orbit,
r3 =
GM s
4π 2
T2 =
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(1.0 s)
4π 2
⇒ r = 1.5 × 106 m
13.53. Model: Assume the solar system is a point particle.
Solve: (a) The radius of the orbit of the solar system in the galaxy is 25,000 light years. This means
r = 25,000 light years = 2500(3.0 × 108 m/s)(365 d/y)(24 h/d)(3600 s/h)(1 y) = 2.36 × 1020 m
T=
2π r 2π (2.36 × 1020 m)
=
= 6.64 × 1015 s = 2.1 × 108 years
v
2.30 × 105 m/s
(b) The number of orbits =
5.0 × 109 years
2.05 × 108 years
(c) Applying Newton’s second law yields
GM g center mss
r
2
=
mssv 2
r
= 24 orbits.
⇒ M g center =
v 2r (2.30 × 105 m/s) 2 (2.36 × 1020 m)
=
= 1.9 × 1041 kg
G
6.67 × 10−11 N ⋅ m 2 /kg 2
(d) The number of stars in the center of the galaxy is
1.87 × 1041 kg
1.99 × 1030 kg
= 9.4 × 1010
13.54. Model: Assume the three stars are spherical masses.
Visualize:
The stars rotate about the center of mass, which is the center of the triangle and equal distance r from all three stars.
The gravitational force between any two stars is the same. On a given star the two forces from the other stars make an
angle of 60°.
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13-26
Chapter 13
Solve: The value of r can be found as follows:
L/2
L
1.0 × 1012 m
=
= 0.577 × 1012 m
= cos(30°) ⇒ r =
r
2cos(30°) 2cos(30°)
The gravitational force between any two stars is
Fg =
GM 2
2
L
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) 2
(1.0 × 1012 m) 2
= 2.64 × 1026 N
The component of this force toward the center is
Fc = Fg cos(30°) = (2.64 × 1026 N)cos(30°) = 2.29 × 1026 N
The net force on a star toward the center is twice this force, and that force equals MRω 2 . This means
⎛ 2π ⎞
2 × 2.29 × 1026 N = MRω 2 = MR ⎜
⎟
⎝ T ⎠
T=
4π 2 MR
4.58 × 10
26
N
=
4π 2 (1.99 × 1030 kg)(0.577 × 1012 m)
4.58 × 1026 N
2
= 3.15 × 108 s = 10 years
13.55. Model: Angular momentum is conserved for a particle following a trajectory of any shape.
Visualize:
For a particle in an elliptical orbit, the particle’s angular momentum is L = mrvt = mrv sin β , where v is the velocity
G
G
tangent to the trajectory and β is the angle between r and v.
Solve: At the distance of closest approach (rmin ) and also at the most distant point, β = 90°. Since there is no tangential force (the only force being the radial force) there is no torque, so angular momentum must be conserved:
mPlutov1rmin = mPlutov2rmax
v2 = v1 (rmin /rmax ) = (6.12 × 103 m/s)((4.43 × 1012 m)/(7.30 × 1012 m) = 3.71 km/s
13.56. Model: Angular momentum is conserved for a particle following a trajectory of any shape.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Newton’s Theory of Gravity
13-27
Visualize:
For a particle in an elliptical orbit, the particle’s angular momentum is L = mrvt = mrv sin β , where v is the velocity
G
G
tangent to the trajectory, and β is the angle between r and v.
Solve: At the distance of closest approach (rmin ) and also at the most distant point, β = 90°. Since there is no tangential force (the only force being the radial force) there is no torque, so angular momentum must be conserved:
mMercuryv1rmin = mMercuryv2rmax
rmin = rmax v2 /v1 =
(6.99 × 1010 m)(38.8 km/s)
= 4.60 × 1010 m
59.0 km/s
13.57. Model: For the sun + comet system, the mechanical energy is conserved.
Visualize:
Solve: The conservation of energy equation K f + U f = Ki + U i gives
1
GM s M c 1
GM s M c
M cv22 −
= M cv12 −
r2
r1
2
2
Using G = 6.67 × 10−11 Nm 2 /kg 2 , M s = 1.99 × 1030 kg, r1 = 8.79 × 1010 m, r2 = 4.50 × 1012 m, and v1 = 54.6 km/s,
we get v2 = 4.49 km/s.
13.58. Model: Model the planet (p) as a spherical mass and the spaceship (s) as a point mass.
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13-28
Chapter 13
Visualize:
Solve: (a) For the circular motion of the spaceship around the planet,
GM p ms
r02
=
mv02
r0
⇒ v0 =
GM p
r0
Immediately after the rockets were fired v1 = v0 /2 and r1 = r0 . Therefore,
v1 =
1 GM p
2
r0
(b) The spaceship’s maximum distance is rmax = r0 . Its minimum distance occurs at the other end of the ellipse. The
energy at the firing point is equal to the energy at the other end of the elliptical trajectory. That is,
GM p ms 1
GM p ms
1
msv12 −
= msv22 −
r1
r2
2
2
Since the angular momentum at these two ends is conserved, we have
mv1r1 = mv2r2
⇒ v2 = v1 (r1/r2 )
With this expression for v2 , the energy equation simplifies to
GM p
1 2 GM p 1 2
v1 −
= v1 (r1/r2 )2 −
r1
r2
2
2
Using r1 = r0 and v1 = v0 /2 =
1 GM p
,
r0
2
1 ⎛ 1 GM p ⎞ GM p 1 ⎛ 1 GM p ⎞ r02 GM p
1
1
r
1
= ⎜
⇒
− = 0 −
⎜
⎟−
⎟ −
r0
r2
2 ⎝ 4 r0 ⎠
2 ⎝ 4 r0 ⎠ r22
8r0 r0 8r22 r2
⎛ 7 ⎞ 2
7
r
1
r0
+ 02 − = 0 ⇒ ⎜
⎟ r2 − r2 + = 0
8r0 8r2 r2
8
r
8
⎝ 0⎠
The solutions are r2 = r0 (the initial distance) and r2 = r0 /7. Thus the minimum distance is rmin = r0 /7.
13.59. Solve: (a) At what distance from the center of Saturn is the acceleration due to gravity the same as on the
surface of the earth?
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Newton’s Theory of Gravity
13-29
(b)
(c) The distance is 6.21 × 107 m. This is 1.06RSaturn
13.60. Solve: (a) A 1000 kg satellite orbits the earth with a speed of 1997 m/s. What is the radius of the orbit?
(b)
(c) The radius of the orbit is
r=
GM E
(vpayload )
2
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
(1997 m/s)
2
= 1.00 × 108 m
13.61. Solve: (a) A 100 kg object is released from rest at an altitude above the moon equal to the moon’s radius. At
what speed does it impact the moon’s surface?
(b)
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13-30
Chapter 13
(c) The speed is v2 = 1680 m/s.
13.62. Solve: (a) Kepler’s third law for circular orbits is
⎛ 4π 2 ⎞ 3
4π 2 32
T2 =⎜
r ⇒ T=
r
⎟
⎜ GM ⎟
GM
⎝
⎠
Letting a =
3
4π 2
, the first satellite obeys T = ar 2 . For the second satellite, which orbits the same mass,
GM
3
⎛ Δr ⎞ 2
T + ΔT = a( r + Δr ) = ar ⎜1 +
⎟
r ⎠
⎝
3
2
Since
3
2
Δr
1, we can use the approximation (1 ± x )n ≈ 1 ± nx.
r
Thus
3 ⎛
3 Δr ⎞
T + ΔT ≈ ar 2 ⎜1 +
⎟
⎝ 2 r ⎠
3
Subtracting the equation T = ar 2 for the first satellite from this,
3 3 ⎛ Δr ⎞
ΔT = ar 2 ⎜ ⎟
2
⎝ r ⎠
Dividing this by the equation for the first satellite,
ΔT 3 Δr
=
T
2 r
(b) The satellites orbit the earth. The fractional difference in their periods is
ΔT 3 1 km
=
= 2.24 × 10−4
T
2 6700 km
After
1
2.24 × 10−4
= 4467 periods they will meet again. For the inner satellite,
T=
4π 2
G (5.98 × 10
3
24
kg)
(6.700 × 106 m) 2
= 5456 s = 1.52 hrs
So the satellites will meet again in 4467 × 1.52 hrs = 6770 hrs = 282 days.
Assess: A communications satellite has an orbital period of around 1.5 h. The surprising length of time between the
two satellites meeting is due to the small differences in their periods.
13.63. Model: Model the earth and sun as spherical masses, the satellite as a point mass. Assume the satellite’s
distance from the earth is very small compared to the earth’s (and satellite’s) distance to the sun.
Visualize:
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Newton’s Theory of Gravity
13-31
Solve: The net force on the satellite is the sum of the gravitational force toward the sun and the gravitational force
toward the earth. This net force is responsible for circular motion around the sun. We want to chose the distance d to
make the period T match the period Te with which the earth orbits the sun. The earth’s orbital period is given by
Te2 = (4π 2 /GM s ) Re3 . Thus
Fnet =
GM s m
r2
−
GM e m
d2
2
= macentripetal =
mv 2 m ⎛ 2π r ⎞
4π 2
GM s
= ⎜
⎟ = mr 2 = mr 3
r
r ⎝ Te ⎠
Te
Re
Using r = Re − d and canceling the Gm term gives
Ms
( Re − d ) 2
Me
−
=
d2
Ms
Re3
( Re − d )
This equation can’t be solved exactly, but we can make use of the fact that d Re to use the binomial approximation. Factor the Re out of the expressions Re − d to get
Ms
2
Re (1 − d/Re )2
−
Me
d
2
=
Ms
Re2
(1 − d/Re )
If we think of d/Re = x 1, we can simplify the first term by using (1 ± x )n ≈ 1 ± nx. Here n = −2, so we get
Ms
Re2
[1 − (−2)d/Re ] −
Me
d
2
=
Ms
Re2
(1 − d/Re ) ⇒
Me
d
2
=3
Ms
Re2
Thus d = [ M e /(3M s )]1/3 Re = 1.50 × 109 m.
Assess: d/Re = 0.010, so our assumption that d/Re 1 is justified.
13.64. Model: The earth and sun are spherical masses. The earth is in a circular orbit around the sun. The projectile is a particle. The effect of the earth’s rotation on the projectile’s velocity is ignored. The earth and sun are so
massive compared to the rocket that they are unaffected by the rocket’s motion (no recoil). The rocket escapes from
the influence of the earth’s gravitation in a distance small in comparison with the distance from the earth to the sun,
so that the change in solar potential of the rocket as it escapes the earth is negligible.
Visualize:
Solve: To escape the sun’s gravitational pull at the earth’s distance from the sun, a projectile must have speed
vescape . The energy conservation equation for the projectile is
M m
1 2
mvescape − G s = 0 J
Re-s
2
vescape = 2G
Ms
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
=
= 4.207 × 104 m/s
Re-s
(1.50 × 1011 m)
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13-32
Chapter 13
The earth’s speed in its orbit is found by considering Equation 13.22:
ve =
GM s
= 2.975 × 104 m/s
Re-s
The projectile must also escape from the earth’s gravitational pull. Far from the earth, the projectile’s total speed is
the sum of its speed relative to the earth far from the influence of earth’s gravity, vp , and the earth’s speed.
vescape = ve + vp
vp = vescape − ve = 4.207 × 104 m/s − 2.975 × 104 m/s = 1.232 × 104 m/s.
To obtain the speed vp far from the earth, the projectile must be launched from the earth with speed vlaunch . Using
energy conservation,
1 2
M m 1
mvlaunch − G e = mvp2
2
Re
2
=
⇒ vlaunch = 2G
2(6.67 × 10211 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
(6.37 × 106 m)
Me
+ vp2
Re
+ (1.232 × 104 m/s) 2
= 1.66 × 104 m/s
Assess: The projectile must attain a speed of 16.6 km/s if launched in the direction of earth’s motion. This is about a
factor of 7 times less than what is required from rest. Note that the speed of the earth in its orbit about the sun ve
differs from the escape speed from the sun vescape by a factor of
2.
13.65. Model: Model the 400 kg satellite and the 100 kg satellite as point masses and model the earth as a spherical
mass. Momentum is conserved during the inelastic collision of the two satellites.
Visualize:
Solve: For the given orbit, r0 = Re + 1 × 106 m = 7.37 × 106 m. The speed of a satellite in this orbit is
v0 =
GM e
= 7357 m/s
r0
The two satellites collide, stick together, and move with velocity v1. The equation for momentum conservation for
the perfectly inelastic collision gives
(400 kg + 100 kg)v1 = (400 kg)(7357 m/s) − (100 kg)(7357 m/s)
v1 = 4414 m/s
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Newton’s Theory of Gravity
13-33
The new satellite’s radius immediately after the collision is still r1 = r0 = 7.37 × 106 m. Now it is moving in an elliptical orbit. We need to determine if the minimum distance r2 is larger or smaller than the earth’s radius
Re = 6.37 × 106 m.
The combined satellites will continue moving in an elliptical orbit. The momentum of the combined satellite is
L = mrv sin β (see Equation 13.26) and is conserved in a trajectory of any shape. The angle β is 90° when
v1 = 4414 m/s and when the satellite is at its closest approach to the earth. From conservation of angular momentum,
we have
r1v1 = r2v2 = (4414 m/s) = 3.253 × 1010 m 2 /s
r2 =
3.253 × 1010 m 2 /s
v2
Using the conservation of energy equation at positions 1 and 2,
1
GM e (500 kg) 1
GM e (500 kg)
(500 kg)(4414 m/s) 2 −
= (500 kg)v22 −
6
2
r2
7.37 × 10 m 2
Using the above expression for r2 , we can simplify the energy equation to
v22 − (2.452 × 104 m/s)v2 + (8.876 × 107 m 2 /s 2 ) = 0 m 2 /s 2
v2 = 20,107 m/s and 4414 m/s
A velocity of 20,107 m/s for v2 yields
r2 =
3.253 × 1010 m 2 /s
= 1.62 × 106 m
20,107 m/s
Since r2 < Re = 6.37 × 106 m, the combined mass of the two satellites will crash into the earth.
13.66. Model: Planet Physics is a spherical mass. The cruise ship is in a circular orbit.
Solve: (a) At the surface, the free-fall acceleration is g = GM/R 2 . From kinematics,
y f = yi + v0Δt − 2 g (Δt ) 2
⇒ 0 m = 0 m + (11 m/s)(2.5 s) − 2 g (2.5 s) 2
g = 2.20 m/s 2
The period of the cruise ship’s orbit is 230 × 60 = 13,800 s. For the circular orbit of the cruise ship,
⎛ 4π 2 ⎞
T2 =⎜
(2 R )3
⎜ GM ⎟⎟
⎝
⎠
R=
⇒
⎛ R2
= R⎜
⎜ GM
32π
⎝
T2
(2.20 m/s 2 )(13,800 s) 2
32π 2
2
⎞
⎛1⎞
⎟⎟ = R ⎜ ⎟
⎝g⎠
⎠
= 1.327 × 106 m.
The mass is thus M = ( R 2 /G ) g = 5.8 × 1022 kg.
(b) From part (a), R = 1.3 × 106 m.
13.67. Model: The moon is a spherical mass. The moon lander is originally in a circular orbit.
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13-34
Chapter 13
Visualize:
Solve: Energy and momentum are conserved between points 1 and 2 in the elliptical orbit. Also, at both points
GM m
β = 90°, so L = mrv sin β = mrv. Let h = 1000 km. The original speed of the lander is v0 =
= 1338 m/s.
Rm + h
Conservation of angular momentum requires mr1v1 = mr2v2
⇒
v2 r1 Rm + h
= =
. The energy conservation equav1 r2
Rm
tion gives
1
M m 1
M m
mv12 − G m = mv22 − G m
2
Rm + h 2
Rm
⎛R +h⎞
Substituting v2 = ⎜ m
⎟ v1,
⎝ Rm ⎠
2
1 2
Mm
1 ⎛ R +h⎞
Mm
= v12 ⎜ m
v1 − G
⎟ −G
2
Rm + h 2 ⎝ Rm ⎠
Rm
⎡ ⎛ R + h ⎞2 ⎤
⎛ 1
1 ⎞
v12 ⎢1 − ⎜ m
−
⎟ ⎥ = 2GM m ⎜
⎟
⎢ ⎝ Rm ⎠ ⎥
⎝ Rm + h Rm ⎠
⎣
⎦
v12 =
2GM m h
1
2GM m ⎛ Rm ⎞
6
2 2
=
⎜
⎟ = 1.392 × 10 m /s
Rm ( Rm + h) [( Rm + h)/Rm ]2 − 1 ( Rm + h) ⎝ 2 Rm + h ⎠
v1 = 1180 m/s
The fractional change in speed required to just graze the moon at point 2 is
1338 m/s − 1180 m/s
= 11.8%
1338 m/s
Assess: A reduction in speed by almost 12% is reasonable.
13.68. Model: Model the earth as a spherical mass and the satellite as a point mass. This is an isolated system, so
mechanical energy is conserved. Also, the angular momentum of the satellite is conserved.
Visualize: Please refer to Figure CP13.68.
Solve: (a) Angular momentum is L = mrv sin β . The angle β = 90° at points 1 and 2, so conservation of angular
momentum requires
mr1v1′ = mr2v2′
⎛r ⎞
⇒ v1′ = ⎜ 2 ⎟ v2′
⎝ r1 ⎠
The energy conservation equation is
1
GMm 1
GMm
m(v2′ )2 −
= m(v1′ ) 2 −
2
r2
2
r1
⎛1 1⎞
⇒ (v2′ ) 2 − (v1′ ) 2 = 2GM ⎜ − ⎟
⎝ r2 r1 ⎠
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Newton’s Theory of Gravity
13-35
Using the angular momentum result for v1′ gives
2
⎡ r2 − r2 ⎤
⎛r ⎞
⎛r −r ⎞
⎛r −r ⎞
(v2′ ) 2 − (v2′ ) 2 ⎜ 2 ⎟ = 2GM ⎜ 1 2 ⎟ ⇒ (v2′ ) 2 ⎢ 1 2 2 ⎥ = 2GM ⎜ 1 2 ⎟
⎝ r1 ⎠
⎝ r1r2 ⎠
⎝ r1r2 ⎠
⎣⎢ r1 ⎦⎥
v2′ =
2GM (r1/r2 )
r1 + r2
⎛r ⎞
2GM (r1/r2 )
and v1′ = ⎜ 2 ⎟ v2′ =
r1 + r2
⎝ r1 ⎠
(b) For the circular orbit,
v1 =
GM
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
= 7730 m/s
r1
(6.37 × 106 m + 3 × 105 m)
For the elliptical orbit,
r1 = Re + 300 km = 6.37 × 106 m + 3 × 105 m = 6.67 × 106 m
r2 = Re + 35,900 km = 6.37 × 106 m + 3.59 × 107 m = 4.23 × 107 m
v1′ =
2GM (r1/r2 )
r1 + r2
⇒ v1′ = 10,160 m/s
1
1
(c) From the work-kinetic energy theorem, W = ΔK = mv1′2 − mv12 = 2.17 × 1010 J
2
2
(d) v′2 =
2GM ( r1/r2 )
r1 + r2
Using the same values of r1 and r2 as in (b), v2′ = 1600 m/s. For the circular orbit,
v2 =
GM
= 3070 m/s
r2
1
1
(e) W = mv22 − mv2′ 2 = 3.43 × 109 J
2
2
(f) The total work done is 2.52 × 1010 J. This is the same as in Example 13.6, but here we’ve learned how the work
has to be divided between the two burns.
13.69. Model: The rod is thin and uniform.
Visualize:
Solve: (a) The rod is not spherical so must be divided into thin sections each dr wide and having mass dm. Since the
rod is uniform,
dm dr
=
M
L
⇒ dm =
M
dr
L
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13-36
Chapter 13
The width of the rod is small enough so that all of dm is distance r away from m. The gravitational potential energy
of dm and m is
dU = −G
m dm
m( M/L) dr
= −G
r
r
The total potential energy of the rod and mass m is found by adding the contributions dU from every point along the
rod in an integral:
U = ∫ dU = −
GMm
L
x+
∫
x−
L
2
dr
GMm ⎛ x + L/2 ⎞
ln ⎜
=−
⎟
r
L
⎝ x − L/2 ⎠
L
2
Note r is increasing with the limits chosen as they are.
(b) The force on m when at x is
dU GMm d ⎡ ⎛
L⎞
L ⎞ ⎤ GMm ⎛ 1
1 ⎞
⎛
=
ln ⎜ x + ⎟ − ln ⎜ x − ⎟ ⎥ =
−
⎜
⎟
dx
L dx ⎣⎢ ⎝
2⎠
2
L
x
L
/2
x
L/2 ⎠
+
−
⎝
⎠⎦
⎝
4
L
⎛
⎞
, x≥
= −GMm ⎜ 2
2⎟
2
⎝ 4x − L ⎠
F =−
Assess: The direction of the force is toward the − x direction, as expected. The force magnitude approaches ∞ as
the mass m approaches the end of the rod, but goes to zero like x −2 as x gets large. This is expected since from far
away the rod looks like a point mass.
13.70. Model: The ring is uniform and is so thin that every point on it may be considered to be the same distance r
from m.
Visualize:
Solve: (a) We must determine the gravitational potential (dU) between m and an arbitrary part of the ring dm, then
add using an integral all the contributions to U. Since the ring is uniform,
dm
dl
=
M 2π r
⇒ dm =
M
dl
2π r
The distance from m to dm is r = x 2 + R 2 . The gravitational potential between m and dm is
dU = −G
m dm
mM
= −G
r
2π r
dl
2
x + R2
The total gravitational potential is
U = ∫ dU = −
GmM
2π r x 2 + R 2
∫
dl
ring
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Newton’s Theory of Gravity
13-37
Note that x and R do not change for any location of dm. The integral is just the length of the ring.
∫
dl = 2π R
ring
Thus
U =−
GmM
x2 + R2
(b) The force on m when at x is
dU
d
⎛ 1⎞
−1
−3
= GmM [( x 2 + R 2 ) 2 ] = GmM ⎜ − ⎟ ( x 2 + R 2 ) 2 (2 x)
dx
dx
⎝ 2⎠
x
= −GmM
3
( x2 + R2 ) 2
Fx = −
Thus the magnitude of the force is
F = GmM
x
2
3
( x + R2 ) 2
.
Assess: The force is zero at the center of the ring. Elsewhere its direction is toward the origin. As x gets large, the
force decreases like x22 This is expected since from far away the ring looks like a point mass.
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14
OSCILLATIONS
Conceptual Questions
14.1. The period of a block oscillating on a spring is T = 2π m/k . We are told that T1 = 2.0 s.
(a) In this case the mass is doubled: m2 = 2m1.
T2 2π m2 /k
m2
2m1
=
=
=
= 2
T1 2π m1/k
m1
m1
So T2 = 2T1 = 2(2.0 s) = 2.8 s.
(b) In this case the spring constant is doubled: k2 = 2k1.
T2 2π m/k2
k
k1
1
=
= 1 =
=
T1 2π m/k1
k2
2k1
2
So T2 = T1/ 2 = (2.0 s)/ 2 = 1.4 s.
(c) The formula for the period does not contain the amplitude; that is, the period is independent of the amplitude.
Changing (in particular, doubling) the amplitude does not affect the period, so the new period is still 2.0 s.
It is equally important to understand what doesn’t appear in a formula. It is quite startling, really, the first time you
realize it, that the amplitude doesn’t affect the period. But this is crucial to the idea of simple harmonic motion. Of
course, if the spring is stretched too far, out of its linear region, then the amplitude would matter.
14.2. The period of a simple pendulum is T = 2π L/g . We are told that T1 = 2.0 s.
(a) In this case the mass is doubled: m2 = 2m1. However, the mass does not appear in the formula for the period of a
pendulum; that is, the period does not depend on the mass. Therefore the period is still 2.0 s.
(b) In this case the length is doubled: L2 = 2 L1.
T2 2π L2 /g
=
=
T1 2π L1/g
L2
2 L1
=
= 2
L1
L1
So T2 = 2T1 = 2(2.0 s) = 2.8 s.
(c) The formula for the period of a simple small-angle pendulum does not contain the amplitude; that is, the period is
independent of the amplitude. Changing (in particular, doubling) the amplitude, as long as it is still small, does not
affect the period, so the new period is still 2.0 s.
It is equally important to understand what doesn’t appear in a formula. It is quite startling, really, the first time you
realize it, that the amplitude (θ max ) doesn’t affect the period. But this is crucial to the idea of simple harmonic
motion. Of course, if the pendulum is swung too far, out of its linear region, then the amplitude would matter. The
amplitude does appear in the formula for a pendulum not restricted to small angles because the small-angle
approximation is not valid; but then the motion is not simple harmonic motion.
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14-1
14-2
Chapter 14
14.3. (a) The amplitude is the maximum displacement obtained. From the graph, A = 10 cm.
(b) From the graph, the period T = 2.0 s. The angular frequency is thus ω =
2π
= 3.1 rad/s.
T
(c) The phase constant specifies at what point the cosine function starts. From the graph, the starting point looks like
1
1
π
A. So at t = 0, A cos φ = A means φ0 = ± ( ±60°). Since the oscillator is moving to the left at t = 0, it is in the
2
2
3
π
upper half of the circular-motion diagram, and we choose φ 0 = + .
3
14.4. (a) A position-vs-time graph plots x(t ) = A cos( wt + φ 0 ). The graph of x(t) starts at
1
A and is increasing. So
2
1
π
π
A = A cos φ0 ⇒ φ0 = ± . We choose φ0 = − since the particle is moving to the right, indicating that it is
2
3
3
in the bottom half of the circular-motion diagram.
at t = 0,
(b) The phase at each point can be determined in the same manner as for part (a). For points 1 and 3, the amplitude is
1
π
again A. At point 1, the particle is moving to the right so φ 1 = − . At point 3, the particle is moving left, so
2
3
π
φ 3 = + . At point 2, the amplitude is A, so cos φ2 = 1 ⇒ φ 2 = 0.
