GY.90707.FM.pgs 6/25/04 9:38 AM Page i
Fall 2004
Dear Student,
This chapter is from the eighth edition of Young and Geller’s College Physics,
scheduled for publication in 2005. We’ve printed this preview booklet to gather
detailed feedback from both students and instructors so that we can publish a
textbook that truly helps you to learn physics and succeed in your course. While
these pages are not yet final, they already reflect the input of many thousands of
students and hundreds of professors who’ve reviewed and class tested the material before.
Hugh Young and Robert Geller are committed to creating a book that provides
exceptionally clear explanations, helps build strong problem-solving skills and
conceptual understanding, and incorporates the latest research into addressing
key areas of student misconception and using well-designed pedagogic figures to
aid understanding. We now need you to tell us: does this book meet its goals?
We would greatly appreciate your comments on the writing, illustrations, and
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Thank you for participating in this student review. Your feedback is invaluable to
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Margot Otway
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GY.90707.FM.pgs 6/25/04 9:38 AM Page ii
About the
Authors
Hugh D. Young is Professor of Physics at Carnegie Mellon University in Pittsburgh, PA. He attended Carnegie Mellon for both undergraduate and graduate
study and earned his Ph.D. in fundamental particle theory under the direction of
the late Richard Cutkosky. He joined the faculty of Carnegie Mellon in 1956, and
has also spent two years as a Visiting Professor at the University of California at
Berkeley.
Hugh's career has centered entirely around undergraduate education. He has
written several undergraduate-level textbooks, and in 1973 he became a coauthor
with Francis Sears and Mark Zemansky for their well-known introductory texts.
In addition to his role on Sears and Zemansky's College Physics, he is currently a
coauthor with Roger Freedman on Sears and Zemansky's University Physics.
Hugh is an enthusiastic skier, climber, and hiker. He also served for several
years as Associate Organist at St. Paul's Cathedral in Pittsburgh, and has played
numerous organ recitals in the Pittsburgh area. Prof. Young and his wife Alice
usually travel extensively in the summer, especially in Europe and in the desert
canyon country of southern Utah.
Robert M. Geller teaches physics at the University of California, Santa Barbara,
where he also obtained his Ph.D. under Robert Antonucci in observational cosmology. Currently, he is involved in two major research projects: a search for
cosmological halos predicted by the Big Bang, and a search for the flares that are
predicted to occur when a supermassive black hole consumes a star.
Rob also has a strong focus on undergraduate education. In 2003, he received
the Distinguished Teaching Award. He trains the graduate student teaching assistants on methods of physics education. He is also a frequent faculty leader for the
UCSB Physics Circus, in which student volunteers perform exciting and thoughtprovoking physics demonstrations to elementary schools.
Rob loves the outdoors. He and his wife Susanne enjoy backpacking along
rivers and fly fishing, usually with rods she has built and flies she has tied. Their
daughter Zoe loves fishing too, but her fish tend to be plastic, and float in the
bathtub.
GY.90707.FM.pgs 6/25/04 9:38 AM Page iii
Young and Geller’s College Physics 8 th Edition
Class Test Student Questionnaire
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a. The solution has Set Up, Solve, and Reflect steps. Does this structure help you learn how to tackle this type of
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GY.90707.FM.pgs 6/25/04 9:38 AM Page iv
7. Read the Do It Yourself 7.1 (Seesaw) on page 25. Did this problem help you bridge the gap from reading the chapter to
doing comparable homework problems on your own?
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a. In diagrams that show a scene, the key object is colored, whereas the rest is in grayscale. Does this practice help you
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9. What did you like most about this chapter? _____________________________________________________________
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GY.90707.FM.pgs 6/25/04 9:38 AM Page v
Brief
Contents
Chapter 1
Models, Measurements, and Vectors
Chapter 2
One-Dimensional Motion
Chapter 3
Two-Dimensional Motion
Chapter 4
Newton’s Laws of Motion
Chapter 5
Applications of Newton’s Laws
Chapter 6
Circular Motion and Gravity
Chapter 7
Work and Energy
Chapter 8
Linear Momentum
Chapter 9
Rotational Kinematics and Energy
Chapter 10
Rotational Equilibrium and Angular Momentum
Chapter 11
Periodic Motion and Elasticity
Chapter 12
Waves and Sound
Chapter 13
Fluids
Chapter 14
Temperature and Heat
Chapter 15
The Ideal Gas Law and
the First Law of Thermodynamics
Chapter 16
The Second Law of Thermodynamics; Entropy
Chapter 17
Electric Charges, Forces, and Fields
Chapter 18
Electric Potential and Electric Energy
Chapter 19
Electric Current and Direct-Current Circuits
Chapter 20
Magnetism
Chapter 21
Faraday’s Law and Inductance
Chapter 22
Alternating Currents
Chapter 23
Electromagnetic Waves
Chapter 24
Light and Geometric Optics
Chapter 25
Lenses and Optical Instruments
Chapter 26
Interference and Diffraction
Chapter 27
Relativistic Mechanics
Chapter 28
Photons, Electrons, and Atoms
Chapter 29
Atomic Structure of Matter
Chapter 30
Nuclear and High-Energy Physics
Appendix A
Mathematics Review
Appendix B
The International System of Units
Appendix C
Metric Prefixes
Appendix D
Periodic Table of the Elements
Appendix E
Properties of Selected Isotopes
Answers to Odd-Numbered Problems
GY.90707.FM.pgs 6/25/04 9:38 AM Page vi
GY.90707.07.pgs 6/25/04 9:41 AM Page 1
7
Work and Energy
T
he previous three chapters developed the
Newtonian account of mechanics, which
provides a complete description of the motion
of an object under the influence of a force. In principle, Newton’s laws could be
used to describe the motion of every molecule in a cup of water. However, doing so
would be impractical, as it would require determining the accelerations of over 1024
particles. To have any chance of describing such complex systems, we must understand another fundamental property of their behavior: energy. Energy is at the core
of every description in physics, from that of the smallest particle to that of the
largest complex system.
We are all familiar with using energy, such as electrical energy or the muscular
energy we expend during physical exertion. In this chapter, we will consider what
energy is and how we can use it to describe an object’s motion. The situations we
will consider can be described either by Newton’s laws or in terms of energy. Later,
when we study fluids, gases, and other systems, it will be necessary to describe
much of their basic behavior in terms of energy. In the big picture, energy is one of
the most important concepts in all of physics.
For an instant, this
skateboarder hangs
in the air. As he
descends, gravity
will increase his
speed. In this chapter we will learn to
analyze such
sequences in terms
of a new quantity,
energy.
7.1 An Overview of Energy
In this opening section, we will tour the main ideas of energy on a qualitative
level. In the rest of the chapter, we will describe the same ideas mathematically
and learn how to use them.
What Is Energy?
Even for physicists, it is difficult to say what energy is. Instead, we will define it
in terms of what it does:
1
GY.90707.07.pgs 6/25/04 9:41 AM Page 2
2
CHAPTER 7 Work and Energy
Nuclear energy in
the Sun's core …
Becomes energy of
the Sun's hot gas …
Becomes energy of
sunlight …
Which is converted by
plants to the chemical
energy of grains and
other foods …
Which you may
consume as calories …
Which can be used
to lift weights …
And the story
goes on.
Figure 7.1 A typical sequence of energy
transfers and transformations—one that sustains life on earth.
Figure 7.2 Elastic potential energy stored
in a stretched rubber band.
Changes in the physical world are caused by transfers and transformations of energy.
In other words, every change in the physical world represents a transfer of energy
from one object to another or a transformation of energy from one form to
another. For instance, when you kick a ball, you transfer energy from your foot to
the ball. The kick results from the muscle cells in your leg transforming chemical
energy to energy of motion.
As the preceding definition implies, energy can take various forms. The
energy of a kicked ball is different from that of an electrical current, boiling
water, or a stretched spring, although all these things have energy because they
can cause physical changes. In this chapter, we will describe just four forms of
energy. Some additional forms will be introduced later in the book. You may
encounter still other forms of energy in other courses.
To give you a sense for energy transfer and transformation, Figure 7.1 shows
how a little of the sun’s nuclear energy can become energy you spend exercising.
As we will learn, this energy is not created in the sun’s core, nor does your exercise “use it up.” Our figure shows just a brief portion in a sequence of energy
transfers and transformations that began with the origin of the universe.
Three Forms of Mechanical Energy
Next, we introduce the three forms of energy that will be described mathematically in this chapter. We start with the form that is easiest to describe: the energy
associated with an object’s motion. It is obvious that a moving object, such as a
kicked ball or a speeding car, has energy: It can cause physical changes by striking things. Energy of motion is called kinetic energy. The faster a given object
moves, the more kinetic energy it has.
To describe the next two forms of energy, we have to introduce the concept of
stored energy. Consider a stretched rubber band (Figure 7.2). You have to expend
muscular energy to stretch it, and as soon as it is released, physical changes will
occur—for instance, it may fly across the room. An unstretched rubber band lacks
this capability to change or cause change. In effect, the act of stretching the rubber
band stores energy in it, and releasing the rubber band allows this stored energy to
transform to other types of energy, such as kinetic energy. We call stored energy
potential energy. Energy that is stored in an elastic object when you stretch, compress, twist, or otherwise deform it is called elastic potential energy.
Now consider the skateboarder in Figure 7.3. When he descends the halfpipe, he speeds up, thus gaining kinetic energy. While he is at the top, he doesn’t
have kinetic energy, but he can gain kinetic energy whenever he lets gravity pull
him downward. Thus, his elevated position represents a form of stored energy,
called gravitational potential energy. Anytime an object can be pulled downward by gravity, it has gravitational potential energy. Moreover, the farther the
object can fall, the greater the gravitational potential energy.
When you climb a ladder, you may be inclined to say that you have
gained gravitational potential energy. However, strictly speaking, the
energy is stored not in you, but in your position relative to the earth. In
more abstract terms, we say that the potential energy is stored in the system consisting of you and the earth. This distinction isn’t very important
for small objects near the earth’s surface, but it becomes important when
you deal with astronomical objects, such as the moon and the earth or the
sun and the planets. N OT E
Here, the skateboarder has gravitational
potential energy because of his position
in Earth's gravitational field.
Gravitational potential
energy in the process of
becoming kinetic energy.
Together, kinetic energy, elastic potential energy, and gravitational potential
energy are called mechanical energy. (Strictly speaking, mechanical energy also
Figure
7.3 Transformation of gravitational potential energy to kinetic energy.
GY.90707.07.pgs 6/25/04 9:41 AM Page 3
7.1 An Overview of Energy
3
Efficient locomotion through springs. The hopping gait of kangaroos is exceptionally
energy efficient. Much of this efficiency comes from the fact that kangaroo’s hind legs are, in
effect, big pogo sticks. The massive tendons you see in the photo are elastic, like stiff rubber
bands. At the end of each hop, the impact stretches these tendons, transforming energy of
motion into elastic potential energy. This stored energy helps to launch the animal on its next
hop, thus transforming back to kinetic energy. Because most of the kinetic energy from each
hop is saved and reused, the kangaroo’s muscles need to add only a little kinetic energy at the
start of each hop.
includes some additional types of energy which we’ll discuss later in the course.
For the purposes of this chapter, however, mechanical energy means kinetic
energy plus elastic and gravitational potential energy.) It is important to realize
that mechanical energy is not itself a form of energy—it is just a collective term
for these forms of energy.
At the start of this section, we said that even physicists have a hard time saying what energy is. You, too, may now feel that the concept is slippery. What is
this entity that goes from object to object and from form to form? Actually,
you’re familiar with something that behaves in very much the same way: money.
Money takes many forms: a coin, a bill, or an electronic number in a bank’s computer, for example. You can store money, or you can spend it to buy things.
Money could be defined as “the ability to buy things.” Thus, money is a pure
abstraction—except that it is also real; you cannot buy anything without it.
Energy, similarly, is both an abstract idea (“the ability to cause physical
changes”) and, within the physical universe, quite real—you can’t cause changes
without it. Table 7.1 sums up this analogy between money and energy.
TABLE 7.1
Stretched sling stores
elastic potential energy
The analogy between energy and money
How money is like energy:
• It can take multiple forms (coins, bills, checks, bank accounts).
• You can transform it (e.g., by cashing a check) or transfer it to others.
• These transfers and transformations do not change its total amount.
• It can be stored or spent.
• It could be defined as “the ability to buy things,” much as energy is “the ability to cause physical
changes.”
How money is not like energy:
• Money can be created or destroyed, whereas energy cannot be.
Conservation of Energy
Now we’re ready for the central fact about energy: Energy can be passed among
objects and can take different forms, but its total amount never changes. Energy
cannot be created or destroyed. Physicists have tested this law innumerable times
and have never seen it violated. We call this law conservation of energy, and it is
one of the core principles of physics.
To explore conservation of energy, let’s look at what happens when we shoot
a rock with a slingshot, as shown in Figure 7.4. For simplicity, we will shoot the
rock straight upward, and we will ignore air resistance. As shown in Figure 7.4a,
the rock sits initially in the pouch of the stretched sling. At this time, the sling’s
elastic bands hold elastic potential energy. When the pouch is released, the sling
propels the rock upward. The sequence in Figure 7.4b begins where the rock has
just lost contact with the pouch. Most of the elastic potential energy that was
stored in the sling has now become kinetic energy of the rock. (A little stays in
the vibrating sling.) As the ball rises, it slows. Why? Because as it rises, it gains
gravitational potential energy, and this energy must come from somewhere. It
comes from the ball’s kinetic energy. The total energy remains the same, but the
(a) Before release
Top of trajectory:
Gravitational potential energy only
v0
S
S
v
S
v
Rock rising:
Kinetic energy
of rock S
gravitational
potential energy
Rock falling:
Gravitational
potential energy S
kinetic energy of
rock
S
Rock released: v
Elastic potential
energy of sling S
kinetic energy of
rock
S
v
(b) After release
Figure
7.4 Energy is conserved as it is first
transferred from a slingshot to a rock and
then transformed between kinetic energy and
gravitational potential energy during the
rock’s flight.
GY.90707.07.pgs 6/25/04 9:41 AM Page 4
4
CHAPTER 7 Work and Energy
ball’s kinetic energy transforms to gravitational potential energy. At the top of its
rise, the rock is briefly motionless (its kinetic energy is zero); all of the energy is
now gravitational potential energy. As the rock falls, the gravitational potential
energy again becomes kinetic energy. Every change in this system represents a
transformation or transfer of energy, but the total energy remains the same.
Typical solid material
The atoms and molecules of a solid can be
thought of as particles vibrating randomly on
springlike bonds. This random vibration is an
example of thermal energy. The stronger the
vibration, the hotter the object.
Figure 7.5 A simple model of thermal
energy in a solid.
Dissipation of Mechanical Energy
In the preceding example, we showed only the conversions from one type of
mechanical energy to another—elastic potential energy becoming kinetic energy
becoming gravitational potential energy. Those are the main energy changes in
this system. But real systems typically include interactions that allow mechanical
energy to transform into nonmechanical types of energy. In the case of the slingshot, for instance, we ignored air drag—but you know that air drag would slow
the rock throughout its trajectory, removing kinetic energy from the system. And
when the rock hits the ground, all of its mechanical energy will vanish—we will
have a rock sitting on the ground. Conservation of energy tells us that this lost
mechanical energy must still exist. So where does it go?
The answer is that most of it goes into slightly warming the air through which
the rock passes, the ground the rock strikes, and the rock itself. To understand how
this warming works, and what form the energy takes, let us go back to our familiar
model of friction: a block kicked across the floor. We know that friction brings the
block to a halt. To find what happened to the block’s kinetic energy, we must look
at the block and floor on a microscopic level. Microscopically, we can think of a
typical solid material as consisting of tiny particles (atoms or molecules) held
together by a web of springlike bonds. Any disturbance will set the particles vibrating randomly, as shown in Figure 7.5. The hotter an object, the more strongly its
atoms or molecules are vibrating. This type of random, atomic-scale vibration is
a form of energy called thermal energy or, sometimes, internal energy.
Now we can understand how friction reduces kinetic energy. From Chapter 5,
we know that friction represents the surfaces of two objects catching and tugging
on each other microscopically. These interactions cause the surface atoms of each
object to vibrate more strongly. The vibrations pass inward; soon, all the atoms
near the surfaces in contact are vibrating more strongly than before. As shown in
Figure 7.6, when friction slows a sliding block, the block’s kinetic energy goes
into the random vibrations of atoms in the block and the floor. Thus, friction
causes an object’s kinetic energy to transform into thermal energy. To experience
this transformation, just rub your palms together hard: Your palms warm up as the
friction between them converts their energy of motion into thermal energy.
Before sliding
Sliding
After sliding
S
fk
Figure 7.6 How friction converts kinetic
energy to thermal energy. The “atomic-scale”
view in the square insets represents a far
higher level of magnification than the smallscale roughness in the middle inset.
At room temperature,
atoms vibrate
moderately.
Kinetic friction due to
small-scale surface
roughness slows block.
Friction has converted the
kinetic energy of the sliding
block to stronger atomic
vibrations (thermal energy).
GY.90707.07.pgs 6/25/04 9:41 AM Page 5
7.2 Work: Transfer of Energy by Forces
How does this description apply to the slingshot example? First, the rock
loses kinetic energy to air drag, which is a form of friction. As in a solid, the
greater the random motion of air molecules, the hotter the air. Figure 7.7 shows
the effect of air drag on a tennis ball. As the ball passes through air, some of its
kinetic energy is transferred to the air, increasing the motion of the air molecules.
The motion of the disturbed air quickly becomes random, and a thermometer
would show that the air has been warmed. Thus, air drag converts some of an
object’s kinetic energy to thermal energy in the air. (The surface of the object also
warms slightly.) When the rock in our example lands, most of its kinetic energy
goes into vibrations in the rock and the ground, which quickly become random
thermal energy. A little of the rock’s kinetic energy goes into sound energy and
perhaps into the kinetic energy of a puff of dust.
Any force, such as friction, that turns mechanical energy into nonmechanical
forms of energy is called a dissipative force, because it dissipates the mechanical
energy in a system.
While we will refer to the dissipation of mechanical energy, physicists and engineers typically just say that a system “loses” mechanical
energy. This doesn’t mean that energy really is lost! Conservation of
energy always holds. However, mechanical energy can be converted (dissipated) to other forms so that the energy is no longer available to influence
the motion of the system. N OT E
5
Undisturbed air ahead
In the ball's wake, the
of the ball. The lines
air's motion is stronger
are thin smoke streams. and more random.
Figure 7.7 The motion of air in the wake of
a tennis ball. This photo is from a wind-tunnel
study of the aerodynamics of tennis balls at
typical court speeds of over 100 mi hr. The
wind tunnel blows air past a fixed ball, but the
effects are exactly the same as those on a ball
moving through still air. The lines in the air
are due to thin smoke streams that are used to
make the air’s motion visible.
/
Let’s summarize the key points we established in this section:
• For an object to change its behavior, it must undergo some transfer or transformation of energy.
• Energy can take different forms. In this chapter, we will concentrate on gravitational potential energy, elastic potential energy, and kinetic energy—collectively
called mechanical energy.
• Energy is conserved. It can be transferred between objects or transformed into
different forms, but its total amount remains the same.
• Mechanical energy can be dissipated to nonmechanical forms of energy. For
instance, friction dissipates kinetic energy, converting it to thermal energy.
7.2 Work: Transfer of Energy by Forces
In the examples we looked at in the preceding section, energy was transferred to
or from objects by the action of forces. In fact,
The transfer of energy by forces is so common that we give it a
name: work.
You’ve spent the last three chapters learning how forces affect an object’s
motion. Thus, it will be no surprise that work—the transfer of energy by forces—
also affects an object’s motion:
When all forces are taken into account, the net work on an object is
only the transfer of energy that results in changing the object’s
speed.
After developing an understanding of work in this section, we’ll be able to
derive the connection between the net work and an object’s speed in the next section. As you can see, the meaning of “work” in physics is more specific than its
use in everyday language—for instance, in reference to homework or a workout.
Low and sleek. Many engineering problems boil down to reducing dissipative forces.
