Chapter 6: Work and Energy
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Why it is impossible to build a perpetual motion machine?
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Why does blood pressure increase when the aorta walls thicken?
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How big is a black hole?
Make sure you know how to:
1. Choose a system and the initial and final states of a physical process.
2. Use Newton’s second law to analyze a physical process.
3. Use kinematics to describe motion.
CO For centuries people have dreamed of making a perpetual motion machine. This sort of
machine would be able to function forever without needing any sort of power source. For
example, imagine operating a laptop computer continuously year after year with no external
power. Claims of successful perpetual motion machines go back as far as the 8th century. In 1775
the Royal Academy of Sciences in Paris decreed that the Academy “will no longer accept or deal
with proposals concerning perpetual motion.” Why did French scientists and mathematicians
reject ideas concerning perpetual motion? Do physics principles rule out the possibility of
constructing a perpetual motion machine? We will begin investigating these questions in this
chapter.
Lead In the last chapter, we developed the idea of a conserved quantity. We also discovered
two examples of conserved quantities: mass and momentum. We learned the conditions under
which the momentum of a system was constant, and how to handle situations in which it wasn’t
(by including external impulses). By making a wise choice of system, we could analyze collisions
between objects without needing to understand the details of the forces they exert on each other.
On the other hand, because momentum is a vector quantity, we have to use vector
component techniques to apply the ideas of momentum and impulse when analyzing physical
processes. Vector equations can often be more complicated than equations involving only scalar
quantities. Also, when the momentum of a system is not constant, dealing with external impulses
involves knowing the time interval during which external forces are exerted, and details of the
magnitude and direction of those forces. It is possible (common actually) that these forces will
not be constant in time, which will require some sort of averaging technique or calculus to
analyze the processes. Because of these complexities, it is worth looking for an additional
conserved quantity that doesn’t have these disadvantages.
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6.1 Work—a new quantity for analyzing physical processes
We have had good success at analyzing a variety of everyday phenomena with the physics
developed in the first five chapters. As just mentioned most of that analysis involved vector
quantities (acceleration, force, impulse and momentum) that relied on applying the component
forms of important principles that involved these quantities (Newton’s second law and the
impulse and momentum principle). Can we analyze such processes using a different type of
thinking (based on new quantities and a new principle) that depends less on vector quantities?
That might make our analysis somewhat easier. Let’s try.
To start, consider several experiments and look for a pattern or perhaps for a new quantity
to help explain what we observe. In each experiment we choose a system of interest, its initial
state, its final state, and the external force that is causing the system to change from its initial state
to its final state. This change involves a displacement of one of the system objects from one
position to another. In the analysis we draw vectors indicating the external force being exerted on
the system and the resulting displacement of the object in the system on which this force was
exerted.
Observational Experiment Table 6.1 Breaking chalk
Observational Experiments
(a) You lift a block from just above a piece of chalk
(initial state) to about 30 cm above the chalk (final state).
The block can fall on the chalk from the initial position or
from the final position.
Analysis
If the block is released from the final
position, it could break the chalk; it could not
if dropped from the initial position. The force
that you exerted on the block, and the block’s
displacement are in the same direction. The
force you exerted caused an increase in the
block’s elevation.
(b) You push a dynamics cart initially at rest (initial state)
until it is moving fast about 2/3 way across track (final
state). The fast moving cart (no longer being pushed) could
collide with a piece of chalk taped to the end of the track.
At rest, the cart cannot break the chalk, but
the fast-moving cart could. The force that
you exerted on the cart, and the cart’s
displacement are in the same direction. The
force you exerted caused the cart’s speed to
increase.
(c) An un-stretched slingshot with a piece of chalk in the
sling (initial state) is pulled back until the slingshot is fully
stretched (final state). If released, the chalk could break
against the wall opposite to it.
The chalk in the un-stretched slingshot would
not hit the wall and break. If released from
the stretched slingshot, the chalk could break
against the wall. The force that you exerted
on the sling and its displacement are in the
same direction. The force you exerted caused
the slingshot to stretch.
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(d) A large heavy box sita on shag carpet (initial state).
You pull the box over a long distance on the carpet (final
state).
The bottom of the box is slightly warmer
than the top after being pulled across the
rough surface. You exerted a force on the box
in the direction of its displacement. This
motion caused the warming.
Patterns
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The external force FYou on Object and the object’s displacement d Object were in the same direction and
caused the system to change so that it gained the potential to do something new:
(a) The block at higher elevation above the earth could break the chalk.
(b) The fast-moving cart could break the chalk.
(c) The stretched slingshot could break the chalk.
(d) The box pulled across the rough surface became warmer.
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An important pattern in all four experiments above is that an external force F was
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exerted on an object in the system, causing a displacement d of the object in the same direction
as the external force. This caused the systems in the first three experiments to gain the ability to
break chalk. In the fourth experiment, the system became warmer due to the external force
pulling the box across a rough surface; again, the external force was in the direction of the system
object’s displacement. Physicists would say that an object in the environment (in this case, you)
did work on the system.
Tip! Notice that Earth was a part of the system in the experiment (a). As a part of the system,
Earth does not do work on the block, as the force it exerts on it is an internal force. However, if
we chose Earth to be an external object, it would do work on the block as it falls down, making it
go faster and faster and thus increasing its ability to break the chalk.
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We say that the system gained energy because of the work done on it by external forces.
There were four types of energy change in systems during these experiments.
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Gravitational potential energy: In experiment (a), the block in the final state was at a higher
elevation with respect to the earth than in the initial state; the energy of the object–Earth
system associated with the elevation of the object above the earth is called gravitational
potential energy.
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Kinetic energy: In experiment (b), the cart was moving faster in the final state than when at
rest in the initial state; the energy associated with the motion of the object is called kinetic
energy.
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Elastic potential energy: In experiment (c), the slingshot in the final state was stretched more
than in the initial state; the energy associated with the degree of stretch is called elastic
potential energy.
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Internal thermal energy: In experiment (d), the bottom surface of the box became warmer as
it was pulled across the rough surface. The energy associated with temperature is called
thermal energy, which is one form of a more general type of energy called internal energy.
In these experiments different types of energy increased due to the work done on the system.
When work causes the energy of a system to increase, we say that positive work is done on the
system.
Negative and zero work
So far we have dealt only with work, which led to an increase in the system’s energy—
positive work. Is it possible to devise a process where an external force causes the energy of a
system to decrease or possibly cause no energy change at all? Let’s try three more observation
experiments to investigate these questions. We analyze them again by drawing vectors
representing: (1) the external force being exerted on the system, and (2) the displacement of the
system object while the external force was exerted
Observational Experiment Table 6.2 More investigations of work.
Observational Experiments
(e) You release a block high above the chalk (initial
state). A friend catches the falling block, slows it
down, and stops it (final state).
Analysis
The block’s potential to break the chalk is
greater in the initial state than in the final state.
The direction of the force exerted by the friend
on the block is opposite the block’s
displacement. The force exerted by your friend
caused the falling block to stop, reducing its
potential to break the chalk.
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(f) A friend pushes lightly on the moving cart (initial
state) opposite the direction of its motion causing it to
stop (final state).
The cart’s potential to break the chalk is greater
in the initial state than in the final state. The
direction of the force exerted by the friend on the
cart is opposite to the cart’s displacement. The
force exerted by your friend caused the moving
cart to stop, reducing its potential to break the
chalk.
(g) Your friend holds a block less than 1 cm above a
piece of chalk (initial state)—so close that the chalk
would not break if the block is released. She slowly
moves the block to the right keeping the block just
above the tabletop until the block hangs less than 1 cm
above a second piece of chalk (final state), which also
would not break if the block is released.
The block’s potential to break the chalk did not
change. The direction of the force exerted by
your friend on the block is perpendicular to the
block’s displacement. The exerted force caused
no change in the block’s potential to break the
chalk.
Patterns
! In (e) and (f) the external force exerted on the system object is opposite its displacement, and the system’s
ability to break the chalk decreases.
! In (g) the external force exerted on the system object is perpendicular to its displacement, and the system’s
ability to break the chalk is unchanged.
In experiments (e) and (f), the external force exerted on the system pointed in the
opposite direction to the displacement of one of the system’s objects, causing a decrease in the
system’s ability to break the chalk. We can say that the external force did negative work on the
system causing its energy to decrease.
In experiment (g), the external force was perpendicular to the displacement of the system
object, which had no effect on the system’s ability to break the chalk. We can say that the
external force did zero work on the system causing no change in the system’s energy.
Formal definition of work
Now that we have a conceptual understanding of positive, negative, and zero work, we
can devise a quantitative definition of work as a physical quantity—an equation that lets us
determine how much work a particular external force does on a system. We will then check to see
if this equation produces results that are consistent with observation experiments (a) through (g).
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Work: The work done by an external force F on a system object while that system object
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undergoes a displacement d is:
Work = W # Fd cos "
! "#$
where F is the magnitude of the force (always positive), d is the magnitude of the
displacement (always positive), and
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is the angle between the direction of F and the
direction of d . The sign of cos " determines the sign of the work (see Fig. 6.1).
The unit of work is the product of the unit of force (newton) and the unit of displacement (meter),
and is called a joule (J):
1 J = 1 N•m
This definition of work assumes the magnitude of the external force is constant, and the angle "
also remains constant.
Figure 6.1 Definition of work
Is this definition consistent with the experiments in Observational Experiment Tables 6.1
and 6.2? The sign of the work depends on the value of cos " %& as both F and d are always
positive values. Consider the circumstances under which the cosine function has positive,
negative, or zero values and if these signs are consistent with experiments (a) through (g) (other
examples of the work done by a particular force during a process are shown in Fig. 6.2).
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! If F and d are parallel, " # 00 , cos 00 # 1.0 , and W # $ Fd . This is consistent with what we
observed in experiments (a) through (d), where the system energy increased (the work done by
the force that the person exerts on a wagon while pulling it up the hill in Fig. 6.2a is positive).
!
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! If F and d point in opposite directions, " # 1800 , cos1800 # %1.0 , and W # % Fd . This is
consistent with what we observed in experiments (e) and (f), where the system energy
decreased (the work done by the force that the person exerts on a wagon while lowering it
down the hill in Fig. 6.2b is negative).
!
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! If F and d point perpendicular to each other, " # 900 , cos 900 # 0 , and W # 0 , consistent
with what we observed in experiment (g) where the system energy did not change (the work
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done by the normal force that the hill’s surface exerts on a wagon wheels as it is pulled up the
hill in Fig. 6.2c is zero).
Figure 6.2 Sign of work depends on relative directions of force and displacement
In each experiment used to help develop the definition of work, we assumed that the
magnitude of the external force exerted on the system object was constant, and that the angle
between the directions of the force and displacement also remained constant. Thus, the above
definition only applies to processes that have these two features. Generalizing Eq. (6.1) so that it
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can handle a non-constant F or " requires the application of calculus.
We also assumed that the displacements of the objects were measured with respect to the
earth. If the object of reference were different, the displacement of the system object might also
be different, and the value of the work will change. When you are calculating work, make sure
you consider the displacement with respect to a particular object of reference.
Tip! An alternative useful way to think about work is as the product of the component of the
external force in the direction of the displacement Fd # F cos" times the magnitude of the
displacement d : W # Fd d . Notice that if the force is perpendicular to the displacement, then
Fd # F cos" # 0 ,and the work is zero.
It is tempting to equate the work done on a system with the force that is exerted on it.
However, in physics, there must be a displacement of a system object in order for an external
force to do work. To get used to this idea, try to think of three situations where the force exerted
on an object is not zero but the work done by this force is zero.
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Quantitative Exercise 6.1 Pushing a bicycle uphill Two friends are cycling up a hill inclined at
80 —steep for bicycle riding. The stronger cyclist exerts a 50-N pushing force parallel to the hill
on his friend’s bicycle while the friend moves a distance of 100 m up the hill. Determine the work
done by the stronger cyclist on the weaker cyclist.
Represent Mathematically Choose the system to be the weaker cyclist. The situation is similar to
that in Fig. 6.2a . The external force that the stronger cyclist exerts on the weaker cyclist is
parallel and upward along the hill and the 100-m displacement of the weaker cyclist is also
parallel and upward along the hill. The work done by the stronger cycles on the weaker cyclist is
W # Fd cos " . The hill is inclined at 80 above the horizontal. Before looking at the “Solve” part,
decide what angle you would insert in this equation.
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Solve and Evaluate Note that the angle between FStrong on Weak and d is 00 and not 80. Thus:
WStrong on Weak # FStrong on Weak d cos " # & 50 N '&100 m ' cos 00 # 5000 N•m = 5000 J.