3
14.5. (a) A velocity-vs-time graph is a plot of v(t ) = − Aω sin (ωt + φ 0 ). At t = 0, the amplitude is
1
1
vmax = Aω
2
2
π
⎛ 1 ⎞ 7π
or − . Since v(t = 0 s) is increasing, we need − sin (ωt + φ0 ) increasing at
= − Aω sin φ 0 , so φ 0 = sin −1 ⎜ − ⎟ =
2
6
6
⎝
⎠
7π
.
t = 0, so we choose φ 0 =
6
1
5π
⎛1⎞ π
(b) The phase is φ = wt + φ 0 . At points 1 and 3, v(t ) = − vmax , so φ = sin21 ⎜ ⎟ = or
. At point 1, we need
2
6
⎝ 2⎠ 6
− sin φ 1, increasing, so φ 1 =
5π
π
. At point 3, − sin φ 3 is decreasing, so φ 3 = . At point 2, the amplitude is vmax , so
6
6
π
− sin φ 2 = 1, thus φ 2 = − .
2
⎛1
⎞
14.6. Energy is transferred back and forth between all potential energy at the extremes ⎜ kA2 ⎟ and all kinetic
⎝2
⎠
⎛1
⎞
energy at the equilibrium point(s) ⎜ mvmax 2 ⎟ . The equation does not say that the particle ever has amplitude A and
⎝2
⎠
speed vmax at the same time. The equation relates expressions for the energy at two different times.
1
2
(b) The turning points occur where the potential energy is equal to the total energy. These points are at x = 14 cm
and 26 cm.
14.7. (a) The equilibrium length is 20 cm since that is where the potential energy U = k ( x − xe ) 2 = 0.
(c) The maximum kinetic energy of 7 J is obtained when the potential energy is minimal, at the equilibrium point.
(d) The total energy will be around 14 J. If a horizontal line is drawn at that energy, it intersects the potential energy
curve at the turning points x = 12 cm and 28 cm.
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Oscillations
14-3
1
2
14.8. One expression for the total energy is E = kA2 , which occurs at the turning points when v(t ) = 0. If the total
energy is doubled,
1
1
2 E = k ( A′) 2 = k ( 2 A) 2
2
2
Thus the new amplitude is A′ = 2 A = 2(20 cm) = 28 cm.
1
2
2
14.9. One expression for the total energy is E = mvmax
, which occurs at the equilibrium point when U = 0. If the
total energy is doubled,
1
1
′ ) 2 = m( 2vmax ) 2
2 E = m(vmax
2
2
′ = 2vmax = 2(20 cm/s) = 28 cm/s
Thus the new maximum speed is vmax
14.10. The time constant τ =
m
decreases as b increases, meaning energy is removed from the oscillator more
b
quickly since E (t ) = E0e −t/τ .
(a) The medium is more resistive.
(b) The oscillations damp out more quickly.
(c) The time constant is decreased.
14.11. (a) T, the period, is the time for each cycle of the motion, the time required for the motion to repeat itself.
τ , the damping time constant, is the time required for the amplitude of a damped oscillator to decrease to e−1 ≈ 37%
of its original value.
(b) τ , the damping time constant, is the time required for the amplitude of a damped oscillator to decrease to
e −1 ≈ 37% of its original value, while t1/2 , the half-life, is the time required for the amplitude of a damped oscillator
to decrease to 50% of its original value.
14.12. Natural frequency is the frequency that an oscillator will oscillate at on its own. You may drive an oscillator
at a frequency other than its natural frequency. For example, if you are pushing a child in a swing, you build up the
amplitude of the oscillation by driving the oscillator at its natural frequency. You can achieve resonance by driving
an oscillator at its natural frequency.
Exercises and Problems
Section 14.1 Simple Harmonic Motion
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence
T=
1
1
=
= 2.27 × 10−3 s = 2.27 ms
f 440 Hz
14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion.
Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark.
33 s
(a) T =
= 3.3 s/oscillation = 3.3 s
10 oscillations
1
1
(b) f = =
= 0.303 Hz ≈ 0.30 Hz
T 3.3 s
(c) ω = 2π f = 2π ( 0.303 Hz ) = 1.90 rad/s
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14-4
Chapter 14
(d) The oscillation from one side to the other is equal to 60 cm − 10 cm = 50 cm = 0.50 m. Thus, the amplitude is
A = 12 (0.50 m) = 0.25 m.
(e) The maximum speed is
⎛ 2π ⎞
vmax = ω A = ⎜
⎟ A = (1.90 rad/s)(0.25 m) = 0.48 m/s
⎝ T ⎠
14.3. Model: The air-track glider attached to a spring is in simple harmonic motion.
Visualize: The position of the glider can be represented as x (t ) = A cos ωt .
Solve: The glider is pulled to the right and released from rest at t = 0 s. It then oscillates with a period T = 2.0 s
and a maximum speed vmax = 40 cm/s = 0.40 m/s.
2π
2π
v
0.40 m/s
= 0.127 m = 12.7 cm = 13 cm
=
= π rad/s ⇒ A = max =
ω
π rad/s
T
2.0 s
(b) The glider’s position at t = 0.25 s is
x0.25 s = (0.127 m)cos[(π rad/s)(0.25 s)] = 0.090 m = 9.0 cm
(a) vmax = ω A and ω =
Section 14.2 Simple Harmonic Motion and Circular Motion
14.4. Model: The oscillation is the result of simple harmonic motion.
Solve: (a) The amplitude A = 20 cm.
(b) The period T = 4.0 s, thus
1
1
=
= 0.25 Hz
T 4.0 s
(c) The position of an object undergoing simple harmonic motion is x(t ) = A cos(ωt + φ 0). At t = 0 s, x0 = 10 cm. Thus,
f =
π
⎛ 10 cm ⎞
−1 ⎛ 1 ⎞
10 cm = (20 cm)cos φ 0 ⇒ φ0 = cos −1 ⎜
⎟ = cos ⎜ ⎟ = ± rad = ±60°.
3
⎝ 20 cm ⎠
⎝2⎠
Because the object is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram and thus
must have a phase constant between π and 2π radians. Therefore, φ 0 = −
π
3
rad = −60°.
14.5. Model: The oscillation is the result of simple harmonic motion.
Solve: (a) The amplitude A = 10 cm.
(b) The time to complete one cycle is the period, hence T = 2.0 s and
1
1
f = =
= 0.50 Hz
T 2. 0 s
(c) The position of an object undergoing simple harmonic motion is x(t ) = A cos(ωt + φ 0). At t = 0 s, x0 = −5 cm, thus
−5 cm = (10 cm)cos[ω (0 s) + φ 0 ]
⇒ cos φ 0 =
−5 cm
1
2π
⎛ 1⎞
= − ⇒ φ 0 = cos −1 ⎜ − ⎟ = ±
rad or ± 120°
10 cm
2
2
3
⎝
⎠
Since the oscillation is originally moving to the left, φ 0 = +120°.
14.6. Visualize: The phase constant 23 π has a plus sign, which implies that the object undergoing simple harmonic
motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left.
Solve: The position of the object is
x(t ) = A cos(ωt + φ 0 ) = A cos(2π ft + φ 0 ) = (4.0 cm)cos[(4π rad/s)t + 23 π rad]
The amplitude is A = 4 cm and the period is T = 1/f = 0.50 s. A phase constant φ 0 = 2π /3 rad = 120° (second
quadrant) means that x starts at − 12 A and is moving to the left (getting more negative).
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Oscillations
14-5
Assess: We can see from the graph that the object starts out moving to the left.
π
implies that the object that undergoes simple harmonic motion is in the
2
lower half of the circular motion diagram. That is, the object is moving to the right.
Solve: The position of the object is given by the equation
14.7. Visualize: A phase constant of −
⎡⎛ π
⎤
⎞ π
x(t ) = A cos(ωt + φ 0) = A cos(2π ft + φ 0) = (8.0 cm)cos ⎢⎜ rad/s ⎟ t − rad ⎥
⎠ 2
⎣⎝ 2
⎦
The amplitude is A = 8.0 cm and the period is T = 1/f = 4.0 s. With φ 0 = −π /2 rad, x starts at 0 cm and is moving
to the right (getting more positive).
Assess: As we see from the graph, the object starts out moving to the right.
14.8. Solve: The position of the object is given by the equation
x(t ) = A cos(ωt + φ 0) = A cos(2π ft + φ 0)
We can find the phase constant φ 0 from the initial condition:
0 cm = (4.0 cm)cos φ 0 ⇒ cos φ 0 = 0 ⇒ φ 0 = cos −1 (0) = ± 12 π rad
Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence,
φ 0 = − 12 π rad. The final result, with f = 4.0 Hz, is
x(t ) = (4.0 cm)cos[(8.0π rad/s)t − 12 π rad]
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14-6
Chapter 14
14.9. Solve: The position of the object is given by the equation
x(t ) = A cos(ωt + φ 0)
The amplitude is A = 8.0 cm. The angular frequency ω = 2π f = 2π ( 0.50 Hz ) = π rad/s. Since at t = 0 it has its most
negative position, it must be about to move to the right, so φ 0 = −π . Thus
x (t ) = (8.0 cm ) cos[(π rad/s)t − π rad]
14.10. Model: The air-track glider is in simple harmonic motion.
Solve: (a) We can find the phase constant from the initial conditions for position and velocity:
x0 = A cos φ 0
v0 x = −ω A sin φ 0
Dividing the second by the first, we see that
sin φ 0
v
= tan φ 0 = − 0 x
cos φ 0
ω x0
The glider starts to the left ( x0 = −5.00 cm) and is moving to the right (v0 x = +36.3 cm/s). With a period of 1.5 s =
3
2
s,
the angular frequency is ω = 2π /T = 43 π rad/s. Thus
⎛
⎞ 1
36.3 cm/s
2
⎟ = 3 π rad (60°) or − 3 π rad ( − 120°)
⎝ (4π /3 rad/s)(−5.00 cm) ⎠
φ 0 = tan −1 ⎜ −
The tangent function repeats every 180°, so there are always two possible values when evaluating the arctan function.
We can distinguish between them because an object with a negative position but moving to the right is in the third
quadrant of the corresponding circular motion. Thus φ 0 = − 23 π rad, or −120°.
(b) At time t, the phase is φ = ω t + φ 0 = ( 43 π rad/s)t − 23 π rad. This gives φ = − 23 π rad, 0 rad, 23 π rad, and 43 π rad at,
respectively, t = 0 s, 0.5 s, 1.0 s, and 1.5 s. This is one period of the motion.
Section 14.3 Energy in Simple Harmonic Motion
Section 14.4 The Dynamics of Simple Harmonic Motion
14.11. Model: The block attached to the spring is in simple harmonic motion.
Solve: The period of an object attached to a spring is
T = 2π
m
= T0 = 2.0 s
k
where m is the mass and k is the spring constant.
(a) For mass = 2m,
T = 2π
(b) For mass
2m
= ( 2)T0 = 2.8 s
k
1 m,
2
T = 2π
1m
2
k
= T0 / 2 = 1.41 s
(c) The period is independent of amplitude. Thus T = T0 = 2.0 s
(d) For a spring constant = 2k ,
T = 2π
m
= T0 / 2 = 1.41 s
2k
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Oscillations
14-7
14.12. Model: The air-track glider attached to a spring is in simple harmonic motion.
Solve: Experimentally, the period is T = (12.0 s)/(10 oscillations) = 1.20 s. Using the formula for the period,
T = 2π
2
2
m
⎛ 2π ⎞
⎛ 2π ⎞
⇒k =⎜
⎟ m=⎜
⎟ (0.200 kg) = 5.48 N/m
k
T
⎝
⎠
⎝ 1.20 s ⎠
14.13. Model: The mass attached to the spring oscillates in simple harmonic motion.
Solve: (a) The period T = 1/f = 1/2.0 Hz = 0.50 s.
(b) The angular frequency ω = 2π f = 2π ( 2.0 Hz ) = 4π rad/s.
(c) Using energy conservation
1 kA2
2
= 12 kx02 + 12 mv02x
Using x0 = 5.0 cm, v0 x = −30 cm/s and k = mω 2 = (0.200 kg)(4π rad/s)2 , we get A = 5.54 cm.
(d) To calculate the phase constant φ 0,
A cos φ 0 = x0 = 5.0 cm
⎛ 5.0 cm ⎞
⇒ φ 0 = cos −1 ⎜
⎟ = 0.45 rad
⎝ 5.54 cm ⎠
(e) The maximum speed is vmax = ω A = (4π rad/s)(5.54 cm) = 70 cm/s.
(f) The maximum acceleration is
amax = ω 2 A = ω (ω A) = (4π rad/s)(70 cm/s) = 8.8 m/s 2
2
(g) The total energy is E = 12 mvmax
= 12 (0.200 kg)(0.70 m/s)2 = 0.049 J.
(h) The position at t = 0.40 s is
x0.4 s = (5.54 cm)cos[(4π rad/s)(0.40 s) + 0.45 rad] = +3.8 cm
14.14. Model: The oscillating mass is in simple harmonic motion.
Solve: (a) The amplitude A = 2.0 cm.
(b) The period is calculated as follows:
2π
2π
= 10 rad/s ⇒ T =
= 0.63 s
10 rad/s
T
(c) The spring constant is calculated as follows:
k
ω=
⇒ k = mω 2 = (0.050 kg)(10 rad/s)2 = 5.0 N/m
m
ω=
(d) The phase constant φ 0 = − 14 π rad.
(e) The initial conditions are obtained from the equations
x(t ) = (2.0 cm)cos(10t − 14 π ) and vx (t ) = −(20.0 cm/s)sin (10t − 14 π )
At t = 0 s, these equations become
x0 = (2.0 cm)cos(− 14 π ) = 1.41 cm and v0 x = −(20 cm/s)sin ( − 14 π ) = 14.1cm/s
In other words, the mass is at +1.41 cm and moving to the right with a velocity of 14.1 cm/s.
(f) The maximum speed is vmax = Aω = (2.0 cm)(10 rad/s) = 20 cm/s.
(g) The total energy E = 12 kA2 = 12 (5.0 N/m)(0.020 m)2 = 1.00 × 10−3 J.
(h) At t = 0.41 s, the velocity is
v0 x = −(20 cm/s)sin[(10 rad/s)(0.40 s) − 14 π ] = 1.46 cm/s
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14-8
Chapter 14
14.15. Model: The block attached to the spring is in simple harmonic motion.
Visualize:
Solve: (a) The conservation of mechanical energy equation K f + U sf = Ki + U si is
1 mv 2
1
2
+ 12 k (Δx) 2 = 12 mv02 + 0 J ⇒ 0 J + 12 kA2 = 12 mv02 + 0 J
m
1.0 kg
v0 =
(0.40 m/s) = 0.10 m = 10 cm
k
16 N/m
(b) We have to find the velocity at a point where x = A/2. The conservation of mechanical energy equation
K 2 + U s2 = Ki + U si is
⇒ A=
2
1 2 1 ⎛ A⎞
1
1
1
1⎛1
1⎛1
⎞ 1
⎞
mv2 + k ⎜ ⎟ = mv02 + 0 J ⇒ mv22 = mv02 − ⎜ kA2 ⎟ = mv02 − ⎜ mv02 ⎟ =
2
2 ⎝2⎠
2
2
2
4⎝ 2
4⎝ 2
⎠ 2
⎠
⇒ v2 =
3⎛ 1 2 ⎞
⎜ mv0 ⎟
4⎝ 2
⎠
3
3
v0 =
(0.40 m/s) = 0.346 m/s
4
4
The velocity is 35 cm/s.
Section 14.5 Vertical Oscillations
14.16. Model: The vertical oscillations constitute simple harmonic motion.
Visualize:
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Oscillations
14-9
Solve: (a) At equilibrium, Newton’s first law applied to the physics book is
( Fsp ) y − mg = 0 N ⇒ − k Δy − mg = 0 N
⇒ k = − mg/Δy = −(0.500 kg)(9.8 m/s 2 )/(−0.20 m) = 24.5 N/m ≈ 25 N/m
(b) To calculate the period:
k
24.5 N/m
2π
2π rad
ω=
=
= 7.0 rad/s and T =
=
= 0.90 s
ω
m
0.500 kg
7.0 rad/s
(c) The maximum speed is
vmax = Aω = ( 0.10 m )( 7.0 rad/s ) = 0.70 m/s
Maximum speed occurs as the book passes through the equilibrium position.
14.17. Model: The vertical oscillations constitute simple harmonic motion.
Visualize:
Solve: The period and angular frequency are
20 s
2π
2π
T=
= 0.6667 s and ω =
=
= 9.425 rad/s
30 oscillations
T
0.6667 s
(a) The mass can be found as follows:
k
k
15 N/m
⇒m= 2 =
= 0.169 kg ≈ 0.17 kg
ω=
m
ω
(9.425 rad/s) 2
(b) The maximum speed vmax = ω A = (9.425 rad/s)(0.060 m) = 0.57 m/s.
14.18. Model: The vertical oscillations constitute simple harmonic motion.
Solve: To find the oscillation frequency using ω = 2π f = k/m , we first need to find the spring constant k. In equilibrium,
the weight mg of the block and the spring force k ΔL are equal and opposite. That is, mg = k ΔL ⇒ k = mg/ΔL. The
frequency of oscillation f is thus given as
f =
1
2π
k
1
=
m 2π
mg/ΔL
1
=
2π
m
g
1
=
ΔL 2π
9.8 m/s 2
= 3.5 Hz
0.020 m
Section 14.6 The Pendulum
14.19. Model: Assume a small angle of oscillation so there is simple harmonic motion.
Solve: The period of the pendulum is
T0 = 2π
L0
= 4.0 s
g
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14-10
Chapter 14
(a) The period is independent of the mass and depends only on the length. Thus T = T0 = 4.0 s.
(b) For a new length L = 2 L0 ,
T = 2π
2 L0
= 2T0 = 5.7 s
g
(c) For a new length L = L0 /2,
L0 /2
1
T0 = 2.8 s
=
g
2
(d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T = 4.0 s.
T = 2π
14.20. Model: Assume the small-angle approximation so there is simple harmonic motion.
Solve: The period is T = 12 s/10 oscillations = 1.20 s and is given by the formula
T = 2π
2
2
L
⎛ T ⎞
⎛ 1.20 s ⎞
2
⇒L=⎜
⎟ g =⎜
⎟ (9.8 m/s ) = 36 cm
2
2
g
π
π
⎝
⎠
⎝
⎠
14.21. Model: Assume a small angle of oscillation so there is simple harmonic motion.
Solve: (a) On the earth the period is
Tearth = 2π
L
1.0 m
= 2π
= 2.0 s
g
9.80 m/s 2
(b) On Venus the acceleration due to gravity is
g Venus =
GM Venus
2
RVenus
=
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(4.88 × 1024 kg)
⇒ TVenus = 2π
6
(6.06 × 10 m)
2
= 8.86 m/s 2
L
1.0 m
= 2π
= 2.1 s
g Venus
8.86 m/s 2
14.22. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoes simple
harmonic motion.
Solve: Using the formula g = GM/R 2 , the periods of the pendulums on the moon and on the earth are
Tearth = 2π
L
L
R2
L
R2
= 2π earth earth and Tmoon = 2π moon moon
g
GM earth
GM moon
Because Tearth = Tmoon ,
2
⎛M
⎞⎛ R
⎞
L
R2
L
R2
2π earth earth = 2π moon moon ⇒ Lmoon = ⎜ moon ⎟⎜ earth ⎟ Learth
GM earth
GM moon
⎝ M earth ⎠⎝ Rmoon ⎠
2
⎛ 7.36 × 1022 kg ⎞⎛ 6.37 × 106 m ⎞
(2.0 m) = 33 cm
=⎜
⎜ 5.98 × 1024 kg ⎟⎜
⎟⎜ 1.74 × 106 m ⎟⎟
⎝
⎠⎝
⎠
14.23. Model: Assume a small angle of oscillation so that the pendulum has simple harmonic motion.
Solve: The time periods of the pendulums on the earth and on Mars are
Tearth = 2π
L
L
and TMars = 2π
g earth
g Mars
Dividing these two equations,
Tearth
=
TMars
2
2
⎛T
⎞
g Mars
⎛ 1.50 s ⎞
2
⇒ g Mars = gearth ⎜ earth ⎟ = (9.8 m/s 2 ) ⎜
⎟ = 3.67 m/s
gearth
T
.
2
45
s
⎝
⎠
⎝ Mars ⎠
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Oscillations
14-11
14.24. Model: This is not a simple pendulum, but is a physical pendulum. Model the bar as thin enough to use
I = 13 ML2 where L is the full length of the rod.
Visualize: For a small-angle physical pendulum T = 2π
I
Mgl
where l is the distance from the pivot to the center of
mass. Note that l = L/2. We seek L.
Solve: Substitute in for I and then solve for L.
T = 2π
1 ML2
I
= 2π 3
= 2π
Mgl
MgL/2
2
2 L2
3
gL
= 2π
2L
3g
2
⎛ T ⎞ 3g 3 g 2 3 (9.8 m/s )
L=⎜
T =
(1.2 s) 2 = 0.54 m = 54 cm
=
⎟
8
π2
⎝ 2π ⎠ 2 8 π 2
Assess: It seems reasonable that a uniform bar 54 cm long would have a period of 1.2 s.
Section 14.7 Damped Oscillations
Section 14.8 Driven Oscillations and Resonance
14.25. Model: The spider is in simple harmonic motion.
Solve: Your tapping is a driving frequency. Largest amplitude at f ext = 1.0 Hz means that this is the resonance
frequency, so f 0 = f ext = 1.0 Hz. That is, the spider’s natural frequency of oscillation f 0 is 1.0 Hz and ω 0 = 2π f 0 =
2π rad/s. We have
ω 0=
k
⇒ k = mω02 = (0.0020 kg)(2π rad/s) 2 = 0.079 N/m
m
14.26. Model: The motion is a damped oscillation.
Solve: The amplitude of the oscillation at time t is given by Equation 14.58: A(t ) = A0e−t/2τ , where τ = m/b is the
time constant. Using x = 0.368 A and t = 10.0 s, we get
−10 s
10.0 s
0.368 A = Ae−10.0 s/2τ ⇒ ln(0.368) =
⇒τ = −
= 5.00 s
2τ
2ln(0.368)
14.27. Model: The motion is a damped oscillation.
Solve: The position of a damped oscillator is x(t ) = Ae− (t/2τ ) cos(ωt + φ 0). The frequency is 1.0 Hz and the damping
time constant τ is 4.0 s. Let us assume φ 0 = 0 rad and A = 1 with arbitrary units. Thus,
x (t ) = e − t/(8.0 s) cos[2π (1.0 Hz )t ] ⇒ x(t ) = e−0.125 t cos(2π t )
where t is in s. Values of x(t) at selected values of t are displayed in the following table:
t(s)
0
0.25
0.50
0.75
1.00
1.25
1.50
1.75
x(t)
1
0
−0.939
0
0.882
0
−0.829
0
t(s)
2.00
2.50
3.00
3.50
4.00
4.50
5.00
5.50
x(t)
0.779
−0.732
0.687
−0.646
0.607
−0.570
0.535
−0.503
t(s)
6.00
6.50
7.00
7.50
8.00
8.50
9.00
9.50
10.00
x(t)
0.472
−0.444
0.417
−0.392
0.368
−0.346
0.325
−0.305
0.286
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14-12
Chapter 14
14.28. Model: The pendulum is a damped oscillator.
Solve: The period of the pendulum and the number of oscillations in 4 hours are calculated as follows:
L
15.0 m
4(3600 s)
= 2π
= 7.773 s ⇒ N osc =
T = 2π
= 1850
2
7.773 s
g
9.8 m/s
The amplitude of the pendulum as a function of time is A(t ) = Ae −bt/ 2m . The exponent of this expression can be
calculated to be
−
bt
(0.010 kg/s)(4 × 3600 s)
=−
= −0.6545
2m
2(110 kg)
We have A(t ) = (1.50 m )e −0.6545 = 0.78 m.
14.29. Model: Assume the eye is a simple driven oscillator.
Visualize: Given the mass and the resonant frequency, we can determine the effective spring constant using the
relationship ω = 2π f = k/m .
Solve: Solving the above expression for the spring constant, obtain
k = (2π f ) 2 m = [2π ( 29 Hz )]2 (7.5 × 10−3 kg) = 249 N/m ≈ 250 N/m
Assess: As spring constants go, this is a fairly large value, however the musculature holding the eyeball in the socket
is strong and hence will have a large effective spring constant.
14.30. Solve: The position and the velocity of a particle in simple harmonic motion are
x (t ) = A cos(ωt + φ 0 ) and vx (t ) = − Aω sin (ωt + φ 0 ) = −vmax sin (ωt + φ 0 )
From the graph, T = 12 s and the angular frequency is
2π
2π π
=
= rad/s
ω=
T 12 s 6
(a) Because vmax = Aω = 60 cm/s, we have
A=
60 cm/s
(b) At t = 0 s,
ω
=
60 cm/s
π /6 rad/s
= 115 cm
v0 x = − Aω sin φ 0 = −30 cm/s ⇒ −( 60 cm/s ) sin φ 0 = −30 cm/s
⇒ φ 0 = sin −1(0.5 rad) = 16 π rad (30°) or 65 π rad (150°)
Because the velocity at t = 0 s is negative and the particle is slowing down, the particle is in the second quadrant of
the circular motion diagram. Thus φ 0 = 56 π rad.
(c) At t = 0 s, x0 = (115 cm)cos( 65 π rad) = −100 cm.
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Oscillations
14-13
14.31. Solve: The position and the velocity of a particle in simple harmonic motion are
x(t ) = A cos(ωt + φ 0) and vx (t ) = − Aω sin(ωt + φ 0) = −vmax sin (ωt + φ 0)
(a) At t = 0 s, the equation for x yields
(−5.0 cm) = (10.0 cm)cos(φ 0) ⇒ φ 0 = cos −1 (−0.5) = ± 23 π rad
Because the particle is moving to the left at t = 0 s, it is in the upper half of the circular motion diagram, and the
phase constant is between 0 and π radians. Thus, φ 0 = 23 π rad.
(b) The period is 4.0 s. At t = 0 s,
⎛ 2π
v0 x = − Aω sin φ 0 = −(10.0 cm) ⎜
⎝ T
⎞ ⎛ 2π
⎟ sin ⎜
⎠ ⎝ 3
⎞
⎟ = −13.6 cm/s
⎠
(c) The maximum speed is
⎛ 2π ⎞
vmax = ω A = ⎜
⎟ (10.0 cm) = 15.7 cm/s
⎝ 4.0 s ⎠
Assess: The negative velocity at t = 0 s is consistent with the position-vs-time graph and the positive sign of the
phase constant.