For instance, a standard bicycle places the
rider in a position which catches a lot of wind,
creating significant air drag. On a recumbent
bicycle, the rider’s low profile causes much
less air drag. The real-life difference is so
great that, in 1933, a second-string racer rode
a recumbent at nearly 30 miles an hour,
smashing by 10% a record almost 20 years
old. After that, recumbent bicycles were
banned from racing. A fairing—an aerodynamic shell—further reduces air drag and thus
increases speed even more; faired recumbent
bicycles can travel 36 mph and can break
65 mph for short distances.
GY.90707.07.pgs 6/25/04 9:41 AM Page 6
6
CHAPTER 7 Work and Energy
F
x
d
• Sled speeds up (gains kinetic energy)
• W 5 Fd (for a force in the direction of the
displacement)
Figure 7.8 The work done on a sled by a
constant force acting in the sled’s direction of
displacement.
S
F
u
x
S
d
• The force component perpendicular to the
displacement does no work.
S
Direction of
displacement
F
F'
i
u 30°
Fi F cos u
i
• The work on the sled is due to the force
component parallel to the displacement:
W Fi d F cos u d.
Any force perpendicular to the displacement of an object transfers
no energy to that object and performs no work on it.
Figure 7.9 The work done on a sled by a
constant force acting at an angle to the sled’s
direction of displacement.
S
F
x
S
d
• The force component perpendicular to
the displacement does no work.
S
F
To explore the idea of work, we will start by looking at the work done on an
object by a single force. We will consider three cases: (1) a force acting in the
same direction as the object’s displacement, (2) a force acting at an angle to the
object’s displacement, and (3) a force with a component directed opposite to
the object’s displacement.
For the first case, consider a dog pulling a sled on frictionless ice for a distance d 5 2.0 m by applying a constant force F 5 50 N parallel to the sled’s
motion (Figure 7.8). The force causes the sled to speed up (gain kinetic energy),
so we know that it transfers energy to the sled. This transfer of energy is the work
W done on the sled. When a constant force F acts in the direction of the
object’s displacement, the work done by the force on the object is given by the
magnitude of the force times the distance the object moves: W 5 Fd. Thus, in
this case, W 5 Fd 5 1 50 N 2 1 2.0 m 2 5 100 N # m.
Notice that the unit of work is the newton-meter (N # m). We call this compound unit a joule (J): 1 N # m 5 1 J. Because work represents a transfer of
energy, the joule is also the SI unit for energy in general. In our example, the dog
transfers 100 J of energy to the sled. We can also say that the dog does 100 J of
work on the sled.
Figure 7.9 shows the same sled being towed by a tall man for 2 m with a force of
50 N. In this case, the force is directed at an angle u 5 30° from the displacement.
However, the force still has a component F 5 F cos u that is parallel to the direction of motion of the sled. This component causes the sled to speed up and thus
transfers energy to it. To calculate work, we use this parallel component of force:
W 5 F d. Thus, W 5 1 F cos u 2 d 5 1 50 N 2 1 2.0 m 2 cos 30° 5 87 J. Notice that
the component of force perpendicular to the displacement, F', has no effect on the
sled’s speed and thus does no work on it. This is an important general point:
We can obtain the same result mathematically. The work done by a force oriented
perpendicular to the displacement (u 5 90°) would be W 5 F d 5F cos 1 90° 2 d.
However, cos 1 90° 2 5 0, so the work would be zero. Because only the parallel
component of a force does work on an object, a force directed at an angle to the
direction of displacement does less work on the object than a force of equal magnitude directed parallel to the displacement.
We can now give a general definition for the work done on an object by a
constant force:
i
Definition of work, W
Work is the transfer of energy to an object by a force. For a constant force, F,
the work is equal to the component of the force parallel to the displacement,
F 5 F cos u, times the magnitude of the displacement, d, where u is the angle
between the force and displacement:
i
W5Fd
F'
i
u 5 150°
(7.1)
Unit: joule, J
Fi 5 F cos u
• The force component parallel and opposite to
the displacement does negative work on the
sled: W 5 Fi d 5 F cos u d , 0.
Figure 7.10 The work done on a grocery
cart by a constant force with a component
opposite to the cart’s direction of
displacement.
Remember that the component of a force is a scalar quantity. Therefore, work is
the product of two scalars and thus is a scalar itself. In fact, energy in general is
always a scalar quantity. (It has no direction.)
Now let’s consider what happens when you pull back on a runaway grocery
cart to slow it down (Figure 7.10). Intuitively, we know that, in slowing the grocery cart, you are removing kinetic energy from it. Again, the force you exert has
a component parallel to the cart’s displacement, and only this component affects
the cart’s motion. This time, however, F points in the direction opposite that of
i
GY.90707.07.pgs 6/25/04 9:41 AM Page 7
7
7.2 Work: Transfer of Energy by Forces
S
F
S
If the force is in the direction of displacement:
• The work on the object is positive (the object speeds up).
• W Fd
F
(a)
S
d
S
S
F
F
F'
u
If the force has a component in the direction of displacement:
• The work on the object is positive (the object speeds up).
• W Fi d 1F cos u 2 d
u
(b)
Fi F cos u
S
d
S
S
F
F
u
F'
u
(c)
Fi F cos u
S
d
If the force has a component opposite to the direction of displacement:
• The work on the object is negative (the object slows down).
• W Fi d 1F cos u 2 d
• Mathematically, W , 0 because F cos u is negative for 90° , u , 270°.
The perpendicular force component F' never does any work on the object!
Figure
7.11 Three cases in which a single constant force does work on an object.
the displacement. Nonetheless, we can still use Equation 7.1 to find the work:
W 5 F d. The angle u must be measured from the direction of displacement, so it
is 150° in this case. Since cos u is negative for 90° , u , 270°, we find that the
work is negative:
i
W 5 F d 5 1 F cos u 2 d
5 1 50 N 2 cos 1 150° 2 1 2.0 m 2 5 287 J
i
Thus,
A force does negative work on an object when it reduces the object’s
kinetic energy (slows the object down). Such a force has a component
opposite to the object’s direction of displacement.
Figure 7.11 summarizes the three cases we’ve examined so far.
Now let us see how to handle multiple forces that do work on the same object,
such as the block shown in Figure 7.12. To find the net work (or total work) done
on the block by all three forces, we canS calculate the work done by each force and
add up the resulting
values. If a force F1 contributes work W1 5 1 F1 2 d, and likeS
S
wise for forces F2, F3, etc., then the net work is
Wnet 5 W1 1 W2 1 W3 1 c5 a Wi
i
As an equivalent method, you could first calculate the net component of force
parallel to the displacement 1 Fnet 2 :
1 Fnet 2 5 1 F1 2 1 1 F2 2 1 c5 F1 cos u1 1 F2 cos u2 1 c
i
i
i
i
i
Then
Wnet 5 1 Fnet 2 d
i
Thus, the net work done on an object is the transfer of only that energy which
results in a change in the object’s speed. (If only one force does work on the
object, then that force is the net force.) If the object’s speed is not changing,
then no net work is being done on it, however many forces act. These points are
illustrated in Examples 7.1 and 7.2.
S
F1
u3
u1
S
F3
S
d
1F2 2 i
u2
1F1 2 i
S
F2
The net work done on the block is the sum of
the work done by the individual forces:
Wnet 5 W1 1 W2 1 W3
5 1F12 i d 1 1F2 2 i d 2 F3d
Figure 7.12 The net work done on a block
by three forces.
GY.90707.07.pgs 6/25/04 9:41 AM Page 8
8
CHAPTER 7 Work and Energy
Dragging a crate
EXAMPLE 7.1
A man drags a crate 4.0 m with a constant force of 50 N. The rope makes an angle of u 5 30° with the ground,
and the force due to friction is 10 N. (a) What is the work on the crate due to each force? (b) What is the net work
done on the crate? (c) What is the work done on the man by the crate?
SOLUTION
Figure 7.13a shows the situation. For problems involving work, you should draw a free-body diagram and show the
displacement of the object, as in Figure 7.13b. Be sure to make
your diagram clear enough to show both the components of force
parallel to the displacement (these forces will contribute to the
work) and the components of force perpendicular to the displacement (these forces will not contribute to the work).
SET U P
Part (a): We can calculate the work done by each force
on the crate; we will call them Wn, Ww, Wman, and Wf . First, note
that the normal force n and weight w 5 mg are perpendicular to
the displacement of the crate and therefore transfer no energy to
it: Wn 5 Ww 5 0.
Now we find Wman and Wf , using W 5 F d, where F 5 F cos u.
The man exerts a force Fman 5 50 N at u 5 30° to the d 5
4.0 m displacement, so his contribution to the work is Wman 5
1 50 N 2 cos 1 30° 2 1 4.0 m 2 5 173 J. The force of friction is given
as fk 5 10 N and is always directed opposite the displacement,
at u 5 180°. With cos 1 180° 2 5 21, Wf 5 1 10 N 2 1 21 2 #
1 4.0 m 2 5 240 J. To summarize, we have
S O LV E
i
i
Wn 5 0 J, Ww 5 0 J, Wman 5 173 J, and Wf 5 240 J
Part (b): To find Wnet, we can add all the individual contributions to the work found in part (a):
Wnet 5 Wn 1 Ww 1 Wman 1 Wf
5 0 J 1 0 J 1 173 J 2 40 J 5 133 J
Alternative Solution: The other way to find Wnet is to use
Wnet 5 1 Fnet 2 d. Looking at Figure 7.13b, we see that
1 Fnet 2 5Fman cos u 2 fk,
so
Wnet 5 1 Fman cos u 2 fk 2 d 5
1 1 50 N 2 cos 1 30° 2 2 10 N 2 1 4.0 m 2 5 133 J, which agrees
with the answer we got with our first method.
i
i
S
S
Part
(c): SBy Newton’s third law, Fcrate on man 5 2Fman on crate (or
S
Fcrate 5 2Fman, in the notation we use for this problem). Therefore, the work done by the crate on the man will be the opposite
of the work done by the man on the crate, so Wcrate 5 2173 J.
The negative sign makes intuitive sense: If the man adds energy
to the crate, the crate must remove that same energy from the
man.
R E F L E C T In part (a), we obtained a negative value for the work
done on the crate by friction: Wf 5 240 J. The negative sign just
means that friction removes energy from the crate. This makes
sense, since we know that friction converts kinetic energy of the
crate into thermal energy. We now have a complete account of
the energy transfer associated with the crate: The man inputs
Wman 5 173 J of energy through his pull on the rope, and 40 J is
lost as thermal energy due to friction, leaving 173 J 2 40 J 5
133 J to increase the crate’s kinetic energy. This result agrees
with our understanding that the net work on an object is only the
transfer of energy that results in changing the object’s speed.
Practice Problem: Suppose the man wants to keep the crate’s
speed constant. If he still exerts a 50 N force, at what angle u
must he pull? Answer: u 5 78°.
It is helpful to indicate the
direction of displacement
on the free-body diagram.
30°
(a) Physical diagram
Figure
(b) Free-body diagram
7.13 A physical diagram of the situation, and the free-body diagram you should draw in solving this
problem.
In the preceding example, a man does 173 J of work on a crate. While this
energy comes from burning calories, it is important to realize that not all the
energy released in a food calorie can be converted into useful work done by the
muscles. In fact, for most sustainable activities (such as running and biking), only
about 20% of caloric energy results in work done through forces exerted by the
muscles. The remainder of the energy eventually goes into thermal energy, which
is why we get warm when we exercise.
GY.90707.07.pgs 6/25/04 9:41 AM Page 9
7.2 Work: Transfer of Energy by Forces
Conceptual
Analysis 7.1
Work and orbital motion
S
v
A communications satellite moves in a circular orbit
at constant speed in response to gravity, as shown in Figure 7.14.
Which of the following statements is correct?
A.
B.
C.
D.
9
The earth does positive work on the satellite.
The earth does negative work on the satellite.
The earth does no work on the satellite.
Once a coordinate system is specified, the work changes sign
every half orbit, so that the average work is zero.
S O L U T I O N In circular gravitational orbits, the force on the
orbiting object is always perpendicular to the object’s velocity, as
indicated in the figure. The displacement of the object is always
along the direction of the velocity, so the force is always perpendicular to the displacement. Therefore, F 5 0, and the work is
zero for any displacement along the orbit. This result still might
seem confusing, since your intuition may correctly guess that
orbiting objects have energy. To clear up the confusion, just
remember that work is the transfer of energy that changes an
i
Satellite
S
Fg
Figure
7.14 A satellite in a circular orbit.
object’s speed. Since the speed of an object is constant in uniform circular motion, W 5 0, and all answers other than choice
C can be rejected. As we’ll see later, energy is required to set an
object in orbital motion, and the orbital system holds onto this
energy, but no additional energy needs to be supplied to maintain
the object in its orbit. Our solar system has been executing orbital
motion for about 5 billion years and does not require an energy
source to maintain its orbits.
In the preceding Conceptual Analysis, we found that W 5 0 because no
energy transfer is required to maintain a circular orbit at constant speed. Now
let’s consider a case where Wnet 5 0, which helps illustrate the meaning of work.
As shown in Figure 7.15a, a woman lifts a barbell of mass m through a height h.
To keep good form, she lifts the weight with a small and constant speed. Because
the barbell’s speed doesn’t change, the net work on the barbell must be zero. Now
let’s look at the work mathematically.
Figure 7.15b shows the work on the barbell done by the woman and done by
gravity.
Since the barbell is lifted at aSconstant speed, there is no acceleration, so
S
S
S
S
gF 5 ma 5 0. Therefore, w 5 2Flift 5 2mg throughout the displacement.
The woman does mgh joules of work on the barbell, and gravity does 2mgh
joules, so the net work is Wnet 5 Wlift 1 Ww 5 mgh 2 mgh 5 0.
When the woman expends her own caloric energy to lift the weight, where
does the energy go? Since the barbell neither speeds up nor slows down, none of
Wlift 5 mgh
Barbell lifted at
contant speed
h
S
Flift
Ww 5 2mgh
Wnet 5 Wlift 1 Ww 5 0
S
w
(a)
(b)
Figure 7.15 The net work done on a barbell
lifted at constant speed.
GY.90707.07.pgs 6/25/04 9:41 AM Page 10
10
CHAPTER 7 Work and Energy
the woman’s energy goes into kinetic energy. Therefore, in lifting the barbell, the
woman’s energy is transferred into gravitational potential energy. The gravitational potential energy is certainly increased, but work on an object is only the
transfer of energy that changes the object’s speed (kinetic energy), so Wnet 5 0.
Sliding down a ramp
EXAMPLE 7.2
A package of mass m is unloaded from a truck with an inclined
ramp, as shown in Figure 7.16. The ramp has rollers that eliminate
friction, and the truck unloads the package at an initial height h.
The ramp is inclined at an angle u. Find an algebraic expression for
the work done on the package during its trip down the ramp.
mass m
h
u
Figure
7.16 A package sliding down a frictionless ramp.
Figure
7.17 What you should draw to find the work done on the
SOLUTION
Figure 7.17 shows how you should
draw the forces on
S
the package in relation to the displacement d along the ramp. Be
sure to make your diagram clear enough to show the components
of force parallel, and those perpendicular, to the displacement of
S
the package. The normal force n is perpendicular to the displacement, so it does no work on the package.
SET U P
S O LV E First, let’s make sure that we avoid a common mistake.
With W 5 F d 5 1 F cos u 2 d in Equation 7.1 and F 5 mg, why
should you reject the following incorrect expression for the
work: W 5 1 mg cos u 2 d? Because the angle u in Equation 7.1
must be the angle between the force and the displacement, yet u
labels the angle of incline in this problem. Since u is a common
symbol, be sure that it represents the correct angle before using it
in a particular equation.
i
The work on the package is done by the component of the
gravitational force parallel to the displacement: F 5 mg sin u.
Thus, W 5 F d gives
i
i
W 5 1 mg sin u 2 d
We aren’t given the distance d in the problem, but we can
remove it algebraically by noting that d sin u 5 h in Figure 7.17.
Therefore,
package.
W 5 1 mg sin u 2 d 5 mgh
R E F L E C T First, let’s identify the forms of energy that were transformed in this example: There was no friction, and all of the
gravitational energy went into kinetic energy. Also, we found that
W 5 mgh, and for a fixed mass, the work done on the package by
gravity depends only on the vertical height h. We discuss the
implications of this fact next.
Starting from height h, would the package in the previous example arrive at
the ground any slower if a longer ramp were used? Work changes an object’s
speed: To see if the package would have a lower speed, we need to find out
whether less work is done on the package by gravity. We found that W 5 mgh. If
a longer ramp is used, d and u in Figure 7.17 change, but the vertical height h
remains the same. Thus, it makes no difference whether the ramp is long or short:
The amount of gravitational potential energy converted to kinetic energy depends
only on the height through which the package moves.
GY.90707.07.pgs 6/25/04 9:41 AM Page 11
7.3 Work and Kinetic Energy
11
Here, we summarize the main points of this section:
• Work is the transfer of energy by forces.
• When all forces are taken into account, the net work on an object is the transfer
of only that energy which results in changing the object’s speed.
• Any force perpendicular to the displacement of an object transfers no energy to
that object and performs no work on the object.
• A force that has a component opposite to an object’s direction of motion
reduces the object’s kinetic energy and does negative work on the object.
7.3 Work and Kinetic Energy
Now it’s time to come up with mathematical descriptions of the three forms of
mechanical energy we study in this chapter. We’ll start with kinetic energy.
To define kinetic energy, we’ll use the fact that an object’s kinetic energy
depends on its speed. Since changes in speed reflect the net work done on an
object, we can define a relation between net work and kinetic energy. Let’s
start by considering an object of mass m that is displaced through a distance d
by a net force Fnet (Figure 7.18). For simplicity, the force acts in the direction
of the displacement. In this case, the net work is Wnet 5 Fnet d. We wish to
express this work in terms of the object’s initial and final speed (before and
after the displacement).
To relate work and speed, we will use a convenient kinematics equation from
Chapter 2: vf2 5 vi2 1 2a 1 x 2 x0 2 , where 1 x 2 x0 2 5 d. As long as we know
the mass of the object, we also know the acceleration from Newton’s second law:
a 5 Fnet m. Then we have
/
vf2 5 vi2 1 2
1 2
Fnet
d
m
To solve this equation for Wnet 5 Fnet d, we subtract vi2 from both sides, then multiply both sides by m 2:
/
m 2
1 vf 2 vi2 2 5 Fnetd
2
This is the net work, which we’ll write as
Wnet 5 12 mvf2 2 12 mvi2
(7.2)
This equation says that the net work is equal to the change in the quantity 12 mv2.
We identify this quantity as the object’s kinetic energy.
Definition of kinetic energy, K
The kinetic energy K of an object is its energy of motion, which depends on
both its mass and its speed:
K 5 12 mv2
(7.3)
Because kinetic energy is just another form of energy, the units of K are
joules (J). Strictly speaking, Equation 7.3 defines the translational kinetic
energy of an object—the energy that pertains to the object’s motion through
vi
vf
m
Fnet
Fnet
d
Figure
7.18 The scenario we use to define
kinetic energy.
GY.90707.07.pgs 6/25/04 9:41 AM Page 12
12
CHAPTER 7 Work and Energy
space. If an object is rotating, then it has rotational kinetic energy, which we will
study in a later chapter.
Remember that energy, including kinetic energy, is a scalar quantity; thus, v in this equation is speed, not velocity. N OT E
Now we can understand Equation 7.2 in terms of kinetic energy:
Work–kinetic energy theorem
The net work on an object is equal to the change in the object’s kinetic
energy:
Wnet 5 Kf 2 Ki 5 12 mvf2 2 12 mvi2
(7.4)
This equation restates what we’ve said in words: The net work on a given object
is the transfer of energy to or from the object that changes the object’s speed. If
you do positive net work on an object (Wnet . 0), the object speeds up (vf . vi),
and its kinetic energy increases. Negative work (Wnet , 0) causes the object to
slow
down, decreasing its kinetic energy. Recall that we used the formula
S
S
F 5 ma to derive Equation 7.4. Thus, the work-kinetic energy theorem is not a
new law of nature, but is simply a consequence of Newton’s second law.