Try It Yourself: You pull a box 20 m up a 100 ramp. The rope is oriented 200 above the surface
of the ramp. The force that the rope exerts on the box is 100 N . (Fig. 6.3) What is the work done
by the rope on the box?
Figure 6.3 Work done by rope on box
Answer: WR on B # (100 N)(20 m)cos 200 # 1880 J. Note that you use the angle between the
displacement (parallel to the ramp) and the force the rope exerts on the box ( 200 above the
ramp).
Tip! Remember, the angle that appears in the definition of work is the angle between the external
force and the displacement of the system object. It is useful when calculating work to draw tailto-tail arrows representing the external force doing the work and the system object displacement.
Then note the angle between the arrows.
Review Question 6.1
Describe two processes where an external force is exerted on a system object and there is no work
done on the system.
6.2 Work-energy bar charts
The observational experiments in the previous section led to the idea that when an external
force is exerted on a system object and the object is displaced from one position to another, work
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ALG
6.1.6
is done on the system. Positive work on a system leads to an energy increase of the system,
negative work to an energy decrease, and zero work to a zero energy change. This is analogous to
the system’s momentum in Chapter 5: when an impulse was exerted on the system, the momentum
of the system changed by an amount equal to the impulse. Thus, it is possible that the energy of a
system is another type of conserved quantity.
So far, we have found that the work done on a system object has resulted in a change of one
or more of these four types of energy in the system:
(a) Gravitational potential energy ( U g ): An Earth-object system’s gravitational potential
energy increases or decreases as the object moves higher or lower relative to the earth.
(b) Kinetic energy (K): When an object moves faster or slower, its kinetic energy increases or
decreases.
(c) Elastic potential energy ( U s ): The energy of an elastic object (for example the coils of a
spring) increases as the elastic object stretches or compresses more.
(d) Internal energy ( U int ): When surfaces of the touching objects rub against each other, they
warm (an increase in the random kinetic energy of the objects).
A system can have one or more of those different energies. Thus we can think of the total energy
U as the sum of all the energies of the system:
Total energy # U # K $ U g $ U s $ U int
(6.2)
As we discussed above, when external objects do work on a system, the system’s energy changes
– analogous to when external impulses cause the system’s momentum to change. However,
unlike momentum which only describes the motion of the objects in the system, the energy of a
system can be in different forms and can change from one form to another (for example the
elastic potential energy of the sling shot changes into the kinetic energy of the chalk in the
slingshot-chalk system as the chalk is released from the slingshot, or when the potential energy of
the elevated block-Earth system changes into kinetic energy when the block falls). Other
examples of energy changing from one form to another are shown in Table 6.3.
Table 6.3 Three examples of system energy changes
Description of process
(1) A person starts at rest at the top
of a smooth water slide and is
moving fast at the bottom.
System and initial-final sketch
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Ug ( K
The system started with
considerable gravitational energy,
which converted to the person’s
kinetic energy as moved down the
slide.
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(2) A fast moving car skids to a
stop on a level road.
(3) A pop-up toy starts with a
compressed spring and when
released pops up to a maximum
height above its starting position.
K ( U internal
The fast moving car with kinetic
energy skidded to a stop
converting its energy to internal
thermal energy due to friction.
Us ( Ug
The compressed spring converted
its elastic energy to the final
gravitational energy at the highest
point above the table.
In the processes in Table 6.3 the energy of the system changes forms. Is it possible to represent
those changes using a bar chart tool similar to the one we used when analyzing the changes in the
momentum of a system? Now that we know some differences between momentum and energy,
we can start adapting the bar charts to energy descriptions keeping in mind that the energy of the
system can change from one form to another.
Qualitative work-energy bar charts
Recall that the impulse-momentum bar charts had three parts. The part on the left was for
the component of the momentum of the system along a particular direction at an initial clock
reading. The part on the right was for the momentum component at the final clock reading. These
two parts were separated by the third part that represented the component in the chosen direction
of any external impulses exerted on the system. The sum of the heights of all the momentum bars
in the initial state plus any impulse bars was equal to the sum of heights of the momentum bars in
the final state. We’ll adapt this bar chart approach for analyzing work-energy processes using the
idea that the energy of a system changes when external forces do work on the system.
We represent each type of energy by a vertical bar (see the skill box that follows). The
height of each bar represents the relative amount of that type of energy in the system. The left
side of the bar chart reflects the initial state energies and the right side section the final state
energies. We will however make an exception for internal energy. As it is very difficult to
quantify how much internal energy a system possesses at a particular moment, we will only
represent the change in the system’s internal energy. We’ll represent this change as a bar on the
right side (final state) of the bar chart.
To represent any work done by external forces we add a shaded section in the middle
similar to the impulse section of an impulse-momentum bar chart. The shading will remind us
that the bars found there do not represent energies that the system has. Instead, they represent
mechanisms by which energy enters or leaves the system due to work done by external objects.
In the skill box, we construct a work–energy bar chart to analyze the following process:
A compressed spring at the bottom of an inclined plane pushes against a cart, causing the cart to
shoot up the inclined plane.
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Skill box: Constructing qualitative work-energy bar charts
1. Sketch the initial-final states.
2. Choose a system.
Initial state
Final state
y
µ=0
0
Earth
Ki Ugi Usi
W
Kf Ugf
Usf 'Uint
0
3. Use initial state sketch to
choose energy types and bars
for the initial state.
4. Use final state sketch to choose
energy types and bars for the final
state.
5. Decide if work is done by an
external object.
At this point, there is no way to know the relative amounts (bar heights) of these different
forms of energy. The bars just remind us to include these types of energy in our energy analysis.
However, if energy is a conserved quantity, then the sum of the energy bar heights on the left plus
the height of the work bar should be equal to the sum of the energy bar heights on the right. For
the spring-cart example, this would mean that the initial elastic energy of the compressed spring
should equal the sum of the heights of the final kinetic energy and the final gravitational potential
energy bars. If external forces do work on the system, then this work plus the initial energy of the
system should equal the final energy of the system:
Ui $ W # U f
where U with the subscript (initial i or final f) is the symbol for the total energy of the system
and W is the symbol for work done by the external forces on the system. If we represent with the
symbols the different types of energy, the above equation becomes:
( K i $ U g i $ U s i ) $ W # ( K f $ U g f $ U s f $ )U int )
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For the process in the skills box, some of these terms are zero:
Us i # K f $Ug f
Testing energy conservation
The above energy conservation idea is a guess we made based on our prior experience
with momentum; this is called reasoning by analogy. To decide if our guess is a good one, we
need to test it by conduct testing experiments.
One qualitative test of this conservation idea involves an experiment called
Galileo’s pendulum (one experiment will not be enough to convince us that the guess is correct,
but at least it will give us some confidence). It involves a swinging pendulum bob and a nail.
Testing Experiment Table 6.4 A qualitative test of energy conservation.
Testing experiment
A pendulum bob attached to a
string swings in a vertical
plane. A nail can be placed at
different locations along the
path of the string.
You pull the bob to the side
so it starts a height h above
the straight down position.
When the bob swings down
and the string meets the nail,
to what maximum height will
the bob rise on the other side?
Prediction
The bob and Earth is the system. The
energy of the system in its initial state (just
before the bob is released) should equal the
energy of the system in its final state (just
as the bob’s swing to the other side stops).
At both of these moments, the kinetic
energy of the system is zero, which means
the system should only have gravitational
potential energy in the initial and final
states. If energy is a conserved quantity,;
then the bob should reach the same height
independent of the nail location.
Outcome
You observe that independent
of the location of the nail, the
bob always reaches the same
height from which it was
released.
Conclusion
We did not disprove the idea of energy conservation with this experiment.
If you perform this experiment you might be surprised by the outcome, although the idea
of energy conservation predicts it.
Tip! For a system to have gravitational potential energy, Earth must be a part of the system. Thus
a single object even when elevated above the ground, does not have gravitational potential energy
if Earth is not in the system. If Earth is not in the system, the gravitational force it exerts on an
object can do work on the object.
Now that we have some confidence in the ideas of energy conservation, let’s apply them
using work-energy bar charts.
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Pole vault
Pole-vaulting originated in Holland. Water removed from marshland left a network of
open drains or canals intersecting each other at right angles. In order to cross these without
getting wet, and to avoid long walks to bridges, a stack of jumping poles was kept at every house
and used for vaulting over the canals. This led to annual vaulting competitions and eventually to
an Olympic event.
Conceptual Exercise 6.2 Pole vaulter A pole vaulter crosses the bar high above the ground.
(Fig. 6.4) Construct two different work–energy bar charts for the process, starting at the highest
point in the jump and ending: (a) just before the jumper reaches the cushion below; and (b) at the
instant the jumper has stopped after sinking into the cushion.
Figure 6.4Pole Vaulter
Sketch and Translate A sketch of the process starting when the vaulter is just above the bar (the
initial state) to the instant the vaulter reaches the cushion (part a) is shown in Fig. 6.5a. Another
sketch of the vaulter just above the bar to the instant he stops after sinking into the cushion (part
b) is shown in Fig. 6.5b. The system includes the vaulter and Earth, but not the cushion. We
choose the zero-point of the vertical axis used to estimate the gravitational potential energy to be
at the final position of the vaulter after he stops while sinking into the cushion.
Figure 6.5 Sketches and bar charts for pole vaulter
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Simplify and Diagram Assume that air resistance can be ignored and that the vaulter has zero
speed (zero kinetic energy) at the top of the flight. We also assume that the internal energy of the
vaulter does not change significantly (there is a small increase in internal energy since the vaulter
feels some discomfort when crashing into the cushion.) In the initial state for each part, the
vaulter–Earth system has positive gravitational potential energy (the vaulter is at a relatively high
vertical position relative to the earth). The bar charts in Fig. 6.5c and d both have an initial
gravitational potential energy bar. When the vaulter reaches the top of the cushion (the final state
for part a), he has much less gravitational potential energy and now has considerable kinetic
energy (the speed of the pole vaulter is great). See the final state bars in Fig. 6.5c. In the final
state for part b, the vaulter has stopped at the origin of the vertical y-axis and has zero
gravitational potential energy and zero kinetic energy (see the lack of final state bars in Fig. 6.5d).
This energy decrease occurred because of the negative work done by the cushion on the vaulter as
he sank into the cushion. The force that the cushion exerted on the vaulter pointed up opposite the
displacement of the vaulter while sinking down into the cushion—the negative work bar in Fig.
6.5d).
Try It Yourself: You throw a ball straight up. Draw a bar chart starting when the ball is at rest in
your hand and ending when the ball is at the very top of its flight Fig. 6.6a). The system is the ball
and Earth (not the hand). Ignore interactions with the air.
Figure 6.6(a) Hand throwing ball to highest point
Answer: See Fig. 6.6b. Note that there is no kinetic energy in the bar chart as the ball was not
moving in the initial state or in the final state. It does not matter that it was moving between those
two states.
Figure 6.6(b) Bar chart for hand throwing ball to highest point
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Tip Note that the amount of gravitational potential energy in a system depends on where the
origin is placed on the vertical y-axis. This placement is arbitrary; the important thing is the
change in position and the corresponding change in gravitational potential energy.
You might be wondering what objects to include in a system and what objects not to
include. Generally, it is preferable to have a larger system so that the process changes occurring
can be included as energy changes in the system rather than the work done by external forces.
Sometimes, our interest is in the force that some other object exerts on an object of interest. In
that case the other object exerting the force is excluded from the system so that we can calculate
the work done by the other object on the system—a calculation that involves the force that the
other object exerts on the system object. In the end the choice of which objects to include in the
system and which not to include is motivated by the goal of the analysis you are doing.
Review Question 6.2
A system can possess energy but not work. Why?
6.3 Quantitative descriptions of kinetic and gravitational potential
energies
The experiment in Testing Experiment Table 6.4 gave us more confidence in the idea of
energy conservation. In the case of Galileo’s pendulum, no external forces did work on the
system, so the energy of the system was constant. Next, we need to develop mathematical
expressions for different types of energies. We can accomplish this by using the idea that the
energy change of the system is exactly the amount of work done on it by external forces. This
will give us a quantitative version of the idea of energy as a new conserved quantity.
Let’s begin with several thought experiments; we analyze situations using physics developed
earlier but do not actually perform the experiments. We use the following:
1. The mathematical expression of energy conservation (work-energy equation): U i $ W # U f .
!
2. The work done by an external force F oriented at an angle " relative to the displacement
!
d of the system object: W # Fd cos " .
3. Newton’s second law and kinematics principles from the first three chapters.
We apply these principles to a series of simple processes. In each process, only one type of
energy changes when an external force is exerted on a system object during its displacement. We
develop an expression for that type of energy change. We focus first on a process in which
gravitational potential energy changes.