14.32. Model: The vertical mass/spring systems are in simple harmonic motion.
Solve: Spring/mass A undergoes three oscillations in 12 s, giving it a period TA = 4.0 s. Spring/mass B undergoes 2
oscillations in 12 s, giving it a period TB = 6.0 s. We have
TA = 2π
⎛ m ⎞⎛ k ⎞ 4.0 s 2
mA
mB
T
⇒ A = ⎜ A ⎟⎜ B ⎟ =
=
and TB = 2π
kA
kB
TB
⎝ mB ⎠⎝ kA ⎠ 6.0 s 3
If mA = mB , then
kB 4
k
9
= ⇒ A = = 2.25
kA 9
kB 4
14.33. Solve: The object’s position as a function of time is x(t ) = A cos(ωt + φ 0). Letting x = 0 m at t = 0 s, gives
0 = A cos φ 0 ⇒ φ 0 = ± 12 π
Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant
between −π and 0 radians. Thus, φ 0 = − 12 π and
x(t ) = A cos(ωt − 12 π ) ⇒ x(t ) = A sin ωt = (0.10 m)sin ( 12 π t )
where we have used A = 0.10 m and
ω=
Let us now find t where x = 0.060 m:
2π 2π rad π
=
= rad/s
T
4.0 s
2
2
⎛π ⎞
⎛ 0.060 m ⎞
0.060 m = (0.10 m)sin ⎜ t ⎟ ⇒ t = sin −1 ⎜
⎟ = 0.41 s
π
⎝2 ⎠
⎝ 0.10 m ⎠
Assess: The answer is reasonable because it is approximately 18 of the period.
14.34. Model: The block attached to the spring is in simple harmonic motion.
Visualize: The position and the velocity of the block are given by the equations
x(t ) = A cos(ωt + φ 0 ) and vx (t ) = − Aω sin (ωt + φ 0 )
Solve: (a) To graph x(t) we need to determine ω , φ 0 , and A. These quantities will be found by using the initial
(t = 0 s) conditions on x(t) and vx (t ). The period is
T = 2π
m
1.0 kg
= 2π
= 1.405 s ≈ 1.4 s
k
20 N/m
⇒ ω=
2π 2π rad
=
= 4.472 rad/s
T 1.405 s
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14-14
Chapter 14
At t = 0 s, x0 = A cos φ 0 and v0 x = − Aω sin φ 0 . Dividing these equations,
tan φ 0 = −
v0 x
ω x0
=−
( −1.0 m/s)
= 1.1181 ⇒ φ 0 = 0.841 rad
(4.472 rad/s)(0.20 m)
From the initial conditions,
2
2
⎛v ⎞
⎛ −1.0 m/s ⎞
A = x02 + ⎜ 0 x ⎟ = (0.20 m) 2 + ⎜
⎟ = 0.30 m
⎝ ω ⎠
⎝ 4.472 rad/s ⎠
14.35. Model: The astronaut attached to the spring is in simple harmonic motion.
Solve: (a) From the graph, T = 3.0 s, so we have
T = 2π
2
2
m
⎛ T ⎞
⎛ 3.0 s ⎞
⇒m=⎜
⎟ k =⎜
⎟ (240 N/m) = 55 kg
k
⎝ 2π ⎠
⎝ 2π ⎠
(b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, A = 12 (0.80 m) = 0.40 m, φ 0 = 0 rad, and
ω=
2π
2π
=
= 2.1 rad/s
T
3.0 s
The equation for the position of the astronaut is
x (t ) = A cos ωt + 1.0 m = (0.4 m) cos[(2.1 rad/s)t ] + 1.0 m
⇒ 1.2 m = (0.4 m)cos[(2.1 rad/s)t ] + 1.0 m ⇒ cos[(2.1 rad/s)t ] = 0.5 ⇒ t = 0.50 s
The equation for the velocity of the astronaut is
vx (t ) = − Aω sin (ωt )
⇒ v0.5 s = −(0.4 m)(2.1 rad/s)sin[(2.1 rad/s)(0.50 s)] = −0.73 m/s
Thus her speed is 0.73 m/s.
14.36. Model: The particle is in simple harmonic motion.
Solve: The equation for the velocity of the particle is
vx (t ) = −(25 cm)(10 rad/s)sin (10 t )
Substituting into K = 2U gives
1 2
⎛1
⎞ 1
mvx (t ) = 2 ⎜ kx 2 (t ) ⎟ ⇒ m[ −(250 cm/s)sin (10 t )]2 = k[(25 cm)cos (10 t )]2
2
⎝2
⎠ 2
⇒
sin 2 (10 t )
2
⎛ k ⎞ (25 cm)
⎛ 1 ⎞ 2
= 2⎜ ⎟
= 2ω 2 ⎜
⎟s
cos (10 t )
⎝ m ⎠ (250 cm/s) 2
⎝ 100 ⎠
2
1
⎛ 1 ⎞ 2
−1
⇒ tan 2 (10 t ) = 2(10 rad/s)2 ⎜
2.0 = 0.096 s
⎟ s = 2.0 ⇒ t = tan
10
⎝ 100 ⎠
14.37. Model: The spring undergoes simple harmonic motion.
Solve: (a) Total energy is E = 12 kA2 . When the displacement is x = 12 A, the potential energy is
U = 12 kx 2 = 12 k
( 12 A)
2
= 14
(
1 kA2
2
)=
1E
4
⇒ K = E − U = 34 E
One quarter of the energy is potential and three-quarters is kinetic.
(b) To have U = 12 E requires
U = 12 kx 2 = 12 E = 12
(
1 kA2
2
)⇒ x =
A
2
14.38. Solve: Average speed is vavg = Δx/Δt . During half a period (Δt = 12 T ), the particle moves from x = − A to
x = + A(Δx = 2 A). Thus
vavg =
Δx 2 A 4 A
4A
2
2
π
=
=
=
= (ω A) = vmax ⇒ vmax = vavg
Δt T/2 T
2π /ω π
π
2
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Oscillations
14-15
14.39. Model: The ball attached to a spring is in simple harmonic motion.
Solve: (a) Let t = 0 s be the instant when x0 = −5.0 cm and v0 = 20 cm/s. The oscillation frequency is
ω=
k
2.5 N/m
=
= 5.0 rad/s
0.100 kg
m
Solving Equation 14.26 for the amplitude gives
2
2
⎛ 20 cm/s ⎞
⎛v ⎞
A = x02 + ⎜ 0 ⎟ = (−5.0 cm) 2 + ⎜
⎟ = 6.4 cm
⎝ω ⎠
⎝ 5.0 rad/s ⎠
(b) The maximum acceleration is amax = ω 2 A = 160 cm/s 2 .
(c) For an oscillator, the acceleration is most positive (a = amax ) when the displacement is most negative
( x = − xmax = − A). So the acceleration is maximum when x = −6.4 cm.
(d) We can use the conservation of energy between x0 = −5.0 cm and x1 = 3.0 cm:
1 mv 2
0
2
+ 12 kx02 = 12 mv12 + 12 kx12 ⇒ v1 = v02 +
k 2
( x0 − x12 ) = 0.283 m/s
m
The speed is 28 cm/s. Because k is known in SI units of N/m, the energy calculation must be done using SI units of
m, m/s, and kg.
14.40. Model: The block on a spring is in simple harmonic motion.
Solve: (a) The position of the block is given by x (t ) = A cos(ωt + φ 0). Because x (t ) = A at t = 0 s, we have
φ 0 = 0 rad, and the position equation becomes x(t ) = A cos ωt. At t = 0.685 s, 3.00 cm = A cos (0.685ω ) and at
t = 0.886 s, −3.00 cm = A cos(0.886ω ). These two equations give
cos (0.685ω ) = − cos (0.886ω ) = cos (π − 0.886ω )
⇒ 0.685ω = π − 0.886ω ⇒ ω = 2.00 rad/s
(b) Substituting into the position equation,
3.00 cm = A cos ((2.00 rad/s)(0.685 s)) = A cos (1.370) = 0.20 A ⇒ A =
3.00 cm
= 15.0 cm
0.20
14.41. Model: The oscillator is in simple harmonic motion. Energy is conserved.
Solve: The energy conservation equation E1 = E2 is
1 mv 2
1
2
+ 12 kx12 = 12 mv22 + 12 kx22
1
1
1
1
(0.30 kg)(0.954 m/s) 2 + k (0.030 m) 2 = (0.30 kg)(0.714 m/s)2 + k (0.060 m)2
2
2
2
2
⇒ k = 44.48 N/m
The total energy of the oscillator is
1
1
1
1
Etotal = mv12 + kx12 = (0.30 kg)(0.954 m/s)2 + (44.48 N/m)(0.030 m) 2 = 0.1565 J
2
2
2
2
2
Because Etotal = 12 mvmax
,
1
2
0.1565 J = (0.300 kg)vmax
⇒ v max = 1.02 m/s
2
Assess: A maximum speed of 1.02 m/s is reasonable.
14.42. Model: The transducer undergoes simple harmonic motion.
Solve: Newton’s second law for the transducer is
Frestoring = ma max ⇒ 40,000 N = (0.10 × 10−3 kg)amax ⇒ amax = 4.0 × 108 m/s 2
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14-16
Chapter 14
Because amax = ω 2 A,
A=
amax
ω2
=
4.0 × 108 m/s 2
[2π (1.0 × 106 Hz )]2
= 1.01 × 10−5 m = 10.1 μ m
(b) The maximum velocity is
vmax = ω A = 2π (1.0 × 106 Hz )(1.01 × 10−5 m ) = 64 m/s
14.43. Model: The block attached to the spring is in simple harmonic motion.
Solve: (a) The frequency is
f =
1
2π
k
1
=
m 2π
2000 N/m
= 3.183 Hz
5.0 kg
The frequency is 3.2 Hz.
(b) From energy conservation,
2
2
⎛v ⎞
⎛ 1.0 m/s ⎞
A = x02 + ⎜ 0 ⎟ = (0.050 m) 2 + ⎜
⎟ = 0.0707 m
⎝ω ⎠
⎝ 2π ⋅ 3.183 Hz ⎠
The amplitude is 7.1 cm.
(c) The total mechanical energy is
E = 12 kA2 = 12 (2000 N/m)(0.0707 m)2 = 5.0 J
14.44. Model: Model this situation as a simple harmonic oscillator without damping.
Visualize: The period is independent of the amplitude. If the oscillator is undamped the amplitude will be constant;
the amplitude is just the arbitrary distance from the equilibrium from which the mass was released. The column of
amplitude data is not needed.
The period is related to the spring constant however: T = 2π
m
.
k
Solve: The equation T 2 = 4π 2 mk leads us to believe that a graph of T 2 vs. m would give a straight line graph whose
slope is 4π 2 /k . Note that the period is 1/10 of the times given in the table.
From the spreadsheet we see that the linear fit is very good and that the slope is 6.0997s 2 /kg.
slope = 4π 2 /k ⇒ k =
4π 2
6.0997s 2 /kg
= 6.5 N/m
Assess: Check the units carefully in the last equation. 6.5 N/m is a reasonable spring constant.
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Oscillations
14-17
14.45. Model: The block undergoes simple harmonic motion.
Visualize:
Solve: (a) The frequency of oscillation is
f =
The frequency is 1.1 Hz.
(b) Using conservation of energy,
1 mv 2
1
2
1
2π
1
k
=
m 2π
10 N/m
= 1.125 Hz
0.20 kg
+ 12 kx12 = 12 mv02 + 12 kx02 , we find
0.20 kg
m 2 2
(v0 − v1 ) = (−0.20 m)2 +
((1.00 m/s) 2 − ( 0.50 m/s )2 )
10 N/m
k
= 0.2345 m or 23 cm
(c) At time t, the displacement is x = A cos (ωt + φ 0). The angular frequency is ω = 2π f = 7.071 rad/s. The amplitude is
x1 = x02 +
2
2
⎛v ⎞
⎛ 1.00 m/s ⎞
A = x02 + ⎜ 0 ⎟ = (−0.20 m) 2 + ⎜
⎟ = 0.245 m
⎝ω ⎠
⎝ 7.071 rad/s ⎠
The phase constant is
⎛ x0 ⎞
−1 ⎛ −0.200 m ⎞
⎟ = cos ⎜
⎟ = ±2.526 rad or ± 145°
⎝ A⎠
⎝ 0.245 m ⎠
A negative displacement (below the equilibrium point) and positive velocity (upward motion) indicate that the
corresponding circular motion is in the third quadrant, so φ 0 = −2.526 rad. Thus at t = 1.0 s,
φ 0 = cos −1 ⎜
x = (0.245 m)cos ((7.071 rad/s)(1.0 s) − 2.526 rad) = −0.0409 m = −4.09 cm
The block is 4.1 cm below the equilibrium point.
14.46. Model: The mass is in simple harmonic motion.
Visualize:
The high point of the oscillation is at the point of release. This conclusion is based on energy conservation. Gravitational
potential energy is converted to the spring’s elastic potential energy as the mass falls and stretches the spring, then the
elastic potential energy is converted 100% back into gravitational potential energy as the mass rises, bringing the mass back
to exactly its starting height. The total displacement of the oscillation—high point to low point—is 20 cm. Because the
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14-18
Chapter 14
oscillations are symmetrical about the equilibrium point, we can deduce that the equilibrium point of the spring is 10 cm
below the point where the mass is released. The mass oscillates about this equilibrium point with an amplitude of 10 cm,
that is, the mass oscillates between 10 cm above and 10 cm below the equilibrium point.
Solve: The equilibrium point is the point where the mass would hang at rest, with Fsp = FG = mg . At the equilibrium
point, the spring is stretched by Δy = 10 cm = 0.10 m. Hooke’s law is Fsp = k Δy, so the equilibrium condition is
k
g 9.8 m/s 2
=
=
= 98 s22
m Δy 0.10 m
The ratio k/m is all we need to find the oscillation frequency:
[ Fsp = k Δy ] = [ FG = mg ] ⇒
f =
1
2π
k
1
98 s −2 = 1.58 Hz ≈ 1.6 Hz
=
m 2π
14.47. Model: The spring is ideal, so the apples undergo simple harmonic motion.
Solve: The spring constant of the scale can be found by considering how far the pan goes down when the apples
are added.
mg
mg
20 N
⇒k =
=
= 222 N/m
ΔL =
ΔL 0.090 m
k
The frequency of oscillation is
1 k
1
222 N/m
f =
=
= 1.66 Hz ≈ 1.7 Hz
2π m 2π (20 N/9.8 m/s 2 )
Assess: An oscillation of fewer than twice per second is reasonable.
14.48. Model: The compact car is in simple harmonic motion.
Solve: (a) The mass on each spring is (1200 kg)/4 = 300 kg. The spring constant can be calculated as follows:
ω2 =
k
⇒ k = mω 2 = m(2π f ) 2 = (300 kg)[2π (2.0 Hz)]2 = 4.74 × 104 N/m
m
The spring constant is 4.7 × 104 N/m.
(b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg. That is,
m = 300 kg + 70 kg = 370 kg. Thus,
f =
ω
1
=
2π 2π
k
1
=
m 2π
4.74 × 104 N/m
= 1.8 Hz
370 kg
Assess: A small frequency change from the additional mass is reasonable because frequency is inversely proportional
to the square root of the mass.
14.49. Model: Assume simple harmonic motion for the two-block system without the upper block slipping. We
will also use the model of static friction between the two blocks.
Visualize:
Solve: The net force on the upper block m1 is the force of static friction due to the lower block m2 . The two blocks
ride together as long as the static friction doesn’t exceed its maximum possible value. The model of static friction
gives the maximum force of static friction as
( fs ) max = μs n = μs (m1g ) = m1amax ⇒ amax = μs g
2
2
amax ω 2 Amax ⎛ 2π ⎞ ⎛ Amax ⎞ ⎛ 2π ⎞ ⎛ 0.40 m ⎞
=
=⎜
⎟=⎜
⎟ ⎜
⎟ ⎜
⎟ = 0.72
g
g
⎝ T ⎠ ⎝ g ⎠ ⎝ 1.5 s ⎠ ⎝ 9.8 m/s 2 ⎠
Assess: Because the period is given, we did not need to use the block masses or the spring constant in our calculation.
⇒ μs =
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Oscillations
14-19
14.50. Model: The DNA and cantilever undergo simple harmonic motion.
Solve: The cantilever has the same spring constant with and without the DNA molecule. The frequency of
oscillation without the DNA is
k
ω1 = 1
M
3
With the DNA, the frequency of oscillation is
ω2 =
k
+m
1M
3
where m is the mass of the DNA.
Divide the two equations, and express ω2 = ω1 − Δω , where Δω = 2πΔf = 2π (50 Hz).
f1
ω1 f1
=
=
=
ω2 f 2 f1 − Δf
(
(
k
1M
3
)
k
+m
1M
3
=
)
1M +m
3
1M
3
Thus
f12
m = 13 M
( 13 M ) = ( f1 − Δf )2 ( 13 M + m )
( f12 − ( f1 − Δf ) 2 )
( f1 − Δf ) 2
⎛
⎞
f12
= 13 M ⎜
−1
⎜ ( f − Δf ) 2 ⎟⎟
⎝ 1
⎠
⎛ Δf ⎞
Since Δf f1, (50 Hz 12 MHz), ( f1 − Δf ) −2 = f1−2 ⎜1 −
⎟
f1 ⎠
⎝
−2
⎛
Δf
≈ f1−2 ⎜ 1 + 2
f1
⎝
⎞
⎟ . Thus
⎠
⎛
⎞
Δf
Δf
m = 13 M ⎜1 + 2
− 1⎟ = 23 M
f1
f1
⎝
⎠
The mass of the cantilever
M = (2300 kg/m3 )(4000 × 10−9 m)(100 × 10−9 m) = 3.68 × 10−16 kg
Thus the mass of the DNA molecule is
2
⎛ 50 Hz ⎞
−21
m = (3.68 × 10−16 kg) ⎜
⎟ = 1.02 × 10 kg
3
⎝ 12 × 106 Hz ⎠
Assess: The mass of the DNA molecule is about 6.2 × 105 atomic mass units, which is reasonable for such a large
molecule.
14.51. Model: Assume that the swinging lamp makes a small angle with the vertical so that there is simple harmonic
motion.
Visualize:
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14-20
Chapter 14
Solve: (a) Using the formula for the period of a pendulum,
T = 2π
2
2
L
⎛ T ⎞
2 ⎛ 5.5 s ⎞
⇒ L = g⎜
⎟ = (9.8 m/s ) ⎜
⎟ = 7.5 m
g
2
π
⎝
⎠
⎝ 2π ⎠
(b) The conservation of mechanical energy equation K 0 + U g0 = K1 + U g1 for the swinging lamp is
1 mv 2
0
2
2
+ mgy0 = 12 mv12 + mgy1 ⇒ 0 J + mgh = 12 mvmax
+0 J
⇒ vmax = 2 gh = 2 g ( L − L cos3°)
= 2(9.8 m/s 2 )(7.5 m)(1 − cos3°) = 0.45 m/s
14.52. Model: Assume that the angle with the vertical that the pendulum makes is small enough so that there is
simple harmonic motion.
Solve: The angle θ made by the string with the vertical as a function of time is
θ (t ) = θ max cos(ωt + φ 0)
The pendulum starts from maximum displacement, thus φ 0 = 0. Thus, θ (t ) = θ max cos ωt . To find the time t when the
pendulum reaches 4.0° on the opposite side:
(−4.0°) = (8.0°) cos ωt ⇒ ωt = cos −1 (−0.5) = 2.094 rad
Using the formula for the angular frequency,
ω=
g
9.8 m/s 2
2.0944 rad 2.094 rad
=
= 3.130 rad/s ⇒ t =
=
= 0.669 s
L
1.0 m
3.130 rad/s
ω
The time t = 0.67 s.
Assess: Because T = 2π /ω = 2.0 s, a value of 0.67 s for the pendulum to cover a little less than half the oscillation is
reasonable.
14.53. Model: Assume the orangutan is a simple small-angle pendulum.
Visualize: One swing of the arm would be half a period of the oscillatory motion. The horizontal distance traveled in
that time would be 2(0.90m)sin (20°) = 0.616m from analysis of a right triangle.
Solve:
T 2π
=
2
2
speed =
L 2π
=
g
2
0.90 m
9.8 m/s 2
= 0.952s
dist 0.616 m
=
= 0.65 m/s
time 0.952 s
Assess: This isn’t very fast, but isn’t out of the reasonable range.
14.54. Model: The mass is a particle and the string is massless.
Solve: Equation 14.51 is
ω=
Mgl
I
The moment of inertia of the mass on a string is I = Ml 2 , where l is the length of the string. Thus
ω=
Mgl
Ml 2
=
g
l
This is Equation 14.48 with L = l .
Assess: Equation 14.48 is really a specific case of the more general physical pendulum described by Equation 14.51.
14.55. Model: The rod is thin and uniform with moment of inertia described in Table 12.2. The clay ball is a
particle located at the end of the rod. The ball and rod together form a physical pendulum. The oscillations are small.
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Oscillations
14-21
Visualize:
Solve: The moment of inertia of the composite pendulum formed by the rod and clay ball is
1
I rod + ball = I rod + I ball = mrod L2 + mball L2
3
1
= (0.200 kg)(0.15 m) 2 + ( 0.020 kg )( 0.15 m )2 = 1.95 × 10−3 kg m 2
3
The center of mass of the rod and ball is located at a distance from the pivot point of
⎛ 0.15 ⎞
(0.200 kg) ⎜
m ⎟ + ( 0.020 kg )( 0.15 m )
⎝ 2
⎠
= 8.18 × 10−2 m
ycm =
(0.200 kg + 0.020 kg)
The frequency of oscillation of a physical pendulum is
f =
1
2π
The period of oscillation T =
Mgl
1
=
I
2π
(0.220 kg)(9.8 m/s 2 )(8.18 × 10−2 m)
1.95 × 10−3 kg m 2
= 1.51 Hz
1
= 0.66 s.
f
14.56. Model: Model the rod as a small-angle physical pendulum without damping.
Visualize: Figure 14.22 in the chapter shows that l is the distance from the pivot to the center of mass; in this case
l = L/4.
Solve: We first need to compute the moment of inertia of a thin rod about an axis 1/4 of the way from one end. Use
the parallel-axis theorem to do so:
I = I c.m. + Md 2 =
2
1
7
⎛L⎞ ⎛ 1 1 ⎞
ML2 + M ⎜ ⎟ = ⎜ + ⎟ ML2 = ML2
12
48
⎝ 4 ⎠ ⎝ 12 16 ⎠
For the physical pendulum
f =
1
2π
Mgl
1
=
I
2π
Mg
( L4 ) =
7 ML2
48
1 12 g 1
=
2π 7 L π
3g
7L
Assess: The period T = 1/f is a little shorter than swinging the rod from the end, which is what we expect.
14.57. Model: Model the sphere as a small-angle physical pendulum without damping.
Visualize: Figure 14.22 in the chapter shows that l is the distance from the pivot to the center of mass; in this case l = R.
Solve: We first need to compute the moment of inertia of a sphere about an axis tangent to the edge. Use the
parallel-axis theorem to do so:
2
7
I = I c.m. + Md 2 = MR 2 + MR 2 = MR 2
5
5
For the physical pendulum
1 Mgl
1
MgR
1 5g
f =
=
=
2π
2π 7 MR 2 2π 7 R
I
5
Assess: The frequency is less than for a simple pendulum of length R as we expect.
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14-22
Chapter 14
14.58. Model: Model this situation as a small-angle physical pendulum without damping. The moment of inertia
for a uniform meter stick pivoted on one end is
1 ML2 .
3
Visualize: The frequency and length of a physical pendulum are related by f =
1
2π
Mgl
,
I
where l = L/2.
Solve: Square both sides of the equation above and substitute in for I.
f2=
1 Mgl
1 Mg L2
3 g 3g 1
=
= 2 = 2
2 I
2 1
2
4π
4π 3 ML 8π L 8π L
This leads us to believe that a graph of f 2 vs. 1/L would produce a straight line whose slope is
3g
.
8π 2
From the spreadsheet we see that the linear fit is excellent and that the slope is 0.3707m/s 2 .
slope =
3g
8π 2
⇒g=
(8π 2 )(0.3707 m/s 2 )
= 9.76 m/s 2
3
Assess: If the uncertainties in our measurements were small enough then our answer would show that g varies
slightly from place to place on the earth.
14.59. Model: Treat the lower leg as a physical pendulum.
Visualize: We can determine the moment of inertia by combining T = 2π I/mgL and T = 1/f .
Solve: Combining the above expressions and solving for the moment of inertia we obtain
I = mgL/(2π f ) 2 = (5.0 kg)(9.80 m/s 2 )(0.18 m)/[2π (1.6 Hz)]2 = 8.7 × 10−2 kg ⋅ m 2
Assess: NASA determines the moment of inertia of the shuttle in a similar manner. It is suspended from a heavy
cable, allowed to oscillate about its vertical axis of symmetry with a very small amplitude, and from the period of
oscillation one may determine the moment of inertia. This arrangement is called a torsion pendulum.
14.60. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion.
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Oscillations
14-23
Visualize:
Let us denote the 250 g and 500 g masses as m1 and m2 , which have initial velocities vi1 and vi2 . After m1 collides
with and sticks to m2 , the two masses move together with velocity vf .
Solve: The momentum conservation equation pf = pi for the completely inelastic collision is (m1 + m2 )vf = m1vi1 + m2vi2.
Substituting the given values,
(0.750 kg)vf = (0.250 kg)(1.20 m/s) + (0.500 kg)(0 m/s) ⇒ vf = 0.400 m/s
We now use the conservation of mechanical energy equation:
( K + U s )compressed = ( K + U s )equilibrium ⇒ 0 J + 12 kA2 = 12 (m1 + m2 )vf2 + 0 J
⇒ A=
m1 + m2
0.750 kg
vf =
(0.400 m/s) = 0.11 m
k
10 N/m
The period is
T = 2π
m1 + m2
0.750 kg
= 2π
= 1.7 s
k
10 N/m
14.61. Model: The block attached to the spring is oscillating in simple harmonic motion.
Solve: (a) Because the frequency of an object in simple harmonic motion is independent of the amplitude and/or the
maximum velocity, the new frequency is equal to the old frequency of 2.0 Hz.