Up to now, we’ve focused on the relation of kinetic energy and speed. However, as Equation 7.3 shows, an object’s energy of motion depends on its mass as
well as its speed: K 5 12 mv2. Intuitively, that makes sense: At a given speed, an
SUV does more damage in an accident than a small car.
Let’s look a little deeper. Suppose it’s your fate to be caught in a parking-lot
fender bender, but you can choose between being hit by a Honda Civic of mass
1300 kg traveling at 2.0 m s or by a Mercedes S600 that has twice the mass, but
is moving at only 1.0 m s. Which will cause less damage to your car?
Off the top of your head, you might think the two choices are even. However,
notice that kinetic energy (K 5 12 mv2) is proportional to mass (K ~ m), but also
proportional to the square of velocity (K ~ v2). Thus, it may be best to choose the
heavier but slower Mercedes. To see, let’s compute the kinetic energies of the two
cars:
/
/
Mercedes: KMercedes 5 12 mv2 5
Honda:
KHonda 5 12 mv2 5
2600 kg 1 1.0 m s 2 2
5 1300 J
2
/
1300 kg 1 2.0 m s 2 2
5 2600 J
2
/
Indeed, the Mercedes has half the kinetic energy of the Honda, so in this example
you should choose to get your fender bent by the heavier, but slower, car.
Conceptual
Analysis 7.2
Work and kinetic energy
Two blocks of ice, one twice as heavy as the other,
are at rest on a frozen lake. A person pushes each block a distance of 5 m. Ignore friction, and assume that an equal force is
exerted on each block. The kinetic energy of the light block after
the push is
A. smaller than the kinetic energy of the heavy block.
B. equal to the kinetic energy of the heavy block.
C. larger than the kinetic energy of the heavy block.
Three forces act on the two blocks: the pull of the
earth, the normal force of the surface, and the push of the person.
The gravitational and normal forces are perpendicular to the displacement, so they transfer no energy to either block. Equation 7.4, Wnet 5 Kf 2 Ki, says that the change in kinetic energy
equals the net work done. The only force doing work on the
blocks is the force from the person, which is the same in both
cases. Since the initial kinetic energy of each block is zero, both
blocks have the same final kinetic energy. Therefore, the correct
answer is B.
SO LU T I O N
GY.90707.07.pgs 6/25/04 9:41 AM Page 13
7.3 Work and Kinetic Energy
13
Slowing a car
EXAMPLE 7.3
/
/
A 1400 kg car is traveling at 30 m s. (a) How much work must be done on the car to brake it to 10 m s?
(b) Where does the kinetic energy go?
SOLUTION
Part (a): In problems involving work and kinetic
energy, it’s helpful to draw a physical diagram of the motion and
identify the initial and final speeds, as in Figure 7.19.
SET U P
Using Equation 7.4, we can solve for the work, which
causes known changes in speed:
S O LV E
Wnet 5 12 mvf2 2 12 mvi2
5 12 1 1400 kg 2 1 10 m s 2 2 2 12 1 1400 kg 2 1 30 m s 2 2
/
/
5 25.6 3 105 J
Part (b): When the net work is negative, as it is here, we know
that the object loses kinetic energy. Where does the kinetic
energy go? Friction between the brake pads and the brake discs
Quantitative
Analysis 7.3
Figure
7.19 What you should draw to find the work done on a
slowing car.
convert the car’s kinetic energy into thermal energy. In the
process, the brake pads heat up to about 500°C. We’ll learn in a
later chapter that 5.6 3 105 J is enough energy to bring one liter
of room-temperature water to a boil.
Stopping distance of a car
Fbrake
Figure 7.20 shows a car carrying out two braking
trials. In each trial, the driver brings the car to a halt by applying
a constant braking force Fbrake. In the first trial, the car is traveling
at speed v1 when the brakes are applied, and it stops in distance
d1. In the second trial, the car is traveling at twice the earlier
speed (v2 5 2v1) and stops in distance d2. What can be said
about the two stopping distances?
A.
B.
C.
D.
d2
d2
d2
d2
d1
Trial 1: Initial velocity v1
Fbrake
5 d1
5 2d1
5 4d1
5 8d1
We follow the method for proportional reasoning
set forth in Section 2.5. First, we must find an equation that
relates the variables given in the problem. We can express the
net work on the car as Wnet 5 Fnet d. This means that, using
Equation 7.4, we can relate the change in kinetic energy of the
car in terms of the net force on the car and the distance through
which it acts:
v1
v2 2v1
d2
SO LU T I O N
1 Fnet 2 d 5 12 mvf2 2 12 mvi2
Trial 2: Initial velocity v2 2v1
Figure
d2
v22
i
In both cases, vf 5 0, and the braking force points opposite to the
displacement, giving 1 Fnet 2 5 2Fbrake. Then Equation 7.4
becomes
i
1 Fbrake 2 d 5
d
1
5
m 5 constant
2Fbrake
vi2
Finally, we equate expressions for the two trials:
5
d1
v12
5 constant
Since the initial velocity for trial 2 is twice that for trial 1,
v2 5 2v1, it follows that
d2
1 2v1 2
1
2
2 mvi
Next, we separate the variables in the preceding equation from
the terms that remain constant during the two trials. Since Fbrake
and m remain the same in both trials, we have
7.20 Two braking trials with different initial speeds.
or
d2 5
2
5
1 2v1 2 2
v12
d1
v12
d1 5 4d1
Thus, choice C is correct. This is why you should be at least four
times farther from the vehicle in front of you when you travel at
60 mi h than when you travel at 30 mi h.
/
/
GY.90707.07.pgs 6/25/04 9:41 AM Page 14
14
CHAPTER 7 Work and Energy
7.4 Work Done by a Varying Force
Object moving from xa to xb in response
to a changing force in the x direction
Fa
So far, we have defined work done by a constant force. But what happens when
you stretch a spring? The farther you stretch it, the harder you have to pull, so the
force is not constant. In this section, we learn how to compute work done by a
force that varies during the displacement of the object it acts upon.
Figure 7.21a shows an object moving in the x direction in response to a
changing force F. In Figure 7.21b, we graph the magnitude of the force as a function of the particle’s position x. To find the work done by this force, we divide the
displacement into short segments Dx1, Dx2, and so on, as in Figure 7.21c. We
approximate the varying force by the average force within each segment: The
force has the average value F1 in segment Dx1, F2 in segment Dx2, and so on. The
work done by the force in the first segment is then F1 Dx1, in the second F2 Dx2,
and so on. The total work is
Fb
x
xa
(a)
xb
F
Fb
Graph of force magnitude as a function
of position
Fa
xa
xb
x
xb – xa
(b)
W 5 F1 Dx1 1 F2 Dx2 1 c
F The height of each strip
represents the average F5
force for that
F4
interval.
F3
F2
F1
xa ∆x1
(c)
∆x2
∆x3
∆x5
∆x4
However, the products F1 Dx1, F2 Dx2, c, are the areas of the vertical strips in
Figure 7.21c, so the total work is represented by the total area of these strips. As
we make the subdivisions smaller and smaller, this total area becomes more and
more nearly equal to the shaded area between the smooth curve and the x-axis in
Figure 7.21b. Thus, we find that
F6
∆x6
xb
On a graph of force as a function of position, the total work done by
the force is represented by the area under the curve between the initial and final positions.
x
Figure
7.21 Force versus displacement for
a varying force.
Fsp
Work Done by a Spring
A spring provides a varying force; the more it is stretched or compressed, the
larger the force it exerts. Let’s apply the preceding rule to find the work done by
the varying force of a spring. To do this, we’ll plot force versus position for a
spring and then calculate the area under the curve we obtain. From Chapter 5, we
Equilibrium position
(at which spring is relaxed)
Fsp
x
x
2x1
x50
2x1
x1
(a) Spring pushes mass to the right from 2x1 to
x 5 0.
Height 5 kx1
2x1
Base 5 x1
x50
Fsp
Fsp 5 2kx
Area 5 work done
by spring between
2x1 and x 5 0
2x1
x1
Work 5 area
7.22 Work done by the spring from
2x 1 to x 5 0.
x50
S
v
5 12 (base)(height) 5 12 (x1)(kx1) 5 12 kx12
Figure
x1
x
(b) Graph of force versus position for this
displacement.
x1
(a) Displacement from x 5 0 to x1 as spring pulls on mass.
Fsp
Fsp52k12x1)5kx1
x50
Speed increasing,
so W . 0
x
Area 5 work done
by spring between
x 5 0 and x1
Speed decreasing,
so W , 0
(b) Force-versus-position graph for the full displacement from 2x1 to x1.
Figure
7.23 Work done as the mass moves from x 5 0 to x 5 x 1 .
GY.90707.07.pgs 6/25/04 9:41 AM Page 15
7.5 Elastic and Gravitational Potential Energy
know that we can approximate the force from a spring with Hooke’s law:
Fsp 5 2kx. In this equation for the spring force, x is the amount the spring is
stretched or compressed from equilibrium, and the equilibrium position is simply
the length of the unstretched spring. We’ll calculate the work done as the spring
in Figure 7.22a pushes the mass from x 5 2x1 to the equilibrium position x 5 0.
Since the spring pushes on the mass and increases its speed, we know that the
work done on the mass is positive. (We could reach the same conclusion from the
fact that the force acts in the direction of the displacement.)
To find the work, we must find the area under the plot of Fsp versus x between
the initial and final positions, as shown in Figure 7.22b. The area between
x 5 2x1 and x 5 0 is easy to find because it forms a right triangle:
area 5 12 1 base 2 1 height 2 . Thus, we find that the work done on the block by the
spring is W 5 12 kx12.
As Figure 7.23a shows, the mass continues to move beyond the equilibrium
point x 5 0. However, the spring now pulls on it, slowing it down. Because the
mass slows down (and because the force acts oppositely to the displacement),
we know that the work done on the mass by the spring during this phase must
be negative. Figure 7.23b shows the full graph of Fsp versus x during both compression and stretching of the spring. We can generalize this result for any
position x:
15
Fhand on spring Fspring on hand
External force (hand)
compressing a spring
x
x50
Fspring on hand Fhand on spring
External force (hand)
stretching a spring
x
x50
For an external force stretching or
compressing a spring:
Whand on spring . 0
Wspring on hand , 0
By Newton's third law,
Fhand on spring 5 2Fspring on hand
Work done by a spring
When the end of a spring moves from position x towards equilibrium at
x 5 0, the work W done by the spring on an object is positive. When the end
of a spring moves away from equilibrium a distance x in either direction
(stretching or compressing), the work done by the spring is negative.
Wby spring 5 6 12 kx 2
Figure
7.24 Work done by an external
force on a spring.
(7.5)
So far we’ve discussed the work done by a spring. Now, when you stretch or
compress a spring, as shown in Figure 7.24, what is the work you do on the
spring? The answer is that your hand always acts in the direction the spring is
displaced, so the work is positive, which makes intuitive sense. (You are putting
energy into the spring.)
7.5 Elastic and Gravitational Potential Energy
As discussed in Section 7.1, potential energy can be thought of as stored energy.
A book sitting on a table has stored gravitational energy, because, if the book is
nudged off the table, gravity will speed it up as it falls. Similarly, a stretched or
compressed spring stores elastic potential energy. In this section, we’ll learn how
to calculate these two forms of potential energy. We will use the symbol U for
potential energy.
Elastic Potential Energy
To develop a mathematical expression for elastic potential energy, we need to
find how work is related to changes in potential energy. Suppose a spring
pushes a mass from a compressed position 2x to the spring’s equilibrium position x 5 0, as shown in Figure 7.25. The spring transfers kinetic energy to the
mass (does work on it). At the same time, as it goes from compressed to
relaxed, the spring loses elastic potential energy. In fact, the motion occurs
because elastic potential energy of the spring is converted to kinetic energy of
the mass.
x
2x
Ui . 0
x50
Uf 5 0
Elastic potential energy of spring:
DUspring 5 Uf 2 Ui
• The spring does positive work on the mass:
Wby spring 5 DKmass . 0.
• At the same time, the spring loses potential
energy: DUspring , 0.
• By conservation of energy, if friction is
absent, the loss of potential energy by the
spring equals the gain in kinetic energy by
the mass: DKmass 5 Wby spring 5 2DUspring.
Figure
7.25 Relation of elastic potential
energy to work.
GY.90707.07.pgs 6/25/04 9:41 AM Page 16
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CHAPTER 7 Work and Energy
Now, remember the law of conservation of energy, which says that the total
amount of energy does not change. Thus, if no friction is present, then the gain in
kinetic energy by the mass must exactly equal the loss in potential energy by the
spring: DKmass 5 2DUspring. Be sure to understand this minus sign; it is very
important in the discussion that follows. Since the change in kinetic energy of the
mass represents the work done on the mass, DKmass 5 Wby spring 5 2DUspring.
Multiplying both sides by 21 gives us an equivalent expression:
DUspring 5 2Wby spring
This is the expression we need in order to find Uspring, and it is valid for any displacement the spring makes.
To find Uspring, let’s consider a mass at the end of a spring moving from the
equilibrium position at xi 5 0 to a coordinate at the positive position xf 5 x. We
know that the spring force pulls back on the mass, in a direction opposite to its
displacement. Thus, from Equation 7.5, the work done by the spring on the mass
is negative:
Wby spring 5 2 12 kx 2
Since DUspring 5 2Wby spring, where DUspring 5 Uf 2 Ui, we have
Uf 2 Ui 5 2 1 2 12 kx 2 2
At the equilibrium position xi 5 0, there is no stretch or compression, so Ui 5 0.
Therefore,
Spring at equilibrium length: Uel 5 0
Uf 5 12 kx 2
x
x50
Spring
compressed
by x
This is the potential energy of a spring, Uspring, or the elastic potential energy Uel,
of any object whose force is given by Hooke’s law, F 5 2kx, where x is the distance by which the object is compressed or stretched from equilibrium.
x
Elastic potential energy
x50
x
The elastic potential energy of a spring (or, indeed, any elastic object)
increases with the distance x the spring is stretched or compressed from its
equilibrium position at x 5 0:
For a compressed or stretched spring,
Uel 5 12 kx2
Spring stretched by x
Uel 5 12 kx 2
(7.6)
x
x50
x
Figure
7.26 Elastic potential energy of a
compressed or stretched spring.
Either stretching or compressing an elastic object causes its elastic potential
energy Uel to increase, as shown in Figure 7.26. When you stretch the object,
x . 0; when you compress it, x , 0. Since we are often interested in changes in
energy, we also have DUel 5 Uel,f 2 Uel,i or DUel 5 12 k 1 xf2 2 xi2 2 .
Nature’s rubber bands. Biologists used to believe that tendons and ligaments were simple
straps, without significant elasticity. We now know that they are often elastic, like very stiff
rubber bands, and that their elasticity is central to their function in the body. The photo shows
the microscopic basis of tendon and ligament elasticity. The fibers you see consist of collagen,
a tough, fibrous protein that is the main constituent of tendon and ligament tissue. Notice that
the collagen fibers are wavy. As the tissue stretches, these wavy fibers straighten out; then they
pull back to their equilibrium wavy configuration when released.
GY.90707.07.pgs 6/25/04 9:41 AM Page 17
17
7.5 Elastic and Gravitational Potential Energy
Quantitative
Analysis 7.4
Stretching a spring
A spring is stretched from x 5 0 to x 5 2a. Is more
energy required to stretch the spring from x 5 0 to x 5 a or from
x 5 a to x 5 2a?
x
x50
S O L U T I O N A pictorial diagram is shown in Figure 7.27a. The
spring is pulled the same distance in each segment of the displacement, yet the spring force increases with distance. Thus, we
expect the energy required to pull the spring through the second
segment to be larger than for the first. We can also use the graph
of the force of the hand pulling on the spring, Fhand, as a function
of position in Figure 7.27b to understand this point. Recall, that
on a graph of force as a function of position, the work done by
the force is represented by the area under the curve. The work
done by the hand in the first segment to pull the spring from
x 5 0 to x 5 a is represented by a single triangle. As you can
see, the area under the curve in the second segment from x 5 a
to x 5 2a is three times larger than in the first segment, so more
energy is required to pull the spring through the second segment.
x5a
x 5 2a
x5a
x 5 2a
(a)
Fhand
x
x50
The area under the curve from x 5 a to x 5 2a
can be broken into three triangles, each equal to
the area from x 5 0 to x 5 a. Thus, it takes three
times more energy to pull the spring through the
second segment than through the first.
(b)
Figure
7.27 (a) A spring pulled through two equal segments.
(b) The area under the force-versus-distance curve during the two
segments of the spring’s displacement.
Gravitational Potential Energy
Now let’s find a mathematical expression for the gravitational potential energy
Ug. By applying conservation of energy to a spring, we found that DUspring 5
2Wby spring. The same reasoning applies to an object moving solely under gravity’s influence: When an object falls, the work done on it by gravity is positive
(the object speeds up), and the change in the potential energy is negative (as the
object descends). To calculate Ug, we’ll compute Wby gravity and then identify Ug
with the use of the formula DUg 5 2Wby gravity.
Suppose a book of mass m is nudged off the end of a table, as shown in Figure 7.28. Gravity does work on the book as the book falls through a vertical distance d 5 yi 2 yf. The force F 5 mg, so the work done by gravity is W 5
F d 5 mg 1 yi 2 yf 2 . Then DUg 5 2Wby gravity gives
DU 5 Uf 2 Ui
y
i
m
i
yi
DUg 5 2mg 1 yi 2 yf 2
Ui
F 5 mg
or
d
Uf 2 Ui 5 mgyf 2 mgyi
Now we can identify the expression for gravitational potential energy:
yf
Gravitational potential energy
Uf
The gravitational potential energy at vertical position y is
Figure
Ug 5 mgy
(7.7)
Since we are often interested in changes in energy, we also have
DUg 5 mg 1 yf 2 yi 2
(7.8)
7.28 The gravitational potential
energy and work done by gravity on a book.
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CHAPTER 7 Work and Energy
Let’s note three important points about gravitational energy and its changes:
• In the preceding derivation, we assumed that the gravitational force is constant
(F 5 mg). Therefore, our result is valid only near the surface of the earth.
• As an object moves, its change in gravitational energy DUg 5 mg 1 yf 2 yi 2 is
independent of any horizontal motion.
• While for any object, Ug 5 mgy depends on where you place the origin y 5 0,
the change in gravitational potential energy DUg 5 mg 1 yf 2 yi 2 depends only
on the vertical distance the object rises or falls. Intuitively, the change in gravitational potential energy of a book dropped 1 meter is the same whether you
drop the book onto the first floor or onto the third floor of a library.
i
EXAMPLE 7.4
y
Climbing El Capitan
One of the world’s most famous climbing rocks is 884 m El Capitan in
Yosemite National Park. One of the few good sleeping ledges during the fourday ascent is the El Cap Tower at about 514 m. Consider a 75.0 kg man climbing the route in Figure 7.29a. (a) What is the man’s change in gravitational
potential energy when he climbs from El Cap Tower to the top? (b) Next, consider the change in gravitational potential energy for the entire route. With
4186 J in a single food calorie, how many food calories would be needed for
the climb if all of the energy went directly into raising the climber’s height?
y2 5 884 m
El
Capitan
y1 5 514 m
El Cap
Tower
y0 5 0
Figure
7.29 (a) The route up El Capitan. (b) The relevant vertical positions.
SOLUTION
Only the vertical displacement along the climb creates a
change in gravitational potential energy. We have labeled the relevant heights in Figure 7.29b, with the origin at the base of the
climb.