Quantitative Description of Gravitational Potential Energy
Imagine that a rope is slowly lifting a heavy box upward at a constant very slow velocity.
(Fig. 6.7a). The rope is attached to a motor above, which is not shown in the figure. If we choose
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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only the box as the system, we can apply Newton’s second law to find the magnitude of the force
that the rope exerts on it. Since the box moves up at constant velocity, the tension force
ALG
6.3.1
!
!
TR on B exerted by the rope on the box is equal in magnitude to the gravitational force FE on B
exerted by Earth on the box (see the force diagram in Fig. 6.7b). Since the magnitude of the
gravitational force is mB g , we find that the magnitude of the tension force for this process
is TR on B # mB g .
Figure 6.7 Lifting a box at negligible constant speed
Now, to derive an expression for gravitational potential energy, we need to change the
system since the box by itself does not have any gravitational potential energy. The new system
will be the box and Earth. The origin of a vertical y-axis will be the ground directly below the
box. The positive direction will be upward. The initial state of the system is when the box is at
position yi and moving upward at a negligible very slow speed v . The final state is when the box
is at position yf moving upward at the same negligible speed v . We will use the idea that energy
is a conserved quantity to derive the mathematical expression for the gravitational potential
energy. The conservation of energy can be written mathematically as:
Ki $ U i $ W # Kf $ U f .
Since the box is traveling at onstant speed, the kinetic energy of the system does not change.
Therefore K i # K f and we can subtract them from both sides of the equation. This leaves us with
Ui $ W # Uf .
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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We can represent this with the bar chart (Fig.6.7c). Notice that here the only energy that changes
is the gravitational potential energy—we ignored the non-changing negligible kinetic energy.
The rope does work on the box lifting the box from vertical position yi to yf :
WR on B # TR on B d cos " # T & y f % yi ' cos 00 # mg & y f % yi ' cos 00
where we used the idea that T # mg for this constant velocity experiment. Substituting this
expression for work into the work energy equation, we get:
U g i $ mg & y f % yi ' cos 00 # U g f
We can now write an expression for the change in the gravitational potential energy of the
system: U g f % U g i # mgy f % mgyi . The change equals the difference in the value of the
physical quantity that is the product of the mass of the object, the gravitational constant and the
height: mgy . This suggests the following definition for gravitational potential energy of a
system.
Gravitational potential energy: The gravitational potential energy of an object–Earth system is:
U g # mgy
(6.3)
where m is the mass of the object, g # 9.8 N kg , and y is the position of the object with
respect to zero of the vertical coordinate system (the origin of the coordinate system is our
choice). The units of gravitational potential energy are kg(N/kg)m = 1 N•m = 1 J (joule), the
same as the unit of work and of every other type of energy.
The unit for work and energy is named for James Joule (1818–1889). Joule was the son
of a prosperous English brewery owner and a student of John Dalton (famous for his work on
atomic theory.) He spent his life (and the fortune inherited from his father) learning how different
forms of energy transform into each other. He was one of three physicists who contributed the
most to our understanding of energy transformation.
Kinetic Energy
Next, let’s use a simple thought experiment and our previous knowledge to determine an
expression for the kinetic energy of a system that consists of a single object. Imagine that with
!
your hand you exert a force FY on C pushing a cart of mass m toward the right on a horizontal
!
frictionless surface (Fig.6.8a). The cart moves at increasing speed during a displacement d . A
bar chart for this process is shown in Fig. 6.8b.
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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Figure 6.8 Work done by hand causes cart’s kinetic energy to increase
Since the cart moves on a horizontal frictionless surface, there is no change in gravitational
potential energy. The kinetic energy changes from the initial state to the final state because of the
work done by the external force exerted by you on the cart. The process can be represented
mathematically as follows:
K i $ WH on C # K f
or
WH on C # K f % K i .
We know that the work in this case equals FH on C d cos 00 # FH on C d . Thus,
FH on C d # K f % K i .
This does not look like a promising result—the kinetic energy change on the right equals
quantities on the left side that do not depend on the mass or speed of the cart. However, we can
use our knowledge of dynamics and kinematics to get a result that does depend on these
properties of the cart. Use Newton’s second law for the horizontal direction to determine the
unbalanced force exerted on the cart, which in this case is the force that the hand exerts on the
cart:
mC aC # FH on C .
Rearrange a kinematics equation from Chapter 1 ( vf2x # vi2x $ 2ax d x ) to get an expression for the
displacement of the cart in terms of its initial and final speeds and its acceleration:
vf 2 % vi 2
d#
.
2aC
Now, insert these expressions for force and displacement into the left side of the equation
FH on C d # K f % K i :
* vf 2 % vi 2 +
vf 2 vi 2
1
1
%
FH on C d # (mC aC ) ,
) # mC vf 2 % mC vi 2
- # mC (
2
2
2
2
. 2aC /
We can now set the result equal to the right part of the same equation:
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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1
1
mC vf 2 % mC vi 2 # K f % K i
2
2
It appears that
1
mC v 2 is an expression for the kinetic energy of the cart.
2
Kinetic energy: The kinetic energy of an object is:
K#
1 2
mv
2
(6.4)
where m is the object’s mass and v is its speed relative to the chosen coordinate system. The
unit of kinetic energy is the joule (J).
Example 6.3 An acorn falls While sitting on the deck behind your house, 5-g acorns are falling
from the trees high above just missing your chair and head. Use the work-energy principle to
estimate how fast one of these acorns is moving just before it reaches your head.
Sketch and Translate First, draw a sketch of the process (Fig. 6.9a). The origin of a vertical y-axis
will be at your head with the positive y-axis pointing up. This axis is used to keep track of the
acorn’s gravitational potential energy. We assume that the acorn is about 20 m above our head as
it begins to fall. The system will be the acorn and Earth. We keep track of kinetic energy and
gravitational potential energy to find the acorn’s speed when reaching the level of the head. The
initial state will be at the instant the acorn leaves the tree. The final state is as it reaches the level
of your head.
Figure 6.9(a) Acorn falls from tree
Simplify and Diagram Since the size of the acorn is much smaller than the distance it travels, it is
reasonable to model it as a point-like object while it falls. The acorn is small and we assume that
the air does not do a significant amount of work on the acorn as it falls. The process is
represented by a bar chart in Fig. 6.9b. In its initial state, the acorn-Earth system has considerable
gravitational potential energy and zero kinetic energy. In its final state, it has zero gravitational
potential energy and considerable kinetic energy. As Earth is in the system, and the acorn’s
interactions with the air are being ignored, there are no external forces doing work on the system.
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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ALG
6.4.3
-6.4.4
AP
5.2;
5.3
Figure 6.9(b) Acorn falls from tree
Represent Mathematically Use the bar chart to apply the work-energy equation:
Ug i $ 0 # K f
( mgyi $ 0 #
1
mv f 2
2
1
( gyi # v f 2
2
( v f # 2( gyi )
The magnitude of the acorn’s speed depends only on the height from which it was dropped,
which we estimated to be 20 m. Note that the acorn’s mass drops out of the calculation.
Solve and Evaluate Our estimate of the final speed of the acorn is:
v f # 2( gyi ) # 2(9.8 m/s 2 )(20 m) # 20 m/s
That’s 45 mph—that seems reasonable based on the sound when it hits the deck.
Try It Yourself: If you throw an acorn upward at a speed vi , what expression can we use to
determine its maximum height above its launching position before it starts descending?
Answer: vi 2 / 2 g .
What if in Example 6.4 we had chosen only the acorn as the system of interest? In that
case the system would not have any gravitational potential energy. Instead, Earth would be an
object in the environment doing positive work on the acorn system. Let’s apply our ideas of work
and energy with this choice of system. Since the acorn is at rest initially, the system has no energy
at all. Earth does work on the system. In the final state just as the acorn reaches the level or out
head, the acorn system has kinetic energy:
0 $W # K f
( Fd cos " #
1 2
mv f
2
( & mg ' yi cos & 00 ' #
1 2
mv f
2
( v f # 2( gyi )
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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That’s the same result we arrived at with the tomato-Earth system. The choice of system did not
affect the result of the analysis.
Review Question 6.3
When we use energy conservation ideas, how do we incorporate the force that Earth exerts on an
object?
6.4 Quantitative Description of Elastic potential energy
In the previous section we succeeded in constructing expressions for: 1) the gravitational
potential energy of an object–Earth system, and 2) the kinetic energy of a system with a moving
object. Our next goal is to construct a mathematical expression for the elastic potential energy
stored in an object when it has been deformed—stretched or compressed. A good example is a
spring. We will follow the same approach to finding the mathematical expression for the elastic
potential energy as we did for gravitational and kinetic energies. We design a simple process in
which an external force stretches or compresses an elastic object. We determine the work done by
this external force, and then apply the work-energy principle and our former knowledge to come
up with an expression for the change in elastic potential energy. However, there is a special
difficulty in applying this method to an elastic object.
When you stretch an object, whether it is a spring, rubber band, bungee cord, or some
other elastic material, the more you stretch it, the greater the force you must exert on it. Thus, we
cannot use the simple expression W # Fd cos " to determine the work done on the spring. That
equation assumes the force being exerted is constant in magnitude, but the force you need to exert
on the spring changes as the spring stretches more. How does the force you exert to stretch the
spring change as the spring stretches?
Hooke’s law
To answer this question we use two springs, a thinner and less stiff spring 1 and thicker
and stiffer spring 2 (their lengths are the same)—see Fig. 6.10a. The springs are attached at the
left end to a rigid object and placed on a smooth surface. We use a scale to pull on the right end
!
of each spring exerting a force F whose magnitude can be measured by the scale. We record the
magnitude of the force F and the distance x that each spring stretches from its un-stretched
position (see example for spring 1 in Fig. 6.10b). The data for both springs are recorded in Table
6.5.
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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Figure 6.10 Measuring spring stretch, x, due to scale pulling force, F
Table 6.5 Spring extension as a result of pulling on its end exerting an increasing force.
Spring 1 distance x
stretched
0.000 m
0.050 m
0.100 m
0.150 m
0.200 m
Force F exerted by the
scale on the spring
0.0 N
1.0 N
2.0 N
3.0 N
4.0 N
Spring 2 distance x
stretched
0.000 m
0.030 m
0.060 m
0.090 m
0.120 m
Force F exerted by the
scale on the spring
0.0 N
1.0 N
2.0 N
3.0 N
4.0 N
A graph of the above data is shown in Fig. 6.11. We use stretch x as an independent
variable and the force F as a dependent variable. The force exerted by the spring scale on each
spring is proportional to the distance that each spring stretches. However, the coefficients of
proportionality are different. The slope of F ( x) graph for spring 2 has a larger slope than the
slope of the graph for spring 1. Spring 2 is thicker and stiffer than spring 1. This agrees with our
everyday experience: thicker springs are usually more difficult to stretch than thinner springs.
This physical property encoded in the slope of the graph is called the spring constant k . Spring
1’s spring constant is 20 N/m ; spring 2’s spring constant is 33 N/m . Another way to interpret
the spring constant is the force that must be exerted on the spring to stretch it 1.0 meter. For
spring 1 this would be 20-N; for spring 2 it would be 33-N.
Figure 6.11 Force needed to stretch springs1 and 2 different distances
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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We discovered that there is a proportional relationship between the external force
exerted on a spring and the distance the spring stretches. This relationship was first reported by
Robert Hooke (1635-1703), one of Isaac Newton’s scientific rivals.
Let’s consider the same process but using your finger as the object of interest. When
your finger stretches a spring in the positive direction, the spring pulls back exerting a force on
the finger in the opposite negative direction (Newton’s third law). You can feel this pull of the
spring on your finger. If an object pushes on a spring compressing it, the spring pushes back
again exerting a force in the opposite direction. In both cases, the magnitude of this force is
proportional to the distance the spring has been stretched/compressed (Fig.6.12).
Figure 6.12 Spring opposes force that stretches or compresses it
Mathematically this elastic force is expressed as what is known as Hooke’s law.
Elastic force (Hooke’s law): If any object causes a spring to stretch or compress, the spring
exerts an elastic force on that object. If the object stretches the spring along the x direction, the
x component of this force is:
FS on O x # – kx
(6.5)
The spring constant k is measured in N/m and is a measure of the stiffness of the spring (or any
elastic object); x is the distance the object has been stretched/compressed (not the total length
of the object). The elastic force exerted by the spring on the object stretching it points in the
direction opposite to the direction it was stretched (or compressed).