(b) The speed v0 of the block just before it is given a blow can be obtained by using the conservation of mechanical
energy equation as follows:
1 kA2
2
2
= 12 mvmax
= 12 mv02
k
A = ω A = (2π f ) A = (2π )(2.0 Hz)(0.02 m) = 0.25 m/s
m
The blow to the block provides an impulse that changes the velocity of the block:
J x = Fx Δt = Δp = mvf − mv0
⇒ v0 =
(−20 N)(1.0 × 10−3 s) = (0.200 kg)vf − (0.200 kg)(0.25 m/s) ⇒ vf = 0.150 m/s
Since vf is the new maximum velocity of the block at the equilibrium position, it is equal to Aω . Thus,
A=
0.150 m/s
ω
=
0.150 m/s
= 0.012 m = 1.19 cm ≈ 1.2 cm
2π (2.0 Hz)
Assess: Because vf is positive, the block continues to move to the right even after the blow.
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14-24
Chapter 14
14.62. Model: Assume the small-angle approximation.
Visualize:
Solve: The tension in the two strings pulls downward at angle θ . Thus Newton’s second law is
∑ Fy = −2T sin θ = ma y
From the geometry of the figure we can see that
sin θ =
y
2
L + y2
If the oscillation is small, then y L and we can approximate sin θ ≈ y/L. Since y/L is tan θ , this approximation
is equivalent to the small-angle approximation sin θ ≈ tan θ if θ 1 rad. With this approximation, Newton’s second
law becomes
−2T sin θ ≈ −
2T
d2y
y = ma y = m 2
L
dt
⇒
d2y
dt
2
=−
2T
y
mL
This is the equation of motion for simple harmonic motion (see Equations 14.32 and 14.46). The constants 2T/mL are
equivalent to k/m in the spring equation or g/L in the pendulum equation. Thus the oscillation frequency is
f =
1
2π
2T
mL
14.63. Solve: The potential energy curve of a simple harmonic oscillator is described by U = 12 k (Δx) 2 , where
Δx = x − x0 is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is x0 = 0.13 nm.
We can find the bond’s spring constant by reading the value of the potential energy U at a displacement Δx and
using the potential energy formula to calculate k.
x (nm)
Δx (nm)
U (J)
k (N/m)
0.11
0.02
0.8 × 10−19 J
400
0.10
0.03
1.9 × 10−19 J
422
0.09
0.04
3.4 × 10−19 J
425
The three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the spring
constant, we can now calculate the oscillation frequency of a hydrogen atom on this “spring” to be
f =
1
2π
k
1
416 N/m
=
= 7.9 × 1013 Hz
m 2π 1.67 × 10−27 kg
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Oscillations
14-25
14.64. Model: Assume that the size of the ice cube is much less than R and that θ is a small angle.
Visualize:
Solve: The ice cube is like an object on an inclined plane. The net force on the ice cube in the tangential direction is
d 2θ
d 2θ
−( FG ) sin θ = ma = mRα = mR 2 ⇒ −mg sin θ = mR 2
dt
dt
where α is the angular acceleration. With the small-angle approximation sin θ ≈ θ , this becomes
d 2θ
g
= − θ = −ω 2θ
R
dt
This is the equation of motion of an object in simple harmonic motion with a period of
2π
R
T=
= 2π
ω
g
2
14.65. Visualize:
Solve: (a) Newton’s second law applied to the penny along the y-axis is
Fnet = n − mg = ma y
G
Fnet is upward at the bottom of the cycle (positive a y ), so n > mg . The speed is maximum when passing through
G
equilibrium, but a y = 0 so n = mg . The critical point is the highest point. Fnet points down and a y is negative. If
a y becomes sufficiently negative, n drops to zero and the penny is no longer in contact with the surface.
(b) When the penny loses contact ( n = 0), the equation for Newton’s law becomes amax = g . For simple harmonic
motion,
g
9.8 m/s 2
=
= 15.65 rad
A
0.040 m
ω 15.65 rad/s
⇒ f =
=
= 2.5 Hz
2π
2π
amax = Aω 2 ⇒ ω =
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14-26
Chapter 14
14.66. Model: The vertical oscillations constitute simple harmonic motion.
Visualize:
Solve: At the equilibrium position, the net force on mass m on Planet X is:
k g
Fnet = k ΔL − mg X = 0 N ⇒ = X
m ΔL
For simple harmonic motion k/m = ω 2 , thus
ω2 =
gX
⇒ω =
ΔL
2
2
g X 2π
⎛ 2π ⎞
⎛ 2π
⎞
2
=
⇒ gX = ⎜
⎟ ΔL = ⎜
⎟ (0.312 m) = 5.86 m/s
T
14
5
s/10
ΔL T
.
⎝
⎠
⎝
⎠
Assess: Because ordinary tasks seemed easier than on earth we expected an answer less than 9.8 m/s 2 .
14.67. Model: The doll’s head is in simple harmonic motion and is damped.
Solve: (a) The oscillation frequency is
1 k
f =
⇒ k = m(2π f )2 = (0.015 kg)(2π ) 2 (4.0 Hz) 2 = 9.475 N/m
2π m
The spring constant is 9.5 N/m.
(b) Using A(t ) = A0e−bt/2 m , we get
(0.5 cm) = (2.0 cm)e−b(4.0 s)/(2×0.015 kg) ⇒ 0.25 = e− (133.3 s/kg)b
⇒ −(133.33 s/kg )b = ln 0.25 ⇒ b = 0.0104 kg/s ≈ 0.010 kg/s
14.68. Model: The oscillator is in simple harmonic motion.
Solve: The maximum displacement at time t of a damped oscillator is
t
⎛ x (t ) ⎞
xmax (t ) = Ae−t/2τ ⇒ − = ln ⎜ max ⎟
2τ
⎝ A ⎠
Using xmax = 0.98 A at t = 0.50 s, we can find the time constant τ to be
0.50 s
= 12.375 s
2ln(0.98)
25 oscillations will be completed at t = 25T = 12.5 s. At that time, the amplitude will be
τ =−
xmax, 12.5 s = (10 cm)e−12.5 s/(2)(12.375 s) = 6.0 cm
14.69. Model: The vertical oscillations are damped and follow simple harmonic motion.
Solve: The position of the ball is given by x(t ) = Ae −(t/2τ ) cos(ωt + φ 0). The amplitude A(t ) = Ae− (t/2τ ) is a function
of time. The angular frequency is
ω=
k
(15.0 N/m)
2π
=
= 5.477 rad/s ⇒ T =
= 1.147 s
ω
m
0.500 kg
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Oscillations
14-27
Because the ball’s amplitude decreases to 3.0 cm from 6.0 cm after 30 oscillations, that is, after 30 × 1.147 s = 34.41 s,
we have
−34.41 s
⇒ τ = 25 s
3.0 cm = (6.0 cm)e−(34.414 s/2τ ) ⇒ 0.50 = e− (34.41 s/ 2τ ) ⇒ ln(0.50) =
2τ
14.70. Model: The motion is a damped oscillation.
Solve: The position of the air-track glider is x(t ) = Ae −(t/2τ ) cos(ωt + φ 0 ), where τ = m/b and
ω=
k
b2
−
m 4m 2
Using A = 0.20 m, φ 0 = 0 rad, and b = 0.015 kg/s,
ω=
4.0 N/m (0.015 kg/s) 2
−
= 16 − 9 × 10−4 rad/s = 4.0 rad/s
0.250 kg 4(0.250 kg) 2
Thus the period is
T=
2π
ω
=
2π rad
= 1.57 s
4.0 rad/s
The amplitude at t = 0 s is x0 = A and the amplitude will be equal to e−1 A at a time given by
1
m
A = Ae− (t/ 2τ ) ⇒ t = 2τ = 2 = 33.3 s
e
b
The number of oscillations in a time of 33.3 s is (33.3 s)/(1.57 s) = 21.
14.71. Model: The oscillator is in simple harmonic motion.
Solve: The maximum displacement, or amplitude, of a damped oscillator decreases as xmax (t ) = Ae−t/2τ , where τ is
the time constant. We know xmax /A = 0.60 at t = 50 s, so we can find τ as follows:
−
t
50 s
⎛ x (t ) ⎞
= ln ⎜ max ⎟ ⇒ τ = −
= 48.9 s
2τ
2ln(0.60)
⎝ A ⎠
Now we can find the time t30 at which xmax /A = 0.30:
⎛ x (t ) ⎞
t30 = −2τ ln ⎜ max ⎟ = −2(48.9 s)ln(0.30) = 118 s
⎝ A ⎠
The undamped oscillator has a frequency f = 2 Hz = 2 oscillations per second. Damping changes the oscillation
frequency slightly, but the text notes that the change is negligible for “light damping.” Damping by air, which allows
the oscillations to continue for well over 100 s, is certainly light damping, so we will use f = 2.0 Hz. Then the
number of oscillations before the spring decays to 30% of its initial amplitude is
N = f ⋅ t30 = (2 oscillations/s) ⋅ (118 s) = 236 oscillations
14.72. Solve: The solution of the equation
d 2x
dt 2
+
b dx k
+ x=0
m dt m
is x(t ) = Ae−bt/2 m cos(ωt + φ 0 ). The first and second derivatives of x(t) are
dx
Ab −bt/2 m
e
=−
cos(ωt + φ 0 ) − Aωe −bt/2 m sin (ωt + φ 0 )
dt
2m
⎤
⎞
d 2 x ⎡⎛ Ab 2
ω Ab
= ⎢⎜ 2 − Aω 2 ⎟ cos(ωt + φ 0 ) +
sin (ωt + φ 0 ) ⎥ e −bt/2 m
2
⎜
⎟
m
dt
⎢⎣⎝ 4m
⎥⎦
⎠
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14-28
Chapter 14
Substituting these expressions into the differential equation, the terms involving sin (ωt + φ 0 ) cancel and we obtain
the simplified result
⎛ −b 2
k⎞
2
⎜⎜ 2 − ω + ⎟⎟ cos(ωt + φ 0 ) = 0
m⎠
⎝ 4m
Because cos (ωt + φ 0 ) is not equal to zero in general,
−b 2
4m 2
− ω2 +
k
k
b2
=0⇒ω =
−
m
m 4m 2
14.73. Model: The two springs obey Hooke’s law.
Visualize:
Solve: There are two restoring forces on the block. If the block’s displacement x is positive, both restoring forces—
one pushing, the other pulling—are directed to the left and have negative values:
( Fnet ) x = ( Fsp 1 ) x + ( Fsp 2 ) x = −k1x − k2 x = −( k1 + k2 ) x = − keff x
where keff = k1 + k2 is the effective spring constant. This means the oscillatory motion of the block under the influence
of the two springs will be the same as if the block were attached to a single spring with spring constant keff . The
frequency of the blocks, therefore, is
f =
1
2π
keff
1
=
2π
m
k1 + k2
k1
k
=
+ 22 =
2
m
4π m 4π m
f12 + f 22
14.74. Model: The two springs obey Hooke’s law. Assume massless springs.
Visualize: Each spring is shown separately. Note that Δx = Δx1 + Δx2 .
Solve: Only spring 2 touches the mass, so the net force on the mass is Fm = F2 on m . Newton’s third law tells us that
F2 on m = Fm on 2 and that F2 on 1 = F1 on 2 . From Fnet = ma, the net force on a massless spring is zero. Thus Fw on 1
= F2 on 1 = k1Δx1 and Fm on 2 = F1 on 2 = k2 Δx2 . Combining these pieces of information,
Fm = k1Δx1 = k2Δx2
The net displacement of the mass is Δx = Δx1 + Δx2 , so
Δx = Δx1 + Δx2 =
Fm Fm ⎛ 1 1 ⎞
k +k
+
= ⎜ + ⎟ Fm = 1 2 Fm
k1 k2 ⎝ k1 k2 ⎠
k1k2
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Oscillations
14-29
Turning this around, the net force on the mass is
Fm =
k1k2
kk
Δx = keff Δx where keff = 1 2
k1 + k2
k1 + k2
keff , the proportionality constant between the force on the mass and the mass’s displacement, is the effective spring
constant. Thus the mass’s angular frequency of oscillation is
keff
1 k1k2
=
m
m k1 + k2
ω=
Using ω12 = k1/m and ω22 = k2 /m for the angular frequencies of either spring acting alone on m, we have
ω=
(k1/m)(k2 /m)
ω12ω22
=
(k1/m) + ( k2 /m)
ω12 + ω22
Since the actual frequency f is simply a multiple of ω , this same relationship holds for f :
f =
f12 f 22
f12 + f 22
14.75. Model: The blocks undergo simple harmonic motion.
Visualize:
The length of the stretched spring due to a block of mass m is ΔL1. In the case of the two-block system, the spring is
further stretched by an amount ΔL2 .
Solve: The equilibrium equations from Newton’s second law for the single-block and double-block systems are
(ΔL1)k = mg and (ΔL1 + ΔL2 )k = (2m) g
Using ΔL2 = 5.0 cm, and subtracting these two equations, gives us
(ΔL1 + ΔL2 )k − ΔL1k = (2m) g − mg ⇒ (0.05 m) k = mg
k 9.8 m/s 2
=
= 196 s −2
m 0.05 m
With both blocks attached, giving total mass 2m, the angular frequency of oscillation is
⇒
ω=
k
1k
1
196 s −2 = 9.90 rad/s
=
=
2m
2m
2
Thus the oscillation frequency is f = ω /2π = 1.6 Hz.
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14-30
Chapter 14
14.76. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonic motion.
Visualize:
Solve: (a) The equation for conservation of energy after the collision is
1 2 1
k
2500 N/m
kA = (mb + mB )vf2 ⇒ vf =
A=
(0.10 m) = 5.0 m/s
2
2
mb + mB
1.010 kg
The momentum conservation equation for the perfectly inelastic collision pafter = pbefore is
(mb + mB )vf = mbvb + mBvB
(1.010 kg)(5.0 m/s) = (0.010 kg)vb + (1.00 kg )( 0 m/s ) ⇒ vb = 5.0 × 102 m/s
(b) No. The oscillation frequency k/(mb + mB ) depends on the masses but not on the speeds.
14.77. Model: The block undergoes simple harmonic motion after sticking to the spring. Energy is conserved
throughout the motion.
Visualize:
It’s essential to carefully visualize the motion. At the highest point of the oscillation the spring is stretched upward.
Solve: We’ve placed the origin of the coordinate system at the equilibrium position, where the block would sit on
the spring at rest. The spring is compressed by ΔL at this point. Balancing the forces requires k ΔL = mg . The
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Oscillations
14-31
angular frequency is w2 = k/m = g/ΔL, so we can find the oscillation frequency by finding ΔL. The block hits the
spring (1) with kinetic energy. At the lowest point (3), kinetic energy and gravitational potential energy have been
transformed into the spring’s elastic energy. Equate the energies at these points:
K1 + U1g = U 3s + U 3g ⇒ 12 mv12 + mg ΔL = 12 k ( ΔL + A)2 + mg (− A)
We’ve used y1 = ΔL as the block hits and y3 = − A at the bottom. The spring has been compressed by Δy = ΔL + A.
Speed v1 is the speed after falling distance h, which from free-fall kinematics is v12 = 2 gh. Substitute this expression
for v12 and mg/ΔL for k, giving
mg
( ΔL + A) 2 + mg (− A)
2(ΔL)
The mg term cancels, and the equation can be rearranged into the quadratic equation
(ΔL) 2 + 2h(ΔL) − A2 = 0
mgh + mg ΔL =
The positive solution is
ΔL = h 2 + A2 − h = (0.030 m) 2 + ( 0.100 m )2 − 0.030 m = 0.0744 m
Now that ΔL is known, we can find
g
9.80 m/s 2
ω
=
= 11.48 rad/s ⇒ f =
= 1.8 Hz
ΔL
0.0744 m
2π
ω=
14.78. Model: Assume that the pendulum is restricted to small angles so sin θ ≈ θ.
Visualize: Because the string is massless the center of mass is at the center of the sphere, so l = L in the formula for
a physical pendulum: T = 2π
I .
Mgl
Solve:
(a) First we use the parallel axis theorem to find I for the sphere around the pivot point at the top (L away):
I = I cm + Md 2 = 52 MR 2 + ML2 .
2 MR 2
5
I
= 2π
Mgl
T = 2π
+ ML2
MgL
This reduces to the traditional simple pendulum formula T = 2π
L
g
= 2π
2 R2
5
+ L2
gL
in the limit L R .
(b)
Treal
=
Tsimple
2π
2 R2
5
+ L2
gL
2π
L
g
=
2 R2
5
L
+ L2
=
2 (0.010 m) 2
5
+ (1.0 m) 2
1.0 m
= 1.00002
Assess: Making the simple pendulum approximation introduces an error of only 0.002% with these numbers.
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14-32
Chapter 14
14.79. Model: Assume a simple system with a massless spring: T = 2π
m
.
k
Visualize: The simplest way to find ΔT is to consider T as a function of m and use calculus to take the differential
dT and consider that a good stand-in for ΔT . Also note that T = 2π
m
k
⇒ k = 4π 2
m
.
T2
Solve:
(a) Re-write T = 2π
m
k
as a function of m.
2π 1/2
m
k
Now take the differential of T with respect to m using the exponent rule and chain rule.
T ( m) =
dT =
Now cancel the 2s and plug in k = 4π 2
2π ⎛ 1 ⎞ −1/2
dm
⎜ ⎟m
k ⎝2⎠
m
.
T2
dT =
π
1
π
dm =
k m
4π 2
m
T2
1
T
dm =
dm
2m
m
Or, in the notation given in the problem:
ΔT =
(b) The new period, T ′ , is given by
T ′ = T + ΔT = T +
T Δm
2 m
T Δm
⎛ 1 Δm ⎞
⎛ 1
⎞
= T ⎜1 +
⎟ = (2.000 s) ⎜1 + (0.001) ⎟ = 2.001s
2 m
⎝ 2 m ⎠
⎝ 2
⎠
Assess: We did not expect a 0.1% change in mass to change the period by much.
14.80. Model: The rod is thin and uniform.
Visualize:
Solve: We must derive our own equation for this combination of a pendulum and spring. For small oscillations,
G
Fs remains horizontal. The net torque around the pivot point is
⎛L⎞
⎝ ⎠
τ net = Iα = − Fs L cosθ − FG ⎜ ⎟ sin θ
2
With α =
d 2θ
1
, FG = mg , Fs = k Δx = kL sin θ , and I = mL2 ,
3
dt
2
d 2θ
dt
2
=−
3k
3g
sin θ cosθ −
sin θ
m
2L
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Oscillations
14-33
1
We can use sin θ cosθ = sin 2θ . For small angles, sin θ ≈ θ and sin 2θ ≈ 2θ . So
2
d 2θ
dt 2
This is the same as Equations 14.32 and 14.46 with
⎛ 3k 3 g ⎞
= −⎜ +
⎟θ
⎝ m 2L ⎠
ω=
3k 3 g
+
m 2L
The frequency of oscillation is thus
f =
1
2π
3(3.0 N/m) 3(9.8 m/s 2 )
+
= 1.73 Hz
(0.200 kg) 2(0.20 m)
1
= 0.58 s.
f
Assess: Fewer than two oscillations per second is reasonable. The rod’s angle from the vertical must be small
enough that sin 2θ ≈ 2θ . This is more restrictive than other examples, which only require that sin θ ≈ θ .
The period T =
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FLUIDS AND ELASTICITY
15
Conceptual Questions
15.1. (a) The density will not change. Although the mass and volume both increase by 23 = 8 times, their ratio is
unchanged.
(b) The volume V has increased by 23 = 8 times, but the mass has not changed, so the new density ρ ′ is
ρ ′ = m/(8V ) = ρ /8. The new density has decreased by a factor of 8.
15.2. The pressure only depends on the depth with respect to the top surface of the liquid. Since points a, b, and c
are all at the same depth (i.e., with respect to point e) the pressure is the same for each, so
pa = pb = pc
15.3. The pressure only depends on the depth from the surface of the liquid. Since point d is the deepest and point e
the highest, then
pd > pf > pe
A point halfway between point e and point b would have a pressure about the same as the pressure at point d.
15.4. (a) The pressure at the bottom of each tank is given by p = ρ gd , so it will be the same because the depth of
the water in each tank is the same. The area of the bottom of tank A is larger than the area of the bottom of tank B.
From p = F/A, we have F = pA, so the FA > FB because the bottom area of tank A is larger.
(b) The pressure at any given depth is the same in both tanks because the water depth is the same. Since the area of
the sides indicated in Figure Q15.4 is the same for each tank, the force on these sides is also the same, so FA = FB .
This makes sense. Since there’s more water in tank A, the total force of the water on the bottom of tank A is larger.
15.5. A floating object displaces it weight in liquid, so placing the boat above point A will increase the depth of the
body of water just enough to compensate for the boat’s weight. Thus, the depth of point B is increased slightly so that
it is at the same pressure as point A, and pA = pB .
While a column above point A contains the ship, it also contains less water than a similar column above point B, but
both columns contain the same weight (or mass if we assume a uniform gravitational field).
15.6. Archimedes’ principle tells us that the displaced liquid has the same weight as the floating object, so
ρ LVL = ρBVB ⇒ ρB = ρ L (VL /VB )
where the subscripts L and B refer to liquid and block, respectively. The ratio VL /VB is the fraction of the block that
is under the liquid, so the densest block is the one for which this fraction is the largest. Thus,
ρA > ρC > ρB
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15-1
15-2
Chapter 15
You’ve heard that only 10% of an iceberg is visible above the surface of the ocean. That means 90% of the iceberg is
below the surface. Therefore the density of ice is 90% the density of seawater. You can verify this by looking in
Table 15.1 and finding on the Web the density of ice: ρice = 917 kg/m3 .
15.7. Archimedes’ principle states that the buoyant force on an object is equal to the weight of the fluid displaced by
the object. Each object displaces exactly the same amount of fluid since each is the same volume. Therefore, the
buoyant force on all three objects is the same, so Fa = Fb = Fc . Note that the buoyant force does not depend on the
mass or location of the object.
15.8. For objects that are completely submerged the buoyant force is proportional to the volume of the object. If the
densities of objects A, B, and C are the same, then the objects with greater mass (A and C) must also occupy a larger
volume. Thus, A and C will experience a larger buoyant force than B.
FA = FC > FB
15.9. The sphere is floating in static equilibrium, so the upward buoyant force exactly equals the sphere’s
weight: FB = w. But according to Archimedes’ principle, FB is the weight of the displaced liquid. That is, the weight
of the missing water in B is exactly matched by the weight of the added ball. Thus, the total weights of both
containers are equal.
15.10. The lower the velocity of fluid, the higher the pressure at a given point in the pipe (Bernoulli’s principle). The
pressure p in the horizontal liquid-containing pipe is p = ps + ρ gd , where d is the depth of the liquid in the vertical
pipes a, b, or c and ps is the gas pressure at the liquid surface. Thus, the vertical pipes with the highest liquid level d
have the lowest surface pressure ps , or the highest gas velocity. Therefore, the gas velocities are vb > va > vc .
15.11. The pressure is reduced at the chimney due to the movement of the wind above (Bernoulli’s principle). Thus,
the air will flow in the window and out the chimney.
Prairie dogs ventilate their burrows this way; a small breeze above their mound lowers the pressure there and allows
the air in the burrow to move between openings of different types or heights.
15.12. Equation 15.34 is F/A = Y (ΔL/L). The second wire is the same material, and has length L′ = 2 L and area
A′ = π (2r )2 = 4π r 2 = 4 A. The force required to stretch it the same length ΔL = 1 mm is
F ′ = A′Y
ΔL
ΔL
ΔL ⎞
⎛
= 4 AY
= 2 ⎜ AY
⎟ = 2F
L′
2L
L ⎠
⎝
So the force required is 4000 N.
15.13. The stress F/A on the wire is proportional to the strain ΔL/L in the elastic region. The breaking point is past
the elastic limit. The elastic limit is reached at a particular strain. Since the wire is the same diameter the area A stays
the same, so an equal force of 5000 N will also take the wire to its elastic limit.
Exercises and Problems
Section 15.1 Fluids
15.1. Solve: The volume of the liquid is
ρ=
m
V
⇒ V=
m
ρ
=
0.055 kg ⎛ 106 mL ⎞
⎜
⎟ = 50 mL
1100 kg/m3 ⎜⎝ m3 ⎟⎠
Assess: The liquid’s density slightly higher than that of water (1000 kg/m3 ), so it is reasonable that it requires
slightly less than 55 mL to get a mass of 55 g.
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Fluids and Elasticity
15-3
15.2. Solve: The volume of the helium gas in container A is equal to the volume of the liquid in container B. That
is, VA = VB . Using the definition of mass density ρ = m/V , this relationship becomes
mA
ρA
=
mB
ρB
⇒
mHe
ρHe
=
7000 mHe
ρB
⇒
ρB = 7000 ρHe = (7000)(0.18 kg/m3 ) = 1260 kg/m3
Referring to Table 15.1, we find that the liquid is glycerine.
15.3. Model: The density of water is 1000 kg/m3 .
Visualize:
Solve: Volume of water in the swimming pool is
V = 6.0 m × 12 m × 3.0 m − 12 (6.0 m × 12 m × 2.0 m) = 144 m3
The mass of water in the swimming pool is
m = ρV = (1000 kg/m3 )(144 m3 ) = 1.4 × 105 kg
15.4. Model: The densities of gasoline and water are given in Table 15.1.
Solve: (a) The total mass is
mtotal = mgasoline + mwater = 0.050 kg + 0.050 kg = 0.100 kg
The total volume is
Vtotal = Vgasoline + Vwater =
mgasoline
ρgasoline
+
mwater
ρwater
=
0.050 kg
680 kg/m
3
+
0.050 kg
1000 kg/m
3
= 1.24 × 10−4 m3
m
0.100 kg
= 8.1 × 102 kg/m3
ρavg = total =
Vtotal 1.24 × 10−4 m3
(b) The average density is calculated as follows:
mtotal = mgasoline + mwater = ρwaterVwater + ρgasolineVgasoline
ρavg =
ρwaterVwater + ρgasolineVgasoline
Vwater + Vgasoline
=
(50 cm3 )(1000 kg/m3 + 680 kg/m3 )
100 cm
3
= 8.4 × 102 kg/m3
Assess: The above average densities are between those of gasoline and water, which is reasonable.