SET U P
S O LV E Part (a): The change in gravitational potential energy
between an initial and final height is given by DUg 5
mg 1 yf 2 yi 2 . In climbing from El Cap Tower at y1 to the top at y2,
the climber experiences a change
DUg 5 mg 1 y2 2 y1 2
5 1 75.0 kg 2 1 9.80 m s2 2 1 884 m 2 514 m 2
5 2.72 3 105 J
/
Part (b): For this part, we must first find the DUg corresponding
to climbing the route from bottom to top. Once we have that
change in energy, we’ll determine how many food calories correspond to an equivalent amount of energy. In climbing from
y0 5 0 m to y2 5 884 m,
(a)
(b)
DUg 5 mg 1 y2 2 y0 2
5 1 75.0 kg 2 1 9.80 m s2 2 1 884 m 2 0 m 2
5 6.50 3 105 J
/
If we multiply this amount of energy by the amount of food calories per joule, we’ll get the corresponding number of food calories:
Number of food calories 5 1 6.50 3 105 J 2
1
1 food calorie
4186 J
5 155 food calories
2
R E F L E C T For part (a), the horizontal displacement during the
climb does not affect DUg. In fact, any path that leads from y1 to
y2 has the same DUg. The purpose of part (b) is to get a sense of
how much energy is contained in food. The 155 calories required
to raise a man this height of over eight football fields is less than
that in a typical candy bar! In reality, most food calories are converted to heat during heavy exertion, and many thousands of
calories are consumed on a climb like this one.
We now have mathematical descriptions for kinetic energy K and potential
energy U, so we can define the total mechanical energy Emech:
Definition of mechanical energy
Emech 5 K 1 U
(7.9)
GY.90707.07.pgs 6/25/04 9:41 AM Page 19
7.5 Elastic and Gravitational Potential Energy
19
For an object with both gravitational and elastic potential energy, U 5
Ug 1 Uel. For a system with more than one object, the kinetic and potential
energies of each object must be added together.
Later, we’ll encounter another form of potential energy associated with the
electric force. The potential energy U in Equation 7.9 is the sum of all forms of
potential energy influencing a system—in this chapter, gravitational and elastic
potential energy.
Test your understanding of a system’s total mechanical energy in the next two
exercises.
Conceptual
Analysis 7.5
Mass-and-pulley system
Two unequal masses are connected by a massless
cord passing over a frictionless pulley, as shown in Figure 7.30.
If you consider the two masses together as a system, which of the
following statements is true about the total gravitational potential
energy Ug and the total kinetic energy K of the system after the
masses are released from rest?
A. Ug increases and K increases.
B. Ug decreases and K increases.
C. Both Ug and K remain constant.
When a system consists of more than one object, the
total K and Ug of the system is the sum of the kinetic and gravitational potential energies for each object. When the objects are
released, the heavy one moves down while the light one moves up.
Since both objects start from rest and end up moving, the
kinetic energy of the system increases. (Remember that kinetic
energy depends on speed, not velocity, so the direction of motion
has no effect on the kinetic energy.)
Now let’s consider the system’s change in gravitational
potential energy. The downward motion of the heavy block
decreases the Ug associated with that block. The upward motion
SO LU T I O N
Quantitative
Analysis 7.6
m
m
M
M
Before release
After release: How do K and Ug
change for the system of the two
masses?
Figure
7.30 The change in K and Ug for two masses connected by
a cord passing over a frictionless pulley.
of the light block increases the associated Ug. However, since Ug
depends directly on mass, the decrease due to the heavy block
dominates, so the total gravitational potential energy of the system decreases. Thus, choice B is correct.
Total mechanical energy
In Figure 7.31, two blocks of mass M and m, where
M . m, are attached to a thin string that passes over a frictionless pulley of negligible mass. The larger mass is resting on a
spring that is compressed a distance d from its equilibrium position. The spring’s compression is maintained by a trigger mechanism. Which of the following formulas represents the total
mechanical energy of the system?
m
M
A. Mgh1 1 mgh2 1 12 kh12
B. Mgh1 1 mgh2 1 12 kd 2
C. Mgh1 2 mgh2 1 12 kd 2
S O L U T I O N The total mechanical energy is the sum of the energies of each part of the system and is given by Emech 5 K 1 U,
where K 5 12 mv2 1 12 Mv2 and U 5 Ug 1 Uel. Using the tabletop
for the reference level, as shown in Figure 7.31, we find that
Ug 5 Mgh1 1 mgh2. Although the compression distance d of the
h2
h1
Figure
7.31 The energy of a spring-and-pulley system.
spring is not labeled in the figure, Uel 5 12 kd 2. With the trigger in
place, neither block has any speed, and K 5 0. Thus, Emech 5
Ug 1 Uel (choice B).
GY.90707.07.pgs 6/25/04 9:42 AM Page 20
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CHAPTER 7 Work and Energy
Conservative Forces
To find the expression for elastic and gravitational potential energy, we used
DUspring 5 2Wby spring and DUg 5 2Wby gravity. Could we similarly find a potential
energy DUfric 5 2Wby fric associated with the work done by friction? Absolutely
not: Work done by friction creates thermal energy, and this dissipated energy can
never be stored as potential energy and returned, in its entirety, to an object. Thus,
no potential energy is associated with friction.
Any force that is associated with potential energy, such as gravitational and
spring forces, is called a conservative force. The work done by a conservative
force can be stored in the form of the associated potential energy, and it can be
regained in its entirety as kinetic energy. As we’ll see in the next section, the term
conservative indicates that this type of force is required for the conservation of
mechanical energy.
Potential energy and work
Potential energy is related to the work Wc done by a conservative force:
DU 5 2Wc
(7.10)
Energy-efficient transport. In Indonesia
and other countries, some women are known
to carry loads of up to 70% of their body
weight in baskets on their heads. Westerners
who attempt to carry a load in this way
expend almost twice as much energy as people who do it normally. This puzzling difference prompted an additional study, which
found that, when Westerners walk, they have
more up-and-down motion than do women
accustomed to carrying baskets. For larger
vertical motion, more work is done by gravity
on the load; thus, more energy is required to
raise the load a higher distance against the
force of gravity.
Recall that the change in potential energy for the rock climber in Example 7.5
is DUg 5 mg 1 y2 2 y1 2 and that this expression depends only on the initial and
final positions of the climber. The meandering path taken by the climber in Figure 7.29 does not have any influence on DUg. The work Wc done by a conservative force is also independent of the path an object takes between two points,
since Wc 5 2DU:
The work done by a conservative force is path independent; that is, it
does not matter which path is taken between the endpoints.
For a closed path, which ends right where it begins,Wc 5 0. To understand this
idea, consider the rock climber again. If he climbs up a distance h and then back
down to the same location, the resulting work on him due to gravity is
2mgh 1 mgh 5 0. Gravity will do equal amounts of positive and negative work
on him during any path that returns him to his original location.
Long path
Short path
Figure
7.32 A box slides on a warehouse
floor along two paths. Work is done on the box
by a nonconservative force of friction and is not
path independent.
Along either path, the nonconservative force of friction does work on the box.
However, friction does more work on the box along the long path than along
the short path.
GY.90707.07.pgs 6/25/04 9:42 AM Page 21
7.6 Conservation of Energy
21
If you think about the energy you would expend in doing work to
climb up a distance h against gravity and then climb back down to your
starting point, you can easily see that you will certainly expend energy to
make the closed path. However, Wc 5 0 pertains to the work done on you
by the conservative force of gravity, and not the calories burned, or the
work done by your muscles, during the climb.
N OT E
The work done on an object by a nonconservative force such as friction is not
path independent: The longer the path taken between two locations, the more
work friction does on the object. This observation is illustrated with a box sliding
on a warehouse floor in Figure 7.32. We discuss nonconservative forces (such as
friction) later.
7.6 Conservation of Energy
Now that we have developed a mathematical description of three forms of
mechanical energy, we can proceed to our main goal, which is to use energy to
describe the behavior of objects in a variety of circumstances. We will use the law
of conservation of energy, which we introduced at the start of the chapter.
Conservation of Mechanical Energy
The law of conservation of energy says that energy can change form, but can
never be created or destroyed. If you imagine an isolated system for which no
energy enters or escapes, the total energy of the system must therefore remain
constant when evaluated at any initial and final time:
1 Etotal 2 f 5 1 Etotal 2 i
This law has been verified on scales as large as those of galaxies and as small as
those of atoms. It applies to every process in the physical world, including all
phenomena in chemistry, geology, and biology.
While conservation of energy applies to all isolated systems when all forms
of energy are taken into account, in this chapter we focus on the conservation of
mechanical energy, where Emech 5 K 1 U (Equation 7.9). If the forces doing
work in a system are conservative, then the system’s mechanical energy will
be conserved. As long as no dissipative forces are present (such as friction or air
drag), nothing will convert any of the system’s mechanical energy to nonmechanical forms of energy. Also, by saying that a system is isolated, we automatically rule out pulls and pushes from outside that would change the system’s
energy.
Conservation of mechanical energy can be written as Emech,f 5 Emech,i. However, it is easiest to solve energy problems when we express the conservation
of mechanical energy in terms of kinetic and potential energy using
Emech 5 K 1 U:
Conservation of mechanical energy
When only conservative forces do work in an isolated system, the system’s total mechanical energy is conserved:
Kf 1 Uf 5 Ki 1 Ui
Unit: joule, J
(7.11)
Catapult legs. To jump, a grasshopper
must straighten its hind legs faster than its leg
muscles can contract. It does so by using a
catapult mechanism. The black swelling on
the first joint of each jumping leg marks a
spot where the insect’s exoskeleton has been
modified into a pair of stiff springs shaped
like flattened C’s. Before a jump, the limb
muscles build up elastic potential energy by
compressing these springs while the joint is
flexed. To jump, the animal releases the
springs, which cause the legs to snap straight.
Such mechanisms rely on conservation of
mechanical energy.
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CHAPTER 7 Work and Energy
Our choice of
final point
y
v50
y2
U
h
U
K
vi
y1
K
In words, this equation for the conservation of mechanical energy says that when
you evaluate the initial mechanical energy of a system (indicated with subscript
“i” in Emech,i 5 Ki 1 Ui), it will equal the final mechanical energy of the system
(indicated with subscript “f” in Emech,f 5 Kf 1 Uf). A conserved quantity is one
that remains constant, so that evaluating it at any time produces the same number.
For this reason, the initial time can be at any point in the system’s motion, the
final time can be at any point in a system’s motion, and the mechanical energy
will be the same at both points. As you will see, the actual choice of which points
to use for initial and final times depends on the question you are trying to answer
about a particular system. Finally, keep in mind that, although we calculate
mechanical energy at different times, there is no variable for time, t, in the equation for conservation.
Now let us use conservation of energy to solve a simple mechanics problem.
The person in Figure 7.33 tosses a ball of mass m 5 0.20 kg straight up with an
initial velocity vi 5 7.7 m s. Assuming no air friction, how high does the ball
rise? The “energy flasks” in the figure show graphically that the total mechanical
energy remains constant throughout the motion, while the amounts of kinetic and
potential energy are constantly changing.
To solve this problem, we will use Equation 7.11. Therefore, we must choose
the initial and final times at which to evaluate the energy. An obvious initial
point is when the ball leaves the girl’s hand, but what should we use for the final
point? Should the final point be when the ball returns to the girl’s hand? The
answer is that while the ball is in motion, you can choose whatever initial and
final times work best for the problem, because the energy is the same at all
times. In this case, we want to know the maximum height reached by the ball, so
we choose that as our final point. At the maximum height, v 5 0, and therefore
Kf 5 0. Now, equating the initial and final energies as in Equation 7.11
(Kf 1 Uf 5 Ki 1 Ui) yields
/
Our choice of
initial point
y50
0 1 mgy2 5 12 mvi2 1 mgy1
Solving for h 5 y2 2 y1, we obtain
mg 1 y2 2 y1 2 5 12 mvi2
Figure
7.33 Using conservation of energy
to find how high a tossed ball rises.
or
h5
1 2
2 vi
g
and finally,
h5
1
2
1 7.7 m / s 2 2
/
9.8 m s2
5 3.0 m
While we could have solved this problem with Newton’s laws, conservation
of energy can be used for problems that would be quite difficult to solve by Newton’s laws. We will work some problems of this type in the pages that follow.
Unless stated otherwise, we assume that friction is negligible, so that mechanical
energy is conserved.
GY.90707.07.pgs 6/25/04 9:42 AM Page 23
7.6 Conservation of Energy
23
Strategy for solving problems using conservation
of mechanical energy
SET U P
1. Determine whether mechanical energy is conserved. Each force that does
work on the system must be conservative (that is, must have an associated
potential energy). Friction in any form dissipates mechanical energy.
2. Choose the initial and final times you will analyze. You might choose different initial and final times for different parts of a problem.
3. Define your coordinate system, particularly the level at which y 5 0.
Ug 5 mgy assumes that the positive direction for y is upward; we suggest
that you use this choice consistently. If there is a spring, it is easiest to
place y 5 0 at the equilibrium position; when this is not possible, use only
the distance from equilibrium in determining the spring’s potential energy.
S O LV E
4. Make a list of the initial and final kinetic and potential energies—that is,
Ki, Kf, Ui, and Uf. In general, some of these will be known and some
unknown. Use algebraic symbols for any unknown quantities.
5. Using Kf 1 Uf 5 Ki 1 Ui, solve for whatever unknown quantity is
required.
EXAMPLE 7.5
Roman catapult
An ancient Roman catapult propels a 10.0 kg
rock at enemy troops, as illustrated in Figure 7.34a. This type of catapult was powered
by elastic sinews or fibers wound around an
axle connected to the catapult’s arm. The
sinews are stretched as the arm is pulled
down.
We can model the loaded catapult as a
compressed spring with k 5 300 N m. In our
model, the force on the rock when the catapult
is triggered is F 5 kx, where x is the amount
by which the catapult arm is pulled back from
its equilibrium position.
Consider the shot shown in Figure 7.34b,
where the catapult arm is pulled back 7.00 m
and then released. The rock hits the hill 20.0
vertical meters above the launch height.
Ignore the fact that the rock gains some gravitational potential energy while still in the catapult’s basket. Find (a) the speed of the rock as
it leaves the catapult’s basket, (b) the total
mechanical energy of the system, and (c) the
speed of the rock when it hits the hill with the
enemy troops.
/
Final point: all
mechanical energy
is kinetic energy K
Initial point: all
mechanical
Uel
energy is elastic
potential energy
S
vcat
x
m 5 10.0 kg
F 5 kx
x50
x 5 7.0 m
Catapult
Spring model representing catapult
(a)
vhill 5 ?
y
Final point
20.0 m
vcat
K
0
Initial point
K
Ug
(b)
Figure
7.34 (a) The catapult modeled as a compressed spring. (b) The catapult shoots a
rock onto a hill.
SOLUTION
S E T U P We’ll place the origin of the y-axis at the point where the
rock leaves the catapult, as shown in Figure 7.34b. The simple
spring model shown in the figure lets us assume that the rock fol-
lows a straight, rather than a curved, trajectory while in the catapult’s basket.
S O LV E Part (a): We can find the launch speed vcat by using
conservation of mechanical energy. In our simplified model, the
Continued
GY.90707.07.pgs 6/25/04 9:42 AM Page 24
24
CHAPTER 7 Work and Energy
catapult acts as a compressed spring, and all the elastic potential
energy is converted to kinetic energy of the rock. (Remember, we
ignore the small change in Ug during launching.) Mathematically,
the system’s initial mechanical energy is entirely elastic potential
energy: Emech,i 5 Uel 5 12 kx 2. For the purpose of finding vcat, we
define the final time as the moment the rock loses contact with
the basket, so that Emech,f 5 K 5 12 mvcat2. The law of conservation
of mechanical energy, Equation 7.11, requires that
Kf 1 Uf 5 Ki 1 Ui
our final time is the moment the rock hits the ground. We begin
by equating the total mechanical energy at that moment to the
total energy at another point in time, where Kf 5 12 mvhill2 and
Uf 5 mgyhill, with yhill 5 20.0 m in the chosen coordinate system.
But what point do we use for the initial energy—the fully loaded
catapult, or perhaps the point when the stone loses contact with
the basket? The answer is that it doesn’t make any difference,
because the total energy is constant. For an initial time when the
rock is sitting in the loaded catapult, Ki 5 0 and Ui 5 12 kx 2. Then
Kf 1 Uf 5 Ki 1 Ui becomes
where Uf 5 Ki 5 0 in this part of the problem. Thus,
1
2
2 mvcat
1
2
2 mvhill
5 12 kx 2
Solving for vhill gives
and it follows that
vcat 5
1 mgyhill 5 0 1 12 kx 2
1 300 N m 2 1 7.00 m 2 2
kx 2
5
5 38.3 m s
Å m
Å
10.0 kg
/
/
vhill 5
Å
1
2
2 kx
2 mgyhill
1
2m
1
1 300 N m 2 1 7.00 m 2 2 2 1 10.0 kg 2 1 9.80 m s2 2 1 20.0 m 2
Part (b): To calculate the total mechanical energy of the system, 5 2
1
Å
we can evaluate either side of Equation 7.11, since each side rep2 1 10.0 kg 2
resents the total mechanical energy at a different time. Evaluating 5 32.8 m s
Emech just as the rock loses contact with the catapult yields
/
/
/
Emech 5 12 mvcat2
5 12 1 10.0 kg 2 1 38.3 m s 2 2 5 7.33 3 103 J
/
Part (c): To find the speed of the rock when it hits the ground,
vhill, we will again use the equation for conservation of energy,
solving it for vhill, which will be the only unknown. This time,
Notice that by using conservation of energy, we can
find vhill without knowing details such as the range of the rock or
the angle at which the rock is fired. Such details would be necessary for a solution found by using Newton’s laws. As you can see
from the algebraic result, vhill depends only on the initial energy
and the height at which the rock lands.
R EF LECT
As noted in the previous example, the speed with which the rock hits the
troops on the hill does not depend on the details of the rock’s trajectory. This is a
consequence of the fact that the work done by conservative forces is independent
of the path taken. Let’s use the special frictionless, air-drag-free roller coaster in
Figure 7.35 to examine the point further. At point A, located at height H, the car
has just begun to roll; its speed (and thus kinetic energy) are negligible. What is
its speed vh at point B with height h? Using conservation of energy,
Kf 1 Uf 5 Ki 1 Ui
where Ki 5 0, Ui 5 mgH, Kf 5 12 mvf2, and Uf 5 mgh. Thus,
1
2
2 mvf
1 mgh 5 0 1 mgH
so that
vf2 5
2
1 mgH 2 mgh 2
m
A
Ug
vh
Ug
K
H
B
h
Figure
7.35 The speed of a roller-coaster
car at a given height on the roller coaster.
K
C
vh
vh
D
vh vh
GY.90707.07.pgs 6/25/04 9:42 AM Page 25
7.6 Conservation of Energy
25
and the speed of the car at point B with height h is
vh 5 "2g 1 H 2 h 2
(7.12)
You can see that the speed vh at point B depends on the change in height,
H 2 h. However, we obtain the same result if we solve for the speed at points C
and D, which are also at height h. This is true even though the car is moving up at
point C, instead of down. To understand this result, note that the car gains additional kinetic energy as it falls below point B. However, by the time the car comes
back up and reaches point C, all of the extra kinetic energy will have turned back
into the gravitational potential energy seen at height h. By the time the cart
reaches point D, there are many exchanges between K and U, but Equation 7.12
still applies.
Of course, this analysis is strictly valid only in the absence of friction. However, in a real roller coaster, energy is added to the system just during the initial
climb. After that, all the coaster’s motion is “coasting,” representing the interchange of K and Ug. Some mechanical energy is lost to friction, but, as the riders
know, plenty remains.
Conceptual
Analysis 7.7
Slide to a stop
Two objects with masses M and m, where M . m,
slide with equal speeds over a horizontal frictionless surface. The
object with mass M has more kinetic energy because it has
greater mass. They encounter a frictionless hill and slow down as
they slide up it. Which object slides higher before coming to
rest?
A. M, because it had the greater kinetic energy.
B. m, because it has less mass.
C. Neither; they both slide to the same height.
Equation 7.11, Kf 1 Uf 5 Ki 1 Ui, says that the
final mechanical energy equals the initial mechanical energy. For
both masses, Ui 5 0 and Kf 5 0, so Uf 5 Ki; that is, the kinetic
SO LU T I O N
DO IT
YOURSELF 7.1
energy of each object transforms into gravitational potential
energy. Thus, for the lighter mass m,
mgh 5 12 mvi2
or
h5
vi2
2g
Since the mass cancels from both sides of the equation, the
height h does not depend on the mass at all. The initial speed is
the same for both objects, so they slide to equal heights (choice
C). Thus, even though object M has a greater initial kinetic
energy than object m, it is precisely this additional energy that is
required to raise M to the same height as m.