Note that the graph lines in Fig. 6.11 had positive slopes and could be described by the
equation FO on S # $ kx . This was the force that the object exerted on the spring and not that the
spring exerted on the object. We will use this latter force shortly to determine an expression for
the elastic potential energy of a stretched or compressed spring.
You may have noticed that if you stretch or compress a spring too much, it becomes
permanently deformed and will not return to its original shape. The maximum distance a spring
can be compressed or stretched without becoming permanently deformed is called the elastic
limit of the spring. Near and beyond the elastic limit, Hooke’s law no longer describes the
behavior of the spring.
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What we have begun investigating here is the physics of elasticity, stresses, and strains.
This physics goes far beyond simple springs and rubber slingshots. It is critical to designing the
suspension systems of vehicles, understanding how buildings will respond to high winds or
earthquakes, and to describing the skeletal and muscular systems of the human body. Whether
you are a structural engineer or physical therapist, the physics of elasticity plays a central role.
Elastic potential energy
Remember that our goal is to develop an expression for the elastic potential energy of an
object that has been deformed (such as a stretched spring.) Hooke’s Law can help accomplish
this. Consider the constant slopes of the lines in the graph shown in Fig. 6.11. While stretching
the spring with your hand from zero stretch ( x # 0 ) to some arbitrary stretch distance x , the
magnitude of the force your hand exerts on the spring changes in a linear fashion from zero when
un-stretched to kx when stretched.
To calculate the work done on the spring by such a variable force, we can replace this
variable force with the average force ( FH on S ) average :
( FH on S )average #
0 $ kx
.
2
The force your hand exerts on the spring is in the same direction as the direction in which the
spring stretches. Thus the work done by this force on the spring to stretch it a distance x is:
1
1
W # ( FH on S )average x # ( kx) x # kx 2 .
2
2
This work equals the change in the spring’s elastic potential energy. Assuming that the elastic
potential energy of the un-stretched spring is zero, the work we calculated above equals the final
elastic potential energy of the stretched spring. We have achieved our goal. Here is a summary of
the result:
Elastic potential energy The elastic potential energy of a spring-like object with a spring
constant k that has been stretched or compressed a distance x from its undisturbed position is:
Us #
1 2
kx
2
(6.6)
Just like any other type of energy, the unit of elastic potential energy is the joule (J).
Tip! The elastic potential energy was derived using Hooke’s law ( FS on O x # % kx ), which applies
only for elastic objects below their elastic limit. Thus, the expression for the elastic potential
energy also assumes the elastic object obeys Hooke’s law.
Example 6.4 Shooting an arrow You are ready to fir an arrow from a bow. You load an arrow
(mass 0.090 kg ) in the bow and pull the bowstring back 0.40 m . The bow has a measured
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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spring constant k = 900 N/m . Determine how fast the arrow will be moving as it leaves the
bow.
Sketch and Translate A sketch of the process is shown in Fig. 6.13a. The system is the bow and
the arrow. The initial state is when the bowstring is pulled back 0.40 m. The final state is when
the string has just relaxed and the arrow has just left the string.
Figure 6.13(a) A bow shoots an arrow
Simplify and Diagram The elastic potential energy of the bowstring varies as you pull it farther
and farther. Since the arrow moves horizontally in the bow, we do not need to keep track of
gravitational potential energy. A bar chart shows that the initial elastic potential energy of the
bow is being transformed into the final kinetic energy of the arrow. (Fig. 6.13b).
Figure 6.13(b)
Represent Mathematically We use the bar chart to apply the work-energy equation:
Ui $ W # U f
( Ki $ U s i $ 0 # K f $ U s f
1
1
( 0 $ kx 2 # mv 2 $ 0
2
2
1
1
( kx 2 # mv 2
2
2
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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If we multiply both sides of the equation by 2, divide by m , and take the square root, we have
v#
k
x.
m
Solve and Evaluate
v#
k
900 N m
x#
& 0.40 m ' # 40 m s
m
0.090 kg
.
This is reasonable for the speed of an arrow fired from a bow. Let’s also check the units of
to make sure they really are equivalent to
k
x
m
m
:
s
N
(m) =
m kg
kg m
m
(m) = .
s m kg
s
2
The fact that units come out as the units of speed is evidence that the mathematics is correct. To
further evaluate if the result is reasonable – see the Try It Yourself example.
Try it yourself: If the arrow were shot vertically, how high would it go?
Answer: 80 m, a reasonable height for the arrow to travel.
Review question 6.4
If the magnitude of the force exerted by the spring on an object is kx, why is it that the work done
to stretch the spring a distance x is not equal to the kx 1 x # kx 2 ?
6.5 Incorporating friction into energy conservation
In nearly every mechanical process that we encounter in daily life, objects exert friction
forces on each other. Sometimes the effect of friction is minimal (for example, in an air hockey
game); but most of the time friction is extremely important (for example, a driver applying the
brakes to avoid a collision). Our next goal is to investigate how we can incorporate friction into
the ideas about energy conservation. Let’s use the ideas of work and energy to analyze the
process of a car skidding to avoid an accident.
Imagine that the car’s brakes have locked, and the tires are skidding on the road surface
(Fig.6.14a). The system of interest is the car. How much
work does the road’s friction force do on the car? There
are three forces exerted on the car by external objects, and
two of them cancel – the gravitational force exerted by
Earth and the normal force exerted by the road surface.
The only force left is the friction force exerted by the road
Figure 6.14(a)
on the car, which slows the car to a stop.
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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ALG
6.3.3
The friction force does work on the car:
Wfriction # f k d cos1800 # – f k d .
The negative work done by the friction force causes the car’s kinetic energy to decrease to zero
(Fig 6.14b). Mathematically,
K i $ (– f k d ) # 0
But we’ve left out a very important feature of friction. If you touched the car’s tire just
after the car came to rest, the tire would be warm to the touch. You would also notice black skid
marks on the road—some of the rubber had been scrapped off of the tire. Thus, the internal
energy of the system increased ( )U internal 2 0 ). There should be a term on the right side of the
above equation indicating this increase in internal energy of the car. But the two terms on the left
side of the above equation add to zero; so we would get 0 # )U internal --not possible.
Figure 6.14(b) A difficulty occurs using the car alone as system for this energy analysis involving friction
This same difficulty occurs with another process. Imagine that you pull a rope attached to
a box (the system object) so that the box moves at constant very slow velocity on a rough carpet
(Fig. 6. 15a). The box is moving so slowly that we will ignore its kinetic energy. A force diagram
for the box is shown in Fig. 6. 15b. The initial state is when the box has only been pulled a short
way across the carpet, and the final state is when it has been pulled a long way. This process can
be represented with a bar chart such as shown in Fig 6.15c. The force exerted by the rope does
positive work on the box system; the friction force does negative work. If the forces have exactly
the same magnitude (which they should be for the box moving with constant velocity), then the
net work done on the system is zero and the energy of the system should not change. However,
again if you touch the bottom of the box at its final posiion, you find that it is warmer than before
you started pulling and the box has scratches on its bottom. Again, the internal energy of the box
increased. We get 0 # )U internal where )U internal is greater than zero. How can we resolve this?
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Figure 6.15 Pulling box across rough carpet
Cars and boxes are not point-like objects
In earlier chapters we often modeled objects as point-like. A point-like object does not
have internal structure and therefore cannot have thermal energy. There’s no way to internally
distinguish a hot object from a cold one. Real objects have bumpy surfaces. When you pull a box
across a rug or when a car skids to a stop, the microscopic bumps on the contacting surfaces hook
into each other. These bumps pull back on each other as the objects move across each other.
Then, they snap back and get warmer—their internal thermal energy increases. Sometimes, the
bumps are actually pulled off the surfaces—also an increase in internal energy. We over-simplify
these situations by only focusing on the mechanical work done by friction—ignoring all of the
complex changes occurring on the surfaces moving across each other. Perhaps there is an
alternative way to incorporate friction into the idea of energy conservation.
The effect of friction as a change in internal energy
To correct this difficulty of having changes in the system’s internal energy when the net
work done on the system is zero, we choose a different system that includes both surfaces that are
in contact; for example, the car and the road, or the box and the carpet. The friction force is then
an internal force and therefore does no work. But, there is a change in the internal energy of the
system caused by friction between the two surfaces. This change of the system is the key to
resolving the difficulty described above.
For the rope pulling the box across the rough carpet at constant velocity, we know that if
the rope pulls horizontally on the box, the magnitude of the force that it exerts on the box TR on B
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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must equal the magnitude of the friction force that carpet exerts on the box f k C on B . But now the
box and surface are both in the system—so the force of the rope is the only external force. Thus
the work on the box-carpet system is:
W # TR on B d cos 00
Substituting TR on B # f k C on B and cos 00 # 1.0 into the above, we get:
W # $ f k C on B d
The only system energy change is its internal energy )U internal . The work-energy
equation for pulling the box across the surface process is:
W # )U internal .
After inserting the expression for the work done on the box and rearranging, we get:
)U internal # $ f k d .
We have constructed an expression for the change in internal energy of a system caused
by the friction force that the two contacting surfaces exert on each other when one object moves a
distance d across each other.
Increase in the system’s internal energy due to friction
)U internal # $ f k d ,
(6.7)
where f k is the magnitude of the average friction force exerted by the surface on the object
moving relative to the surface, and d is the distance that the object moves across that surface.
The increase in internal energy is shared between the moving object and the surface.
The above approach for including friction into the idea of energy conservation produces
the same result as calculating the work done by friction when we ignored internal energy changes.
In this new approach, there is an increase in internal thermal energy )U internal # $ f k d in a
system that includes the two surfaces rubbing against each other (this goes on the right side of the
work-energy principle); in the work done by friction approach, the negative work done by friction
Wfriction # – f k d is included if one of the surfaces is not in the system (this term goes on the left
side of the work-energy principle). So mathematically, they have the same effect. For the two
surfaces in the system approach, we can see why a skidding tire gets warmer and why there might
be structural changes; in the latter one surface work done by friction approach, the change in the
internal energy of the rubbing surface is a mystery. In this book we prefer to include friction in
our energy analysis by including both surfaces in the system and considering the increase in
internal energy caused by friction.
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Example 6.5 Skidding to a stop While driving your Chevy Malibu, a Hyundai Sonata crosses
the road at an intersection in front of you. To avoid a collision, you apply the brakes, leaving
24-m skid marks on the road while stopping. A police officer observes the near collision and
gives you a speeding ticket, claiming that you were exceeding the 35-mph speed limit. He
estimates your car’s mass to be 1390 kg and that the coefficient of kinetic friction between your
Malibu tires and this particular road is about 0.70 . Is the speeding ticket deserved?
Sketch and Translate A sketch of the process is shown in Fig. 6.16a. We choose the Malibu and
the road surface as the system. We need to decide if your car was traveling faster than 35 mph at
the instant you applied the brakes.
Figure 6.16(a) A car skids to a stop
Simplify and Diagram Assume that the process occurs on a horizontal level road and neglect
interactions with the air (in fact, it is significant here but for now we will ignore it since we do not
yet have a way of incorporating it quantitatively). The initial state is just before the brakes are
applied. The final state is just after the Malibu has come to rest. The energy bar chart in Fig.
6.16b represents the process. In the initial state, the system has kinetic energy. In the final state,
the system has no kinetic energy, and has increased internal energy due to friction.
Figure 6.16(b)
Represent Mathematically Convert the bar chart into an equation:
K i $ 0 # U int f
1 2
mvi # f k R on C d
2
1
( mvi2 # ( 3k N R on C )d
2
(
The magnitude of the upward normal force N R on C that the road exerts on the car equals the
magnitude of the downward gravitational force FE on C # mC g that Earth exerts on the car. Thus
the above becomes:
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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1
mC vi2 # 3k mC gd
2
Solve and Evaluate Rearranging the above and canceling the car mass, we get;
* 1 mph +
vi # 2 3 k gd # 2 & 0.70 '& 9.8 N kg '& 24 m ' # (18.1 m/s) ,
- # 40 mph
. 0.45 m s /
It looks like you deserve the ticket. We ignored the resistive drag force that air exerts on the car,
which could be significant for a car traveling at 40 mph . Thus, air resistance helps the car slow
down, which means you actually were traveling faster than 40 mph .
Try It Yourself: Imagine the same situation as above, only you are driving a 2000-kg minivan.
Will the answer for the initial speed increase or decrease?
Answer: The speed does not change.
Review question 6.5
Why, when friction cannot be neglected, is it useful to include both surfaces in the system when
analyzing processes using the energy approach?