Section 15.2 Pressure
15.5. Model: The density of sea water is 1030 kg/m3 .
Solve: The pressure below sea level can be found from Equation 15.5 as follows:
p = p0 + ρ gd = 1.013 × 105 Pa + (1030 kg/m3 )(9.81 m/s 2 )(1.1 × 104 m)
= 1.013 × 105 Pa + 1.1103 × 108 Pa = 1.1113 × 108 Pa = 1.1× 103 atm
where we have used the conversion 1 atm = 1.013 × 105 Pa.
Assess: The pressure deep in the ocean is very large.
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15-4
Chapter 15
15.6. Visualize:
Solve: The pressure at the bottom of the vat is p = p0 + ρ gd = 1.3 atm. Substituting into this equation gives
1.013 × 105 Pa + ρ (9.8 m/s 2 )(2.0 m) = (1.3)(1.013 × 105 ) Pa ⇒
ρ = 1550.5 kg/m3
The mass of the liquid in the vat is
m = ρV = ρπ (0.50 m) 2 d = (1550.5 kg/m3 )π (0.50 m) 2 (2.0 m) = 2.4 × 103 kg
15.7. Model: The density of water is 1000 kg/m3 and the density of ethyl alcohol is 790 kg/m3 .
Solve: (a) The volume of water that has the same mass as 8.0 m3 of ethyl alcohol is
Vwater =
mwater
ρwater
=
malcohol
ρwater
=
ρalcoholValcohol ⎛ 790 kg/m3 ⎞
(8.0 m3 ) = 6.3 m3
=⎜
⎜ 1000 kg/m3 ⎟⎟
ρwater
⎝
⎠
(b) The pressure at the bottom of the cubic tank is p = p0 + ρwater gd:
p = 1.013 × 105 Pa + (1000 kg/m3 )(9.8 m/s 2 )(6.3)1/ 3 = 1.2 × 105 Pa
where we have used the relation d = (Vwater )1/ 3 .
15.8. Model: The density of oil is ρoil = 900 kg/m3 and the density of water is ρwater = 1000 kg/m3.
Visualize:
Solve: The pressure at the bottom of the oil layer is p1 = p0 + ρoil gd1, and the pressure at the bottom of the water layer is
p2 = p1 + ρwater gd 2 = p0 + ρoil gd1 + ρwater gd 2
5
p2 = (1.013 × 10 Pa) + (900 kg/m3 )(9.8 m/s 2 )(0.5 m) + (1000 kg/m3 )(9.8 m/s 2 )(1.2 m) = 1.2 × 105 Pa
Assess: A pressure of 1.2 × 105 Pa = 1.2 atm is reasonable.
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Fluids and Elasticity
15-5
15.9. Model: The density of seawater is ρseawater = 1030 kg/m 2 .
Visualize:
Solve: The pressure outside the submarine’s window is pout = p0 + ρseawater gd , where d is the maximum safe depth
for the window to withstand a force F. This force is F/A = pout − pin , where A is the area of the window. With
pin = p0 , we simplify the pressure equation to
pout − p0 =
d=
F
= ρseawater gd
A
⇒ d=
F
Aρseawater g
1.0 × 106 N
2
π (0.10 m) (1030 kg/m 2 )(9.8 m/s 2 )
= 3.2 km
Assess: A force of 1.0 × 106 N corresponds to a pressure of
ρ=
F 1.0 × 106 N
=
= 314 atm
A π (0.10 m)2
A depth of 3 km is therefore reasonable.
15.10. Visualize:
We assume that the seal is at a radius of 5 cm. Outside the seal, atmospheric pressure presses on both sides of the
cover and the forces cancel. Thus, only the 10-cm-diameter opening inside the seal is relevant, not the 20 cm
diameter of the cover.
Solve: Within the 10 cm diameter area where the pressures differ,
Fto left = patmos A
Fto right = pgas A
where A = π r 2 = 7.85 × 10−3 m 2 is the area of the opening. The difference between the forces is
Fto left − Fto right = ( patmos − pgas ) A = (101,300 Pa − 20,000 Pa)(7.85 × 10−3 m 2 ) = 0.64 kN
Normally, the rubber seal exerts a 0.64 kN force to the right to balance the air pressure force. To pull the cover off,
an external force must pull to the right with a force > 0.64 kN.
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15-6
Chapter 15
Section 15.3 Measuring and Using Pressure
15.11. Model: The density of water is ρ = 1000 kg/m3.
Visualize: Please refer to Figure 15.16.
Solve: From the figure and the equation for hydrostatic pressure, we have
p0 + ρ gh = patmos
Using p0 = 0 atm, and patmos = 1.013 × 105 Pa, we get
0 Pa + (1000 kg/m3 )(9.81 m/s 2 ) h = 1.013 × 105 Pa
⇒ h = 10.3 m
Assess: This large value of h is due to water having a much smaller density than mercury.
15.12. Model: Assume that the oil is incompressible. Its density is 900 kg/m3 .
Visualize: Please refer to Figure 15.18. Because the liquid is incompressible, the volume displaced in the left
cylinder of the hydraulic lift is equal to the volume displaced in the right cylinder.
Solve: Equating the two volumes,
2
A1d1 = A2d 2 ⇒ (π r12 )d1 = (π r22 ) d 2
2
⎛r ⎞
⎛ 0.040 m ⎞
⇒ d1 = ⎜ 2 ⎟ d 2 = ⎜
⎟ (0.20 m) = 3.2 m
⎝ 0.010 m ⎠
⎝ r1 ⎠
15.13. Model: Assume that the vacuum cleaner can create zero pressure.
Visualize:
Solve: The gravitational force on the dog is balanced by the force resulting from the pressure difference between the
atmosphere and the vacuum ( phose = 0) in the hose. The force applied by the hose is
F = ( patmos − phose ) A = patmos A = mg
A=
(10 kg)(9.8 m/s 2 )
1.013 × 105 Pa
= 9.7 × 10−4 m 2
Since A = π ( d/2) 2 , the diameter of the hose is d = 2 A/π = 0.035 m = 3.5 cm.
Section 15.4 Buoyancy
15.14. Model: The buoyant force on the sphere is given by Archimedes’ principle.
Visualize:
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Fluids and Elasticity
15-7
Solve: The sphere is in static equilibrium because it is neutrally buoyant. That is,
∑ Fy = FB − FG = 0 N ⇒ ρlVl g − ms g = 0 N
The sphere displaces a volume of liquid equal to its own volume, Vl = Vs , so
ρl =
ms
ms
0.0893 kg
=
=
= 7.9 × 102 kg/m3
3
3
4
4
Vs
r
(0
.
030
m)
π
π
3 s
3
A density of 790 kg/m3 in Table 15.1 identifies the liquid as ethyl alcohol.
Assess: If the density of the fluid and an object are equal, we have neutral buoyancy.
15.15. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
Vcyl is the volume of the cylinder and Vw is the volume of the water displaced by the cylinder. Note that the volume
displaced is only from the part of the cylinder that is immersed in water.
Solve: The cylinder is in static equilibrium, so FB = FG . The buoyant force is the weight ρwVw g of the displaced
water. Thus
FB = ρ wVw g = FG = mg = ρcylVcyl g
ρcyl = (1000 kg/m3 )
⇒ ρ wVw = ρcylVcyl
⇒
ρcyl = ρ w
Vw
Vcyl
A(0.040 m)
= 6.7 × 102 kg/m3
A(0.060 m)
where A is the cross-sectional area of the cylinder.
Assess: ρcyl < ρ w for a cylinder floating in water is an expected result.
15.16. Model: The buoyant force on the sphere is given by Archimedes’ principle.
Visualize:
Solve: The sphere is in static equilibrium. The free-body diagram on the sphere shows that
1
4
∑ Fy = FB − T − FG = 0 N ⇒ FB = T + FG = FG + FG = FG
3
3
4
3
3
ρ wVsphere g = ρsphereVsphere g ⇒ ρsphere = ρ w = (1000 kg/m3 ) = 750 kg/m3
3
4
4
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15-8
Chapter 15
15.17. Model: The buoyant force on the rock is given by Archimedes’ principle.
Visualize:
Solve: Because the rock is in static equilibrium, Newton’s first law gives
Fnet = T + FB − ( FG ) rock = 0 N
⎞
1
1
ρ
⎛1
⎞
⎛
⎞
⎛
⎞⎛ m g ⎞ ⎛
T = ρ rockVrock g − ρ water ⎜ Vrock ⎟ g = ⎜ ρ rock − ρ water ⎟Vrock g = ⎜ ρrock − ρ water ⎟ ⎜ rock ⎟ = ⎜1 − water ⎟ mrock g
2
2
2
2
ρ
ρ
⎝
⎠
⎝
⎠
⎝
⎠ ⎝ rock ⎠ ⎝
rock ⎠
Using ρ rock = 4800 kg/m3 and mrock = 5.0 kg, we get T = 44 N.
15.18. Model: The buoyant force on the aluminum block is given by Archimedes’ principle. The density of
aluminum and ethyl alcohol are ρ Al = 2700 kg/m3 and ρethyl alcohol = 790 kg/m3.
Visualize:
The buoyant force FB and the tension due to the string act vertically up, and the gravitational force on the aluminum block
acts vertically down. The block is submerged, so the volume of displaced fluid equals VAl , the volume of the block.
Solve: The aluminum block is in static equilibrium, so
∑ Fy = FB + T − FG = 0 N ⇒ ρf VAl g + T − ρ AlVAl g = 0 N ⇒ T = VAl g ( ρ Al − ρf )
T = (100 × 10−6 m3 )(9.81 m/s2 )(2700 kg/m3 − 790 kg/m3 ) = 1.9 N
where we have used the conversion 100 cm3 = 100 × (10−2 m)3 = 10−4 m3 .
Assess: The gravitational force on the aluminum block is ρAlVAl g = 2.65 N. A similar order of magnitude for T is
reasonable.
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Fluids and Elasticity
15-9
15.19. Model: The buoyant force on the steel cylinder is given by Archimedes’ principle.
Visualize:
The length of the cylinder above the surface of mercury is d.
Solve: The cylinder is in static equilibrium with FB = FG . Thus
FB = ρHgVHg g = FG = mg = ρcylVcyl g
d = 0.20 m −
⇒ ρHgVHg = ρcylVcyl
⇒ ρHg A(0.20 m − d ) = ρcyl A(0.20 m)
⎛
ρcyl
7900 kg/m3 ⎞
(0.20 m) = (0.20 m) ⎜1 −
= 0.084 m = 8.4 cm
⎜ 13,600 kg/m3 ⎟⎟
ρHg
⎝
⎠
That is, the length of the cylinder above the surface of the mercury is 8.4 cm.
15.20. Model: The buoyant force is determined by Archimedes’ principle. Ignore any compression the air in the
beach ball may undergo as a result of submersion.
Solve: The mass of the beach ball is negligible, so the force needed to push it below the water is equal to the buoyant force.
⎛4
⎞
⎡4
⎤
FB = pw ⎜ π R3 ⎟ g = (1000 kg/m3 ) ⎢ π (0.30 m)3 ⎥ (9.8 m/s 2 ) = 1.1 kN
⎝3
⎠
⎣3
⎦
Assess: It would take a 113 kg (250 lb) person to push the ball below the water. Two people together could do it.
This seems about right.
15.21. Model: The buoyant force on the sphere is given by Archimedes’ principle.
Visualize:
Solve: For the Styrofoam sphere and the mass not to sink, the sphere must be completely submerged and the buoyant
force FB must be equal to the sum of the gravitational force on the Styrofoam sphere and the attached mass.
Neglecting the volume of the hanging mass, the volume of displaced water equals the volume of the sphere, so
FB = ρwaterVwater g = (1000 kg/m3 )
( 43π ) (0.25 m)3 (9.8 m/s2 ) = 641.4 N
( FG )Styrofoam = ρStyrofoamVStyrofoam g = (150 kg/m3 ) ⎡ 43 π (0.25 m)3 ⎤ (9.8 m/s 2 ) = 96.2 N
⎣
⎦
Because ( FG )Styrofoam + mg = FB ,
m=
FB − ( FG )Styrofoam
g
=
641.4 N − 96.2 N
9.8 m/s 2
= 55.6 kg
To two significant figures, the mass is 56 kg.
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15-10
Chapter 15
Section 15.5 Fluid Dynamics
15.22. Model: Treat the water as an ideal fluid. The hose is a flow tube, so the equation of continuity applies.
Solve: The volume flow rate is
Q=
600 L 600 × 10−3 m3
=
= 1.25 × 10−3 m3/s
8.0 min
8.0 × 60 s
Using the definition Q = vA = vπ (d/2)2 , we get
d=
4Q
4(1.25 × 10−3 m3/s)
=
= 2.0 cm
πv
π (4.0 m/s)
Assess: This is a reasonable diameter for a typical garden hose.
15.23. Model: Treat the water as an ideal fluid. The pipe itself is a flow tube, so the equation of continuity applies.
Visualize:
Note that A1, A2 , and A3 and v1, v2 , and v3 are the cross-sectional areas and the speeds in the first, second, and third
segments of the pipe, respectively.
Solve: (a) The equation of continuity is
A1v1 = A2v2 = A3v3
⇒ π r12v1 = π r22v2 = π r32v3
⇒ r12v1 = r22v2 = r32v3
(0.0050 m) 2 (4.0 m/s) = (0.010 m) 2 v2 = (0.0025 m) 2 v3
2
⎛ 0.0050 m ⎞
v2 = ⎜
⎟ (4.0 m/s) = 1.0 m/s
⎝ 0.010 m ⎠
2
⎛ 0.0050 m ⎞
v3 = ⎜
⎟ (4.0 m/s) = 16 m/s
⎝ 0.0025 m ⎠
(b) The volume flow rate through the pipe is
Q = A1v1 = π (0.0050 m) 2 (4.0 m/s) = 3.1 × 10−4 m3/s
15.24. Model: Treat the water as an ideal fluid so that the flow in the tube follows the continuity equation.
Visualize:
Solve: The equation of continuity is v0 A0 = v1 A1, where A0 = L2 and A1 = π
⎛L⎞
v0 L2 = v1π ⎜ ⎟
⎝2⎠
2
( 12 L )
2
. The above equation simplifies to
⎛4⎞
⇒ v1 = ⎜ ⎟ v0 = 1.27v0
⎝π ⎠
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Fluids and Elasticity
15-11
15.25. Model: Treat the oil as an ideal fluid obeying Bernoulli’s equation. Consider the path connecting point 1 in
the lower pipe with point 2 in the upper pipe a streamline.
Visualize: Please refer to Figure EX15.25.
Solve: Bernoulli’s equation is
p2 + 12 ρ v22 + ρ gy2 = p1 + 12 ρ v12 + ρ gy1 ⇒
Using p1 = 200 kPa = 2.0 × 105 Pa,
ρ = 900 kg/m3 ,
p2 = p1 + 12 ρ (v12 − v22 ) + ρ g ( y1 − y2 )
y2 − y1 = 10.0 m,
v1 = 2.0 m/s, and v2 = 3.0 m/s, we get
5
p2 = 1.096 × 10 Pa = 110 kPa.
Section 15.6 Elasticity
15.26. Model: Turning the tuning screws on a guitar string creates tensile stress in the string.
Solve: The tensile stress in the string is given by T/A, where T is the tension in the string and A is the cross-sectional
area of the string. From the definition of Young’s modulus,
T/A
T ⎛L⎞
Y=
⇒ ΔL = ⎜ ⎟
ΔL/L
A⎝Y ⎠
Using T = 2000 N, L = 0.80 m, A = π (0.00050 m) 2 , and Y = 20 × 1010 N/m 2 (from Table 15.3), we obtain
ΔL = 0.010 m = 1.0 cm.
Assess: 1.0 cm is a large stretch for a length of 80 cm, but 2000 N is a large tension.
15.27. Model: The dangling mountain climber creates tensile stress in the rope.
Solve: Young’s modulus for the rope is
Y=
F/A stress
=
ΔL/L strain
The tensile stress is
(70 kg)(9.8 m/s 2 )
π (0.0050 m)
2
= 8.734 × 106 Pa
and the strain is 0.080 m/50 m = 0.00160. Dividing the two quantities yields Y = 5.5 × 109 N/m 2 .
15.28. Model: The hanging mass creates tensile stress in the wire.
Solve: The force (F) pulling on the wire, which is simply the gravitational force (mg) on the hanging mass, produces
tensile stress given by F/A, where A is the cross-sectional area of the wire. The fractional change in the wire’s length
is ΔL/L = 0.010. From the definition of Young’s modulus, we have
mg/A
(π r 2 )Y ( ΔL/L) π (0.50 × 10−3 m) 2 (7 × 1010 N/m 2 )(0.010)
Y=
⇒ m=
=
= 60 kg
ΔL/L
g
9.8 m/s 2
15.29. Model: The load supported by a concrete column creates compressive stress in the concrete column.
Solve: The gravitational force on the load produces tensile stress given by F/A, where A is the cross-sectional area of
the concrete column and F equals the gravitational force on the load. From the definition of Young’s modulus,
Y=
F/A
ΔL/L
2
3.0 m
⎛ F ⎞⎛ L ⎞ ⎡ (200,000 kg)(9.8 m/s ) ⎤ ⎛
⎞
⇒ ΔL = ⎜ ⎟⎜ ⎟ = ⎢
= 1 mm
⎥⎜
2
10
2⎟
π (0.25 m)
⎝ A ⎠⎝ Y ⎠ ⎢⎣
⎥⎦ ⎝ 3 × 10 N/m ⎠
Assess: A compression of 1.0 mm of the concrete column by a load of approximately 200 tons is reasonable.
15.30. Model: Water is almost incompressible and it applies a volume stress.
Solve: (a) The pressure at a depth of 5000 m in the ocean is
p = p0 + ρsea water g (5000 m) = 1.013 × 105 Pa + (1030 kg/m3 )(9.8 m/s 2 )(5000 m) = 5.057 × 107 Pa = 5.1 × 107 Pa
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15-12
Chapter 15
(b) Using the bulk modulus of water,
ΔV
p
5.057 × 107 Pa
=2 =2
= 20.025
V
B
0.2 × 1010 Pa
(c) The volume of a mass of water decreases from V to 0.975V. Thus the water’s density increases from ρ to
ρ /0.975. The new density is
ρ5000 m =
1030 kg/m3
= 1056 kg/m3
0.975
15.31. Solve: The pressure p at depth d in a fluid is p = p0 + ρ gd . Using 1.29 kg/m3 for the density of air,
pbottom = ptop + ρair gd
pbottom − ptop = (1.29 kg/m3 )(9.8 m/s 2 )(16 m) = 202 Pa = 1.99 × 10−3 atm
⇒
Assuming pbottom = 1 atm,
pbottom − ptop
pbottom
=
1.99 × 10−3 atm
= 0.20,
1 atm
15.32. Model: We assume that there is a perfect vacuum inside the cylinders with p = 0 Pa. We also assume that
the atmospheric pressure in the room is 1 atm.
Visualize: Please refer to Figure P15.32.
Solve: (a) The flat end of each cylinder has an area A = π r 2 = π (0.30 m) 2 = 0.283 m 2 . The force on each end is thus
Fatm = p0 A = (1.013 × 105 Pa)(0.283 m 2 ) = 2.86 × 104 N
To two significant figures, the force on each end is 2.9 × 104 N.
(b) The net vertical force on the lower cylinder when it is on the verge of being pulled apart is
∑ Fy = Fatm − ( FG ) players = 0 N ⇒ ( FG )players = Fatm = 2.86 × 104 N
number of players =
2.86 × 104 N
(100 kg)(9.8 m/s 2 )
= 29.2
That is, 30 players are needed to pull the two cylinders apart.
15.33. Model: Assume that the oil is incompressible and its density is 900 kg/m3 .
Visualize: Please refer to Figure P15.33.
Solve: (a) The pressure at depth d in a fluid is p = p0 + ρ gd . Here, pressure p0 at the top of the fluid is due both to
the atmosphere and to the gravitational force on the piston. That is, p0 = patm + ( FG ) p /A. At point A,
pA = patm +
( FG ) P
+ ρ g (1.00 m − 0.30 m)
A
= 1.013 × 105 Pa +
(10 kg)(9.8 m/s 2 )
π (0.020 m) 2
+ (900 kg/m3 )(9.8 m/s 2 )(0.70 m) = 185,460 Pa
FA = pA A = (185,460 Pa)π (0.10 m) 2 = 5.8 kN
(b) In the same way,
( FG )P
+ ρ g (1.30 m) = 190,752 Pa ⇒
A
Assess: FB is larger than FA , because pB is larger than pA .
pB = patm +
FB = 6.0 kN
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Fluids and Elasticity
15-13
15.34. Model: Assume that blood is incompressible.
Solve: When lying down, the pressure at both the heart and the brain is at 118 mm Hg (gauge pressure) or 118 mm
Hg+ 760 mm Hg = 878 mm Hg absolute pressure. Upon standing up, the pressure in the heart remains at 878 mm Hg,
but the pressure in the brain is reduced according to the hydrostatic pressure equation:
pH = pB + ρ gh
where B and H refer to brain and heart, respectively. Inserting h = 0.40 m and solving for pB gives
⎛ 760 mm Hg ⎞
pB = pH − ρ gh = 878 mm Hg − (1060 kg/m3 )(9.8 m/s 2 )(0.40 m) ⎜
⎟ = 847 mm Hg
⎝ 101.3 kPa ⎠
In gauge pressure, the blood pressure in the brain is pB = 847 mm Hg − 760 mm Hg = 87 mm Hg.
Assess: This blood pressure is below the 90 mm Hg that can cause fainting, which explains why people whose blood
vessels do not constrict to compensate for this effect feel faint when standing from a prone position.
15.35. Model: The tire flattens until the pressure force against the ground balances the upward normal force of the
ground on the tire.
Solve: The area of the tire in contact with the road is A = (0.15 m)(0.13 m) = 0.0195 m 2 . The normal force on each
tire is
n=
FG (1500 kg)(9.8 m/s 2 )
=
= 3675 N
4
4
Thus, the pressure inside each tire is
n
3675 N
14.7 psi
pinside = =
= 188,500 Pa = 1.86 atm ×
= 27 psi
2
A 0.0195 m
1 atm
15.36. Visualize:
Solve: (a) Because the patient’s blood pressure is 140/100, the minimum fluid pressure needs to be 100 mm of Hg
above atmospheric pressure. Since 760 mm of Hg is equivalent to 1 atm and 1 atm is equivalent to 1.013 × 105 Pa,
the minimum pressure is 100 mm = 1.333 × 104 Pa. The excess pressure in the fluid is due to force F pushing on the
internal 6.0-mm-diameter piston that presses against the liquid. Thus, the minimum force the nurse needs to apply to
the syringe is
F = fluid pressure × area of plunger = (1.333 × 104 Pa) ⎡π (0.0030 m)2 ⎤ = 0.38 N
⎣
⎦
(b) The flow rate is Q = vA, where v is the flow speed of the medicine and A is the cross-sectional area of the
needle. Thus,
Q 2.0 × 10−6 m3/2.0 s
v= =
= 20 m/s
A π (0.125 × 10−3 m) 2
Assess: Note that the pressure in the fluid is due to F that is not dependent on the size of the plunger pad. Also note
that the syringe is not drawn to scale.
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15-14
Chapter 15
15.37. Solve: The fact that atmospheric pressure at sea level is 101.3 kPa = 101,300 N/m2 means that the weight of
the atmosphere over each square meter of surface is 101,300 N. Thus the mass of air over each square meter is m =
(101,300 N)/g = (101,300 N)/(9.80 m/s 2 ) = 10,340 kg per m 2 . Multiplying by the earth’s surface area will give the
total mass. Using Re = 6.27 × 106 m for the earth’s radius, the total mass of the atmosphere is
M air = Aearth m = (4π Re2 ) m = 4π (6.37 × 106 m) 2 (10,340 kg/m 2 ) = 5.27 × 1018 kg
15.38. Visualize: Let d be the atmosphere’s thickness, p the atmospheric pressure on the earth’s surface, and
p0 (= 0 atm) the pressure beyond the earth’s atmosphere.
Solve: The pressure at a depth d in a fluid is p = p0 + ρ gd . This equation becomes
1 atm = 0 atm + ρair gd
⇒ d=
1 atm
1.013 × 105 Pa
=
= 8.0 km
ρair g (1.3 kg/m3 )(9.8 m/s 2 )
15.39. Model: Oil is incompressible and has a density 900 kg/m3 .
Visualize: Please refer to Figure P15.39.
Solve: (a) The pressure at point A, which is 0.50 m below the open oil surface, is
pA = p0 + ρoil g (1.00 m − 0.50 m) = 101,300 Pa + (900 kg/m3 )(9.8 m/s 2 )(0.50 m) = 106 kPa
(b) The pressure difference between A and B is
pB − pA = ( p0 + ρ gd B ) − ( p0 + ρ gd A ) = ρ g (d B − d A ) = (900 kg/m3 )(9.8 m/s 2 )(0.50 m) = 4.4 kPa
Pressure depends only on depth, and C is the same depth as B. Thus pC − pA = 4.4 kPa also, even though C isn’t
directly under A.
15.40. Model: Assume that oil is incompressible and its density is ρ = 900 kg/m3 .
Visualize: Please refer to Figure P15.40.
Solve: (a) The hydraulic lift is in equilibrium and the pistons on the left and the right are at the same level.
Therefore, Equation 15.10 simplifies to
Fleft piston Fright piston
( FG )student ( FG )elephant
=
⇒
=
Aleft piston Aright piston
π (rstudent )2 π (relephant )2
⎛ (F )
⎞
(70 kg) g
rstudent = ⎜ G student ⎟ (relephant ) =
(1.0 m) = 0.2415 m
⎜ ( FG )elephant ⎟
(1200 kg) g
⎝
⎠
The diameter of the piston the student is standing on is therefore 2 × 0.2415 m = 0.48 m.