Seesaw
Two children sit at rest on a seesaw, at equal distances from the fulcrum, with their feet on the ground as shown
in Figure 7.36. The seesaw is initially level and 0.25 m above the
ground. We will neglect its mass. The child on the right has twice
the mass of the child on the left. After both children lift their feet,
the side with the heavier child falls to the ground, and the lighter
child is inadvertently launched into the air. What is the lighter
child’s launch speed?
/
Answer: 1.3 m s. Get as far as you can using the hints given next,
and use only the hints you need.
At what speed is the
smaller child launched
into the air?
h 5 0.25 m
Before
Figure
After
7.36 Two children on a seesaw.
Continued
GY.90707.07.pgs 6/25/04 9:42 AM Page 26
26
CHAPTER 7 Work and Energy
HINTS
S O LV E
SET U P
4. What is the initial energy? Initially, the seesaw is at rest, and
both masses are at the same height.
1. What is the relevant energy equation? There is no friction,
so mechanical energy is conserved. The mechanical energy
evaluated at any two times will be equal.
2. When should I evaluate the initial and final energies? It is
easiest to take the initial energy to be the energy when the seesaw is level, and the final energy to be the energy when the
right end hits the ground. It will also be helpful to use a subscript of 1 for the child on the left and a subscript of 2 for the
child on the right, so that m2 5 2m1.
5. What is the final energy? Each child may have a kinetic
energy term and a potential energy term, so the final energy
may have up to four terms. You don’t know the velocity yet,
but that’s what you will solve for. To find the height of m1,
inspect the geometry of the situation in Figure 7.36.
6. How are v1 and v2 related? Since the children are at equal
distances from the center of the seesaw, v2 5 v1.
3. Where should I choose y 5 0? For the hints that follow, we
choose the ground to be y 5 0. Setting the initial height of the
seesaw as y 5 0 is an equally good choice.
Recall that when a system includes more than one object, the total mechanical
energy is the sum of the kinetic and potential energies of each object. Test your
understanding of this point in the next example.
Quantitative
Analysis 7.8
System with two masses
Figure 7.37 shows a system in which a heavy object
of mass M is connected to a lighter object of mass m by a thin
string passing over a frictionless pulley of negligible mass. The
heavier object is given an initial push that sends it moving
upward. Gravity brings it to a halt after it has risen a distance h; it
then falls back downward. Which of the following equations
could you use to find h?
M
h
m
M
A. 1 M 1 m 2 gh 5 12 1 M 1 m 2 v2
B. 1 M 2 m 2 gh 5 12 1 M 1 m 2 v2
C. 1 M 2 m 2 gh 5 12 1 M 2 m 2 v2
m
We begin by using Equation 7.11, Kf 1 Uf 5
Ki 1 Ui. The final configuration has the masses at rest, which
means that Kf 5 0. Equation 7.11 can now be written as
Uf 2 Ui 5 Ki. In words, this means that the initial kinetic energy
of the system changes into an increase in gravitational energy.
Choice C can be rejected because it implies that the kinetic energy
of object m is negative (2 12 mv2), but kinetic energy can never be
negative. For the two objects, Ki 5 12 1 M 1 m 2 v2. The large mass
rises so that it gains gravitational potential energy equal to Mgh,
while the small mass moves downward, with its gravitational
SO LU T I O N
Before push
Maximum rise
Figure
7.37 Behavior of two masses on a pulley when the larger
mass is given an upward push.
energy decreasing by mgh. Thus, the system’s increase in gravitational potential energy is Uf 2 Ui 5 1 M 2 m 2 gh, and choice B
is correct.
Now let’s consider a situation involving both elastic and gravitational potential
energy.
EXAMPLE 7.6
Climber’s fall
A 60.0 kg climber has moved 4.00 m above her last anchor point, with 2.00 m of slack in the rope, when
she suddenly falls. She falls 10.0 m before the slack runs out. Fortunately, rock climbing ropes are designed
to stretch, so that once the slack has run out, the deceleration is gradual, reducing injury. While stretching,
the rope acts like a spring with k 5 1125 N m. For this fall, how far does the rope stretch?
/
Continued
GY.90707.07.pgs 6/25/04 9:42 AM Page 27
27
7.6 Conservation of Energy
y
Mass (climber)
y1 4.00 m
1 Climber begins
fall
2.00 m
slack
y0
Spring
(rope)
2 Climber reaches
end of slack
3 Rope reaches
Amount
rope
stretches
maximum stretch
y2 6.00 m
ys y3y2
y3
Pan (end of slack)
1 Climber
2 Climber reaches
begins to fall
(a) Climber's fall
3 Rope reaches
end of slack
maximum stretch
(b) Spring model of climber's fall
Figure 7.38 (a) A climber falls. Her rope stretches to reduce the acceleration as she is brought to a halt. (b) A spring
model to analyze the energy in the fall.
SOLUTION
so that
S E T U P Figure 7.38a diagrams the climber’s fall, and Figure 7.38b shows how we can model the fall as that of a mass (the
climber) and a spring (the rope). The 10.0 m between y2 and y1
represents the climber’s free fall. (Eight meters represents the
climber falling from 4.00 m above the anchor to 4.00 m below,
and the remaining 2 meters come from the slack).
In this problem, the origin of the coordinate system ( y 5 0)
is not at the equilibrium position (y 5 y2) of the spring used to
model the stretching rope. Therefore, we need to express the
elastic potential energy in terms of the distance the rope or spring
is stretched from the equilibrium position. Using the symbol ys
for the stretch from equilibrium, the maximum stretch of the rope
at the bottom of the fall is ys 5 y3 2 y2.
S O LV E Our strategy is to break the motion into two parts. First,
we’ll find the climber’s speed and kinetic energy at the moment
she runs out of slack. Then, we’ll see how much this kinetic energy
K will stretch the rope. In addition to K, Ug also helps to stretch the
rope. For the first part, we’ll apply conservation of mechanical
energy between points y2 and y1, and instead of writing Emech,f 5
Emech,i, in Equation 7.11, we’ll write Emech,2 5 Emech,1. We have
K2 1 U2 5 K1 1 U1
where K2 5 12 m 1 v2 2 2, U2 5 mgy2, K1 5 0, and U1 5 mgy1. Then
1
2m
1 v2 2 2 1 mgy2 5 0 1 mgy1
and, solving for v2, we obtain
1
2m
1 v2 2 2 5 mgy1 2 mgy2
or
v2 5 "2g 1 y1 2 y2 2
v2 5 "2 1 9.80 m s2 2 1 4.00 m 2 1 26.00 m 2 2 5 14.0 m s
/
/
A mistake commonly made at this point is to use v2 to equate the
kinetic energy of the falling climber at height y2 to the elastic
potential energy gained when the rope is stretched to its maximum at y3: Uel,3 5 K2. You can reject this approach when you
recognize that gravitational potential energy also helps stretch
the rope. Therefore,
K3 1 U3 5 K2 1 U2
where U 5 Ug 1 Uel and K3 5 0. Then, using ys 5 y3 2 y2 for
the stretch of the rope, we get
0 1 mgy3 1 12 kys2 5 12 mv22 1 mgy2
Rearranging terms yields
1
2k
1 y3 2 y2 2 2 1 mg 1 y3 2 y2 2 5 12 m 1 v2 2 2
which is a quadratic equation in ys:
1
2
2 kys
1 mgys 2 12 m 1 v2 2 2 5 0
From Appendix A, the solution is ys 5 1 2b 6 "b 2 2 4ac 2 2a,
where a 5 12 k 5 562.5 N m, b 5 mg 5 588.0 N, and c 5
2 12 m 1 v2 2 2 5 25880 N # m. The two solutions of this equation
are ys 5 23.79 m and ys 5 12.75 m. As is usual for a quadratic
equation, you have to decide which solution is physically plausible. With the coordinate system shown in Figure 7.38, we can see
that ys 5 y3 2 y2 , 0, so our answer is ys 5 23.79 m.
/
/
Alternative Solution: It’s worth pointing out that there’s an easier way to solve this problem. Since energy is conserved throughout the fall, we can equate the energy at the bottom of the
stretched rope to the energy at the beginning of the climber’s fall:
Emech,3 5 Emech,1. Doing so gives
Continued
GY.90707.07.pgs 6/25/04 9:42 AM Page 28
28
CHAPTER 7 Work and Energy
0 1 mgy3 1 12 k 1 y3 2 y2 2 2 5 0 1 mgy1
which, with 1 y3 2 y2 2 5 1 y3 2 2 2y3y2 1 1 y2 2 , becomes
2
1
2k
2
2
1 y3 2 2 1 1 mg 2 ky2 2 y3 1 1 12 k 1 y2 2 2 2 mgy1 2 5 0
In this quadratic equation, everything is known except for y3,
which is found by solving the equation. The physical solution is
y3 5 29.79 m, and the rope stretches a distance ys 5 y3 2 y2 5
23.79 m.
As the alternative solution illustrates, you can apply
conservation of energy at any two times, and some choices make
the problem easier to solve than others.
R EF LECT
7.7 Nonconservative Forces: Energy Conservation
with Friction
An assumption we make in applying the principle of conservation of mechanical
energy is that no frictional forces are present. When frictional forces act within a
system, mechanical energy is converted into thermal energy. Thus, mechanical
energy is not conserved in such systems, and friction is called a nonconservative
force.
When friction is present, we can adapt conservation of energy so that it is still
useful for solving problems. Mathematically, we must relate changes in the
mechanical energy to the work done by a nonconservative force. Doing this
requires a short derivation beginning with the work–kinetic energy theorem,
which is valid even in the presence of friction. From Equation 7.4,
Wnet 5 Kf 2 Ki 5 DK
and if we write Wc and Wnc for work done by conservative and nonconservative
forces respectively, we have
Wc 1 Wnc 5 DK
Using Wc 5 2DU (Equation 7.10) to express the work done by a conservative
force, we have
Wnc 5 DK 1 DU
With Emech 5 K 1 U, the change in mechanical energy can be written as
DEmech 5 DK 1 DU. Therefore, Wnc 5 DEmech, or Wnc 5 Emech,f 2 Emech,i.
Conservation of energy with nonconservative forces
For work Wnc done on an object by a nonconservative force,
Wnc 5 Emech,f 2 Emech,i
(7.13)
This result simply says that the energy transferred to or from an object by nonconservative forces will change mechanical energy by the same amount.
Now let’s focus on a familiar dissipative force: friction. Friction forces
always act directly opposite to the direction of displacement. Thus, in Wnc 5 F d,
F is always negative. Therefore, if a constant force of kinetic friction, fk, is present, Wnc 5 2fkd. The minus sign tells us that energy is transferred away from the
object. For work done by a nonconservative force of friction in Equation 7.13,
i
i
Emech,f 5 Emech,i 2 fkd
(7.14)
which simply states that the final mechanical energy equals the initial mechanical
energy minus the energy dissipated through friction. We use this equation in the
next example.
GY.90707.07.pgs 6/25/04 9:42 AM Page 29
29
7.7 Nonconservative Forces: Energy Conservation with Friction
“Boardslide” down the rail
EXAMPLE 7.7
m 5 70.0 kg
y
A skateboarder of mass 70.0 kg rides a stairway handrail as shown in vi 5
Figure 7.39a. He lands onto the top of the rail at a point 2.25 m 2.00 m/s
/
above the ground with a speed of v 5 2.00 m s. A constant frictional force with magnitude fk 5 255 N acting between the skateboard and rail limits the skateboarder’s speed. (a) What is his speed
as he slides off the rail at a height of 1.25 m? (b) By what fraction
does the friction reduce his speed?
yi 5 2.25 m
vf 5 ?
yf 5 1.25 m
d5
1.00 m
1.00 m
sin 30°
30°
30°
Figure 7.39 A skateboarder slides down a rail; a nonconservative friction force acts to slow him down.
SOLUTION
S E T U P A pictorial representation is shown in Figure 7.39a,
together with a coordinate system.
S O LV E Part (a): We know from Equation 7.14 that friction
reduces the mechanical energy of the system according to Emech,f 5
Emech,i 2 fkd. We are asked to find the final speed, which will
come from the kinetic energy contribution to Emech,f. Let’s write
the equation and see what we need in order to find vf:
1
2
2 mvf
1 mgyf 5
1
2
2 mvi
1 mgyi 2 fkd
We already know every variable in this problem, except for the
distance d that the skateboard slides (grinds) along the handrail.
In Figure 7.39b, we can see that d 5 1 1 m 2 1 sin 30° 2 5 2 m.
Now we solve for vf:
/
vf2 5
2 1 2
1 mv 1 mg 1 yi 2 yf 2 2 fkd 2
m 2 i
y50
(a)
(b)
Part (b): To find the fraction by which friction reduces the
skateboarder’s final speed, we must first find the speed, say, v2,
he would have in the absence of friction. In this case, Emech,f 5
Emech,i, so we have
1
2
2 mv2
1 mgyf 5 12 mvi2 1 mgyi
from which it follows that
v2 5 " 1 2.00 m s 2 2 1 2 1 9.80 m s2 2 1 1.00 m s 2 5 4.86 m s
/
/
/
/
/
Then friction reduces the final speed by the fraction vf v2 5
1 3.00 m s 2 1 4.86 m s 2 5 0.62, or 62%.
/ /
/
R E F L E C T From Emech,f 5 Emech,i 2 fk d, we can see that the
change in energy depends only on the total energy dissipated by
friction between yi and yf. Therefore, twice the friction for only
half the 2 m distance would result in the same final speed. This is
true regardless of whether the hypothetically increased friction is
applied during the first or second half of the slide.
/
Multiplying through by 2 m before taking the square root yields
vf 5 " 1
v2i
1 2g 1 yi 2 yf 2 2 1 2 m 2 fkd 2
/
so that
vf 5 " 1 2.00 m s 2 2 1 2 1 9.80 m s2 2 1 1.00 m 2 2 3 2 1 70.0 kg 2 4 1 255 N 2 1 2.00 m 2
5 3.00 m s
/
/
/
/
Water slide
DO IT
YOURSELF 7.2
Figure 7.40 shows a child riding down a 3.0 m water
slide that is inclined at an angle of 20° with the horizontal and that
ends 1.0 m above the water surface. The child rides an inflatable
tube; the combined mass of the child and the tube is 50 kg. The tube
has negligible friction when it is inflated, but halfway down the slide
it pops, and the child continues sliding under a constant friction force
of 40 N. What is the child’s speed when she reaches the water?
3.0
m
20°
1.0 m
/
Answer: 6.1 m s. Get as far as you can using the hints given next,
and use only the hints you need.
Figure
7.40 A child rides down a water slide.
Continued
GY.90707.07.pgs 7/12/04 11:50 AM Page 30
30
CHAPTER 7 Work and Energy
HINTS
S O LV E
SET U P
1. What is the relevant energy equation? Due to friction,
mechanical energy is not conserved. In this case, we have
Emech,f 5 Emech,i 2 fkd, where the last term is the work done
by friction.
2. When should I evaluate the initial and final energies? It is
easiest to evaluate the initial energy at the top of the slide and
the final energy when the child reaches the water.
3. Where should I choose y 5 0? It is easiest to choose y 5 0
either at the water level or at the top of the slide. Both choices
work equally well, although the latter means that the water
level is at a negative height.
4. What are the initial kinetic and potential energies? The
child starts from rest, and the initial height can be found using
trigonometry.
5. What are the final kinetic and potential energies? The final
kinetic and potential energies are found at the moment the
child hits the water. The final speed is the variable you want to
solve for.
6. What is the work done by friction? The friction force acts
only over half the length of the slide.
*7.8 Mechanical Power
Your performance as a runner is limited not by the total force your leg muscles
can generate, but by how much work they can do over a sustained period. For this
and other applications, we need the concept of power, which expresses the rate at
which energy is transferred or transformed. In this chapter, we will deal specifically with average mechanical power, which is the average rate at which a system (such as your running body) does work:
Average mechanical power, P
For work W performed during the time interval Dt, the average mechanical
power is
P5
W
Dt
(7.15)
/
unit: J s 5 watt, W
/
For power, we define a new unit, the watt, W, where 1 W 5 1 J s. Don’t confuse the symbol for watt with work, W 5 F d. Because the joule is the unit for
either work or energy in general, the watt can measure either the rate at which
energy of any sort is transferred or transformed (power) or the rate at which work
is done (mechanical power). You’re probably most familiar with watts as a power
rating for lightbulbs (often 60 to 100 W). The wattage of a lightbulb tells you the
rate at which the bulb will transform electrical energy to the radiant energy of
light.
Another unit of power you may have heard of is horsepower, hp, where
1 hp 5 746 W. The automotive industry uses this term frequently, but without a
consistent conversion to watts. Nonetheless, scientists always use 746 W for 1 hp.
For top atheletes, the maximum sustainable mechanical power that can be produced is around 13 hp. This is why human-powered flight is so difficult: Even the
best human-powered plane designs require about 13 hp to fly.
Let’s get a physical feeling for the mechanical power generated in a common
activity. Riding a stationary bike at the gym with moderate effort requires about
P 5 150 W of mechanical power. The physical stress you feel comes from burning calories, but caloric energy gets converted to both work and heat. In other
i
Measuring calories burned. Animals use
oxygen to burn calories. Therefore, to measure how fast calories are burned during a
given activity, scientists measure the rate at
which the animal consumes oxygen. Here,
the oxygen consumption of a pacing llama is
being measured.
GY.90707.07.pgs 6/25/04 9:42 AM Page 31
*7.8 Mechanical Power
31
words, not all of the expended energy goes into work, and this is important to
keep in mind when you consider mechanical power. In fact, only about 20% of
the energy used gets converted into work during most types of sustained physical
exercise (walking, running, rowing, cycling, etc.). The remainder of the energy is
expended internally and raises your body temperature, which causes you to
sweat. Therefore, on the stationary bike, to produce P 5 150 W of mechanical
power, about five times this power, or 750 W, must be supplied by calories. (The
factor of 5 comes from 5 calories burned for each 1 calorie of work, and 51 is
20%.) 750 W is about 1 hp, but keep in mind that the 13 hp maximum a human can
sustain refers to the mechanical power produced, not calories burned.
Treadmill
EXAMPLE 7.8
/
A 70.0 kg man runs at 8.00 km hr (about 5 mph)
for 30.0 min on a treadmill inclined u 5 5.00°
above horizontal. (a) What is the mechanical
power delivered during the exercise? (b) If only
20% of caloric energy goes into work, how many
calories does the man burn during 15.0 min of
exercise? (Use 1 food cal, C 5 4186 J.)
Figure 7.41 Exercise on a treadmill can
be modeled as exercise on an actual hill.
d 5 4.00 km
u 5 5.00°
(a) A run on a treadmill.
SOLUTION
h 5 349 m
(b) The treadmill run modeled as a
run up a 4 km, 5° hill.
or
S E T U P Figure 7.41a shows the runner. It might seem at first
that the work done on him as he runs is zero. After all, neither
his height nor his speed changes. However, his activity is
equivalent to running up an inclined road for 30.0 min at
8.00 km hr. When you run uphill, you move yourself upward;
on an angled treadmill, you maintain position on a descending
surface by repeatedly moving upward and then getting pulled
back down. Figure 7.41b models the runner’s activity as an
ordinary uphill run. The running distance would be d 5 vDt, or
d 5 1 8.00 km hr 2 3 1 0.50 hr 2 5 4.00 km.
/
/
Part (a): To find the power, we’ll analyze the inclined
road in Figure 7.41b. For the power P 5 W Dt, we need the work
W 5 F d. Finding the work here is similar to finding the work in
the case of a package sliding down the ramp in Example 7.2. To
run at constant speed, the man must push off the treadmill with
an average force of magnitude Fman 5 mg sin u. Then the work
done by the man is W 5 Fman d 5 mgd sin u. Recognizing that
h 5 d sin u, we have
S O LV E
/
P5
Part (b): First, let’s find the total energy required to produce
133 W for 15 min. Then we’ll calculate how many calories must
be burned to produce this amount of energy. For a 20% caloriesto-work efficiency, the total calories burned is five times greater
than the work.