6.6 The generalized work-energy principle
Since the beginning of the chapter we have developed the ideas of work and energy and
used those ideas to analyze physical processes. We developed mathematical expressions for the
different types of energy. We found that the energy of a system changes by exactly the amount of
work done on the system. This also means that the energy of an isolated system is constant. Thus
we now have three conserved quantities: mass, momentum and energy1. The conservation of
ALG
energy is represented mathematically as an equation we call the generalized work-energy
6.3.46.3.5
equation.
Generalized work-energy equation: The sum of the initial energies of a system plus the work
done on the system by external forces equals the sum of the final energies of the system:
Ui $ W # U f
(6.8)
where U # K $ U g $ U s $ U int $ "
The above equation is similar to the generalized impulse-momentum principle, except energy can
change forms while momentum only refers to the motion of the system. As new forms of energy
are investigated in later chapters, they will be included in the above generalized work-energy
principle. Let’s test the generalized work-energy equation one more time.
1
We will learn later that the mass and energy are not conserved independently of each other.
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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Testing Experiment Table 6.6 Testing the generalized work–energy equation.
Testing experiment
You have a Hot Wheels track and a car.
You can tilt the track at different angles
with the bottoms oriented horizontally at
the edge of a table. Where should you
release a Hot Wheels car on the track so the
car always lands the same distance from the
table independently of how the track is
tilted?
Prediction based on the generalized
work-energy equation
For the car to land on the floor at the
same distance from the table’s edge,
the car needs to have the same
horizontal velocity, and therefore the
same kinetic energy when leaving the
track. To get the same kinetic energy at
the bottom of the track, it must have
the same initial gravitational energy
when it starts. This means that no
matter what the incline angle of the
track, the car should start at the same
vertical elevation.
Outcome
When released
from the same
vertical height
with respect to the
table on the tracks
tilted at various
angles, the car
lands the same
distance from the
table.
U gi = K f
mgyi #
1
2
mvf 2 .
For each experiment vf is the same if
yi # h is the same.
Conclusion
The outcome of the experiment matches the prediction for this experiment. So, rather than disprove the
generalized work–energy equation, we have found another supporting experiment. Our confidence in the
work and energy ideas increases.
The generalized work-energy equation gives insight into why perpetual motion machines
very likely cannot exist. By definition, a perpetual machine is a device which produces more
energy than is absorbed in its operation; it violates the principle of conservation of energy.
Alternatively, the term refers to a mechanical device that once set in motion, continues to do
useful work without an input of energy. This too is impossible because of friction processes; the
mechanical energy of the system is transformed to internal energy, which cannot be converted
back to a useful form of energy.
Review question 6.6
Two football players are running toward each other at high speeds. They collide and stop. Where
did their kinetic energy go? What happened to their momentum?
6.7 Skills for analyzing processes using the work–energy principle
In this section, we will apply our problem-solving strategy to analyze work–energy
processes. The general strategy for analyzing such processes is described below and illustrated for
Example 6.7.
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ALG
6.3.6
Example 6.6 An elevator slows to a stop A 1000-kg elevator is moving up toward the top floor
of the Grandover resort hotel. While moving up at speed 4.0 m/s, it starts slowing and stops in 6.0
m. Determine the magnitude of the tension force that the cable exerts on the elevator ( TC on El )
while it is stopping.
Sketch and Translate
! Sketch initial and final states of the
process.
! Include a coordinate system with the
vertical y-axis with a zero point at an
object of reference to use in keeping
track of the gravitational potential
energy.
! Label the sketch with the known
information and the unknown quantity.
! Choose the system of interest. It is
often easier to include more objects in
the system to minimize the number of
external forces that might do work on
the system. Sometimes, you wish to
determine a particular force. In those
cases it is usually best to consider the
object that exerts this force as an
external object that does work on the
system.
Simplify and Diagram
! Decide what simplifications you can
make in the problem situation
regarding the objects, interactions, and
processes.
! Decide which energy types are
changing.
! Decide if there are external objects
doing work. If necessary, isolate a
moving object and draw a force
diagram for it to answer the above
question.
! Use the initial–final sketch to help
draw a work-energy bar chart. While
making the bar chart, you will have to
include work bars (if needed) and
initial and final energy bars and for the
types of energy that are changing.
The elevator and Earth are in the system. We exclude the cable
because we are interested in finding the magnitude of the force it
exerts on the elevator.
! Assume that the cable exerts a constant force on the elevator.
! In this case, we will keep track of kinetic energy (since the
elevator’s speed changes), and the gravitational potential energy
(since the elevator’s vertical position changes).
! The tension force exerted by the cable on the elevator , does
positive work (the tension force points up, and the displacement
of the elevator points up). See the diagram at the right.
! The origin of the coordinate system is at the initial position of
the elevator with the positive direction pointing up.
! In its initial state the system has kinetic energy, but no
gravitational potential energy. In its final state the system has
only gravitational potential energy. A bar chart for the process is
shown at the right.
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ALG
6.4.16.4.13
Represent Mathematically
! Convert the bar chart into a
mathematical description of the
process. Each bar in the chart will
show up as a single term in the
equation.
Ui $ W # U f
1
2
0
( ) mvi $ TC on El ( y f – 0) cos 0 # mgy f
2
1
2
( ) mvi $ TC on El y f # mgy f
2
Solve and Evaluate
mv 2
TC on El # mg –
! Use the mathematical description of
2yf
the process to solve for the desired
unknown quantity.
2
(1000 kg)(4.0 m/s)
! Evaluate the result. Does it have the
# (1000 kg)(9.8 N/kg) –
= 8500 N .
2(6.0 m)
correct units? Is its magnitude
reasonable? Do the limiting cases ! The result has the correct units. The force that the cable exerts is
make sense?
less than the 9800-N force that Earth exerts. This is reasonable
since the elevator slows down and the net force exerted on it
should point down.
! Limiting case: If the elevator slowed down over a much longer
distance (yf = very large number instead of 6.0 m), then the force
would be closer to 9800N.
Try It Yourself: Solve the same problem using Newton’s second law and kinematics.
Answer: 8500 N.
Notice that in the Try It Yourself activity at the end of the last example, you obtained the
same result for this problem when using Newton’s second law and kinematics. Analyzing
situations using the energy approach is often easier – as you can see by comparing the lengths of
the solutions for the previous problem if using an energy approach or a Newton’s second
law/kinematics approach. Another reason for learning the work-energy approach is that it allows
us to understand many microscopic processes better—such as climate temperature changes and
fluid flow (we will learn about these in Chapters 11 and 12.) The next example illustrates the
relative ease of the energy approach on a problem we addressed earlier using Newton’s second
law and kinematics—the human cannon ball in Example 3.10. There we used Newton’s laws and
kinematics to analyze the launch of a human cannon ball. Let’s analyze a similar process using
the ideas of work and energy.
Example 6.7 Return to the human cannon-ball You need to buy a spring with an appropriate
spring constant in order to launch a 60-kg human so that he leaves the cannon moving at a speed
of 15 m/s. This spring will be compressed 3.0 m from its natural length when it is ready to launch
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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the person. The cannon is oriented at an angle of 370 above the horizontal. What spring constant
should the spring have so that the cannon functions as desired?
Sketch and Translate A sketch of the process is shown in.Fig.6.17a. The system is the person, the
cannon (with the spring), and Earth. The initial state is just before the cannon is fired. The final
state is when the person is leaving the end of the barrel of the cannon. All motion is with respect
to the earth. Choose the origin of the vertical y-axis at the initial position of the person.
Figure 6.17(a) Launching a human cannonball
Simplify and Diagram Assume that the spring obeys Hooke’s law and neglect the relatively
small amount of friction between the person and the inside barrel walls of the cannon, and
between the person and the air. We need to keep track of kinetic energy (the person’s speed
changes), gravitational potential energy (Earth is in the system and the person’s vertical position
changes relative to the vertical y-axis), and elastic potential energy changes (the spring
compression changes relative to a special x-axis used to keep track of elastic energy). There are
no external forces being exerted on the system, so the total energy of the system is constant. The
bar chart shown in Fig. 6.17b represents the process.
Figure 6.17(b)
Represent Mathematically Using the bar chart to help write the work-energy equation for this
process:
Ui $ W # U f
( U s ,i $ 0 # K f $ U g , f
(
1 2 1 2
kxi # mv f $ mgy f
2
2
Solve and Evaluate Dividing both parts of the equation by (1/2)x2 we get:
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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1
2( mv f 2 $ mgxi sin 37 0 )
k# 2
xi 2
41
5
2 6 (60 kg)(15 m/s) 2 $ (60 kg)(9.8 N/kg)(3.0 m)sin 37o 7
2
9 # 1740 N m : 1700 N m
# 8
2
(3.0 m)
.
The units of the answer are correct for a spring constant. The spring constant is always a positive
number and we obtained a positive number – both of those checks are important when we
evaluate the answer. The next step is to evaluate the magnitude of the result. The value of k is
quite large, which means this is a stiff spring, which makes sense given that it is launching a
person.
Try It Yourself: What should the spring constant of a spring be if there is a 100-N friction force
exerted on the human cannon ball while he is moving up the barrel?
Answer: 1770 N/m.
In Example 3.8, we used the force–kinematics framework to analyze the crash of
champion parachutist Michael Holmes into some shrubbery when his parachute did not open.
Let’s analyze a similar process using an energy approach.
Example 6.8 Landing in a snow bank In 1955, during an airborne parachute jump exercise
AP
in Alaska, a paratrooper survived a 370-m fall into a snow bank when his parachute failed to
5.5-5.7
open. His fall left a 1.07-m crater in the snow. Estimate the average force that the snow
exerted on the trooper while stopping his fall. The paratrooper and his gear had an estimated total
mass of 90 kg, and just before he contacted the snow, his speed was about 54 m/s (120 mph).
Sketch and Translate A sketch of the process is shown in Fig. 6.18a. We choose the origin to be
at the final position of the paratrooper. The y-axis points upward; the object of reference is the
earth. The initial state is just before the paratrooper touches the snow. The final state is just after
the paratrooper comes to rest in the snow. The system includes the paratrooper and Earth, but not
the snow. This will make the upward force the snow exerts on the paratrooper an external force
that does negative work (the force points up, but the paratrooper travels downward while sinking
into the snow).
Figure 6.18(a) Paratrooper lands in snow
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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Simplify and Diagram Assume that the upward force the snow exerts is constant since we are
interested only in the average force the snow exerts. We keep track of kinetic energy (the
paratrooper’s speed changes) and gravitational potential energy (the paratrooper’s vertical
position changes). A bar chart for the process is shown in Fig. 6.18b.
Figure 6.18(b)
Represent Mathematically We convert the bar chart into the application of the generalized workenergy equation:
1
mvi 2 $ mgyi $ FS on P d cos1800 # 0
2
Solve and Evaluate To determine the force exerted by the snow on the paratrooper FS on P, divide
by d cos1800 and rearrange the terms to get:
FS on P
1
mvi 2 $ mgd
2
#
d
1
(90 kg)(54 m/s) 2 $ (90 kg)(9.8 N/kg)(1.07 m)
#2
# 1.2 x 105 N
(1.07 m)
This is an average force of about 12 tons! The trooper survived with only a broken clavicle.
Compare this solution to that in Example 3.8—is the energy approach easier?
Try It Yourself: How big would the stopping force exerted on the paratrooper be if there was no
snow bank and he landed on the hard ground with a stopping distance of about 0.01 m?
Answer: 1.3 x 107 N.
High systolic blood pressure
We have used the ideas of work and energy to quantitatively examine several real world
phenomena. The work–energy approach also allows us to understand important qualitative
phenomena, such as the relationship between blood pressure and the properties of arteries in the
human body.
Conceptual Exercise 6.9 Stretching the aorta Each heartbeat, the left ventricle of your heart
pumps about 80 cm3 of blood into the aorta. This pumping action occurs during a very short
time interval, about 0.13 s . The blood stretches the aorta walls to accommodate the volume of
blood. During the next 0.4 s or so, the walls of the aorta apply pressure on the blood and moves
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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the blood out of the aorta into the rest of the circulatory system. Thus, the elastic walls of the
aorta serve as an intermediary pump between the heart’s left ventricle and the rest of the
circulatory system. (a) Represent this process with a qualitative work–energy bar chart. Choose
the system to be the aorta and the 80-cm3 of blood that is being pumped. Choose the initial state
to be just before the left ventricle contracts. Choose the final state to be when the blood is in the
stretched aorta and moving out into the rest of the circulatory system. (b) Modify the work-energy
bar chart for a person with hardened and thickened artery walls.
Sketch and Translate A sketch of the process is shown in Fig. 6.19a. The system and initial and
final states have been chosen for us in the task description.