(b) With the second student added, the pistons are no longer at the same height. Therefore, Equation 15.10 gives
Fright piston Fleft piston
=
− ρ gh
Aright piston Aleft piston
With Fleft piston = ms g + (70 kg)g and h = 0.35 m, we can solve for the second students mass ms :
ms =
=
melephant Aleft piston
Aright piston
− 70 kg + ρ hAleft piston
(1200 kg)(0.2415 m) 2
(1.0 m) 2
− 70 kg + (900 kg/m3 )(0.35 m)π (0.2415 m) 2 = 58 kg
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Fluids and Elasticity
15-15
15.41. Model: Assume that the oil is incompressible and its density is 900 kg/m3 .
Visualize:
Solve: The pressures p1 and p2 are equal. Thus,
p0 +
F1
F
= p0 + 2 + ρ gh ⇒
A1
A2
F1 F2
=
+ ρ gh
A1 A2
With F1 − m1g , F2 = 4m2 g , A1 = π r12 , and A2 = π r2 2 , we have
m1g
π r12
=
4m2 g
π r22
1/ 2
⎛ 4m g ⎞
+ ρ gh ⇒ r2 = ⎜ 2 ⎟
⎝ π ⎠
⎛ m1g
⎞
⎜⎜ 2 − ρ gh ⎟⎟
⎝ πr1
⎠
−1/2
Using m1 = 55 kg, m2 = 110 kg, r1 = 0.08 m, ρ = 900 kg/m3 , and h = 1.0 m, the calculation yields r2 = 0.276 m.
The diameter is 55 cm.
Assess: Both pistons are too small to hold the people as shown, but the ideas are correct.
15.42. Model: Water and mercury are incompressible and immiscible liquids.
Visualize:
The water in the left arm floats on top of the mercury and presses the mercury down from its initial level. Because
points 1 and 2 are level with each other and the fluid is in static equilibrium, the pressure at these two points must be
equal. If the pressures were not equal, the pressure difference would cause the fluid to flow, violating the assumption
of static equilibrium.
Solve: The pressure at point 1 is due to water of depth d w = 10 cm:
p1 = patmos + ρw gd w
Because mercury is incompressible, the mercury in the left arm goes down a distance h while the mercury in the right
arm goes up a distance h. Thus, the pressure at point 2 is due to mercury of depth d Hg = 2h:
p2 = patmos + ρHg gd Hg = patmos + 2 ρHg gh
Equating p1 and p2 gives
patmos + ρw gd w = patmos + 2 ρHg gh ⇒ h =
3
1 ρw
⎛ 1 ⎞ ⎛ 1000 kg/m ⎞
dw = ⎜ ⎟ ⎜
⎟ (10 cm) = 3.7 mm
2 ρHg
⎝ 2 ⎠ ⎜⎝ 13,600 kg/m3 ⎟⎠
The mercury in the right arm rises 3.7 mm above its initial level.
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15-16
Chapter 15
15.43. Model: Glycerin and ethyl alcohol are incompressible and do not mix.
Visualize:
Solve: The alcohol in the left arm floats on top of the denser glycerin and presses the glycerin down distance h from
its initial level. This causes the glycerin to rise distance h in the right arm. Points 1 and 2 are level with each other
and the fluids are in static equilibrium, so the pressures at these two points must be equal:
p1 = p2
⇒
h=
p0 + ρeth gd eth = p0 + ρgly gdgly
⇒
ρeth g (20 cm) = ρgly g (2h)
3
1 ρeth
⎛ 1 ⎞ ⎛ 790 kg/m ⎞
(20 cm) = ⎜ ⎟ ⎜
⎟ (20 cm) = 6.27 cm
2 ρgly
⎝ 2 ⎠ ⎜⎝ 1260 kg/m3 ⎟⎠
You can see from the figure that the difference between the top surfaces of the fluids is
Δy = 20 cm − 2h = 20 cm − 2(6.27 cm) = 7.46 cm ≈ 7.5 cm
15.44. Model: Assume the liquid is incompressible.
Visualize:
Solve: (a) Can 2 has moved down with respect to Can 1 since the water level in Can 2 has risen. Since the total
volume of water stays constant, the water level in Can 1 has fallen by the same amount. The water level is equalized
in the two cans at the middle of the height change, so the change in height of the water is half the relative change in
height of the cans. Can 2 has moved relative to Can 1 (6.5 cm − 5.0 cm) × 2 = 3.0 cm down.
(b) The water level in Can 1 has fallen by the same amount. The new level is 5.0 − 1.5 cm = 3.5 cm.
Assess: The two cans are an inexpensive method of measuring relative changes in height.
15.45. Model: The water is in hydrostatic equilibrium.
Visualize:
Solve: (a) The force of the liquid on the bottom is the pressure p = ρ gD times the area A = WL. This gives
F = pA = ρ gDWL
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Fluids and Elasticity
15-17
(b) The pressure on the front panel of the aquarium can be found by integrating the force on a thin strip dy (see
figure above). Using F = pA, we find
dF = ρ gyLdy
D
F = Ñρ gyLdy = 12 ρ gD 2 L
0
(c) For a 100-cm-long, 35-cm-wide, 40-cm-deep aquarium filled with water, the force from the liquid on the front
window is
F = 12 (1000 kg/m3 )(9.8 m/s 2 )(0.40 m) 2 (1.0 m) = 784 N ≈ 0.78 kN
The force on the bottom is
F = (1000 kg/m3 )(9.8 m/s 2 )(0.40 m)(0.35 m)(1.0 m) = 1372 N ≈ 1.4 kN
15.46. Visualize:
The figure shows a small column of air of thickness dz, of cross-sectional area A = 1 m 2 , and of density ρ ( z ). The
column is at a height z above the surface of the earth.
Solve: (a) The atmospheric pressure at sea level is 1.013 × 105 Pa . That is, the weight of the air column with a 1 m 2
cross section is 1.013 × 105 N. Consider the weight of a 1 m 2 slice of thickness dz at a height z. This slice has
volume dV = Adz = (1 m 2 )dz , so its weight is dw = ( ρ dV ) g = ρ g (1 m 2 )dz = ρ0e − z/z0 g (1 m 2 )dz. The total weight of
the 1 m 2 column is found by adding all the dw. Integrating from z = 0 to z = ∞,
∞
w = ∫ ρ0 g (1 m 2 )e − z/z0 dz
0
∞
= ⎡ − ρ0 g (1 m 2 ) z0 ⎤ ⎡e − z/z0 ⎤
⎣
⎦⎣
⎦0
= ρ0 g (1 m 2 ) z0
Because w = 101,300 N = ρ0 g (1 m 2 ) z0 ,
z0 =
101,300 N
(1.28 kg/m3 )(9.8 m/s 2 )(1.0 m 2 )
= 8.1 × 103 m
(b) Using the density at sea level from Table 15.1,
ρ = (1.28 kg/m3 )e− z/ (8.08×10
3
m)
= (1.28 kg/m3 )e −1600 m/(8.08×10
3
m)
= 1.05 kg/m3
This is 82% of ρ0 .
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15-18
Chapter 15
15.47. Model: Ignore the change in water density over lengths comparable to the size of the fish.
Visualize:
Solve: The buoyant force is the force of gravity on the volume of water displaced by the fish times, or
FB = ρ w (VF + Vb ) g
where the subscripts w, F, and b indicate water, fish, and bladder, respectively. The force due to gravity on the fish is
FG = ρ FVF g + ρairVb g
For neutral buoyancy, the two forces must have equal magnitude. Equating them and solving for the Vb gives
ρ w (VF + Vb ) g = ρ FVF g + ρairVb g
Vb =
ρw − ρF
1000 kg/m3 − 1080 kg/m3
VF =
VF = 0.0801VF
ρair − ρ w
1.19 kg/m3 − 1000 kg/m3
so the fish needs to increase its volume by 8.01%.
15.48. Model: The buoyant force on the ceramic statue is given by Archimedes’ principle.
Visualize:
Solve: The gravitational force on the statue is the 28.4 N registered on the scale in air. In water, the gravitational
force on the statue is balanced by the sum of the buoyant force FB and the spring’s force on the statue; that is,
( FG )statue = FB + Fspring on statue
ρstatue =
⇒ 28.4 N = ρ wVstatue g + 17.0 N ⇒ Vstatue =
11.4 N mstatue
=
g ρw
ρstatue
( mstatue g ) ρ w (28.4 N)(1000 kg/m3 )
=
= 2.49 × 103 kg/m3
(11.4 N)
(11.4 N)
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Fluids and Elasticity
15-19
15.49. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
Solve: (a) Initially, as it floats, the cylinder is in static equilibrium, with the buoyant force balancing the
gravitational force on the cylinder. The volume of displaced liquid is Ah, so
FB = ρliq ( Ah) g = FG
Force F pushes the cylinder down distance x, so the submerged length is h + x and the volume of displaced liquid is
A(h + x). The cylinder is again in equilibrium, but now the buoyant force balances both the gravitational force and
force F. Thus
FB = ρliq [ A(h + x)]g = FG + F
Since ρliq ( Ah) g = FG , we’re left with
F = ρliq Agx
(b) The amount of work dW done by force F to push the cylinder from x to x + dx is dW = Fdx = ( ρliq Agx)dx. To
push the cylinder from xi = 0 m to xf = 10 cm = 0.10 m requires work
xf
xf
W = Ñ Fdx = ρliq Ag Ñ xdx = 12 ρliq Ag ( xi2 − xf2 )
xi
=
1 (1000
2
xi
3
kg/m )π (0.020 m)2 (9.8 m/s 2 )(0.10 m) 2 = 0.62 J
15.50. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
Let d1 be the length of the cylinder in the less-dense liquid with density ρ1, and d 2 be the length of the cylinder in
the more-dense liquid with density p2 .
Solve: The cylinder is in static equilibrium, so
∑ Fy = FB1 + FB2 − FG = 0 N ⇒
ρ1 ( Ad1 ) g + ρ 2 ( Ad 2 ) g = ρ A(d1 + d 2 ) g
ρ1
ρ
d1 + d 2 =
(d1 + d 2 )
ρ2
ρ2
Since l = d1 + d 2
⇒ d1 = l − d 2 , we can simplify the above equation to obtain
⎛ ρ − ρ1 ⎞
d2 = ⎜
⎟l
⎝ ρ 2 − ρ1 ⎠
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15-20
Chapter 15
That is, the fraction of the cylinder in the more dense liquid is f = ( ρ − ρ1 ) / ( ρ 2 − ρ1 ).
Assess: As expected f = 0 when ρ is equal to ρ1, and f = 1 when ρ = ρ 2 .
15.51. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
Solve: The tube is in static equilibrium, so
∑ Fy = FB − ( FG ) tube − ( FG ) Pb = 0 N ⇒
ρliquid =
ρliquid A(0.25 m) g = (0.030 kg) g + (0.250 kg) g
(0.280 kg)
(0.280 kg)
=
= 8.9 × 102 kg/m3
A(0.25 m) π (0.020 m) 2 (0.25 m)
Assess: This is a reasonable value for a liquid.
15.52. Model: Archimedes’ Principle determines the buoyant force.
Visualize:
Solve: The plastic hemisphere will hold the most weight when its rim is at the surface of the water. The buoyant
force balances the gravitational force on the bowl and rock.
∑ Fy = FB − ( FG ) rock − ( FG ) bowl = 0 N
Thus
⎛ 1 ⎞⎡4
⎤
mg = pwVbowl g − m bowl g = (1000 kg/m3 ) ⎜ ⎟ ⎢ π (0.040 m)3 ⎥ (9.8 m/s 2 ) − (0.021 kg)(9.8 m/s 2 ) ⇒ m = 0.16 kg
2
3
⎝ ⎠⎣
⎦
Assess: Putting a rock as big as 160 g in an 8-cm-diameter bowl before it sinks is reasonable.
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Fluids and Elasticity
15-21
15.53. Model: The buoyant force is determined by Archimedes’ principle. The spring is ideal.
Visualize:
Solve: The spring is stretched by the same amount that the cylinder is submerged. The buoyant force and spring
force balance the gravitational force on the cylinder.
∑ Fy = FB + FS − mg = 0 N
pw Ayg + ky = mg
y=
mg
(1.0 kg)(9.8 m/s 2 )
=
3
pw Ag + k (1000 kg/m )π (0.025 m) 2 (9.8 m/s 2 ) + 35 N/m
= 0.181 m = 18 cm
Assess: This is difficult to assess because we don’t know the height h of the cylinder and can’t calculate it without
the density of the metal material.
15.54. Model: The buoyant force is determined by Archimedes’ principle. The cross-sectional area of the boat is
denoted by A.
Visualize:
Solve: For the boat to have neutral buoyancy, the buoyant force must have the same magnitude as the force due to
gravity:
FB = Fg ⇒ ρ dAg = ( M + nm) g
where ρ is the density of the liquid, M is the boat’s mass, and n is an integer. Rearranging this expression gives
nm = ρ dA − M
If we plot the mass nm added to the boat as a function of the depth d, the y-intercept will be the negative of the boat’s
mass M and the slope s will be related to the liquid’s density as s = ρ /A.
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15-22
Chapter 15
From the fit to the data we find M = 29 g and s = 2.667 kg/m. This gives a density of
s
2.667 kg/m
ρ= =
= 1067 kg/m3 = 1.1 × 103 kg/m3
A 25 × 10−4 m 2
Assess: The mass of the boat is some 6 times greater than the mass of the blocks, which seems reasonable. The
density of the liquid is slightly greater than the density of water, which is also a reasonable result.
15.55. Model: The buoyant force on the can is given by Archimedes’ principle.
Visualize:
The length of the can above the water level is d, the length of the can is L, and the cross-sectional area of the can is A.
Solve: The can is in static equilibrium, so
∑ Fy = FB − ( FG )can − ( FG ) water = 0 N 12 ρ water A( L − d ) g = (0.020 kg) g + mwater g
The mass of the water in the can is
⎛V
mwater = ρ water ⎜ can
⎝ 2
⎞
3 355 × 10
⎟ = (1000 kg/m )
2
⎠
−6
m3
= 0.1775 kg
ρ water A( L − d ) = 0.020 kg + 0.1775 kg = 0.1975 kg ⇒ d − L = 2
0.1975 kg
= 0.0654 m
ρ water A
Because Vcan = π (0.031 m) 2 L = 355 × 10−6 m3 , L = 0.1176 m. Using this value of L, we get d = 0.0522 m ≈ 5.2 cm.
Assess: d/L = 5.22 cm/ 11.76 cm = 0.444, thus 44.4% of the length of the can is above the water surface. This is
reasonable.
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Fluids and Elasticity
15-23
15.56. Model: The buoyant force on the boat is given by Archimedes’ principle.
Visualize:
The minimum height of the boat that will enable the boat to float in perfectly calm water is h.
Solve: The boat barely floats if the water comes completely to the top of the sides. In this case, the volume of
displaced water is the volume of the boat. Archimedes’ principle in equation form is FB = ρ wVboat g . For the boat to
float FB = ( FG ) boat . Let us first calculate the gravitational force on the boat:
( FG ) boat = ( FG )bottom + 2( FG )side 1 + 2( FG )side 2 , where
( FG ) bottom = ρsteelVbottom g = (7900 kg/m3 )(5.0 × 10 × 0.020 m3 ) g = 7900g N
( FG )side 1 = ρsteelVside 1g = (7900 kg/m3 )(5.0 × h × 0.0050 m3 ) g = 197.5gh N
( FG )side 2 = ρsteelVside 2 g = (7900 kg/m3 )(10 × h × 0.0050 m3 ) g = 395gh
( FG ) boat = [7900g + 2(197.5gh) + 2(395gh)] N = (7900+1185h) g N
Going back to the Archimedes’ equation and remembering that h is in meters, we obtain
ρ wVboat g = (7900 + 1185h) g N ⇒ (1000)[10 × 5.0 × (h + 0.020)] = 7900 + 1185h
h = 14 cm
15.57. Model: The two pipes are identical.
Visualize:
Solve: The water speed is the same in both pipes. The flow rate is
Q = 3.0 × 106 L/min = 2(vA)
Q
v=
=
2A
⎛ 1 ⎞
(3.0 × 106 )(10−3 ) ⎜ ⎟ m3/s
⎝ 60 ⎠
= 3.5 m/s
2π (1.5 m)2
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15-24
Chapter 15
15.58. Model: Treat the liquid as an ideal liquid obeying Bernoulli’s equation (Equation 15.27).
Visualize:
Solve: (a) Bernoulli’s equation gives
p0 + 12 ρ v02 + ρ gy0 = p1 + 12 ρ v12 + ρ gy1
Because the pipe expands equally in the positive and negative y directions, the contributions from the gravitationalpotential-energy term cancel by symmetry, so Bernoulli’s equation reduces to
p0 + 12 ρ v02 = p1 + 12 ρ v12
Expressing p1 in terms of p0 gives
p1 = p0 + 12 ρ (v02 − v12 )
Finally, we can express the velocity v1 in terms of v0 using the continuity equation (Equation 15.17):
v0 A0 = v1 A1 ⇒ v1 = v0
d02
d12
Inserting this into the expression for p0 gives
⎛ d4 ⎞
p1 = p0 + 12 ρ v02 ⎜1 − 04 ⎟
⎜ d ⎟
1 ⎠
⎝
(b) Inserting the given values gives
⎡ ⎛ 16.8 cm ⎞4 ⎤
p1 = 50 kPa + 12 (1000 kg/m3 )(10.0 m/s) 2 ⎢1 − ⎜
⎟ ⎥ = 50 kPa + 25.1 kPa = 75 kPa
⎢ ⎝ 20.0 cm ⎠ ⎥
⎣
⎦
Assess: We find that the pressure increases where the velocity has decreased, which is in accordance with
Bernoulli’s equation.
15.59. Solve: Treat the sap as incompressible and assume sap is almost entirely made of water.
Solve: To replace the water lost to the atmosphere, the vessels must supply 110 g/h of sap. The sap provided by each
vessel is ρ vπ (d/2)2 , so for N = 2000 sap vessels, the flow rate in each vessel is
2
4(0.110 kg/h)(1 h/ 3600 s)
⎛d ⎞
N ρ vπ ⎜ ⎟ = 0.110 kg/h ⇒ v =
= 187 nm/s
π (1040 kg/m3 )(2000)(100 × 10−6 m)2
⎝2⎠
15.60. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins in the bigger size
pipe and ends at the exit of the narrower pipe.
Visualize: Please see Figure P15.60. Let point 1 be beneath the standing column and point 2 be where the water
exits the pipe.
Solve: (a) The pressure of the water as it exits into the air is p2 = patmos .
(b) Bernoulli’s equation, Equation 15.26, relates the pressure, water speed, and heights at points 1 and 2:
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ p1 − p2 = 12 ρ (v22 − v12 ) + ρ g ( y2 − y1 )
From the continuity equation,
v1 A1 = v2 A2 = (4 m/s)(5 × 10−4 m 2 ) ⇒ v1 (10 × 10−4 m 2 ) = 20 × 10−4 m3/s ⇒ v1 = 2.0 m/s
Substituting into Bernoulli’s equation,
p1 − p2 = p1 − patmos = 12 (1000 kg/m3 )[(4.0 m/s) 2 − (2.0 m/s) 2 ] + (1000 kg/m3 )(9.8 m/s)(4.0 m)
= 6000 Pa + 39,200 Pa = 45 kPa
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Fluids and Elasticity
15-25
But p1 − p2 = ρ gh, where h is the height of the standing water column. Thus
h=
45 × 103 Pa
(1000 kg/m3 )(9.8 m/s 2 )
= 4.6 m
15.61. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins at the faucet and
continues down the stream.
Visualize:
The pressure at point 1 is p1 and the pressure at point 2 is p2 . Both p1 and p2 are atmospheric pressure. The
velocity and the area at point 1 are v1 and A1 and they are v2 and A2 at point 2. Let d be the distance of point 2
below point 1.
Solve: The flow rate is
2.0 × 1000 × 10−6 m3
2.0 × 10−4 m3/s
= 2.0 × 10−4 m3/s ⇒ v1 =
= 1.0 m/s
10 s
π (0.0080 m)2
Bernoulli’s equation at points 1 and 2 is
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ ρ gd = 12 ρ (v22 − v12 )
Q = v1 A =
From the continuity equation,
v1 A1 = v2 A2 ⇒ (1.0 m/s)π (8.0 × 10−3 m)2 = v2π (5.0 × 10−3 m) 2
⇒ v2 = 2.56 m/s
Going back to Bernoulli’s equation, we have
gd = 12 [(2.56 m/s) 2 − (1.0 m/s) 2 ] ⇒ d = 0.283 m ≈ 28 cm
15.62. Model: Treat the air as an ideal fluid obeying Bernoulli’s equation.
Solve: (a) The pressure above the roof is lower due to the higher velocity of the air.
(b) Bernoulli’s equation, with yinside ≈ youtside , is
pinside = poutside + 12 ρair v 2
2
⇒ Δp =
1
1
⎛ 130 × 1000 m ⎞
ρair v 2 = (1.28 kg/m3 ) ⎜
⎟ = 835 Pa
2
2
⎝ 3600 s ⎠
The pressure difference is 0.83 kPa
(c) The force on the roof is (Δp ) A = (835 Pa)(6.0 m × 15.0 m) = 7.5 × 104 N. The roof will blow up, because pressure
inside the house is greater than pressure on the top of the roof.
15.63. Model: The ideal fluid obeys Bernoulli’s equation.
Visualize: Please refer to Figure P15.63. There is a streamline connecting point 1 in the wider pipe on the left with
point 2 in the narrower pipe on the right. The air speeds at points 1 and 2 are v1 and v2 and the cross-sectional area
of the pipes at these points are A1 and A2 . Points 1 and 2 are at the same height, so y1 = y2 .
Solve: The volume flow rate is Q = A1v1 = A2v2 = 1200 × 10−6 m3/s. Thus
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15-26
Chapter 15
v2 =
1200 × 10−6 m3/s
π (0.0020 m)2
= 95.49 m/s
v1 =
1200 × 10−6 m3/s
π (0.010 m)2
= 3.82 m/s
Now we can use Bernoulli’s equation to connect points 1 and 2:
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2
p1 − p2 = 12 ρ (v22 − v12 ) + ρ g ( y2 − y1 ) = 12 (1.28 kg/m3 )[(95.49 m/s)2 − (3.82 m/s)2 ] + 0 Pa = 5.83 kPa
Because the pressure above the mercury surface in the right tube is p2 and in the left tube is p1, the difference in the
pressures p1 and p2 is ρ Hg gh. That is,
p1 − p2 = 5.83 kPa = ρ Hg gh ⇒ h =
5.83 × 103 Pa
(13,600 kg/m3 )(9.8 m/s 2 )
= 4.4 cm
15.64. Model: The ideal fluid (i.e., air) obeys Bernoulli’s equation.
Visualize: Please refer to Figure P15.64. There is a streamline connecting points 1 and 2. The air speeds at points 1
and 2 are v1 and v2 , and the cross-sectional areas of the pipes at these points are A1 and A2 . Points 1 and 2 are at
the same height, so y1 = y2 .
Solve: (a) The height of the mercury is 10 cm. So, the pressure at point 2 is larger than at point 1 by
ρ Hg g (0.10 m) = (13,600 kg/m3 )(9.8 m/s 2 )(0.10 m) = 13,328 Pa ⇒
p2 = p1 + 13,328 Pa
Using Bernoulli’s equation,
p1 + 12 ρair v12 + ρair gy1 = p2 + 12 ρair v22 + ρair gy2
v12 − v22 =
2( p2 − p1 )
⇒
p2 − p1 = 12 ρair (v12 − v22 )
2(13,328 Pa)
= 20,825 m 2 /s 2
(1.28 kg/m3 )
From the continuity equation, we can obtain another equation connecting v1 and v2:
ρair
A1v1 = A2v2
⇒ v1 =
=
A2
π (0.0050 m)2
v2 =
v2 = 25v2
A1
π (0.0010 m)2
Substituting v1 = 25v2 in the Bernoulli equation, we get
(25 v2 ) 2 − v22 = 20,825 m 2 /s 2
⇒ v2 = 5.78 m/s
2
Thus v2 = 5.8 m/s and v1 = 25v2 = 1.4 × 10 m/s.
(b) The volume flow rate A2v2 = π (0.0050 m) 2 (5.78 m/s) = 4.5 × 10−3 m3/s.
15.65. Model: Treat the water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline from the top
of the water to the hole.
Visualize: The top of the water (at y1 = h) and the hole (at y2 = y ) are at atmospheric pressure. The speed of the
water at the top is zero because the tank is kept filled.
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Fluids and Elasticity
15-27
Solve: (a) Bernoulli’s equation connecting the two points gives
0 + ρ gh = 12 ρ v 2 + ρ gy ⇒ v = 2 g (h − y )
(b) For a particle shot horizontally from a height y with speed v, the range can be found using kinematic equations.
For the y-motion, using t0 = 0 s, we have
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2
⇒ 0 m = y + (0 m/s)t1 + 12 ( − g )t12 ⇒ t1 = 2 y/g
For the x-motion,
x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2
⇒
x = 0 m + vt1 + 0 m ⇒
x = v 2 y/g
(c) Combining the results of (a) and (b), we obtain
x = 2 g (h − y ) 2 y/g = 4 y (h − y )
To find the maximum range relative to the vertical height,
dx
1
=0 ⇒
[4( h − y ) − 4 y ] = 0 ⇒
dy
4 y (h − y )
y=
h
2
With y = 12 h, the maximum range is
h⎞
⎛ h ⎞⎛
xmax = 4 ⎜ ⎟⎜ h − ⎟ = h
2⎠
⎝ 2 ⎠⎝
15.66. Model: Treat the water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline from the top
of the water to the hole.
Visualize: The top of the water and the hole are at atmospheric pressure. The speed of the water at the top is not zero
because the tank is emptying.