To find the energy required to produce 133 W for 15 min, or
Dt 5 15 min 1 60 s min 2 5 900 s, recall that the transferred
energy is work. We can find the work from
/
W 5 PDt
5 1 133 J s 2 1 900 s 2 5 1.20 3 105 J
i
h 5 d sin u 5 v 1 Dt 2 sin u
5 1 4.00 3 103 m 2 1 sin 5.00° 2 5 349 m
and W 5 mgh.
We can see that the work done by the man simply depends on
the height h he rises against the pull of gravity along the hypothetical inclined road.
The man does this work for 30 min, or Dt 5 1800 s. Therefore, the power generated by the man is
P5
mgh
W
5
Dt
Dt
1 70.0 kg 2 1 9.80 m / s2 2 1 349 m 2
5 133 W
1800 s
/
1
Converting this to calories yields
1 1.20 3 105 J 2
2
1 food cal
5 28.7 food cal
4186 J
These 28.7 food cal went into work, and required five times as
many total calories burned. Therefore, during 15 min of this
workout, 5 3 1 28.7 2 5 144 food cal are burned. This is similar
to the amount of calories in a typical exercise bar.
Notice that, because W 5 mgh, no net work is done on
a person running horizontally at constant speed. Why does a runner still expend so much energy? We will discuss that next.
R EF LECT
Practice Problem: How much power must the man produce to
run at 8.00 km h at 5.00°during a 20 min workout? Answer: 133 W.
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CHAPTER 7 Work and Energy
At the end of the previous example, we pointed out that no net work is done
on you when you run on level ground at constant speed. Running horizontally
still burns plenty of calories, so where does the energy go? When you run on flat
ground, your legs repeatedly launch you upward, doing work to raise your body
against gravity. This work converts some of the calories you burn into gravitational potential energy. When you land from a stride, some of the energy is dissipated as thermal energy. Thus, the up-and-down motion of running burns
calories. This is why running in place or jumping rope burns an amount of calories similar to that burned in regular running.
The biomechanics of running requires most animals (including humans) to
move up and down. To decrease energy losses in this movement, ligaments in
legs have evolved to be elastic, so that they act partially like springs. Thus, running is similar to hopping on a pogo stick, and the calories burned in up-anddown motion largely compensate for the portion of elastic energy that dissipates
into thermal energy. In other words, the biomechanics of running requires vertical motion, and elastic ligaments reduce the energy required for such motion.
Now let’s look at another method for finding the power produced by the
treadmill runner in Example 7.9. From Equation 7.15, we have
P5
mgh
W
5
Dt
Dt
Earlier, we found that h 5 v 1 Dt 2 sin u, so we can write the power as
P5
mgv 1 Dt 2 sin u
5 1 mg sin u 2 v
Dt
We can identify mg sin u as F and write
i
P 5 F v.
(7.16)
i
This result, which is valid for any constant net force F that gives an object a
speed v, gives you another way to evaluate power. Clearly, when the force and
speed are constant, the power will be constant as well.
As an example using P 5 F v, consider a woman on a bike riding at
v 5 6.0 m s (about 13 mph) on flat road. To maintain this constant speed, she must
produce a force equal in magnitude to the air drag, which is about Fair 525 N. Thus,
the mechanical power she must produce is P 5 Fv 5 1 25 N 2 1 6.0 m s 2 5150 W.
So far, we have been talking about mechanical power, or the rate at which
work is performed. However, earlier we stated that power also can mean the rate
at which any form of energy is transferred or transformed. We mentioned that this
broader definition applies to lightbulbs, which don’t perform any work, but
instead transform electrical energy to radiant energy. The most general way to
think of power is
i
i
/
/
P5
Quantitative
Analysis 7.9
Energy vs. power: 10,000 W camera flash!
The flash on a pocket camera can be quite bright.
Using nothing more than the battery, the flash can have a power
of 10,000 W. That’s the same power you’d get from 100 lightbulbs emitting 100 W each! How can a camera do this with a
small battery?
S O L U T I O N In this example, we’re considering power in the
more general sense of Equation 7.17 and the transformation of
energy transferred or transformed
elapsed time
(7.17)
electrical energy to radiant energy. As expressed in that equation,
the power of a camera flash will be the amount of energy transformed, divided by the time over which the transformation takes
place. If a fixed quantity of energy is delivered over a very short
time, the power will be very large. In a camera, the battery
spends about 1 s “charging up,” or preparing 10 J of energy to be
delivered to the flash. When a snapshot is made, the flash uses
this 10 J in about 0.001 s. Therefore, P 5 1 10 J 2 1 0.001 s 2 5
10,000 W.
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33
Summary
SUMMARY
Work Done by a Constant Force
Work done by a single force:
(Section 7.2) The work W done by a force on an object is the transfer of
W 5 F d 5 1F cos u2d
energy by that force to or from the object. For a constant force, W 5 F d
W.0
W,0
(Equation 7.1), where F is the component of the force parallel to the
S
S
F
F F'
object’s displacement of magnitude d. If F points opposite to the displace- F'
u
ment, then W , 0, and the force removes energy from the object by
u
opposing its displacement. When multiple forces do work on an object,
F 5 F cos u
F 5 F cos u
the total or net work is the sum of the work due to each force.
Work done by
multiple forces:
Wnet5W11W21c
i
i
i
S
F1
i
i
v 5 1.0 m/s
Mercedes, mass 2600 kg
K 5 2 mv 5 2600 J
Potential Energy
(Section 7.4) Potential energy can be thought of as stored energy.
When a spring is stretched or compressed a distance x from its
equilibrium length, the spring stores elastic potential energy in
accordance with the equation Uel 5 12 kx 2 (Equation 7.6). A mass
x50
at height y above a chosen origin y 5 0 has a gravitational potenx
x
tial energy Ug 5 mgy (Equation 7.7). Potential energy is associElastic
potential
energy:
For
a
ated only with conservative forces, not with nonconservative
spring
stretched
or
compressed
forces such as friction.
K5
1
2
mv2 5 1300 J
y
Ug,i 5 mgyi
yi
m
F 5 mg
x
Ug,f 5 mgyf
yf
Gravitational potential energy:
For an object at vertical position y,
by a distance x from equilibrium,
1
2
—
2 kx
u2
S
F2
i
Work and Kinetic Energy
v 5 2.0 m/s
1
2
(Section 7.3) The kinetic energy of an object is K 5 2 mv (Equation 7.3). The net work on an object changes its kinetic energy:
Wnet 5 Kf 2 Ki (Equation 7.4). An object speeds up when Honda, mass 1300 kg
Wnet . 0 and slows down when Wnet , 0.
1
2
Uel 5
u1
Ug 5 mgy
For a change in vertical position,
DUg 5 Ug,f 2 Ug,i
Conservation of Energy
(Sections 7.6 and 7.7) When all forms of energy are taken into account, the energy of an isolated system is conserved and therefore remains constant. If only conservative forces act in a system, then
the system’s mechanical energy will be conserved. The mechanical energy can be written as
Emech 5 K 1 U (Equation 7.9), and this sum of energies remains constant in time when conserved:
Kf 1 Uf 5 Ki 1 Ui (Equation 7.11). When frictional forces act in a system, some or all of the initial mechanical energy is dissipated away, reducing the final mechanical energy by
Emech,f 5 Emech,i 2 fkd (Equation 7.14).
y
v50
y2
h
Ug
K
vi
y1
y50
Power
Power is the rate at which energy is transferred or transformed. For
mechanical power, the energy must be in the form of work: P 5 W Dt (Equation 7.15).
We can also talk about power more generally as the rate at which any energy is transferred or transformed, as in Equation 7.17. A very large amount of power can be
obtained when a given amount of energy is transferred over a very short period of time.
(Section 7.8)
/
Ug
K
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CHAPTER 7 Work and Energy
Conceptual Questions
1. How does friction produce thermal energy? Explain what is
happening at the molecular level to produce this form of energy.
2. True or false? If hydrogen molecules and oxygen molecules
have the same kinetic energy, then they have the same speed.
Explain your answer.
3. What is wrong with the following statement? “If a stone is
thrown into the air at an angle u above the horizontal, the ycomponent of its kinetic energy is converted to potential
energy at its highest point.” Explain your reasoning.
4. An elevator is hoisted by its cables at constant speed. Is the
total work done on the elevator positive, negative, or zero?
Explain your reasoning.
5. A rope tied to an object is pulled, causing the object to accelerate. According to Newton’s third law, the object pulls back on
the rope with an equal and opposite force. Is the total net work
done on the object then zero? If so, how can the object’s
kinetic energy change?
6. If a projectile is fired upward at various angles above the horizontal, it has the same initial kinetic energy in each case. Why
does it not then rise to the same maximum height in each case?
7. When you use a jack to lift a car, the force you exert on the jack
is much less than the weight of the car. Does this mean that less
work is done on the car than if the car were lifted directly?
8. Using simple observations, how can you tell that (a) sound has
energy and (b) light has energy? (Hint: Can you find ways in
which sound and light produce more familiar forms of energy,
such as kinetic energy or thermal energy?)
9. A compressed spring is clamped in its compressed position
and is then dissolved in acid. What becomes of the spring’s
potential energy?
10. You bounce on a trampoline, going a little higher with each
bounce. Explain how you increase your total mechanical energy.
11. When you jump from the ground into the air, where does your
kinetic energy come from? What force does work on you to lift
you into the air? Could you jump if you were floating in outer
space?
12. Hydroelectric energy comes from gravity pulling down water
through dams in rivers. Explain how such energy is really just
a form of solar energy by tracing how the sun’s energy is able
to get the water from the ocean to the reservoir behind the dam.
13. Does the kinetic energy of a car change more when the car
speeds up from 10 mph to 15 mph or from 15 mph to 20 mph?
14. True or false? If a force is in the negative direction, then the
work it does is negative. Explain your reasoning.
15. You often hear that the energy we use on the earth comes from
the sun. Explain how this is true. For example, how does the
kinetic energy of a running person come from the sun?
16. On your electrical “power” bill, you are charged for kilowatthours (kWh). Is this really a power bill? What kind of quantity
is the kilowatt-hour?
Multiple-Choice Problems
1. A spiral spring is compressed so as to add U units of potential
energy to it. When this spring is allowed to elongate by twothirds of the distance it was compressed, its remaining potential energy in the same units will be (see Section 2.5 for a
review of proportional reasoning)
A. 2U 3
B. 4U 9
C. U 3
D. U 9
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/
/
/
2. A block slides a distance d down a frictionless plane and then
comes to a stop after sliding a distance s across a rough horizontal plane, as shown in the accompanying figure. What fraction of the distance s does the block slide before its speed is
reduced to one-third of the maximum speed it had at the bottom of the ramp?
A. s 3
B. 2s 3
C. s 9
D. 8s 9
/
/
/
d
Smoo
/
s
th
Rough
Figure for Multiple-Choice 2
3. You slam on the brakes of your car in a panic and skid a distance d on a straight and level road. If you had been traveling
twice as fast, what distance would the car have skidded under
the same conditions?
A. 4d
B. 2d
C. "2d
D. d 2
4. You wish to accelerate your car from rest at a constant acceleration. Assume that there is negligible air drag. To create a constant acceleration, the car’s engine must
A. maintain a constant power output.
B. develop ever-decreasing power.
C. develop ever-increasing power.
5. Consider two frictionless inclined planes with the same vertical height. Plane 1 makes an angle of 30° with the horizontal,
and plane 2 makes an angle of 60° with the horizontal. Mass
m1 is placed at the top of plane 1, and mass m2 is placed at the
top of plane 2. Both masses are released at the same time. At
the bottom, which mass is going faster?
A. m1
B. m2
C. Neither; they both have the same speed at the bottom.
6. A brick is dropped from the top of a building through the air
(friction is present) to the ground below. How does the brick’s
kinetic energy (K ) just before striking the ground compare
with its gravitational potential energy (Ug) at the top of the
building? Set y 5 0 at the ground level.
A. K is equal to Ug.
B. K is greater than Ug.
C. K is less than Ug.
7. Which of the following statements about work is or are true?
(More than one statement may be true.)
A. Negative net work done on an object always reduces the
object’s kinetic energy.
B. If the work done on an object by a force is zero, then either
the force or the displacement must have zero magnitude.
C. If a force acts downward, it does negative work.
D. The formula W 5 Fd cos u can be used only if the force is
constant over the distance d.
8. A 10 kg stone and a 100 kg stone are released from rest at the
same height above the ground. There is no appreciable air
drag. Which of the following statements is or are true? (More
than one statement may be true.)
A. Both stones have the same initial gravitational potential
energy.
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GY.90707.07.pgs 6/25/04 9:42 AM Page 35
Multiple-Choice Problems
9.
10.
11.
12.
B. Both stones will have the same acceleration as they fall.
C. Both stones will have the same speed when they reach the
ground.
D. Both stones will have the same kinetic energy when they
reach the ground.
E. Both stones will reach the ground at the same time.
Two identical objects are pressed against two different springs
so that each spring stores 50 J of potential energy. The objects
are then released from rest. One spring is quite stiff (hard to
compress), while the other one is quite flexible (easy to compress). Which of the following statements is or are true? (More
than one statement may be true.)
A. Both objects will have the same maximum speed after
being released.
B. The object pressed against the stiff spring will gain more
kinetic energy than the other object.
C. Both springs are initially compressed by the same amount.
D. The stiff spring has a larger spring constant than the flexible spring.
E. The flexible spring must have been compressed more than
the stiff spring.
Two objects with different masses each have 100 J of gravitational potential energy. They are released and fall to the
ground. Which of the following statements is or are true?
(More than one statement may be true.)
A. Both objects are released from the same height.
B. Both objects will have the same kinetic energy when they
reach the ground.
C. Both objects will have the same speed when they reach the
ground.
D. Both objects will accelerate toward the ground at the same
rate.
E. Both objects will reach the ground at the same time.
Two objects with different masses are launched vertically into
the air by identical springs. The two springs are compressed by
the same amount before launching. Which of the following
statements is or are true? (More than one statement may be
true.)
A. Both masses will reach the same maximum height.
B. Both masses leave the spring with the same energy.
C. Both masses leave the spring with the same speed.
D. Both masses leave the spring with the same kinetic energy.
E. The lighter mass will gain more gravitational potential
energy than the heavier mass.
Two objects with unequal masses are released from rest from
the same height. They slide without friction down a slope and
then encounter a rough horizontal region, as shown in the
accompanying figure. The coefficient of kinetic friction in the
rough region is the same for both masses. Which of the following statements is or are true? (More than one statement may be
true.)
A. Both masses start out with the same gravitational potential
energy.
B. Both objects have the same speed when they reach the base
of the slope.
C. Both masses have the same kinetic energy at the bottom of
the slope.
D. Both masses travel the same distance on the rough horizontal surface before stopping.
E. Both masses will generate the same amount of thermal
energy due to friction on the rough surface.
35
Smo
oth
Rough
Figure for Multiple-Choice 12
13. A 1 g grasshopper and a 0.1 g cricket leap vertically into the
air with the same initial speed in the absence of air drag.
Which of the following statements must be true? (More than
one statement may be true.)
A. The grasshopper goes 10 times higher than the cricket.
B. Both will reach the same height.
C. At the highest point, both will have the same change in
gravitational potential energy.
D. At the highest point, the grasshopper’s change in gravitational potential energy will be 10 times more than the
cricket’s.
E. The grasshopper and the cricket have the same initial
kinetic energy.
14. Two balls having different masses reach the same height when
shot into the air from the ground. If there is no air drag, which
of the following statements must be true? (More than one
statement may be true.)
A. Both balls left the ground with the same speed.
B. Both balls left the ground with the same kinetic energy.
C. Both balls will have the same gravitational potential energy
at the highest point.
D. The heavier ball must have left the ground with a greater
speed than the lighter ball.
E. Both balls have no acceleration at their highest point.
15. The stone in the accompanying figure can be carried from the
bottom to the top of a cliff by various paths. Which path
requires more work?
A. AC
B. ABC
C. The work is the same for both paths.
C
Cliff
A
B
Figure for Multiple-Choice 15
Problems
7.2 Work: Transfer of Energy by Forces
1. A fisherman reels in 12.0 m of line while landing a fish, using
a constant forward pull of 25.0 N. How much work does the
tension in the line do on the fish?
2. A tennis player hits a 58.0 g tennis ball so that it goes straight
up and reaches a maximum height of 6.17 m. How much work
does gravity do on the ball on the way up? On the way down?
3. A boat with a horizontal tow rope pulls a water skier. She skis
off to the side, so the rope makes an angle of 15.0° with the
GY.90707.07.pgs 6/25/04 9:42 AM Page 36
36
4.
5.
6.
7.
8.
9.
10.
11.
CHAPTER 7 Work and Energy
forward direction of motion. If the tension in the rope is
180 N, how much work does the rope do on the skier during a
forward displacement of 300.0 m?
A constant horizontal pull of 8.50 N drags a box along a horizontal floor through a distance of 17.4 m. (a) How much
work does the pull do on the box? (b) Suppose that the same
pull is exerted at an angle above the horizontal. If this pull
now does 65.0 J of work on the box while pulling it through
the same distance, what angle does the force make with the
horizontal?
You push your physics book 1.50 m along a horizontal tabletop with a horizontal push of 2.40 N while the opposing force
of friction is 0.600 N. How much work does each of the following forces do on the book? (a) your 2.40 N push, (b) the
friction force, (c) the normal force from the table, and (d) gravity? (e) What is the net work done on the book?
A 128.0 N carton is pulled up a very smooth baggage ramp
inclined at 30.0° above the horizontal by a rope exerting a
72.0 N pull parallel to the ramp’s surface. If the carton travels
5.20 m along the surface of the ramp, calculate the work done
on it by (a) the rope, (b) gravity, and (c) the normal force of the
ramp. (d) What is the net work done on the carton? (e) Suppose that the rope is angled at 50.0° above the horizontal,
instead of being parallel to the ramp’s surface. How much
work does the rope do on the carton in this case?
A factory worker moves a 30.0 kg crate a distance of 4.5 m
along a level floor at constant velocity by pushing horizontally
on it. The coefficient of kinetic friction between the crate and
the floor is 0.25. (a) What magnitude of force must the worker
apply? (b) How much work is done on the crate by the
worker’s push? (c) How much work is done on the crate by
friction? (d) How much work is done by the normal force? By
gravity? (e) What is the net work done on the crate?
An 8.00 kg package in a mail-sorting room slides 2.00 m down
a chute that is inclined at 53.0° below the horizontal. The coefficient of kinetic friction between the package and the chute’s
surface is 0.40. Calculate the work done on the package by
(a) friction (b) gravity, and (c) the normal force. (d) What is
the net work done on the package?
A 715 N person walks up three flights of stairs, covering a
total vertical distance of 10.5 m. (a) If, as is typical, only 20%
of the caloric (food) energy is converted to work by the muscles, how many joules and food calories of energy did the person use? (One food calorie 5 4186 J.) (b) What happened to
the other 80% of the food energy?
Assume that the earth and the moon have circular orbits
(which is almost true) and that the periods of their orbits are
365 days and 28 days, respectively. (a) How much work does
the earth do on the moon in one day? (b) How much work does
the sun do on the earth in one month?
A boxed 10.0 kg computer monitor is dragged by friction
5.50 m up along the moving surface of a conveyor belt inclined
at an angle of 36.9° above the horizontal. If the monitor’s
speed is a constant 2.10 cm s, how much work is done on the
monitor by (a) friction, (b) gravity, and (c) the normal force of
the conveyor belt?
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13.
/
14.
BIO
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15.
BIO
16.
BIO
17.
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18.
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19.
/
20.
21.
/
22.
7.3 Work and Kinetic Energy
12. It takes 4.186 J of energy to raise the temperature of 1.0 g of
water by 1.0°C. (a) How fast would a 2.0 g cricket have to
jump to have that much kinetic energy? (b) How fast would a
4.0 g cricket have to jump to have the same amount of kinetic
energy?