Figure 6.19(a) Left ventricle pumps blood into aorta
Simplify and Diagram We keep track of the kinetic energy (the blood speed changes) and the
elastic potential energy (the aorta wall stretches). We will ignore the slight increase in the vertical
elevation of the blood. Since the left ventricle is not in the system, it will be exerting an external
force on the blood. The left ventricle is doing positive work since it pushes the blood upward in
the direction of the blood’s displacement. The work–energy bar chart in Fig. 6.19b represents this
process for an aorta with flexible walls. If the person has stiff, thick arteries (atherosclerosis),
then more energy than normal is required to stretch the aorta walls, resulting in high blood
pressure. This process is represented by the bar chart in Fig. 6.19c. In this case, the so-called
systolic pressure (the higher number of the two that comprises blood pressure) will be higher than
it would be for a healthy heart.
Figure 6.19(b)(c)
Bungee Jumping
In the 1950s, David Attenborough and a BBC film crew filmed the first recorded bungee
jumpers (called “land divers”) on Pentecost Island in Vanuatu in the South Pacific. As a test of
their courage, young men jumped from tall wooden platforms with vines tied to their ankles.
Years later, on April 1, 1979, four members of the Oxford University Dangerous Sport Club
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made the first modern bungee jump. They each jumped from the 76-m-high (250 ft) Clifton
Suspension Bridge in Bristol, England while tied to the bridge with a single rubber bungee cord.
Let’s analyze the Clifton Suspension Bridge jump using work-energy principles.
Example 6.10 Bungee jumping: We estimate that the Oxford team used a 40-m -long bungee
cord that had stretched another 35 m when the jumper was at the very lowest point in the jump.
We estimate that the jumper’s mass is 70 kg . Imagine that your job was to buy a bungee cord
that would provide a safe jump with the above specifications. Specifically, you need to decide the
spring constant k of the cord you need to buy.
Sketch and Translate A sketch of the process is shown in Fig. 6.20a. The initial state is just before
the jumper jumps, and the final state is when the cord is fully stretched and the jumper is
momentarily at rest at the lowest position. The motion is with respect to the earth. We choose a
coordinate system with the positive y-direction pointing up and the origin at the jumper’s final
position. The system is the jumper, the cord, and Earth.
Figure 6.20(a) Bungee jump
Simplify and Diagram We keep track of gravitational potential energy (the diver’s elevation
changes) and elastic potential energy (the cord stretch changes). We do not need to keep track of
kinetic energy since in both the initial and final states, the system has no kinetic energy. Assume
the bungee cord obeys Hooke’s law, and that its mass is small compared to the mass of the
jumper (meaning, we will not include the gravitational potential energy of the cord). There are no
external forces doing work on the system; thus the energy of the system is constant. The energy
bar chart in Fig. 6.20b represents the process. The initial gravitational potential energy of the
system is transformed completely into the elastic potential energy of the stretched bungee cord.
Figure 6.20(b)
Represent Mathematically We apply the generalized work-energy equation with one term for
each bar in the bar chart:
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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Ug i # Us f
( mgyi #
1 2
kx
2
Solve and Evaluate Solving the above for the spring constant gives:
k#
2mgyi
x2
The length of the cord is L = 40 m and the cord stretches xi # 35 m . Thus, the distance of the
person’s initial position yi above the final y f # 0 position is:
yi # L $ x f # 40 m $ 35 m # 75 m .
This means that the spring constant has a value:
k#
2mgyi 2 & 70 kg '& 9.8 N kg '& 75 m '
#
# 84 N m
2
x2
& 35 m '
The units for the spring constant are the correct and the magnitude is reasonable.
Try It Yourself: Suppose the bungee cord used by the Oxford University Dangerous Sport Club in
the last jump had a force constant of 40 N/m. How long should the un-stretched cord be so the
total distance of the jump remained 75 m?
Answer: The cord should be only 24 m long and will stretch 51 m during the jump—a much more
easily stretched cord.
Review Question 6.7
Why is it important to choose the system of interest before attempting to analyze a process?
6.8 Collisions
In Chapter 5, we used impulse and momentum principles to analyze collisions. An
example of a collision is a baseball hit by a bat (Fig. 6.21). Notice how the ball compresses
during the first half of the collision; then decompresses during the second half of the collision.
We learned that the forces that the two colliding objects exert on each other during the collision
are complicated, non-constant, and are exerted for a very brief
time interval—roughly 1 millisecond for the baseball/bat
collision. In Chapter 5 we learned that we could use the idea
of a conserved quantity to analyze these relatively complex
collisions. Even if complicated things are happening, we can
use the idea of a conserved quantity to successfully make a
prediction about the outcome of the collision.
Figure 6.21 Bat hits ball
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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Analyzing collisions using momentum and energy principles
We have already analyzed collisions using momentum principles. Can we learn anything
new by analyzing them using energy principles? Three different observational experiments
involving collisions are shown in Table 6.7. In each case a 1.0-kg object attached to the end of a
string (the bob of a pendulum) swings down and hits a 4.0-kg wheeled cart at the lowest point of
its swing. In each experiment, the pendulum bob and cart start at the same initial positions but the
compositions of the pendulum bob and cart are varied. After each collision, the cart moves at a
nearly constant speed due to the smoothness of the surface on which it rolls. We use momentum
and energy principles to analyze the results of the experiments. In all cases, the system is the
pendulum bob and the cart. The initial state of the system is the moment just before the collision.
The final state is the moment just after the collision ends. The momentum of the system is in the
positive x-direction. We keep track of the kinetic energy of both the pendulum bob and the cart.
In some of the experiments one or more objects in the system are deformed—we will discuss
internal energy changes when finished. We do not keep track of gravitational potential energy
since the vertical position of the system objects does not change between the initial and final
states.
Observational Experiment Table 6.7 Analyzing energy and momentum during collisions.
Observational experiment
(1) In the first experiment the 1.0-kg ball
(object 1, the bob) and the 4.0-kg cart
(object 2) are both metal. The ball swings
and hits the cart. Their masses and
velocity components just before and just
after the collision are:
m1 # 1.0 kg; v1ix # 10 m/s
v1 fx # –6.0 m/s
m2 # 4.0 kg; v2 ix # 0 m/s
Analysis
Momentum:
Before collision:
(1.0 kg)(+10 m/s) + (4.0 kg) 0 = +10 kg • m/s
After collision:
(1.0 kg)( – 6.0 m/s) + (4.0 kg)(+4.0 m/s) = +10 kg • m/s
Kinetic energy:
Before collision:
2
2
(1/2)(1.0 kg)(+10 m/s) + (1/2)(4.0 kg) 0 = 50 J
After collision:
2
2
(1/2)(1.0 kg)( – 6.0 m/s) + (1/2)(4.0 kg)(+4.0 m/s) = 50 J
v2 fx # 4.0 m/s
(2) In this experiment instead of the metal
ball, we have a 1.0-kg sand-filled balloon.
It swings down and hits a flimsy 4.0-kg
cardboard cart. After the collision, the
damaged cart moves across the table at
constant speed. The side of the balloon
Momentum:
Before collision:
(1.0 kg)(+10 m/s) + (4.0 kg) 0 = +10 kg • m/s
After collision:
(1.0 kg)( – 4.2 m/s) + (4.0 kg)(+3.55 m/s) = +10 kg • m/s
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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that hit the cart is flattened in the
collision.
m1 # 1.0 kg; v1ix =10 m/s
v1 fx = - 4.2 m/s
m2 # 4.0 kg; v2 ix # 0 m/s
Kinetic energy:
Before collision:
2
2
(1/2)(1.0 kg)(+10 m/s) + (1/2)(4.0 kg) 0 = 50 J
After
collision:
2
2
(1/2)(1.0 kg)( – 4.2 m/s) + (1/2)(4.0 kg)(+3.55 m/s) = 34 J
v2 fx # 3.55 m/s
(3) In this experiment the 1.0-kg sandfilled balloon is covered with Velcro. It
swings down and sticks to a flimsy
4.0-kg Velcro covered cardboard cart.
The string holding the balloon is cut by a
razor blade immediately after the balloon
contacts the cart. The damaged cart and
flattened balloon move off together across
the table.
m1 # 1.0 kg; v1ix # 10 m/s
v1 fx # 2.0 m/s
Momentum:
Before collision:
(1.0 kg)(+10 m/s) + (4.0 kg) 0 = +10 kg • m/s
After collision:
(1.0 kg)(+2.0 m/s) + (4.0 kg)(+2.0 m/s) = +10 kg • m/s
Kinetic energy:
Before collision:
2
2
(1/2)(1.0 kg)(+10 m/s) + (1/2)(1.0 kg) 0 = 50 J
After collision:
2
2
(1/2)(1.0 kg)(+2.0 m/s) + (1/2)(4.0 kg)(+2.0 m/s) = 10 J
m2 # 4.0 kg; v2 ix # 0 m/s
v2 fx # 2.0 m/s
Patterns
Two important patterns emerge from the data collected from these three different collisions.
! First, the momentum of the system is constant in all three experiments.
! Second, the kinetic energy of the system seems only to be constant when no damage is done to the
system objects during the collision (Experiment 1 but not in Experiments 2 and 3).
We can understand the first pattern in Table 6.7 using our knowledge of impulsemomentum. The x-component of the net force exerted on the system in all three cases was zero;
hence the x-component of momentum should be constant.
What about the second pattern? In experiment 1, the system objects were very rigid; but
in experiments 2 and 3, they were more fragile and as a result were damaged during the collision.
Using the data we can determine the amount of energy that was transformed from kinetic energy
to internal energy. But it is extremely difficult to be able to predict this amount ahead of time.
Unfortunately this means that in collisions where any deformation of the system objects occurs,
we cannot make predictions about the amount of kinetic energy converted to internal energy.
However, we now know that even in collisions where the system objects become damaged, the
momentum of the system still remains constant (assuming the impulse exerted on the system is
zero.) So, even though the work–energy principle is less useful in these types of collisions, the
impulse-momentum principle is still very useful.
Types of collisions
The experiments in Observational Experiment Table 6.7 are examples of the three
general collision categories: elastic collisions (experiment 1), inelastic collisions (experiment 2),
and totally inelastic collisions (experiment 3).
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Table 6.8 Types of collisions.
Elastic collisions
Both the momentum and kinetic
energy of the system are constant.
The internal energy of the system
does not change. The colliding
objects never stick together.
Elastic collisions are very rare in
nature, and never happen for
macroscopic collisions. Some
collisions (such as in
Observational Experiment Table
6.7 Experiment 1) are almost
elastic. Collisions between atoms
or sub-atomic particles are
extremely close to being perfectly
elastic.
Inelastic collisions
The momentum of the system is
constant but the kinetic energy is
not. The colliding objects do not
stick together. Internal energy
increases during the collisions.
Inelastic collisions are very
common in daily life: for
example, a volleyball bouncing
off your arms, or jumping on a
trampoline.
Totally inelastic collisions
These are inelastic collisions in
which the colliding objects stick
together. Typically a large
fraction of the kinetic energy of
the system is transformed into
internal energy in this type of
collision. Totally inelastic
collisions are common in daily
life; for example, catching a
football, or jumping into the back
of a moving pickup truck.
Measuring the speed of a fast moving projectile
In Chapter 5, we encountered a device that
measures the speed of fast projectiles, such as arrows
or bullets. Here we analyze a variation of such a
device, known as a ballistic pendulum. The huge
ballistic pendulum shown in Fig. 6.22 includes a
cannon that launches a massive ball into the opening
of a large metal block. The block with the cannon ball
swings up and stops at a height above its starting
position. That height depends on the speed of the ball
before it entered the block. By measuring the height
that the block with the ball swings up, the initial speed
of the ball can be determined.
Figure 6.22 Very large ballistic pendulum
Example 6.11 A ballistic pendulum A gun fires a 10-g bullet. Imagine that you place the gun a
few centimeters from a 1.0 kg wooden block hanging at the end of strings. You fire the gun, and
the bullet embeds in the block and swings upward to a height of 0.20 m . What was the speed of
the bullet when leaving the gun?
Sketch and Translate A sketch of the process is shown in Fig. 6.23a. We analyze the situation in
two parts. The first part is the collision of the bullet with the wood block, a totally inelastic
collision as the bullet combines with the block. The second part is the swinging of the block
upward to its maximum height. The system for analysis in the part I collision is the bullet and the
block. During the part I collision, the kinetic energy of the system is not constant, but the
momentum is. So, we use momentum constancy to analyze the collision. The initial state is the
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moment just before the collision begins, and the final state is the moment just after over the bullet
joins the block. In the initial state, only the bullet has momentum. In the final state, both the bullet
and the wood block have momentum, and their velocities are equal. A horizontal x-axis is used to
indicate the known information and the unknown quantity.