Solve: (a) Bernoulli’s equation connecting the two points gives
2
2
patm + ρ gh + 12 ρ vtop
= patm + 12 ρ vhole
+ ρ g (h − d )
The continuity equation relates the flow rates at the surface and at the hole:
⎛r⎞
⇒ vtop = vhole ⎜ ⎟
⎝R⎠
Inserting this into Bernoulli’s equation and solving for vhole gives
2
vtop Atop = vhole Ahole
vhole =
2 gd
1 − (r/R )4
(b) Inserting the given values, we find that the water level will initial at the following rate:
2
2
⎛ 2.0 × 10−3 m ⎞
2(9.8 m/s 2 )(1.0 m)
=
= 1.1 mm/min
⎜
⎟
⎟ 1 − [(2.0 × 1023 m) /(1.0 m)]4
1 − ( r/R ) 4 ⎜⎝ 1.0 m
⎠
Assess: For the given dimensions, the result seems reasonable.
⎛r⎞ ⎛r⎞
vtop = vhole ⎜ ⎟ = ⎜ ⎟
⎝R⎠ ⎝R⎠
2 gd
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15-28
Chapter 15
15.67. Model: The aquarium creates tensile stress.
Solve: Weight of the aquarium is
FG = mg = ρ waterVg = (1000 kg/m3 )(10 m3 )(9.8 m/s 2 ) = 9.8 × 104 N
where we have used the conversion 1 L = 1023 m3 . The weight supported by each wood post is
1 (9.8 × 104
4
N) =
2.45 × 104 N. The cross-sectional area of each post is A = (0.040 m) 2 = 1.6 × 10−3 m 2 . Young’s modulus for the
wood is
Y = 1 × 1010 N/m 2 =
ΔL =
F/A
FL
=
ΔL/L AΔL
FL
(2.45 × 104 N)(0.80 m)
=
= 1.23 × 10−3 m = 1 mm
AY (1.6 × 10−3 m 2 )(1 × 1010 N/m 2 )
Assess: A compression of 1 mm due to a weight of 2.45 × 104 N is reasonable.
15.68. Model: The weight of the person creates tensile stress in the disk cartilage.
Solve: The cross-sectional area of each disk of cartilage is π r 2 . Inserting this into Equation 15.34 gives
ΔL
F
=Y
A
L
ΔL =
FL mgL (33 kg)(9.8 m/s 2 )(0.0050 m)
=
=
= 1.3 mm
YA
YA (1.0 × 106 N/m 2 )π (0.020 m) 2
Assess: The cartilage experiences a significant compressive axial strain (~25%).
15.69. Model: Pressure applies a volume stress to water in the cylinder.
Solve: The volume strain of water due to the pressure applied is
ΔV
p
2 × 106 Pa
=− =−
= −1 × 10−3
V
B
0.2 × 1010 Pa
ΔV = V ′ − V = −(1 × 10−3 )(1.30 m3 ) = −1.30 × 10−3 m3 ≈ −1 L
As the safety plug on the top of the cylinder bursts, the water comes back to atmospheric pressure. To the precision
of the data (one significant figure), one liter of water comes out.
15.70. Model: Air is an ideal gas and obeys Boyle’s law.
Visualize: Please refer to Figure CP15.70. The quantity h is the length of the air column when the mercury fills the
cylinder to the top. A is the cross-sectional area of the cylinder.
Solve: For the column of air, Boyle’s law is p0V0 = p1V1, where p0 and V0 are the pressure and volume before any
mercury is poured, and p1 and V1 are the pressure and volume when mercury fills the cylinder above the air. Using
p1 = p0 + ρ Hg g (1.0 m − h), Boyle’s law becomes
p0V0 = [ p0 + ρ Hg g (1.0 m − h)]V1 ⇒
p0 (1.0 m − h) = ρ Hg g (1.0 m − h) h ⇒ h =
p0 A(1.0 m) = [ p0 + ρ Hg g (1.0 m − h)] Ah
p0
ρ Hg g
=
1.013 × 105 Pa
(13,600 kg/m3 )(9.8 m/s 2 )
= 0.76 m = 76 cm
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Fluids and Elasticity
15-29
15.71. Model: The buoyant force on the cone is given by Archimedes’ principle.
Visualize:
Solve: It may seem like we need the formula for the volume of a cone. You can use that formula if you know it, but
it isn’t essential. The volume is clearly the area of the base multiplied by the height multiplied by some constant.
That is, the cone shown above has V = aAl where a is some constant. But the radius of the base is r = l tan α , where
α is the angle of the apex of the cone, and A = π r 2 , making A proportional to l 2 . Thus the volume of a cone of
height l is V = cl 3 , where c is a constant. Because the cone is floating in static equilibrium, we must have FB = FG .
The cone’s density is ρ0 , so the gravitational force on it is FG = ρ0Vg = ρ0cl 3 g . The buoyant force is the
gravitational force on the displaced fluid. The volume of displaced fluid is the full volume of the cone minus the
volume of the cone of height h above the water, or Vdisp = cl 3 − ch3. Thus FB = ρf Vdisp g = ρf c(l 3 − h3 ) g , and the
equilibrium condition is
FB = FG
⇒
3
3
3
ρf c(l − h ) g = ρ0cl g ⇒
⎛ h3 ⎞
ρf ⎜ 1 − 3 ⎟ = ρ0
⎜
l ⎟⎠
⎝
1/3
⇒
h ⎛ ρ0 ⎞
= ⎜1 −
⎟
l ⎝ ρf ⎠
15.72. Model: The grinding wheel is a uniform disk. We will use the model of kinetic friction and hydrostatics.
Visualize: Please refer to Figure CP15.72.
Solve: This is a three-part problem. First find the desired angular acceleration, then use that to find the force applied
by each brake pad, then finally the needed oil pressure. The angular acceleration required to stop the wheel is found
using rotational kinematics.
Δω ωf − ω i 0 rad/s − 900 × 2π × (1/ 60) rad/s
α=
=
=
= −18.85 rad/s 2
Δt
Δt
5.0 s
Each brake pad applies a frictional force f k = μ k n to the wheel. The normal force is equal to the force applied by the
piston by Newton’s third law. Rotational dynamics can be used to find the magnitude of the force. f k is applied 12 cm
from the rotation axis on both sides of the disk.
τ net = Iα
⎛1
⎞
22 f k (0.12 m) = ⎜ MR 2 ⎟α
2
⎝
⎠
MR 2α
(15 kg)(0.13 m) 2 (−18.85 rad/s 2 )
=
= 16.6 N
24 μ k (0.12 m)
24(0.60)(0.12 m)
The oil pressure required to generate this much force at each brake pad is
n=
p=
F
16.6 N
=
= 53 kPa
A π (0.010 m) 2
relative to atmospheric pressure. The absolute pressure is 53 kPa + 101.3 kPa = 154.3 kPa = 1.5 × 105 Pa to two
significant figures.
Assess: The required oil pressure is about half an atmosphere, which is quite reasonable.
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15-30
Chapter 15
15.73. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
A is a cross-sectional area of the cylinder.
Solve: (a) The gravitational force on the cylinder is FG = ρ0 ( Al ) g and the weight of the displaced fluid is
FB = ρf ( Ah) g . In static equilibrium, FB = FG and we can write
ρf Ahg = ρ0 Alg ⇒ h = ( ρ0 / ρf )l
(b) Now the volume of the displaced liquid is A(h − y ). Applying Newton’s second law in the y-direction gives
∑ Fy = − FG + FB = − ρ0 Alg + ρf A(h − y ) g
Using ρf Ahg = ρ0 Alg from part (a), we find
( Fnet ) y = − ρf Agy
(c) The result in part (b) is F = − ky , where k = ρf Ag . This is Hooke’s law.
(d) Since the cylinder’s equation of motion is determined by Hooke’s law, the angular frequency for the resulting
simple harmonic motion is ω = k/m , and the period is
T=
2π
ω
= T = 2π
m
m
ρ0 Al
h
= 2π
= 2π
= 2π
k
ρf Ag
ρf Ag
g
where we have used the expression for h from part (a).
(e) The oscillation period for the 100-m-tall iceberg ( pice = 917 kg/m3 ) in sea water is
T = 2π
h
ρ0l
(917 kg/m3 )(100 m)
= 2π
= 2π
= 18.9 s
g
ρf g
(1030 kg/m3 )(9.80 m/s 2 )
15.74. Model: A streamline connects every point on the surface of the liquid to a point in the drain. The drain
diameter is much smaller than the tank diameter (r R ).
Visualize:
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Fluids and Elasticity
15-31
Solve: The pressures at the surface and drain (points 1 and 2) are equal to one atmosphere. When the liquid is a
depth y, Bernoulli’s principle connecting points 1 and 2 is
1
1
p1 + ρ v12 + ρ gy = p2 + ρ v22 + ρ g (0)
2
2
v22 = v12 + 2 gy
The flow rate through the drain is the same as through a horizontal layer in the tank:
A
Q = v1 A1 = v2 A2 ⇒ v1 = v2 2
A1
Thus the velocity of the liquid through the drain is
2
v22
⎛ A ⎞
2 gy
= ⎜ v2 2 ⎟ + 2 gy ⇒ v2 =
A
1 − ( r/R ) 4
1⎠
⎝
Since r R, v2 ≈ 2 gy , and the flow rate through the drain is
Q ≈ π r 2 2 gy
The flow rate gives the volume of liquid flowing out per unit time. The inverse gives the time needed for a unit
volume of liquid to flow out. For the volume of liquid to decrease by dV requires a time
dV −(π R 2dy )
− R 2 dy
dt =
= 2
= 2
Q π r 2 gy r 2 g y
Note that dV < 0 implies the volume of liquid is decreasing. Integrating both sides from the initial condition
(t = 0, y = d ) to the final condition (t , y = 0) yields
t=−
R2
r
2
0
∫y
2g
d
− 12
dy = −
2R2
r
2
2g
y
1
2
0
d
=
R2
r
2
2d
g
Assess: The time for the tank to drain depends on the ratio of the cross-sectional areas of the tank to drain, which
makes sense, as well as on the strength of the acceleration due to gravity. Note that, if g were larger (say we were on
Jupiter), the time to drain would be shorter, whereas if the tank were taller (larger d), the time would be longer.
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A MACROSCOPIC DESCRIPTION OF MATTER
16
Conceptual Questions
16.1. Convert them to the absolute Kelvin scale to compare.
T1 = 0 K
T2 = 0°C = 273 K because TK = TC = 273
5
T3 = 0°F = 255 K because TC = (TF − 32) = −17.8°C ⇒ TK = 255 K
9
So T1 < T3 < T2 .
16.2. We must use the absolute temperature scale when we double the temperature.
10°C = 283 K
2 ⋅ 283 K = 566 K = 293°C
16.3. (a) Yes. Because the water-ice phase boundary has a negative slope down to the triple point and the solid-gas
phase boundary has a positive slope up to the triple point, ice does not exist for temperatures greater than the triple
point.
(b) No. The positive slope of the solid-vapor boundary begins at the point (0 K, 0 atm). This means that as long as
the pressure is low enough, water can always exist in the vapor phase.
16.4. The pressure must be the same on both sides of the piston; were that not the case then there would be a greater
force on one side and the net force would cause the piston to accelerate.
16.5. (a) The ideal-gas law says
⎛pV ⎞
⎛ 3 p ⎞⎛ 2V ⎞
p1V1 p2V2
=
⇒ T2 = ⎜ 2 2 ⎟ T1 = ⎜ 1 ⎟⎜ 1 ⎟ T1 = 6T1
T1
T2
⎝ p1V1 ⎠
⎝ p1 ⎠⎝ V1 ⎠
⎛1 ⎞
⎛ p2V2 ⎞
⎛ 3 p1 ⎞ ⎜ 2 V1 ⎟
3
(b) T2 = ⎜
⎟ T1 = T1.
⎟ T1 = ⎜
⎟⎜
p
V
p
V
2
⎝ 11⎠
⎝ 1 ⎠⎜ 1 ⎟
⎝
⎠
16.6. (a) Because the container is sealed the number of moles in the container remains constant.
(b)
n1
p n
p
3 p1 n2 3 n1
= 1 2= 2 =
=
.
V1 RT1 V2 RT2 R (2T1 ) V2 2 V1
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16-1
16-2
Chapter 16
16.7. It is likely that the air pressure in the underwater apartment is higher than normal atmospheric air pressure, so
the freezing point of water would be lower than the value at the surface, and the boiling point would be higher than
the value at the surface. This is because the solid-liquid transition line in the phase diagram has a negative slope, but
the liquid-gas transition line has a positive slope.
16.8. (a) As the piston squeezes the sample the pressure increases. To see what happens refer to the phase diagram
in Figure 16.4. As the pressure on the gas increases from that initial point on the phase diagram it passes from the gas
region to the solid region and eventually to the liquid region.
(b) When the initial temperature is warmer (+0.02°C) the sample never passes through the solid state, but goes
directly from the gas state to the liquid state.
16.9. (a) The ideal-gas law says
⎛ V ⎞⎛ T ⎞
⎛ V ⎞⎛ 3T ⎞
p1V1 p2V2
3
=
⇒ p2 = ⎜ 1 ⎟⎜ 2 ⎟ p1 = ⎜ 1 ⎟⎜ 1 ⎟ p1 = p1
T1
T2
2
⎝ V2 ⎠⎝ T1 ⎠
⎝ 2V1 ⎠⎝ T1 ⎠
⎛
⎞
⎜ V1 ⎟ ⎛ 3T1 ⎞
(b) p2 = ⎜
⎟⎜
⎟ p1 = 6 p1
⎜ 1 V1 ⎟ ⎝ T1 ⎠
⎝2 ⎠
16.10. We want T2 /T1.
p1V1 p2V2
T
pV
⇒ 2 = 2 2 but p2 = p1, and from the figure V2 = 3V1, so
=
T1
T2
T1
p1V1
T2 3V1
=
=3
T1 V1
16.11. The figure shows that V2 = V1. We also know p1 = 2 atm, T1 = 300 K, and T2 = 1200 K.
⎛ V ⎞⎛ T ⎞
⎛ V ⎞ ⎛ 1200 K ⎞
p1V1 p2V2
=
⇒ p2 = ⎜ 1 ⎟⎜ 2 ⎟ p1 = ⎜ 1 ⎟ ⎜
⎟ (2 atm) = 8 atm
T1
T2
V
T
⎝ 2 ⎠⎝ 1 ⎠
⎝ V1 ⎠ ⎝ 300 K ⎠
16.12. Part (a) of the figure represents a constant pressure, or isobaric, expansion of the gas. Part (b) represents a
constant volume reduction of pressure of the gas. During part (b), the temperature also decreases. Part (c) represents
a decrease in volume along with an increase in pressure. However, part (c) is not isothermal since the graph is a
straight line. Isothermal processes are hyperbolas on pV diagrams.
A correct diagram would look like the following figure.
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A Macroscopic Description of Matter
16-3
Exercises and Problems
Section 16.1 Solids, Liquids, and Gases
16.1. Model: Recall the density of water is 1000 kg/m3.
Solve: The mass of gold mAu = ρ AuVAu = (19,300 kg/m3 )(100 × 10−6 m3 ) = 1.93 kg. For water to have the same mass
its volume must be
Vwater =
mwater
ρ water
=
1.93 kg
1000 kg/m3
= 0.00193 m3 ≈ 1900 cm3
Assess: Since the lead is 19.3 times as dense we expect the water to take 19.3 times the volume.
16.2. Model: Assume the nucleus is spherical.
Solve: The volume of the uranium nucleus is
V = 43 π R3 = 43 π (7.5 × 10−15 m)3 = 1.767 × 10−42 m3
The density of the uranium nucleus is
ρ nucleus =
mnucleus
4.0 × 10−25 kg
=
= 2.3 × 1017 kg/m3
Vnucleus 1.767 × 10−42 m3
Assess: This density is extremely large compared to the typical density of materials.
16.3. Model: The volume of a hollow sphere is
V=
4π 3
(rout − rin3 )
3
Solve: We are given m = 0.690 kg, rout = 0.050 m, and we know that for aluminum ρ = 2700 kg/m3.
Solve the above equation for rin .
3
rin = 3 rout
−
3
= 3 rout
−
V
4π
3
m/ρ
4π
3
= 3 0.050 m3 −
0.690 kg/2700 kg/m3
4π
3
= 0.040 m
So the inner diameter is 8.0 cm.
Assess: We are happy that the inner diameter is less than the outer diameter, and in a reasonable range.
16.4. Solve: The volume of the aluminum cube is 10−3 m3 and its mass is
mAl = ρ AlVAl = (2700 kg/m3 )(1.0 × 10−3 m3 ) = 2.7 kg
The volume of the copper sphere with this mass is
4π
m
2.7 kg
= 3.027 × 10−4 m3
VCu =
(rCu )3 = Cu =
3
ρCu 8920 kg/m3
1
⇒ rCu
⎡ 3(3.027 × 10−4 m3 ) ⎤ 3
=⎢
⎥ = 0.042 m
4π
⎣⎢
⎦⎥
The diameter of the copper sphere is 0.0833 m = 8.3 cm.
Assess: The diameter of the sphere is a little less than the length of the cube, and this is reasonable considering the
density of copper is greater than the density of aluminum.
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16-4
Chapter 16
Section 16.2 Atoms and Moles
16.5. Solve: The volume of the aluminum cube V = 8.0 × 10−6 m3 and its mass is
M = ρV = (2700 kg/m3 )(8.0 × 10−6 m3 ) = 0.0216 kg = 21.6 g
One mole of aluminum ( 27Al ) has a mass of 27 g. The number of atoms is
⎛ 6.02 × 1023 atoms ⎞ ⎛ 1 mol ⎞
23
N =⎜
⎟⎟ ⎜
⎟ (21.6 g) = 4.8 × 10 atoms
⎜
1
mol
27
g
⎠
⎝
⎠⎝
Assess: This is almost one mole of atoms, which is a reasonable value.
16.6. Solve: The volume of the copper cube is 8.0 × 10−6 m3 and its mass is
M = ρV = (8920 kg/m3 )(8.0 × 10−6 m3 ) = 0.07136 kg = 71.36 g
Because the atomic mass number of Cu is 64, one mole of Cu has a mass of 64 g. The number of moles in the cube is
⎛ 1 mol ⎞
n=⎜
⎟ (71.36 g) = 1.1 mol
⎝ 64 g ⎠
Assess: This answer is in the same ballpark as the previous exercise.
16.7. Solve: (a) The number density is defined as N/V , where N is the number of particles occupying a volume V.
Because Al has a mass density of 2700 kg/m3 , a volume of 1 m3 has a mass of 2700 kg. We also know that the
molar mass of Al is 27 g/mol or 0.027 kg/mol. So, the number of moles in a mass of 2700 kg is
⎛ 1 mol ⎞
5
n = (2700 kg) ⎜
⎟ = 1.00 × 10 mol
⎝ 0.027 kg ⎠
The number of Al atoms in 1.00 × 105 mols is
N = nN A = (1.00 × 105 mol)(6.02 × 1023 atoms/mol) = 6.02 × 10 28 atoms
Thus, the number density is
N 6.02 × 1028 atoms
=
= 6.02 × 1028 atoms/m3
V
1 m3
(b) Pb has a mass of 11,300 kg in a volume of 1 m3. Since the atomic mass number of Pb is 207, the number of
moles in 11,300 kg is
⎛ 1 mole ⎞
n = (11,300 kg) ⎜
⎟
⎝ 0.207 kg ⎠
The number of Pb atoms is thus N = nN A , and hence the number density is
N nN A ⎛ 11,300 kg ⎞ ⎛
atoms
23 atoms ⎞ (1 mol)
=
=⎜
= 3.28 × 1028
⎟⎜ 6.02 × 10
⎟
3
V
V
0.207
kg
mol
m3
⎠ 1m
⎝
⎠⎝
Assess: We expected to get very large numbers like this.
16.8. Solve: The mass density is ρ = M/V . The mass M of the sample is related to the number of atoms N and the
mass m of each atom by M = Nm. Combining these, the atomic mass is
m=
M ρV
ρ
1750 kg/m3
=
=
=
= 3.986 × 10−26 kg/atom
N
N
N/V 4.39 × 1028 atoms/m3
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A Macroscopic Description of Matter
16-5
the atomic mass in m = A u, where A is the atomic mass number. Thus
A=
m 3.986 × 10−26 kg
=
= 24
1 u 1.661 × 10−27 kg
The element’s atomic mass number is 24.
Assess: This is a reasonable answer for an isotope of neon (although neon is a gas at normal temperatures), sodium,
or magnesium.
16.9. Solve: The mass of mercury is
⎛ 10−6 m3 ⎞
= 0.136 kg = 136 g
M = ρV = (13,600 kg/m3 )(10 cm3 ) ⎜
⎜ 1 cm3 ⎟⎟
⎝
⎠
and the number of moles is
M
0.136 g
n=
=
= 0.6766 mol
M mol 201 g/mol
The mass of aluminum with 0.6766 mol of Al is
⎛ 27 g ⎞
M = (0.6766 mol) M mol = (0.6766 mol) ⎜
⎟ = 18.27 g = 0.01827 kg
⎝ mol ⎠
This mass M of aluminum corresponds to a volume of
M 0.01827 kg
=
= 6.8 × 10−6 m3 = 6.8 cm3
V=
ρ 2700 kg/m3
Assess: We expected an answer in the same order of magnitude. The size of atoms doesn’t vary as much as the
density of atoms from element to element.
4
3
Visualize: We want to know D = 2r. Because the atomic mass number of gold is 197, 1.0 mol of gold has a mass of
16.10. Model: Assume the gold is shaped into a solid sphere of volume V = π r 3.
197 g or 0.197 kg. Table 16.1 gives ρ = 19,300 kg/m3.
Solve:
4
3 M
3
0.197 kg
V = π r 3 = M/ρ ⇒ r = 3
=3
= 0.0135 m = 1.35 cm
3
4π ρ
4π 19,300 kg/m3
D = 2r = 2(1.35 cm) = 2.7 cm.
Assess: This is about the right size for a chunk that contains one mole of material.
Section 16.3 Temperature
Section 16.4 Phase Changes
16.11. Solve: The lowest temperature is
TF = 95 TC + 32° ⇒ −127°F = 95 TC + 32° ⇒ TC = −88°C ⇒ Tk = (−88.3 + 273) K = 185 K
In the same way, the highest temperature is
136°F = 95 TC + 32° ⇒ TC = 58°C = 331 K
Assess: On the absolute scale the highest recorded temperature is not quite twice the lowest.
16.12. Solve: Let TF = TC = T :
TF = 95 TC + 32° ⇒ T = 95 T + 32° ⇒ T = −40°
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16-6
Chapter 16
That is, the Fahrenheit and the Celsius scales give the same numerical value at −40°.
Assess: It is usually unnecessary to specify the scale when the temperature is reported as −40°.
16.13. Model: A temperature scale is a linear scale.
Solve: (a) We need a conversion formula for °C to °Z, analogous to the conversion of °C to °F. Since temperature
scales are linear, TC = aTZ + b, where a and b are constants to be determined. We know the boiling point of liquid
nitrogen is 0°Z and −196°C. Similarly, the melting point of iron is 1000°Z and 1538°C. Thus
−196 = 0a + b
1538 = 1000a + b
From the first, b = −196°. Then from the second, a = (1538 + 196)/1000 = 1734/1000. Thus the conversion is TC =
(1734/1000)TZ − 196°. Since the boiling point of water is TC = 100°C, its temperature in °Z is
⎛ 1000 ⎞
TZ = ⎜
⎟ (100° + 196°) = 171°Z
⎝ 1734 ⎠
(b) A temperature Tz = 500°Z is
⎛ 1734 ⎞
TC = ⎜
⎟ 500° − 196° = 671°C = 944 K
⎝ 1000 ⎠
16.14. Solve: (a) The triple point of water is T = 0.01°C and p = 0.006 atm. Thus
TF = 95 TC + 32° = 95 (0.01) + 32 = 32.02°F
1.013 × 105 Pa
= 608 Pa
1 atm
(b) The triple point of carbon dioxide is T = −56°C and p = 5 atm. Thus
p = 0.006 atm ×
TF = 95 TC + 32° = 95 ( −56°) + 32° = −68.8°F
⎛ 1.013 × 105 Pa ⎞
5
p = 5.0 atm = (5.0 atm) ⎜
⎟⎟ = 5.06 × 10 Pa
⎜
1 atm
⎝
⎠
Section 16.5 Ideal Gases
16.15. Model: Treat the nitrogen gas in the closed cylinder as an ideal gas.
Solve: (a) The density before and after the compression are ρ before = m1/V1 and ρafter = m2 /V2 . Noting that m1 = m2
and V2 = 12 V1,
m V1
ρafter
=
= 2 ⇒ ρafter = 2 ρ before
ρ before V2 m
The mass density has changed by a factor of 2.
(b) The number of atoms in the gas is unchanged, implying that the number of moles in the gas remains the same;
hence the number density is unchanged.
16.16. Model: Treat the gas in the container as an ideal gas.
Solve: From the ideal-gas law pV = nRT , the pressure of the gas is
p=
nRT (3.0 mol)(8.31 J/mol K)[(273 − 120) K]
=
= 1.9 × 106 Pa = 19 atm
3
−3
V
(2.0 × 10 m )
Assess: 19 atm is a high pressure, but not unreasonable.
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A Macroscopic Description of Matter
16-7
16.17. Model: Treat the gas as an ideal gas in the sealed container.
Solve: (a) For p1 = p0 , the before-and-after relationship of an ideal gas in a sealed container is
V1 V0
T
473 K
= ⇒ V1 = V0 1 = V0
= 1.27V0
T1 T0
T0
373 K
where T0 = (273 + 100) K and T1 = (273 + 200) K.
(b) When the Kelvin temperature is doubled, T1 = 2T0 = 2(373 K) = 746 K and the above equation becomes
V1 = V0
T1
746 K
= V0
= 2V0
T0
373 K
Assess: When we use the Kelvin scale we expect the volume to double when the temperature doubles (if the pressure
is kept constant).
16.18. Model: Treat the gas in the sealed container as an ideal gas.
Solve: (a) From the ideal gas law equation pV = nRT , the volume V of the container is
V=
nRT (2.0 mol)(8.31 J/mol K)[(273 + 30 K)]
=
= 0.050 m3
p
1.013 × 105 Pa
Note that pressure must be in Pa in the ideal-gas law.
(b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is
p1V p2V
T
(273 + 130) K
=
⇒ p2 = p1 2 = (1.0 atm)
= 1.3 atm
T1
T2
T1
(273 + 30) K
Note that gas-law calculations must use T in kelvins.