A bullet is fired into a large stationary absorber and comes to
rest. Temperature measurements of the absorber show that it
received 1960 J of energy from the bullet, and high-speed photos of the bullet show that it was moving at 965 m s just as it
struck the absorber. What is the mass of the bullet?
Animal energy. Adult cheetahs, the fastest of the great cats,
have a mass of about 70 kg and have been clocked at up to
72 mph (32 m s). (a) How many joules of kinetic energy does
such a swift cheetah have? (b) By what factor would its kinetic
energy change if its speed were doubled?
A racing dog is initially running at 10.0 m s, but is slowing
down. (a) How fast is the dog moving when its kinetic energy
has been reduced by half? (b) By what fraction has its kinetic
energy been reduced when its speed has been reduced by half?
If a running house cat has 10.0 J of kinetic energy at speed v,
(a) At what speed (in terms of v) will she have 20.0 J of kinetic
energy? (b) What would her kinetic energy be if she ran half as
fast as the speed in part (a)?
A 0.145 kg baseball leaves a pitcher’s hand at a speed of
32.0 m s. If air drag is negligible, how much work has the
pitcher done on the ball by throwing it?
A 1.50 kg book is sliding along a rough horizontal surface. At
point A it is moving at 3.21 m s, and at point B it has slowed
to 1.25 m s. (a) How much work was done on the book
between A and B? (b) If 20.750 J of work is done on the book
from B to C, how fast is it moving at point C? (c) How fast
would it be moving at C if 10.750 J of work were done on it
from B to C?
Stopping distance of a car. The driver of an 1800 kg car
(including passengers) traveling at 23.0 m s slams on the
brakes, locking the wheels on the dry pavement. The coefficient of kinetic friction between rubber and dry concrete is
typically 0.700. (a) Use the work–energy principle to calculate
how far the car will travel before stopping. (b) How far would
the car travel if it were going twice as fast? (c) What happened
to the car’s original kinetic energy?
Everyday kinetic energies. Make reasonable estimates of the
quantities necessary to calculate the following kinetic energies: (a) Approximately how many joules of kinetic energy
does a walking adult have? What about a jogging adult?
(b) Approximately how many joules of kinetic energy does a
large car going at freeway speeds have? (c) If you drop a 1 kg
weight (about 2 lb) from shoulder height, approximately how
many joules of kinetic energy will the weight have when it
reaches the ground?
You throw a 20 N rock into the air from ground level and
observe that, when it is 15.0 m high, it is traveling upward at
25.0 m s. Use the work–energy principle to find (a) the rock’s
speed just as it left the ground and (b) the maximum height the
rock will reach.
A 0.420 kg soccer ball is initially moving at 2.00 m s. A soccer player kicks the ball, exerting a constant 40.0 N force in
the same direction as the ball’s motion. Over what distance
must her foot be in contact with the ball to increase the ball’s
speed to 6.00 m s?
A 61 kg skier on level snow coasts 184 m to a stop from a
speed of 12.0 m s. (a) Use the work–energy principle to find
23.
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GY.90707.07.pgs 6/25/04 9:42 AM Page 37
Problems
the coefficient of kinetic friction between the skis and the
snow. (b) Suppose a 75 kg skier with twice the starting speed
coasted the same distance before stopping. Find the coefficient of kinetic friction between that skier’s skis and the
snow.
24. Other metric energy units. Besides the joule, several other
units of energy are commonly used in the metric system. One of
these is the calorie, which is defined as the amount of heat
energy required to raise the temperature of 1.0 g of water by
1.0°C; 1.00 cal is equal to 4.186 J. (Note: Do not confuse this
calorie with the food calorie, which is a thousand of the calories
defined here.) Another unit is the erg; just as the joule is defined
as kg # m2 s2, the erg is defined as g # cm2 s2. A third energy unit
is the kilowatt-hour (kW # h), which is the energy produced
when 1.0 kilowatt of power (1 kW 5 1000 J s) acts for an hour.
(a) By converting units, show that 1 J 5 107 ergs. (b) How
many joules and how many calories are there in 1.00 kW # h?
(c) If a 0.250 mg flea jumps at 2.0 m s, how many ergs of
kinetic energy does it have? (d) How many kW # h and how
many ergs are required to raise the temperature of 1.0 g of water
by 1.0°C?
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/
/
7.4 Work Done by a Varying Force
25. To stretch a certain spring by 2.5 cm from its equilibrium position requires 8.0 J of work. (a) What is the force constant of
this spring? (b) What was the maximum force required to
stretch it by that distance?
26. A spring is 17.0 cm long when it is lying on a table. One end is
then attached to a hook and the other end is pulled by a force
of 25.0 N, causing the spring to stretch to a length of 19.2 cm.
(a) What is the force constant of this spring? (b) How much
work was required to stretch the spring from 17.0 cm to
19.2 cm? (c) How long will the spring be if the 25 N force is
replaced by a 50 N force?
27. A spring of force constant 300.0 N m and unstretched length
0.240 m is stretched by two 15.0 N forces pulling in opposite
directions at opposite ends of the spring. How long will the
spring now be, and how much work was required to stretch it
that distance?
28. An unstretched spring has a force constant of 1200 N m. How
large a force and how much work are required to stretch the
spring: (a) by 1.0 m from its unstretched length, and (b) by
1.0 m beyond the length reached in part (a)?
29. The graph in the accompanying figure shows the magnitude of
the force exerted by a given spring as a function of the distance
/
/
x the spring is stretched. How much work does it take to stretch
this spring: (a) a distance of 5.0 cm, starting with it
unstretched, and (b) from x 5 2.0 cm to x 5 7.0 cm?
7.5 Elastic and Gravitational Potential Energy
30. A force of 800.0 N stretches a certain spring by 0.200 m from
its equilibrium position. (a) What is the force constant of this
spring? (b) How much elastic potential energy is stored in the
spring when it is: (i) stretched 0.300 m from its equilibrium
position and (ii) compressed by 0.300 m from its equilibrium
position? (c) How much work was done in stretching the
spring by the original 0.200 m?
31. Tendons. Tendons are strong elastic fibers that attach muscles
BIO to bones. To a reasonable approximation, they obey Hooke’s
law. In laboratory tests on a particular tendon, it was found
that, when a 250 g weight was hung from it, the tendon
stretched 1.23 cm. (a) Find the force constant of this tendon in
N m. (b) Because of its thickness, the maximum tension this
tendon can support without rupturing is 138 N. By how much
can the tendon stretch without rupturing, and how much
energy is stored in it at that point?
32. A certain spring stores 10.0 J of potential energy when it is
stretched by 2.00 cm from its equilibrium position. (a) How
much potential energy would the spring store if it were
stretched an additional 2.00 cm? (b) How much potential
energy would it store if it were compressed by 2.00 cm from its
equilibrium position? (c) How far from the equilibrium position
would you have to stretch the string to store 20.0 J of potential
energy? (d) What is the force constant of this spring?
33. In designing a machine part, you need a spring that is 8.50 cm
long when no forces act on it and that will store 15.0 J of
energy when it is compressed by 1.20 cm from its equilibrium
position. (a) What should be the force constant of this spring?
(b) Can the spring store 850 J by compression?
34. Measurements on the magnitude of the restoring force F of a
15.0-cm-long spring as a function of the distance x it is
stretched from its equilibrium position yield the graph shown
in the accompanying figure. Use numerical information from
this graph to answer the following questions. (a) Does this
spring obey Hooke’s law? How do you know? (b) What is the
force constant of this spring? (c) How much force does it take
to stretch this spring 8.0 cm from its equilibrium position, and
how much energy is stored in the spring at that point?
/
F (N)
240
300
200
160
120
80
250
40
F (N)
350
37
x (cm)
200
1
2
3 4 5
6
7 8 9 10 11
150
100
50
x (cm)
1 2
3 4 5
6 7
Figure for Problem 29
8
Figure for Problem 34
35. How high can we jump? The maximum height a typical
BIO human can jump from a crouched start is about 60 cm. How
much gravitational potential energy does a 72 kg person gain
in such a jump? Where does this energy come from?
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38
CHAPTER 7 Work and Energy
36. A 575 N woman climbs a staircase that rises at 53° above the
/
41. An exercise program. A 75 kg person is put on an exercise
BIO horizontal and is 4.75 m long. Her speed is a constant 45 cm s.
BIO program by a physical therapist, the goal being to burn up
(a) Is the given weight a reasonable one for an adult woman?
(b) How much gravitational potential energy has she gained by
climbing the stairs? (c) How much work has gravity done on
her as she climbed the stairs?
37. The accompanying figure shows a graph of the gravitational
potential energy of an object as a function of its height above
the ground, at which y 5 0. Use information obtained from
this graph to find the object’s mass.
500 food calories in each daily session. Recall that human
muscles are about 20% efficient in converting the energy they
use up into mechanical energy. The exercise program consists
of a set of consecutive high jumps, each one 50 cm into the air
(which is pretty good for a human) and lasting 2.0 s, on the
average. How many jumps should the person do per session,
and how much time should be set aside for each session? Do
you think that this is a physically reasonable exercise session?
7.6 Conservation of Energy
Ug(J)
500
400
300
200
100
h (m)
1
2
3
4 5
6
Figure for Problem 37
38. Interplanetary potential energy. A hammer dropped on earth
has an initial gravitational potential energy of 60.0 J. (a) What
would be the initial gravitational potential energy of the hammer on the moon (where the acceleration due to gravity is
1.67 m s2) if it is released from the same height in both
places? (b) From what height (compared with the height on
earth) would you have to drop the hammer on Mars (where the
acceleration due to gravity is 3.7 m s2) for it to have the same
initial amount of potential energy that it had on earth?
39. Food calories. The food calorie, equal to 4186 J, is a measure
BIO of how much energy is released when food is metabolized by
the body. A certain brand of fruit-and-cereal bar contains
140 food calories per bar. (a) If a 65 kg hiker eats one of these
bars, how high a mountain must he climb to “work off” the
calories, assuming that all the food energy goes only into his
gravitational potential energy? (b) If, as is typical, only 20% of
the food calories go into mechanical energy, what would be
the answer to part (a)? (Note: In this and all other problems,
we are assuming that 100% of the food calories that are eaten
are absorbed and used by the body. This is actually not true. A
person’s “metabolic efficiency” is the percentage of calories
eaten that are actually used; the rest are eliminated by the
body. Metabolic efficiency varies considerably from person to
person.)
40. A good workout. You overindulged on a delicious dessert, so
BIO you plan to work off the extra calories at the gym. To accomplish this, you decide to do a series of arm raises holding a
5.0 kg weight in one hand. The distance from your elbow to
the weight is 35 cm, and in each arm raise you start with your
arm horizontal and pivot it until it is vertical. Assume that the
weight of your arm is small enough compared with the weight
you are lifting that you can ignore it. As is typical, your muscles are 20% efficient in converting the food energy they use
up into mechanical energy, with the rest going into heat. If
your dessert contained 350 food calories, how many arm raises
must you do to work off these calories? Is it realistic to do
them all in one session?
/
/
42. Tall Pacific Coast redwood trees (Sequoia sempervirens) can
reach heights of around 100 m. If air drag is negligibly small,
how fast is a sequoia cone moving when it reaches the ground
if it dropped from the top of a 100 m tree?
43. The total height of Yosemite Falls is 2425 ft. (a) How many
joules of gravitational potential energy does each kilogram of
water at the top of this waterfall have compared with each
kilogram of water at the foot of the falls? (b) Find the kinetic
energy and speed of each kilogram of water as it reaches the
base of the waterfall, assuming that there are no losses due to
friction with the air or rocks and that the mass of water had
negligible vertical speed at the top. How fast (in m s and mph)
would a 70 kg person have to run to have that much kinetic
energy? (c) How high would Yosemite Falls have to be so that
each kilogram of water at the base had twice the kinetic energy
you found in part (b); twice the speed you found in part (b)?
44. The speed of hailstones. Although the altitude may vary considerably, hailstones sometimes originate around 500 m (about
1500 ft) above the ground. (a) Neglecting air drag, how fast
will these hailstones be moving when they reach the ground,
assuming that they started from rest? Express your answer in
m s and in mph. (b) From your own experience, are hailstones
actually falling that fast when they reach the ground? Why not?
What has happened to most of their initial potential energy?
45. Pebbles of weight w are launched from the edge of a vertical
cliff of height h at speed v0. How fast (in terms of the quantities just given) will these pebbles be moving when they reach
the ground if they are launched (a) straight up, (b) straight
down, (c) horizontally away from the cliff, and (d) at an angle
u above the horizontal? (e) How would the answers to the previous parts change if the pebbles weighed twice as much?
46. Volcanoes on Io. Io, a satellite of Jupiter, is the most
volcanically active moon or
planet in the solar system. It
has volcanoes that send
plumes of matter over 500
km high (see the accompanying figure). Due to the satellite’s small mass, the
acceleration due to gravity
on Io is only 1.81 m s2, and Figure for Problem 46
Io has no appreciable atmosphere. Assume that there is no variation in gravity over the distance traveled. (a) What must be the speed of material just as it
leaves the volcano to reach an altitude of 500 km? (b) If the
gravitational potential energy is zero at the surface, how much
potential energy does a 25 kg fragment have at its maximum
/
/
/
GY.90707.07.pgs 6/25/04 9:42 AM Page 39
Problems
47.
BIO
48.
49.
height on Io? How much gravitational potential energy would it
have if it were at the same height above earth?
Human energy vs. insect energy. For its size, the common
flea is one of the most accomplished jumpers in the animal
world. A 2.0-mm-long, 0.50 mg critter can reach a height of
20 cm in a single leap. (a) Neglecting air drag, what is the takeoff speed of such a flea? (b) Calculate the kinetic energy of this
flea at takeoff and its kinetic energy per kilogram of mass. (c) If
a 65 kg, 2.0-m-tall human could jump to the same height compared with his length as the flea jumps compared with its
length, how high could he jump, and what takeoff speed would
he need? (d) In fact, most humans can jump no more than 60 cm
from a crouched start. What is the kinetic energy per kilogram
of mass at takeoff for such a 65 kg person? (e) Where does the
flea store the energy that allows it to make such a sudden leap?
A 25 kg child plays on a swing having support ropes that are
2.20 m long. A friend pulls her back until the ropes are 42°
from the vertical and releases her from rest. (a) How much
potential energy does the child have just as she is released,
compared with her potential energy at the bottom of the
swing? (b) How fast will she be moving at the bottom of the
swing? (c) How much work does the tension in the ropes do as
the child swings from the initial position to the bottom?
Two stones of different masses are thrown straight upward,
one on earth and one on the moon (where gravity produces an
acceleration of 1.67 m s2). Both reach the same height. If the
stone on the moon had an initial speed v, what was the initial
speed of the stone on earth, in terms of v?
A slingshot obeying Hooke’s law is used to launch pebbles
vertically into the air. You observe that if you pull a pebble
back 2.0 cm against the elastic band, the pebble goes 6.0 m
high. (a) Assuming that air drag is negligible, how high will
the pebble go if you pull it back 4.0 cm instead? (b) How far
must you pull it back so it will reach 12 m? (c) If you pull a
pebble that is twice as heavy back 2.0 cm, how high will it go?
When a piece of wood is pressed against a spring and compresses the spring by 5.0 cm, the wood gains a maximum
kinetic energy K when it is released. How much kinetic energy
(in terms of K) would the piece of wood gain if the spring were
compressed 10.0 cm instead?
A 1.5 kg box moves back and forth on a horizontal frictionless
surface between two different springs, as shown in the accompanying figure. The box is initially pressed against the stronger
spring, compressing it 4.0 cm, and then is released from rest.
(a) By how much will the box compress the weaker spring?
(b) What is the maximum speed the box will reach?
/
50.
51.
52.
k 32 N/cm
k 16 N/cm
Box
Smooth surface
Figure for Problem 52
53. A 12.0 N package of whole wheat flour is suddenly placed on
the pan of a scale such as you find in grocery stores. The pan is
supported from below by a vertical spring of force constant
325 N m. If the pan has negligible weight, find the maximum
distance the spring will be compressed if no energy is dissipated by friction.
/
39
54. A basket of negligible weight hangs from a vertical spring
scale of force constant 1500 N m. (a) If you suddenly put a
3.0 kg adobe brick in the basket, find the maximum distance
that the spring will stretch. (b) If, instead, you release the brick
from 1.0 m above the basket, by how much will the spring
stretch at its maximum elongation?
/
7.7 Nonconservative Forces: Energy Conservation
with Friction
55. A 1.50 kg brick is sliding along on a rough horizontal surface
at 13.0 m s. If the brick stops in 4.80 s, how much mechanical
energy is lost, and what happens to this energy?
56. A fun-loving 11.4 kg otter slides up a hill and then back down
to the same place. If she starts up at 5.75 m s and returns at
3.75 m s, how much mechanical energy did she lose on the
hill, and what happened to that energy?
57. A 12.0 g plastic ball is dropped from a height of 2.50 m and is
moving at 3.20 m s just before it hits the floor. How much
mechanical energy was lost during the ball’s fall?
58. Old Faithful. Old Faithful, the most famous geyser in Yellowstone National Park, shoots water approximately 61 m into the
air. The water has been measured to leave the geyser at a speed
of 50 m s. (a) How many joules of mechanical energy are lost
by each kilogram of water on the way up? (b) What happens to
the “lost” energy? Is it really lost? To what other forms of
energy is it converted?
59. While a roofer is working on a roof that slants at 36° above the
horizontal, he accidentally nudges his 85.0 N toolbox, causing
it to start sliding downward, starting from rest. If it starts
4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the
kinetic friction force on it is 22.0 N?
60. A 1.1 kg mass hangs from a spring of force constant
400.0 N m. The mass is then pulled down 13.0 cm from the
equilibrium position and released. At the end of five complete
cycles of vibration, the mass reaches only 10.0 cm from the
equilibrium position. (a) How much mechanical energy is lost
during these five cycles? (b) What percentage of the mechanical energy is lost during the five cycles?
61. A loaded 375 kg toboggan is traveling on smooth horizontal
snow at 4.5 m s when it suddenly comes to a rough region.
The region is 7.0 m long and reduces the toboggan’s speed by
1.5 m s. (a) What average friction force did the rough region
exert on the toboggan? (b) By what percent did the rough
region reduce the toboggan’s (i) kinetic energy and (ii) speed?
62. A 62.0 kg skier is moving at 6.50 m s on a frictionless, horizontal snow-covered plateau when she encounters a rough
patch 3.50 m long. The coefficient of kinetic friction between
this patch and her skis is 0.300. After crossing the rough patch
and returning to friction-free snow, she skis down an icy,
frictionless hill 2.50 m high. (a) How fast is the skier moving
when she gets to the bottom of the hill? (b) How much thermal
energy was generated in crossing on the rough patch?
/
/
/
/
/
/
/
/
/
*7.8 Mechanical Power
63. (a) How many joules of energy does a 100 watt lightbulb use
every hour? (b) How fast would a 70 kg person have to run to
have that amount of kinetic energy? Is it possible for a person
to run that fast? (c) How high a tree would a 70 kg person
have to climb to have the same amount of gravitational poten-
GY.90707.07.pgs 6/25/04 9:42 AM Page 40
40
64.
CHAPTER 7 Work and Energy
tial energy relative to the ground? Are there any trees that
tall?
The engine of a motorboat delivers 30.0 kW to the propeller
while the boat is moving at 15.0 m s. What would be the tension in the towline if the boat were being towed at the same
speed?
At 7.35 cents per kilowatt-hour, (a) what does it cost to operate
a 10.0 hp motor for 8.00 hr? (b) What does it cost to leave a
75 W light burning 24 hours a day?
A tandem (two-person) bicycle team must overcome a force of
165 N to maintain a speed of 9.00 m s. Find the power required
per rider, assuming that each contributes equally.
A ski tow operates on a 15.0° slope of length 300 m. The rope
moves at 12.0 km h and provides power for 50 riders at one
time, with an average mass per rider of 70.0 kg. Calculate the
power required to operate the tow.
U.S. power use. The total consumption of electrical energy in
the United States is about 1.0 3 1019 joules per year. (a) Express
this rate in watts and kilowatts. (b) If the U.S. population is
about 260 million people, what is the average rate of electrical
energy consumption per person?