During part II, the block–bullet swings upward to its maximum height. We choose the
bullet, block, and Earth as the system. The initial state is just after the collision is over, and the
final state is when the block reaches its maximum height. In this case, the tension force exerted
by the string on the system does no work (it always points perpendicular to the direction of
motion of the block-bullet), and the energy of the system is constant. The initial kinetic energy of
the bullet and block is converted into the final gravitational potential energy of the system at the
end of the swing. The two parts are analyzed separately and then combined to determine the
bullet’s speed before hitting the block.
Figure 6.23(a) Analysis of ballistic pendulum
Simplify and Diagram Part I: The force that the block exerts on the bullet and the force that the
bullet exerts on the block are internal forces that do not change the momentum of the system. The
momentum bar chart in Fig. 6.23b represents part I. The motion occurs along the x-axis; the bar
chart is for the x-component of momentum. We use subscripts b for the bullet and B for the block
of wood; when they join together, the subscript is bB.
Part II: The string does no work on the block, as it is perpendicular to the block’s motion at every
instant. The energy bar chart in Fig. 6.24c represents part II of the process.
Figure 6.23(b)(c)
Represent Mathematically Apply momentum constancy to part I and energy constancy to part II.
For part I, the initial x-component of momentum is pb i x $ 0 # mvbi ; the final component is
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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pbB ix # & m $ M ' vbB i . For part II, the initial kinetic energy of the system transforms into the
system’s gravitational potential energy.
mvb i # & mb $ M B ' vbB i
Part I:
Part II:
1
& mb $ M B ' vbB2 i # & mb $ M B ' gybB f
2
where mb is the mass of the bullet, M B is the mass of the block, vb i is the initial speed of the
bullet, vbB i is the speed of the bullet + block immediately after the collision, and ybB f is the ycoordinate of the bullet + block at its highest point.
We wish to determine vb i . From the part I equation, we get:
vb i #
* M +
mb $ M B
vbB i # ,1 $ B - vbB i
mb
mb /
.
.
We don’t know vbB i . We can get this speed from the part II equation:
1
& m $ M ' vbB i 2 # & m $ M ' gybB f
2
( vbB i # 2 gybB f
Now we combine these two equations to eliminate vbB i and solve for vb i :
* M+
* M+
vb i # ,1 $ - vbB i # ,1 $ - 2 gybB f
m/
m/
.
.
.
Solve and Evaluate We can now insert the known information into the above to determine the
bullet’s speed as it left the gun:
*
1.0 kg +
vb i # ,1 $
- 2 & 9.8 N kg '& 0.20 m ' # 200 m s
. 0.01 kg /
That’s close to 450 mph —very fast but reasonable for a bullet`` fired from a gun.
Try It Yourself: Determine the initial kinetic energy in this example, the final gravitational
potential energy of the block-bullet system, and the increase in internal energy of the system.
Answers: K i # 200 J; U gf # 2 J; and )U int f # 198 J .
Review Question 6.8
Imagine that a collision occurs. Before the event, you measured the masses of the two objects and
the velocities of the objects both before and after the collision. Describe how you could use this
data to determine which type of collision had occurred.
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6.9 Power
Jose is rushing up the stairs to the 4th floor of a building to meet with his professor. On
the way up, he passes Nick, who is approximately the same mass. Nick is walking casually. When
Jose reaches the 4th floor, he is out of breath, so he rests for a minute. Nick finally reaches the 4th
floor but is not out of breath. From an energy point of view, this seems odd. If you analyze the
situation choosing Jose and Earth as the system, and Nick and Earth as another system, you
would determine that both systems gained the same amount of gravitational potential energy
reaching the 4th floor. Why was it more difficult for Jose to reach the 4th floor than for Nick?
To climb the stairs, both Jose and Nick converted the some amount of internal energy in
their bodies into gravitational potential energy. The amount of internal energy transformed into
gravitational energy might have been the same in both cases; but the rate at which that
transformation occurred was not. Jose converted the energy at a faster rate. The rate at which the
transformation occurs is called the power.
Power The power of a process is the amount of some type of energy transformation )U
during a process divided by the time interval 't for the process to occur:
Power # P #
)U
)t
(6.9)
If the process involves external forces doing work, then power can also be defined as the
magnitude of the work W done on the system divided by the time interval 't needed to do
the work:
Power # P #
W
)t
(6.10)
The SI unit of power is the watt (W), where 1 watt is 1 joule/second (1 W = 1 J/s).
One example of power is the output of a light bulb. For example, a 60-watt light bulb
converts electrical energy into light and thermal energy at a rate of 60 J each second. Another
example of power is a cyclist pedaling a bicycle at moderate speed. A cyclist in good shape will
transform about 400–500 joules of internal chemical energy each second ( 400-500 W ) into
kinetic, gravitational potential, and thermal energies.
Another power unit is the horsepower (hp): 1 hp = 746 W . It’s most often used to
describe the power rating of engines or other machines. A 0.5-hp gasoline engine transforms the
internal energy of the fuel into other forms of energy at a rate of 0.5 ; 746 W = 373 W , or
373 J/s .
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Example 6.12 Lifting weights Xueli is doing what is called a dead lift. She lifts a 30-pound
barbell ( 13.6 kg ) from the floor to the level of her waist (a vertical distance of 1.0 m ) in
0.80 s . Determine the power during the lift.
Sketch and Translate First, sketch the process (Fig. 6.24a). The system is the barbell and Earth.
The initial state is just before Xueli starts lifting. The final state is just after Xueli is finished
lifting. A vertical y-axis is used to indicate the values of the known quantities.
Figure 6.24(a) Power while doing a dead lift
Simplify and Diagram We assume that Xueli lifts the barbell so slowly that kinetic energy is zero
during the process. She does work on the barbell, causing the system’s gravitational potential
energy to increase (the barbell’s vertical y-coordinate position increases), but no change in kinetic
energy (it’s zero in the initial and final states). The origin is at the initial position of the barbell.
Since the barbell is moving at a small constant velocity, the external force exerted by Xueli on the
system is very nearly constant and equal to the gravitational force exerted by Earth on the barbell.
An energy bar chart for the process is shown in Fig. 6.24b.
Figure 6.25(b)
Represent Mathematically and Solve The power of this process is
P#
W
Fd cos "
mgd cos "
#
#
)t
)t
)t
Solve and Evaluate
P#
mgd cos " &13.6 kg '& 9.8 N kg '&1.0 m ' cos 00
#
# 170 W
)t
0.80 s
This is a reasonable power for a process – if you use an exercise machine that displays the power
output, compare what you can achieve with this number.
Try It Yourself: Xueli performs an overhead press—lifting the same barbell from her shoulders to
above her head. Determine the power involved in this process. The length of her arm is
approximately 40 cm . It takes her 1.0 s to lift the bar.
Answer: About 50 W .
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Example 6.13 Power and driving A 1400-kg car is traveling on a level road at a constant speed
of 27 m/s (60 mph). The drag force exerted by the air on the car, and the rolling friction force
exerted by the road on the car add to a net force of 680 N pointing opposite the direction of
motion of the car. (a) Determine the power due to the work done by this opposing force during
this process. (b) The car instead drives up a 4.00 incline at the same speed. Determine the power
due to the work of the opposing friction-like forces and the force exerted on the car by Earth
during this process. Express both results in watts and in horsepower.
Sketch and Translate First, sketch each process (Fig. 6.25a and b). Choose the car alone as the
system—we want to determine the work done by the Earth, road, and surrounding air on the car.
The initial state is the moment the car passes position x on an x-axis and the final state is a short
time interval )t later when the car has had a displacement )x parallel to the road.
Figure 6.25(a)(b) Power used by car to overcome resistive forces
Simplify and Diagram For part (a) two forces oppose the car’s motion and point in the negative x-
!
direction: air friction and rolling friction ( FAir + Rolling on C ). They are combined in one arrow in the
force diagram in Fig. 6.25c. We need to determine the magnitude of work per unit time done by
those forces. The gravitational force exerted by Earth on the car points perpendicular to the car’s
motion and so does no work.
For part (b) while the car is moving uphill, air friction and rolling resistance still exert
opposing forces now parallel to the inclined road. In addition Earth exerts a gravitational force on
the car that has a component ( FE on C x # mC g sin 40 ) that also opposes the motion along the
inclined road (see the force diagram in Fig. 6.25d). Thus, we will want to determine the
magnitude of the work done per unit time by these three forces.
Figure 6.26(c)(d)
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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Represent Mathematically To determine the power for each process, we will determine the work
done during displacement )x and divide by the time interval )t needed to complete that
displacement. The power is the magnitude of that ratio. For part (a),
P#
( FAir + Rolling on C ))x cos1800
)x
W
#
# %(680 N)
# (680 N)v .
)t
)t
)t
The magnitude of the power for the part (b) process when the car is going up the hill is:
P#
$ mg sin 40 ))x cos1800
(F
W
)x
# Air + Rolling on C
# %(680 N + 957 N)
# &1637 N ' v
)t
)t
)t
Solve and Evaluate Substitute in the above to determine the power in each case. For part (a):
P # (680 N)v # & 680 N '& 27 m s ' # 1.8 ; 104 W # 24 hp
That’s a relatively small power. Let’s see what changes when the car is climbing a hill.
The magnitude of the power for the part (b) process when the car is going up the hill is:
P # %(680 N + 957 N)
)x
# (%1637 N)v # &1637 N '& 27 m s ' # 4.4 ; 104 W # 59 hp .
)t
The power involved in this process is more than twice as large as when the car was moving on a
level road.
Try It Yourself: What is the average power of the process when the car is moving on the
horizontal surface, as in case (a), but accelerates from 20 to 27 m/s in 5 seconds?
Answer: P = 8.0 ; 104 W or 107 hp .
Review Question 6.9
Jim (mass 80 kg) moves on rollerblades on a smooth linoleum floor a distance of 4.0 m in 5.0 s.
Determine the power of this process.
6.10 Improving our model of gravitational potential energy
So far in our investigation of energy, we have made a major assumption involving
gravity: the gravitational force exerted by Earth on an object is constant. Specifically,
FE on O # mg
where m is the mass of the object and g # 9.8 N kg # 9.8 m s 2 is the gravitational constant
for Earth. We know from our study of gravitation at the end of Chapter 4 that this assumption is
only appropriate when the object is located near Earth’s surface, and Earth’s motion towards the
object can be ignored. Since we built our model of gravitational potential energy on this
assumption, it means whenever we use U g # mgy , we are making this assumption. It also means
that if we want to apply work-energy principles to objects not near the surface of Earth, we need
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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to improve on that model. We will do this by analyzing a large-scale version of the discussion
from the beginning of Section 6.3 that led to our original expression for gravitational potential
energy.
Imagine that a ‘space elevator’ has been built to transport supplies (typically of mass
1000 kg) from the surface of Earth up to the International Space Station (ISS). The elevator
moves at constant speed, except for the brief acceleration and deceleration periods at the
beginning and end of its motion, which last for a very small fraction of the total time interval the
elevator is in motion. How much work must be done to lift the supplies from the surface to the
ISS?
First sketch the process (Fig. 6.26a). The initial state is the moment just before the
supplies leave the surface. The final state is the moment just after they arrive at the ISS. We
choose Earth and the supplies as the system. The force that the elevator cable exerts on the
supplies is an external force that does positive work on the system.
Figure 6.26 A cable lifts supplies to space station
We keep track of gravitational potential energy only (it is the only type of energy that
changes between the initial and final states.) Since the supplies are moving at constant velocity,
the force exerted by the elevator cable on the supplies is equal in magnitude to the gravitational
force exerted by Earth on the supplies (Fig 6.26b). Consider the earth to be the object of
reference and the origin of the coordinate system at the center of the Earth. The process can be
described mathematically as follows:
Ug i $W # Ug f
( W # Ug f %Ug i
At this point it would be tempting to say that the work done by the elevator cable on the system is
W # Fd cos " , where F is the constant force that the cable exerts on the supplies. However, as
the supplies attain higher and higher altitudes, the force exerted by the elevator cable decreases—
the Earth exerts a weaker and weaker force on the supplies.
We mentioned earlier in the chapter that determining the work done by a variable force
requires the application of calculus. Specifically, we need to integrate Newton’s law of universal
gravitation along the path taken by the supplies as they are lifted from the surface of the earth up
to the ISS. The details of how to do that is beyond the level of this textbook, so we will just quote
the result:
*
M E mS + *
M E mS +
W # , %G
- % , %G
RE $ hISS / .
RE /
.
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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where M E is the mass of Earth, RE is the radius of Earth, mS is the mass of the supplies, and
hISS is the altitude of the ISS. Before you move on, check whether this complicated equation
makes sense to you (for example, check the units). The work is written as the difference in two
quantities – the latter is the quantity that describes the initial state and the former – the final state.