16.19. Model: We’ll assume that air is an ideal gas so we can use the ideal gas law, pV = nRT .
Visualize: We are given V = 5.0 L = 0.0050 m3 , p = 1 atm = 101.3 kPa, and T = 37°C = 310 K.
Also recall that R = 8.31 J/(mol ⋅ K) and oxygen makes up 20% of the air.
Solve: Solve Equation 12.12 for n, the number of moles of air.
n=
pV (101.3 kPa)(0.0050 m3 )
=
= 0.20 mol
RT (8.31 J/(mol ⋅ K))(310 K)
Multiply the number of moles of air by 20% to get the number of moles of oxygen: (0.20 mol)(0.20) = (0.040 mol)
(6.02 × 1023 molecules/mol) = 2.4 × 1022 molecules of oxygen.
Assess: The answer is a small number of moles of oxygen, but a large number of molecules of oxygen.
16.20. Model: Treat the oxygen gas in the cylinder as an ideal gas.
Solve: (a) The number of moles of oxygen is
M
50 g
n=
=
= 1.563 mol ≈ 1.6 mol
M mol 32 g/mol
(b) The number of molecules is
N = nN A = (1.563 mol)(6.02 × 1023 mol−1 ) = 9.41 × 1023 ≈ 9.4 × 1023
(c) The volume of the cylinder V = π r 2 L = π (0.10 m)2 (0.40 m) = 1.257 × 10−2 m3. Thus,
N
9.41 × 1023
=
= 7.5 × 1025 m −3
V 1.257 × 10−2 m3
(d) From the ideal-gas law pV = nRT we can calculate the absolute pressure to be
p=
nRT (1.563 mol)(8.31 J/mol K)(293 K)
=
= 303 kPa
V
1.257 × 10−2 m3
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16-8
Chapter 16
where we used T = 20°C = 293 K. But a pressure gauge reads gauge pressure:
pg = p − 1 atm = 303 kPa − 101 kPa = 202 kPa ≈ 200 kPa
16.21. Model: Treat the neon gas in the sealed cylinder as an ideal gas.
Solve: The volume of the cylinder is V = π r 2h = π (0.05 m)2 (0.30 m) = 2.356 × 10−3 m3. The gauge pressure of the
gas is 120 psi ×
1 atm 1.013 × 105 Pa
×
= 8.269 × 105 Pa, so the absolute pressure of the gas is 8.269 × 105 Pa +
14.7 psi
1 atm
1.013 × 105 Pa = 9.282 × 105 Pa. The temperature of the gas is T = (273 + 30) K = 303 K. The number of moles of the
gas in the cylinder is
n=
pV (9.282 × 105 Pa)(2.356 × 10−3 m3 )
=
= 0.869 mol
RT
(8.31 J/mol K)(303 K)
The mass of the helium is
M = nM mol = (0.869 mol)(20 g/mol) = 17.37 g = 17.37 × 10−3 kg
The mass density is
ρ=
M 17.37 × 10−3 kg
=
= 7.4 kg/m3
V 2.356 × 10−3 m3
Section 16.6 Ideal-Gas Processes
16.22. Model: The gas is assumed to be ideal and it expands isothermally.
Solve: (a) Isothermal expansion means the temperature stays unchanged. That is T2 = T1.
(b) The before-and-after relationship of an ideal gas under isothermal conditions is
⎛V ⎞ p
p1V1 p2V2
V
=
⇒ p2 = p1 1 = p1 ⎜ 1 ⎟ = 1
T1
T1
V2
⎝ 2V1 ⎠ 2
16.23. Model: The gas is assumed to be ideal.
Solve: (a) Isochoric means the volume stays unchanged. That is V2 = V1.
(b) The before-and-after relationship of an ideal gas under isochoric conditions is
⎛1 ⎞
⎜ p1 ⎟ T
p1V1 p2V2
p2
=
⇒ T2 = T1
= T1 ⎜ 3 ⎟ = 1
3
T1
T1
p1
p
⎜⎜ 1 ⎟⎟
⎝
⎠
16.24. Model: The rigid sphere’s volume does not change, so this is an isochoric process. The air is assumed to be
an ideal gas.
Solve: (a) When the valve is closed, the air inside is at p1 = 1 atm and T1 = 100°C. The before-and-after relationship
of an ideal gas in the closed sphere (constant volume) is
⎛T ⎞
p1V p2V
(273 + 0) K
=
⇒ p2 = p1 ⎜ 2 ⎟ = (1.0 atm)
= 0.73 atm
T1
T2
T
(273
+ 100) K
⎝ 1⎠
(b) Dry ice is CO 2 . From Figure 16.4, we can see that the solid-gas transition line gives a temperature of −78°C
when p = 1 atm. Cooling the sphere to −78°C gives
⎛T ⎞
(273 − 78) K
p2 = p1 ⎜ 2 ⎟ = (1.0 atm)
= 0.52 atm
373 K
⎝ T1 ⎠
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A Macroscopic Description of Matter
16-9
16.25. Model: The gas is assumed to be ideal, and since the container is rigid V2 = V1.
Solve: Convert both temperatures to the Kelvin scale.
p1V1 p2V2
T
⎛ 253 K ⎞
=
⇒ p2 = p1 2 = 3 atm ⎜
⎟ = 2.6 atm
T1
T1
T1
⎝ 293 K ⎠
Assess: On the absolute scale the temperature only went up a little bit, so we expect the pressure to rise a little bit.
16.26. Model: Assume the gas is an ideal gas.
Visualize: The pressure in the gas exerts exactly enough upward force to balance the weight of the piston plus the
pressure of the air above the piston. The initial temperature is Ti = 303°C + 273 = 576 K.
Solve: (a) The pressure of the gas in the cylinder is
p = 1 atm +
F
mg
(20 kg)(9.80 m/s 2 )
= 101300 Pa + 2 = 101300 Pa +
= 105600 Pa
A
πr
π (0.12 m) 2
(b) The pressure is determined only by the weight of the piston. It does not change with temperature. The temperature
change is an isobaric process, and the pressure remains 105600 Pa.
For an isobaric process,
Vi Vf
hA h A
h h
=
⇒ i = f
⇒ i= f
Ti Tf
Ti
Tf
Ti Tf
where V = hA is the volume of a cylinder of cross section area A. The final piston height is
T
(15 + 273) K
hf = hi f = 82 cm ×
= 42 cm
Ti
576 K
Assess: The temperature decreased dramatically, so a large decrease in piston height is to be expected.
16.27. Model: In an isochoric process, the volume of the container stays unchanged. Argon gas in the container is
assumed to be an ideal gas.
Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 × 10−6 m3 , and T1 = 20°C = 293 K.
This produces a pressure
nRT (0.1 mol)(8.31 J/mol K)(293 K)
=
= 4.87 × 106 Pa = 4870 kPa ≈ 4900 kPa
p1 =
V1
50 × 10−6 m3
An ideal gas process has p2V2 /T2 = p1V1/T1. Isochoric heating to a final temperature T2 = 300°C = 573 K has
V2 = V1, so the final pressure is
V1 T2
573
p1 = 1 ×
× 4870 kPa = 9520 kPa ≈ 9500 kPa
293
V2 T1
Note that it is essential to express temperatures in kelvins.
(b)
p2 =
Assess: All isochoric processes will be a straight vertical line on a pV diagram.
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16-10
Chapter 16
16.28. Model: The isobaric heating means that the pressure of the argon gas stays unchanged.
Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 × 10−6 m3 , and T1 = 20°C = 293 K.
This produces a pressure
nRT1 (0.1 mol)(8.31 J/mol K)(293 K)
=
= 4.87 × 106 Pa = 4870 kPa ≈ 4900 kPa
p1 =
V1
50 × 10−6 m3
An ideal gas process has p2V2 /T2 = p1V1/T1. Isobaric heating to a final temperature T2 = 300°C = 573 K has p2 = p1,
so the final volume is
V2 =
p1 T2
573
V1 = 1 ×
× 50 cm3 = 97.8 cm3 ≈ 98 cm3
p2 T1
293
(b)
Assess: All isobaric processes will be a straight horizontal line on a pV diagram.
16.29. Model: In an isothermal expansion, the temperature stays the same. The argon gas in the container is
assumed to be an ideal gas.
Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 × 10−6 m3 , and T1 = 20°C = 293 K.
This produces a pressure
p1 =
nRT1 (0.1 mol)(8.31 J/mol K)(293 K)
=
= 4.87 × 106 Pa = 12.02 atm ≈ 12 atm
V1
50 × 10−6 Pa
An ideal-gas process obeys p2V2 /T2 = p1V1/T1. Isothermal expansion to V2 = 200 cm3 gives a final pressure
p2 =
T2 V1
200
p1 = 1 ×
× 12.02 atm = 48 atm
T1 V2
50
(b)
16.30. Model: Assume the gas to be an ideal gas.
Solve: (a) Because the volume stays unchanged, the process is isochoric.
(b) The ideal-gas law p1V1 = nRT1 gives
T1 =
p1V1 (1 × 1.013 × 105 Pa)(200 × 10−6 m3 )
=
= 609.5 K = 336°C
nR
(0.0040 mol)(8.31 J/mol K)
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A Macroscopic Description of Matter
16-11
The final temperature T2 is calculated as follows for an isochoric process:
p1 p2
p
⎛ 3 atm ⎞
=
⇒ T2 = T1 2 = (609.5 K) ⎜
⎟ = 1829 K = 1556°C
T1 T2
p1
⎝ 1 atm ⎠
Assess: All straight vertical lines on a pV diagram represent isochoric processes.
16.31. Model: Assume that the gas is ideal.
Solve: (a) Because the process is at a constant pressure, it is isobaric.
(b) For an ideal gas at constant pressure,
V2 V1
V
300 cm3
= ⇒ T2 = T1 2 = [(273 + 900) K]
= 3519 K = 3246°C
T2 T1
V1
100 cm3
(c) Using the ideal-gas law p2V2 = nRT2 ,
n=
p2V2 (3 × 1.013 × 105 Pa)(300 × 10−6 m3 )
=
= 3.18 × 10−3 mol
RT2
(8.31 J/mol K)(3519 K)
Assess: All straight horizontal lines on a pV diagram represent isobaric processes.
16.32. Model: Assume that the gas is an ideal gas.
Solve: (a) The graph shows that the pressure is inversely proportional to the volume. The process is isothermal.
(b) From the ideal-gas law,
T1 =
p1V1 (3 × 1.013 × 105 Pa)(400 × 1026 m3 )
=
= 731 K = 458°C
nR
(0.020 mol)(8.31 J/mol K)
T2 is also −90°C, because the process is isothermal.
(c) The before-and-after relationship of an ideal gas under isothermal conditions is
p
⎛ 3 atm ⎞
3
p1V1 = p2V2 ⇒ V2 = V1 1 = (400 cm3 ) ⎜
⎟ = 1200 cm
p2
⎝ 1 atm ⎠
16.33. Visualize:
Solve: Suppose we have a 1 m × 1 m × 1 m block of copper of mass M containing N atoms. The atoms are spaced a
distance L apart along all three axes of the cube. There are N x atoms along the x-edge of the cube, N y atoms along
the y-edge, and N z atoms along the z-edge. The total number of atoms is N = N x N y N z . If L is expressed in
meters, then the number of atoms along the x-edge is N x = (1 m)/L. Thus,
1/3
⎛ 1 m3 ⎞
1 m 1 m 1 m 1 m3
N=
×
×
= 3 ⇒L=⎜
⎜ N ⎟⎟
L
L
L
L
⎝
⎠
This relates the spacing between atoms to the number of atoms in a 1-meter cube. The mass of the large cube of
copper is
M = ρCuV = (8920 kg/m3 )(1 m3 ) = 8920 kg
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16-12
Chapter 16
But M = mN , where m = 64 u = 64 × (1.661 × 10−27 kg) is the mass of an individual copper atom. Thus,
N=
M
8920 kg
=
= 8.39 × 1028 atoms
m 64 × (1.661 × 10−27 kg)
1/3
⎛ 1 m3 ⎞
⇒L=⎜
= 2.28 × 10−10 m = 0.228 nm
⎜ 8.39 × 1028 ⎟⎟
⎝
⎠
Assess: This is a reasonable interatomic spacing in a crystal lattice.
16.34. Solve: The volume of the cube associated with each atom is
Vatom = (0.227 × 10−9 m)3 = 1.1697 × 10−29 m3
The volume of a mole of atoms is
V = Vatom N A = (1.1697 × 10−29 m3 )(6.02 × 1023 mol −1 ) = 7.0416 × 10 −6 m3/mol
Thus, the mass of a mole of atoms is
M mol = V ρ = (7.0416 m3/mol)(7950 kg/m3 ) = 0.056 kg/mol = 56 g/mol
The atomic mass number of the element is 56.
Assess: This is likely the most common isotope of iron.
16.35. Model: Assume the density of the liquid water is
1.0 g
.
1.0 cm3
Visualize: There are 10 protons in each water molecule. 1 mol of water molecules has a mass of 18 g.
Solve:
23
⎛ 1.0 g ⎞ ⎛ 1 mol ⎞ ⎛ 6.02 × 10 molecules ⎞ ⎛ 10 protons ⎞
26
1.0 L = 1000 cm3 ⎜
⎜
⎟⎟ ⎜
⎟
⎟⎜
⎟ = 3.3 × 10 protons
1 mol
1
molecule
⎝ 1.0 cm3 ⎠ ⎝ 18 g ⎠ ⎜⎝
⎝
⎠
⎠
Assess: This is an unimaginably large number, but reasonable considering the data.
16.36. Model: Air is an ideal gas. The pressure at sea level is 1 atm.
Solve: From the ideal gas law pV = NkBT , the number density is
N
p
1.013 × 105 Pa
=
=
= 2.5 × 1025 molecules/m3
V kBT (1.38 × 10−23 J/K)(293 K)
Assess: Each cubic meter of air holds an unimaginably large number of molecules.
16.37. Model: Assume the gas in the solar corona is an ideal gas.
Solve: The number density of particles in the solar corona is N/V . Using the ideal-gas equation,
pV = NkBT ⇒
N
p
=
V kBT
(0.03 Pa)
N
=
= 1.1 × 1015 particles/m3
−
V (1.38 × 10 23 J/K)(2 × 106 K)
Assess: This answer is a lot smaller than the one in the previous problem.
16.38. Model: Assume that the gas in the vacuum chamber is an ideal gas.
Solve: (a) The fraction is
pvacuum chamber 1.0 × 10−10 mm of Hg
=
= 1.3 × 10−13
patmosphere
760 mm of Hg
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A Macroscopic Description of Matter
16-13
(b) The volume of the chamber V = π (0.20 m)2 (0.30 m) = 0.03770 m3. From the ideal-gas equation pV = NkBT ,
the number of molecules of gas in the chamber is
N=
pV (1.32 × 10−13 )(1.013 × 105 Pa)(0.03770 m3 )
=
= 1.2 × 1011 molecules
kBT
(1.38 × 10−23 J/K)(293 K)
16.39. Model: Model the oxygen molecules as an ideal gas.
⎛ 1 mol ⎞
2
Visualize: Do some preliminary calculations. n = 100 mg O 2 ⎜
⎟ = 0.003125 mol. The area of the cap is A = π r =
32
g
⎝
⎠
π (0.030 m) 2 = 0.002827 m 2 . The volume of the cylinder is V = π r 2 L = π (0.030 m) 2 (0.10 m) = 0.0002827 m3.
The pressure inside the cylinder will be the outside pressure minus the force needed to remove the cap divided by the
area of the cap. p = 100 kpa −
184 N
0.002827 m 2
= 34.9 × 103 Pa.
Solve: Solve the ideal gas law for temperature.
pV (34.9 × 103 Pa)(0.0002827 m3 )
=
= 380 K = 107°C
nR (0.003125 mol)(8.31 J/mol ⋅ K)
Assess: The answer is much hotter than room temperature, but the pressure was pretty low.
T=
16.40. Model: Assume the nebula gas is ideal.
Visualize: Use pV = NkBT . We are given N/V = 100 atoms/cm3 = 1 × 108 atoms/m3.
Solve:
NkBT
= (1.0 × 108 atoms/m3 )(1.38 × 10−23 J/K)(7500 K) = 1.0 × 10−11 Pa = 1.0 × 10−16 atm
V
Assess: This is much lower pressure than the best vacuum we can achieve in a laboratory, but it is a higher pressure
than in non-nebula space.
p=
16.41. Model: Assume the air is pure N 2 , with a molar mass of M mol = 28 g/mol.
Visualize: We will use pV = nRT both before and after. Our intermediate goal is n1 − n2 . We are given T2 = T1 =
T = 20°C + 273 = 293 K and V2 = V1 = V = π r 2C = π (0.011 m 2 )(2.0 m) = 7.60 × 10−4 m3.
Solve: We convert the gauge pressures to absolute pressures with p = pg + 1 atm = pg + 14.7 psi. p1 = 110 psi +
14.7 psi = 124.7 psi = 860 kPa. p2 = 80 psi + 14.7 psi = 94.7 psi = 653 kPa.
n1 − n2 =
=
p1V1 p2V2
−
RT1 RT2
V
( p1 − p2 )
RT
7.60 × 10−4 m3
(860 kPa − 653 kPa)
(8.31 J/mol ⋅ K)(293 K)
= 0.0646 mol
Thus 0.0646 mol of N 2 was lost; this is
=
⎛ 28 g ⎞
0.0646 mol ⎜
⎟ = 1.8 g
⎝ 1 mol ⎠
Assess: The result seems to be a reasonable number.
16.42. Model: The carbon dioxide in the cube is an ideal gas.
Solve: Using the ideal gas equation and n = M/M mol ,
pV = nRT ⇒ V =
nRT
MRT
=
p
pM mol
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16-14
Chapter 16
The molar mass of CO 2 is 44 g/mol or 0.044 kg/mol. Thus,
V=
(10,000 kg)(8.31 J/mol K)(273 K)
(1.013 × 105 Pa)(0.044 kg/mol)
= 5090 m3
The length of the cube is L = (V )1/3 = 17.2 m.
Assess: This is sobering when you multiply it by the number of people in the industrialized world. Good thing plants
take up CO 2 in large quantities.
16.43. Model: Assume the gas is an ideal gas.
Solve: The before-and-after relationship of an ideal gas is
1 p
3
p0V0 p2V2
p V
0 3V0
=
⇒ T2 = T0 2 2 = T0 2
= T0
T0
T2
p0 V0
p0 V0 2
16.44. Model: Model the neon as an ideal gas.
Visualize: Because the cylinder is rigid V doesn’t change. The ideal gas law requires that everything be in SI units.
2.0 L = 0.0020 m3.
Solve: From the ideal gas law pV = nRT we solve for p as a function of T. p =
of pressure vs. temperature would be a straight line whose slope is
nR
.
V
( nRV )T . This leads us to believe that a graph
Make sure the given data is restated in SI units.
From the graph we see that the fit is reasonably good and that the slope is 2074.6 Pa/K.
nR
slope × V (2074.6 Pa/K)(0.0020 m3 )
⇒ n=
=
= 0.50 mol
8.31 J/mol ⋅ K
V
R
For noon, 1 mole has a mass of 20 g.
slope =
⎛ 20 g ⎞
0.50 mol ⎜
⎟ = 10 g
⎝ 1 mol ⎠
Assess: The units check out. We are just a bit bothered that the intercept of the line isn’t as close to zero as we would
like; there appears to be a bit of systematic error in all of the measurements, but that doesn’t affect the slope.
16.45. Model: Assume the trapped air to be an ideal gas.
Visualize:
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A Macroscopic Description of Matter
16-15
Initially, as the pipe is touched to the water surface and the gas inside is thus closed off from the air, the pressure
p1 = patmos = 1 atm and the volume is V1 = L1 A, where A is the cross-sectional area of the pipe. By pushing the pipe
in slowly, the gas temperature in the pipe remains the same as the water temperature. Thus, this is an isothermal
compression of the gas with T2 = T1.
Solve: From the ideal-gas law,
p2V2 = p1V1 ⇒ p2 L2 A = p1L1 A ⇒ p2 L2 = patmos L1 ⇒ p2 = patmos ( L1/L2 )
As the pipe is pushed down, the increasing water pressure pushes water up into the pipe, compressing the air. In
equilibrium, the pressure at points a and b, along a horizontal line, must be equal. (This is like the barometer. If the
pressures at a and b weren’t equal, the pressure difference would cause the liquid level in the pipe to move up or
down.) The pressure at point a is just the gas pressure inside the pipe: pa = p2 . The pressure at point b is the pressure
at depth L2 in water: pb = patmos + ρ gL2 . Equating these gives
p2 = patmos = ρ gL2
Substituting the expression for p2 from the ideal-gas equation above, the pressure equation becomes
patmos L1
L2
= patmos + ρ gL2 ⇒ ρ gL22 + patmos L2 − patmos L1 = 0
This is a quadratic equation for L2 with solutions
L2 =
− patmos ± ( patmos ) 2 + 4 ρ gpatmos L1
2ρ g
Length has to be a positive quantity, so the one physically acceptable solution is
L2 =
−101,300 Pa + (101,300 Pa) 2 + 4(1000 kg/m3 )(9.8 m/s 2 )(101,300 Pa)(3.0 m)
2(1000 kg/m3 )(9.8 m/s 2 )
= 2.4 m
16.46. Model: Assume that the steam is an ideal gas.
Solve: (a) The volume of water is
V=
M
ρ
=
nM mol
ρ
=
pV M mol 50(1.013 × 105 Pa)(5.0 m3 )(0.018 kg/mol)
=
= 0.0815 m3 = 81.5 L ≈ 82 L
RT ρ
(8.31 J/mol K)(673 K)(1000 kg/m3 )
(b) Using the before-and-after relationship of an ideal gas,
p2V2 p1V1
T p
⎛ (273 + 150) K ⎞⎛ 50 atm ⎞
3
3
3
=
⇒ V2 = 2 1 V1 = ⎜
⎟⎜
⎟ (5.0 m ) = 78.6 m ≈ 79 m
T2
T1
T1 p2
673
K
2.0
atm
⎝
⎠⎝
⎠
16.47. Model: We assume that the volume of the tire and that of the air in the tire is constant.
Solve: A gauge pressure of 30 psi corresponds to an absolute pressure of (30 psi) + (14.7 psi) = 44.7 psi. Using the
before-and-after relationship of an ideal gas for an isochoric (constant volume) process,
p1 p2
T
⎛ 273 + 45 ⎞
=
⇒ p2 = 2 p1 = ⎜
⎟ (44.7 psi) = 49.4 psi
T1 T2
T1
⎝ 273 + 15 ⎠
Your tire gauge will read a gauge pressure pg = 49.4 psi − 14.7 psi = 34.7 psi. ≈ 35 psi.
16.48. Model: The air is assumed to be an ideal gas.
Solve: At 20°C and 1 atm pressure, the number of moles in the container is
n1 =
p1V1 (1.013 × 105 Pa)(10−3 m3 )
=
= 0.0416 mol
RT1
(8.31 J/mol K)(293 K)
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16-16
Chapter 16
At 100°C and 1 atm pressure, the number of moles is
n2 =
p2V2 (1.013 × 105 Pa)(10−3 m3 )
=
= 0.0327 mol
RT2
(8.31 J/mol K)(373 K)
When heated, the pressure will rise as the number of moles remains n1. When opened, the pressure drops to 1 atm as
gas escapes. Thus, the number of moles of air that escape as the container is opened is n1 − n2 = 0.0416 mol −
0.0327 mol = 0.0089 mol.
16.49. Model: The gas’s temperature doesn’t change, so this is an isothermal compression.
Solve: At the surface, the pressure inside the cylinder must exactly equal the outside air pressure of 1.0 atm. If the
pressures were not equal, a net force would push the piston in or pull it out until the pressures balanced and
equilibrium was achieved. As the cylinder is pulled underwater, the increasing water pressure pushes the piston in
and compresses the gas. Equilibrium at depth d requires that the gas pressure inside the cylinder equal the water
pressure pwater = p0 + ρ gd , where p0 = 1.0 atm is the pressure at the surface. As long as the cylinder moves slowly,
the gas will stay at the same temperature as the surrounding water. The value of T is not important; all we need to
know is that the compression is isothermal. In that case, because T2 /T1 = 1,
V2 = V1
T2 p1
p
p0
= V1 1 = V1
T1 p2
p2
p0 + ρ gd
The initial pressure p0 must be in SI units: p0 = 1.0 atm = 1.013 × 105 Pa. Then a straightforward computation gives
V2 = 155 cm3.
Assess: V2 is less than V1. This is expected because the gas is being compressed.
16.50. Model: Assume the gas in the manometer is an ideal gas.
Solve: In the ice-water mixture the pressure is
p1 = patoms + ρ Hg g (0.120 m)
5
= 1.013 × 10 Pa + (13,600 kg/m3 )(9.8 m/s 2 )(0.120 m) = 1.173 × 105 Pa
In the freezer the pressure is
p2 = patm + ρ Hg g (0.030 m)
5
= 1.013 × 10 Pa + (13,600 kg/m3 )(9.8 m/s 2 )(0.030 m) = 1.053 × 105 Pa
Assume that a drop in length of 90 mm produces a very small change in gas volume compared with the total volume
of the gas cell. This means the volume of the chamber can be considered constant. Hence,
⎛p ⎞
p1 p2
1.053 × 105 Pa
=
⇒ T2 = T1 ⎜ 2 ⎟ = (273 K)
= 245 K = −28°C
T1 T2
1.173 × 105 Pa
⎝ p1 ⎠
Assess: This is a reasonable temperature for an industrial freezer.
16.51. Model: The air in the closed section of the U-tube is an ideal gas.
Visualize: The length of the tube is l = 1.0 m and its cross-sectional area is A.
Solve: Initially, the pressure of the air in the tube is p1 = patmos and its volume is V1 = Al. After the mercury is
poured in, compressing the air, the air-pressure force supports the weight of the mercury. Thus the compressed
pressure equals the pressure at the bottom of the column: p2 = patmos + ρ gL. The volume of the compressed air is
V2 = A(l − L). Because the mercury is poured in slowly, we will assume that the gas remains in thermal equilibrium
with the surrounding air, so T2 = T1. In an isothermal process, pressure and volume are related by
p1V1 = patmos Al = p2V2 = ( patmos + ρ gL ) A(l − L)
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