Solar energy. The sun transfers energy to the earth by radiation
at a rate of approximately 1.0 kW per square meter of surface.
(a) If this energy could be collected and converted to electrical
energy with 25% efficiency, how large an area (in square kilometers) would be required to collect the electrical energy used
by the United States? (See the previous problem.) (b) If the
solar collectors were arranged in a square array, what would be
the length of its sides in kilometers and in miles? Does an array
of these dimensions seem technologically feasible?
A 20.0 kg rock slides on a rough horizontal surface at 8.00 m s
and eventually stops due to friction. The coefficient of kinetic
friction between the rock and the surface is 0.200. What average thermal power is produced as the rock stops?
Horsepower. In the English system of units, power is expressed as horsepower (hp) instead of watts, where 1.00 hp 5
746 W. Horsepower is often used for motors and automobiles. (a) What should be the power rating in horsepower of a
100 W lightbulb? (b) How many watts does a 75 hp motor
produce? (c) Electrical resistors are rated in watts to indicate
how much power they can tolerate without burning up. If a
resistor is rated at 2.00 W, what should be its rating in horsepower? (d) Our sun radiates 3.92 3 1026 J of energy each
second. What is its horsepower rating? (e) How many kilowatt-hours of energy do you use when you run a 25 hp motor
for 90 minutes?
Maximum sustainable human power. The maximum sustainable mechanical power a human can produce is about 13 hp.
How many food calories can a human burn up in an hour by
exercising at this rate? (Remember that only 20% of the food
energy used goes into mechanical energy.)
Should you walk or run? It is 5.0 km from your home to the
physics lab. As part of your physical fitness program, you
could run that distance at 10 km hr (which uses up energy at
the rate of 700 W), or you could walk it leisurely at 3.0 km hr
(which uses energy at 290 W). Which choice would burn up
more energy, and how much energy (in joules) would it burn?
Why is it that the more intense exercise actually burns up less
energy than the less intense one?
/
65.
66.
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67.
/
68.
69.
70.
71.
72.
BIO
73.
BIO
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/
/
74. Human power. Time yourself while running up a flight of
steps, and compute your maximum power in horsepower. Are
you stronger than a horse?
75. The power of the human heart. The human heart is a powerBIO ful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume that the work done by
the heart is equal to the work required to lift that amount of
blood a height equal to that of the average American female,
approximately 1.63 m. The density of blood is 1050 kg m3.
(a) How much work does the heart do in a day? (b) What is the
heart’s power output in watts? (c) In fact, the heart puts out
more power than you found in part (b). Why? What other
forms of energy does it give the blood?
76. Hydroelectric power. The Grand Coulee Dam is 1270 m wide
and 170 m high. The electrical power output from the generators at the base of the dam is approximately 2000 MW.
(a) How many cubic meters per second of water must flow
from the top of the dam to produce this amount of power if all
the initial gravitational potential energy is converted to electrical energy? (Each cubic meter of water has a mass of 1000 kg.)
(b) What would be the answer to part (a) if the conversion
process to electrical energy were only 92% efficient?
/
General Problems
77. Bumper guards. You are asked to design spring bumpers for
the walls of a parking garage. A freely rolling 1200 kg car
moving at 0.65 m s is to compress the spring no more than
7.0 cm before stopping. (a) What should be the force constant
of the spring, and what is the maximum amount of energy that
gets stored in it? (b) If the springs that are actually delivered
have the proper force constant but can become compressed by
only 5.0 cm, what is the maximum speed of the given car for
which they will provide adequate protection?
78. Human terminal velocity. By landing properly and on soft
ground (and by being lucky!), humans have survived falls
from airplanes when, for example, a parachute failed to open,
with astonishingly little injury. Without a parachute, a typical
human eventually reaches a terminal velocity of about 62 m s.
Suppose the fall is from an airplane 1000 m high. (a) How fast
would a person be falling when he reached the ground if there
were no air drag? (b) If a 70 kg person reaches the ground
traveling at the terminal velocity of 62 m s, how much
mechanical energy was lost during the fall? What happened to
that energy?
79. A 0.800 kg ball is tied to the end of a string 1.60 m long and
swung in a vertical circle. (a) Starting anywhere in the path,
calculate the net work done on this ball during one complete
cycle by (i) the tension in the string and (ii) gravity. (b) Repeat
part (a) for motion along the semicircle from the lowest to the
highest point on the path.
80. Mountain climbing! A 75.0 kg mountain climber is holding his
60.0 kg partner over a cliff when he suddenly steps on frictionless ice at the horizontal top of the cliff, as shown in the accompanying figure. The rope has negligible mass and is held
horizontally by the climber. There is no appreciable friction at
the icy edge of the cliff. Use energy methods to calculate
the speed of the climbers after the lower one has descended
1.50 m starting from rest.
/
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GY.90707.07.pgs 6/25/04 9:42 AM Page 41
General Problems
41
87. A pump is required to lift 750 liters of water per minute from a
well 14.0 m deep and eject it with a speed of 18.0 m s. How
much work per minute does the pump do?
88. A 350 kg roller coaster starts from rest at point A and slides
down the frictionless loop-the-loop shown in the accompanying figure. (a) How fast is this roller coaster moving at point B?
(b) How hard does it press against the track at point B?
75.0 kg
/
A
4.00 m
B
25.0 m
60.0 kg
12.0 m
Figure for Problem 80
3.00 m
81. More mountain climbing! What would be the speed of the
climbers in the previous problem after they had moved 1.50 m
if there were friction between the upper climber and the ice
with a coefficient of kinetic friction of 0.250?
82. Suppose the worker in Problem 7 pushes downward at an
angle of 30° below the horizontal. (a) What magnitude of
force must the worker apply to move the crate at constant
velocity? (b) How much work is done on the crate by this
force when the crate is pushed a distance of 4.5 m? (c) How
much work is done on the crate by friction during the displacement of the crate? (d) How much work is done by the
normal force of the floor? By gravity? (e) What is the net work
done on the crate?
83. A 50 g object and a 500 g object are launched by pressing
them against identical springs and compressing both springs
by the same amount. (a) Which object has more initial potential energy? (b) Which object gains more kinetic energy after
being released? (c) Which object gains more speed after being
released? (d) Which object is acted upon by a greater initial
force from the spring? (e) Which object has a greater acceleration the instant after it is released? In each case, explain the
reasoning behind your answer.
84. On an essentially frictionless horizontal ice-skating rink, a
skater moving at 3.0 m s encounters a rough patch that
reduces her speed by 45% due to a friction force that is 25% of
her weight. Use the work–energy principle to find the length of
the rough patch.
85. Pendulum. A small 0.12 kg metal ball is tied to a very light
(essentially massless) string 0.80 m long to form a pendulum
that is then set swinging by releasing the ball from rest when
the string makes a 45° angle with the vertical. Air drag and
other forms of friction are negligible. What is the speed of the
ball when the string passes through its vertical position, and
what is the tension in the string at that instant?
86. Pendulum. A pendulum consists of a small 4.5 N metal weight
attached to a very light 1.75 m string. The weight is released
from rest when the string makes an angle of 62.0° with the vertical. At the end of the first swing, the string reaches an angle of
only 56.5°. (a) How much mechanical energy has the pendulum
lost during this swing? (b) Where did the energy go? Explain
physically how it could have been converted to other forms of
energy.
/
Figure for Problem 88
89. Automobile air-bag safety. An
automobile air bag cushions
the force on the driver in a
head-on collision by absorbing
her energy before she hits the
steering wheel. Such a bag can
be modeled as an elastic force,
similar to that produced by a
spring. (a) Use energy conservation to show that the effec
tive force constant k of the air Figure for Problem 89
2
2
bag is k 5 mv0 xmax , where m
is the mass of the driver, v0 is the speed of the car (and driver)
at the instant of the accident, and xmax is the maximum distance
the bag gets compressed, which, in a severe accident, would be
the distance from the driver’s body to the steering wheel.
(b) Show that the maximum force the air bag would exert on
the driver is Fmax 5 mv02 xmax. (c) Now let’s put in some realistic numbers. Experimental tests have shown that injury occurs
when a force density greater than 5.0 3 105 N m2 acts on
human tissue. (The total force is this force density times the
area over which it acts.) As the accompanying figure shows,
the force of the air bag acts mostly on the upper front half of
the driver’s body, over an area of about 2500 cm2. (Check
your own body to see if this is reasonable.) Use this value to
calculate the total force on the driver’s body at the threshold
of injury. (d) Use your results to calculate the effective force
constant k of the air bag and the maximum speed for which
the bag will prevent injury to a 65 kg driver if she is 30 cm
from the steering wheel at the instant of impact. Express the
speed in m s and mph. (e) How could you design a safer air
bag for higher speed collisions? What things could you alter
to do this? Would it be safe to make a stiffer air bag by inflating it more? Explain your reasoning.
90. On a smooth horizontal laboratory floor, a 15.0 N scientific
instrument is attached to a light (that is, essentially massless)
spring having a force constant of 75 N cm. The other end of
the spring is attached to the wall. (a) If the instrument is pulled
1.20 cm from its equilibrium position and released from rest,
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91.
BIO
92.
BIO
CHAPTER 7 Work and Energy
find its maximum speed and its maximum acceleration. (b) Do
the maximum speed and acceleration occur at the same time?
If not, when in the motion does each of them occur?
Bone fractures. The maximum energy that a bone can support
without breaking depends on its characteristics, such as its
cross-sectional area and its elasticity. For healthy human leg
bones of approximately 6.0 cm2 cross-sectional area, this
energy has been experimentally measured to be about 200 J.
(a) From approximately what maximum height could a 60 kg
person jump and land rigidly upright on both feet without
breaking his legs? (b) You are probably surprised at how small
the answer to part (a) is. People obviously jump from much
greater heights without breaking their legs. How can that be?
What else absorbs the energy when they jump from greater
heights? (Hint: How did the person in part (a) land? How do
people normally land when they jump from greater heights?)
(c) In light of your answers to parts (a) and (b), what might be
some of the reasons that older people are much more prone than
younger ones to bone fractures from simple falls (such as a fall
in the shower)?
Whiplash injuries. When a car is hit from behind, its passengers undergo sudden forward acceleration, which can cause a
severe neck injury, known as whiplash. During normal acceleration, the neck muscles play a large role in accelerating the
head so that the bones are not injured. But during a very sudden
acceleration the muscles do not react immediately, because
they are flexible, so most of the accelerating force is provided
by the neck bones. Experimental tests have shown that these
bones will fracture if they absorb more than 8.0 J of energy.
(See the discussion in the previous problem.) (a) If a car waiting
at a stoplight is rear-ended in a collision that lasts for 10.0 ms,
what is the greatest speed this car and its driver can reach without breaking neck bones if his head has a mass of 5.0 kg (which
is about right for a 70 kg person)? Express your answer in m s
and mph. (b) What is the acceleration of the passengers during
the collision in part (a), and how large a force is acting to accelerate their heads? Express the acceleration in m s2 and in g’s.
A 250 g object on a smooth, horizontal lab table is pushed
against a spring of force constant 35 N cm and then released.
Just before the object is released, the spring is compressed
12.0 cm. How fast is the object moving when it has gained half
of the spring’s original stored energy?
In the accompanying figure, a 5.0 N ball is fastened to two
identical springs that pull in opposite directions and that can
slide on a perfectly smooth horizontal surface. The ball is initially stationary and is then moved 3.0 cm to the right and
released from rest. (a) Find the magnitude and direction of the
net force on the ball just after it is released. (b) What is the
potential energy of the system just after the ball is released?
(c) Find the maximum speed of the ball.
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93.
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94.
Ball
k 5 12 N/cm
k 5 12 N/cm
Smooth surface
Figure for Problem 94
95. Automobile accident analysis. In an auto accident, a car hit a
pedestrian and the driver then slammed on the brakes to stop
the car. During the subsequent trial, the driver’s lawyer
claimed that the driver was obeying the posted 35 mph speed
limit, but that the limit was too high to enable him to see and
react to the pedestrian in time. You have been called as the
state’s expert witness. In your investigation of the accident site,
you make the following measurements: The skid marks made
while the brakes were applied were 280 ft long, and the tread
on the tires produced a coefficient of kinetic friction of 0.30
with the road. (a) In your testimony in court, will you say that
the driver was obeying the posted speed limit? You must be
able to back up your answer with clear numerical reasoning
during cross-examination. (b) If the driver’s speeding ticket is
$10 for each mile per hour he was driving above the posted
speed limit, would he have to pay a ticket, and if so, how much
would it be?
96. Bouncing ball. A rubber ball is dropped from an initial
height h. After each bounce, the ball returns to 75% of its previous height. What percent of its maximum potential energy,
maximum kinetic energy, and maximum speed does the ball
lose as a result of the first bounce?
97. Riding a loop-the-loop. A
A
car in an amusement park ride
travels without friction along h
the track shown in the accompanying figure, starting from
rest at point A. If the loop the
car is currently on has a radius Figure for Problem 97
of 20.0 m, find the minimum
height h so that the car will not fall off the track at the top of the
circular part of the loop.
Wood
98. A 2.0 kg piece of wood slides
on the surface shown in the
Rough bottom
accompanying figure. All parts
of the surface are perfectly Figure for Problem 98
smooth, except for a 30-m-long
rough segment at the bottom, where the coefficient of kinetic
friction with the wood is 0.20. The wood starts from rest 4.0 m
above the bottom. (a) Where will the wood eventually come to
rest? (b) How much work is done by friction by the time the
wood stops?
99. A 68 kg skier approaches the foot of a hill with a speed of
15 m s. The surface of this hill slopes up at 40.0° above the
horizontal and has coefficients of static and kinetic friction of
0.75 and 0.25, respectively, with the skis. (a) Use energy conservation to find the maximum height above the foot of the
hill that the skier will reach. (b) Will the skier remain at rest
once she stops, or will she begin to slide down the hill? Prove
your answer.
100. Energy requirements of the body. A 70 kg human uses
BIO energy at the rate of 80 J s, on average, for just resting and
sleeping. When the person is engaged in more strenuous
activities, the rate can be much higher. (a) If the individual
did nothing but rest, how many food calories per day would
she or he have to eat to make up for those used up? (b) In
what forms is energy used when a person is resting or sleeping? In other words, what happens to those 80 J s? (Hint:
What kinds of energy, mechanical and otherwise, do our body
components have?) (c) If an average person rested and did
other low-level activity for 16 hours (which consumes 80 J s)
and did light activity on the job for 8 hours (which consumes
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General Problems
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101.
200 J s), how many calories would she or he have to consume per day to make up for the energy used up?
From the top edge of a horizontal plateau, a 175.0 N bag of
mountain-climbing gear is launched horizontally by pressing
it against a spring of force constant 385.0 N cm. Just before
the launch, the spring is compressed 10.0 cm from its equilibrium position. Once the bag is free of the spring, it goes over
a vertical cliff 3.45 m high and lands in the snow below. How
far from the foot of the cliff does the bag land, and how fast is
it moving when it gets there?
Two paint buckets are connected
by a lightweight rope passing
over a pulley of negligible mass
and friction. (a) As shown in the
accompanying figure, the system
12.0 kg
is released from rest with the
12.0 kg bucket 2.00 m above the
floor. Use energy conservation
2.00 m
to find the speed with which this
bucket strikes the floor. (b) Sup4.0 kg
pose the pulley had appreciable
mass but no friction. Would the
bucket’s speed be greater than, Figure for Problem 102
less than, or the same as you
found in part (a)? Explain your reasoning in terms of energy.
(a) Use data from Appendix F to calculate the kinetic energy
of the earth as it moves around the sun and the kinetic energy
of the moon as it moves around the earth. (b) Express these
kinetic energies in units of megatons of TNT. (One ton of
TNT releases 4.184 3 109 J of energy.)
A ball is thrown upward with an initial velocity of 15 m s at
an angle of 60.0° above the horizontal. Use energy conservation to find the ball’s greatest height above the ground.
Gasoline energy. When it is burned, 1 gallon of gasoline produces 1.3 3 108 J of energy. A 1500 kg car accelerates from
rest to 37 m s in 10.0 s. The engine of this car is only 15%
efficient (which is typical), meaning that only 15% of the
combustion energy goes into accelerating the car. The rest
goes into things like heating up the engine and exhaust gas.
(a) How many gallons of gasoline does the car use during its
acceleration? (b) How many such accelerations will it take to
burn 1 gallon of gasoline?
Mass extinctions. One of the greatest mass extinctions
occurred about 65 million years ago, when, along with many
other life-forms, the dinosaurs went extinct. Most geologists
and paleontologists agree that this event was caused when a
large asteroid hit the earth. Scientists estimate that this asteroid was about 10 km in diameter and that it would have been
traveling at least as fast as 11 km s. The density of asteroid
material is about 3.5 g cm3, on the average. (a) What would
be the approximate mass of the asteroid, assuming it to be
spherical? (b) How much kinetic energy would the asteroid
have delivered to the earth? (c) In order to put the amount of
energy you found in part (b) in perspective, consider the following: One of the largest nuclear weapons ever tested by the
U. S. (the Castle/Bravo bomb) was a 15 megaton bomb,
meaning that it released the energy of 15 million tons of
TNT. One ton of TNT releases 4.184 3 109 J. How many
Castle/Bravo bombs would it take to produce the energy of
the asteroid that did in the dinosaurs?
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102.
103.
104.
105.
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43
107. Avoiding mass extinctions. It has been suggested that we
can protect the earth from devastating asteroidal impacts such
as the one discussed in the previous problem by using nuclear
devices to alter the orbits of such asteroids around the sun so
that they will miss our planet. If this is done very far from
earth, it is necessary to move them only a few centimeters to
spare the earth a mass extinction. How much energy would it
take to move the asteroid that has been implicated in the
dinosaur extinction by a few centimeters? (See the previous
problem.) To make the calculation reasonable, assume that
we need to exert a force on the asteroid that will accelerate it
uniformly from rest through a distance of 5.0 cm in 0.50 s.
The energy we must give to the asteroid is the added kinetic
energy from this motion. To see if it is feasible to do this,
how many 1.0 megaton bombs would it take to accomplish
the task?
108. In an experiment, a ball is dropped from various heights and
its kinetic energy is measured just as it reaches the floor. A
graph of the kinetic energy of the ball as a function of the
height h from which it is dropped is shown in the accompanying figure. Use information obtained from the graph to
answer the following questions: (a) How many newtons does
the ball weigh? (b) From what height should the ball be
dropped so it will be moving at 4.0 m s when it reaches the
floor? (c) Sketch a clear graph of the ball’s potential energy
as a function of h.
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K (J)
200
100
h (m)
1
2 3 4 5
Figure for Problem 108
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106.
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109. A graph of the potential energy stored in a tendon as a function of the square of the distance x it has stretched from its
equilibrium position is shown in the accompanying figure.
(a) Does the tendon obey Hooke’s law? Explain your reasoning. (b) What is the force constant of the tendon? (c) How far
must the tendon be stretched from its equilibrium position to
store 10.0 J of energy? (d) What force is necessary to hold the
tendon in place in part (c)? (e) Sketch a clear graph of the
potential energy stored in the tendon as a function of x.
U (J)
8
7
6
5
4
3
2
1
x 2 (m2)
0.020
0.040
Figure for Problem 109
0.060
GY.90707.07.pgs 6/25/04 9:42 AM Page 44
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CHAPTER 7 Work and Energy
110. The accompanying figure shows the kinetic energy of an
object as a function of the square of its speed. (a) What is the
mass of the object? (b) If the object is moving at 3.0 m s,
what is its kinetic energy? (c) How fast must the object move
to have a kinetic energy of 5.0 J? (d) Sketch a graph of the
object’s kinetic energy as a function of its speed.
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111. A sled with rider having a combined mass of 125 kg travels
over the perfectly smooth icy hill shown in the accompanying
figure. How far does the sled land from the foot of the cliff?
K (J)
11.0 m Cliff
12
22.5 m/s
10
8
6
4
2
v
1 2 3 4
2
Figure for Problem 111
(m2/s2)
5
Figure for Problem 110
Answers to odd-numbered questions and problems
can be found at www.aw-bc.com/info/young8e.