If we compare the above result with W # U g , f % U g ,i , we see that each component is an
expression for the gravitational potential energy of the Earth-object system for a particular state:
Gravitational potential energy of a system consisting of Earth and any object
U g # %G
M E mO
rE <O
(6.11)
where M E is the mass of Earth ( 5.98 ; 1024 kg ), mO is the mass of the object, rE <O is the
distance from the center of Earth to the center of the object, and G # 6.67 ; 10%11 N 1 m 2 kg 2 is
Newton’s universal gravitational constant.
The gravitational potential energy is a negative number. Note that the cable did positive
work on the system while pulling the supplies away from Earth. When the object is infinitely far
away, the gravitational potential energy is zero. The only way to add positive energy to a system
and have it become zero is if it started with negative energy; for example,
%5 $ 5 # 0; –100 $ 100 # 0 . We can now represent the process of pulling an object from the
surface of the Earth to infinity with a work–energy bar chart (Fig. 6.27). The initial state is when
the object is near the earth, the final state is when it is infinitely far away.
Figure 6.27 Bar chart representing work done to take an object from near Earth to infinity
Now we can determine the amount of work needed to raise the supplies to the ISS.
*
M E mS + *
M E mS +
W # , %G
- % , %G
RE $ hISS / .
RE /
.
*
1
1 +
# %GM E mS ,
%
. RE $ hISS RE /
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
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# % & 6.67 ; 10%11 N 1 m 2 kg 2 '& 5.98 ;1024 kg ' &1000 kg '
1
1
*
+
%
,
6
5
6
. 6.37 ; 10 m $ 3.50 ; 10 m 6.37 ; 10 m /
# 3.26 ;109 J
Let’s compare this with what we would have calculated had we used our original
expression for gravitational potential energy. Choose the zero-level at the surface of Earth.
W # ms gy f % ms gyi # ms gy f % 0
# &1000 kg '& 9.8 N kg ' & 3.50 ; 105 m ' # 3.43 ; 109 J
This differs by only about 5 percent from the more accurate result. We might expect the
gravitational force exerted on the station to be much weaker at the altitude of the ISS compared
with at the earth’s surface. But remember, situations where U g # mgy is reasonable are when the
distance above the surface of Earth is a small fraction of the radius of Earth. The altitude of the
ISS (350 km) is a small fraction of the radius of the Earth (6371 km); so U g # mgy is still
reasonably accurate.
Escape speed
You are used to thinking about the gravitational force that the Earth exerts on you when
you are standing or jumping on its surface. The best Olympic high jumpers can leap over bars that
are about 8 feet (2.5 m) above the earth’s surface. We can use energy principles to estimate the
jumper’s speed when leaving the ground in order to attain that height. To do this, we would
choose the jumper and Earth as the system, and the zero level of gravitational potential energy at
ground level. We model the jumper as a point-like object. This is a questionable assumption since
the jumper bends his legs to jump, and rotates in the air as he passes over the bar. So, this is an
estimate. The kinetic energy of the jumper transforms into the gravitational potential energy of
the system.
1 2
mv # mgy
2
( v # 2 gy # 2 & 9.8 N kg '& 2.5 m ' # 7.0 m s
What if the high jumper attempted the jump on the Moon? How high could he jump?
Using Newton’s law of Universal Gravitation (Chapter 4), we find that the gravitational constant
of objects near the Moon’s surface is g m # 1.6 N kg . Using the same energy ideas, we can
estimate how high the jumper could jump on the Moon.
1 2
mv # mg m y
2
& 7.0 m s ' # 15.3 m
v2
(y#
#
2 g m 2 &1.6 N kg '
2
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That’s almost 50 feet!
You might be wondering if it is possible to jump off a celestial body entirely. By
“entirely” we mean jumping up and never coming down! Another way to ask this question is:
what is the minimum speed you would need in order to leave the surface and never fall back
down? This speed is called escape speed.
Example 6.14 Escape speed What vertical speed must a jumper have in order to leave the
surface of a planet and never come back down?
Sketch and Translate First draw a sketch of the process (see Fig. 6.28a). The initial state will be
the instant after the jumper’s feet leave the surface. The final state will be when the jumper has
traveled far enough away from the planet to no longer feel the effects of its gravity; we have to go
to r # = for the gravitational force exerted on the jumper to be zero. Choose the system to be the
jumper and the planet.
Figure 6.28(a) Person tries to jump fast enough to escape a planet
Simplify and Diagram We keep track of gravitational potential energy and kinetic energy and
need to use our improved Eq. (6.16) expression for gravitational potential energy, since the
jumper will end far from the planet’s surface. An energy bar chart for the process is shown in Fig.
6.28b. In the initial state, the system has both kinetic and gravitational potential energy. In the
final state, the jumper will not have kinetic energy. (We are interested in the minimum speed the
jumper needs to never come down. If he had some leftover kinetic energy after getting far away
from the planet, then he could have left the planet’s surface traveling a little slower). The system
also will have zero gravitational potential energy (the jumper is very far from the planet at
r # = ). We assume that no other external forces affect the system.
Figure 6.28(b)
Represent Mathematically Using the generalized work–energy equation and the bar chart:
Ui $ W # U f
( Ki $ U g i $ 0 # K f $ U g f
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(
*
mm +
1
mJ v 2 $ , %G P J - $ 0 # 0 $ 0
2
rP /
.
where mP is the mass of the planet, mJ is the mass of the jumper, rP is the radius of the planet,
and G # 6.67 ;10%11 N 1 m 2 kg 2 is the gravitational constant.
Solve and Evaluate Solving for the speed of the jumper
v#
2GmP
rP
For the Moon, the escape speed is
v#
2 & 6.67 ;10%11 N 1 m 2 kg 2 '& 7.35 ; 1022 kg '
1.74 ; 106 m
# 2370 m s
No high jumper could ever achieve that!
Try It Yourself: What is the escape speed of a particle near the surface of the Sun? The mass of
the Sun is 2 x 1030 kg and its radius is 700,000 km.
Answer: 616 km/s.
In the above example, we derived an expression for escape speed—the speed that an
object needs in order to escape completely from an object of mass m and radius r:
v#
2Gm
r
(6.12)
If the object has less speed, it is still possible for it to escape. For example, if there were a road
built to the Moon, you could bicycle your way to the Moon. Escape speed is the speed of an
object that is not interacting with any other objects and is able to completely escape the planet or
star from its surface. For Earth this escape speed is
v#
2GmE
2 ; (6.67 Nm 2 / kg 2 )5.97 ; 1024 kg
#
# 11200 m/s # 11.2 km s
rE
6.40 ;106 km
Tip: Notice that the escape speed does not depend on the mass of the escaping object – a tiny
speck of dust and a huge boulder would need the same initial speed to leave Earth.
Why Earth’s atmosphere has oxygen but not hydrogen
The concept of escape speed has been important for the development of life. All of the
molecules that make up the atmosphere have a particular average kinetic energy (we will
investigate this statement when we study the physics of gases later on in the book). Because of
this, molecules with smaller masses will have larger average speeds. During Earth’s formation,
the less massive atoms and molecules, such as helium (He) and hydrogen (H) had higher average
speeds and therefore a greater likelihood of escaping Earth’s gravitational pull. More massive
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molecules, such as oxygen (O2), nitrogen (N2), carbon dioxide (CO2), and water vapor (H2O) had
a much lower likelihood of escaping. The presence of these heavier molecules is necessary for
animal and plant life. Imagine what would have happened if Earth were smaller (resulting in a
smaller escape velocity), had a lower density (also resulting in a smaller escape velocity), or was
closer to the Sun (resulting in higher temperatures, which we will learn, means higher average
kinetic energy for the gas molecules in the atmosphere). All of these would result in the oxygen,
nitrogen, carbon dioxide, and water vapor having a much greater chance of escaping into space.
On Earth, the conditions were just right for complex life to arise.
Dark Stars
Equation (6.13) for the escape speed suggests something fairly amazing. If the mass of a
planet was large enough and/or the radius of the planet was small enough, the escape speed could
be made arbitrarily large. What if the planet’s escape speed were greater than light speed
( c # 3.00 ; 108 m s )? What would this planet look like? Consider that light emanating from a
planet is the Sun’s light reflected by the planet’s surface. But, if the escape speed from this planet
was greater than light speed, then no light from its surface could ever reach your eyes. This planet
would be completely dark.
Let’s imagine what would happen if Earth started compressing and shrinking. How small
would Earth have to be for its escape speed to be greater than light speed? We use Eq. (6.13) with
v # c to answer this question.
v#c#
2Gm p
(6.13)
rp
or
( rp #
2Gm p
c2
#
2 & 6.67 ; 10%11 N 1 m 2 kg 2 '& 5.98 ; 1024 kg '
& 3.00 ;10
8
m s'
2
# 8.86 mm
Earth would have to be smaller than 9 millimeters! It’s difficult to imagine Earth compressed to
the size of a marble. Equation (6.14) was first constructed by the brilliant astronomer PierreSimon Laplace. He also coined the term ‘dark star’.
Quantitative Exercise 6.15 Sun as a dark star? How small would our Sun need to be in order
to become a dark star?
Represent Mathematically The mass of the Sun is 1.99 ; 1030 kg . All we need to do is use Eq.
(6.14) to find the radius of this dark star:
rSun #
2GmSun
.
c2
Solve and Evaluate
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rSun #
2 & 6.67 ;10%11 N 1 m 2 kg 2 '&1.99 ; 1030 kg '
& 3.00 ;10
8
m s'
2
# 2.95 ;103 m : 3 km
So, if the Sun collapsed to smaller than 3 km in radius, it would become a dark star. Could this
happen? We will return to this question in later chapters.
Try It Yourself: Estimate the size to which a human would need to shrink to become invisible in
the same sense that a dark star is invisible.
Answer: About 10-25 m
We’ve been talking about objects whose escape speed is larger than light speed.
Physicists once thought that since light has zero mass, the gravitational force exerted on it would
always be zero. Recall that in Chapter 4, we discussed that the law of Universal Gravitation
cannot account for the orbit of Mercury. At the beginning of the 20th century, Albert Einstein’s
theory of General Relativity improved greatly on Newton’s Law of Universal Gravitation,
providing a conceptual picture of gravitational interactions. According to General Relativity, light
is affected by gravity. Amazingly, the theory gives the same final prediction for the size of an
object from the surface of which no light can escape as is provided by the Newtonian theory. In
General Relativity, these objects are known as black holes, a term coined in 1967 by John
Wheeler, one of the most influential physicists of the 20th century in the field of General
Relativity.
Review Question 6.10
Why is the gravitational potential energy of two large bodies (for example, the Sun and Earth)
negative?
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Summary
Words
Pictorial and physical
representations
Work (W) Work is a way to change the
energy of a system when an external force
of magnitude F is exerted on an object in
the system as it undergoes a displacement
of magnitude d. The work depends on the
!
angle " between the directions of F and
!
d . It. is scalar quantity (6.1)
Kinetic energy (K) The energy of an
object of mass m moving at speed v. It is
a scalar quantity.
Gravitational potential energy ( U g ) The
energy that a system has due to the
gravitational interaction of its objects. It
is a scalar quantity. A single object
cannot have gravitational potential
energy.
Elastic potential energy ( U s ) The energy
of a stretched or compressed elastic object
(e.g., coils of a spring or stretched bow
string).
Internal energy ( U int ) The random
kinetic energy (thermal energy) and
chemical energy of a system of particles.
The internal energy of rubbing surfaces in
a system changes in proportion to the
friction force between the surfaces and
the displacement across each other.
Total energy (U) The sum of all the
energies of the system.
Generalized work-energy equation The
energy of a system changes due to work
of the external forces done on it. Internal
forces do not change the energy of the
system. Energy is a conserved quantity
similar to momentum. A bar chart
represents such work-energy processes.
Collisions There are three basic types of
collision:
! Elastic: momentum and kinetic energy
are conserved—no internal energy
produced.
! Inelastic: momentum conserved but not
kinetic energy—internal energy
produced.
! Totally inelastic: Same as above and
the colliding objects stick together.
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6
Mathematical representations
W # Fd cos "
K#
1
2
mv
2
U g # mgy
(near Earth’s
surface)
U g # %G
mA mB
rAB (general
expression)
Us #
1 2
kx
2
)U int # f k d (change
caused by friction)
U # K $ U g $ U s $ U int $ "
Ui $ W # U f
All collisions:
!
!
>pi # >p f
Elastic collisions only:
>K i # > K f
For other collisions,
)U int 2 0
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