Chapter 6: Work and Energy -...

Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 6 6-1

! Why it is impossible to build a perpetual motion machine?

! Why does blood pressure increase when the aorta walls thicken?

! 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.

Chapter 6: Work and Energy


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 Analysis

(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.

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.


(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

The external force You on ObjectF!

and the object’s displacement Objectd!

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.

An important pattern in all four experiments above is that an external force F!

was

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.


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.

! 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.

! 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.

! 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.

! 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 Analysis

(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).

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.


(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).


Work: The work done by an external force F!

on a system object while that system object

undergoes a displacement d!

is:

Work = cos W Fd "# ! "#$

where F is the magnitude of the force (always positive), d is the magnitude of the

displacement (always positive), and 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).

! If F!

and d!

are parallel, 00" # ,

0cos0 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).

! If F!

and d!

point in opposite directions, 0180" # ,

0cos180 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).

! If F!

and d!

point perpendicular to each other, 090" # ,

0cos90 0# , and 0W # , consistent

with what we observed in experiment (g) where the system energy did not change (the work


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

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 cosdF F "# times the magnitude of the

displacement d : dW F d# . Notice that if the force is perpendicular to the displacement, then

cos 0dF F "# #,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.


Quantitative Exercise 6.1 Pushing a bicycle uphill Two friends are cycling up a hill inclined at

08 —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

cosW Fd "# . The hill is inclined at 08 above the horizontal. Before looking at the “Solve” part,

decide what angle you would insert in this equation.

Solve and Evaluate Note that the angle between Strong on WeakF!

and d!

is 00 and not 80. Thus:

& '& ' 0

Strong on Weak Strong on Weak cos 50 N 100 m cos 0 5000 N•m = 5000 J.W F d "# # #

Try It Yourself: You pull a box 20 m up a 010 ramp. The rope is oriented

020 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: 0

R on B (100 N)(20 m)cos 20 1880 J.W # # Note that you use the angle between the

displacement (parallel to the ramp) and the force the rope exerts on the box (020 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 tail-

to-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

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 ( gU ): 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 ( sU ): 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 ( intU ): 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:

intTotal energy g sU K U U U# # $ $ $ (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 System and initial-final sketch Energy changes

(1) A person starts at rest at the top

of a smooth water slide and is

moving fast at the bottom.

gU K(

The system started with

considerable gravitational energy,

which converted to the person’s

kinetic energy as moved down the

slide.


(2) A fast moving car skids to a

stop on a level road.

internalK U(

The fast moving car with kinetic

energy skidded to a stop

converting its energy to internal

thermal energy due to friction.

(3) A pop-up toy starts with a

compressed spring and when

released pops up to a maximum

height above its starting position.

s gU U(

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.


Skill box: Constructing qualitative work-energy bar charts

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:

i fU W U$ #

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:

int( ) ( )i g i s i f g f s fK U U W K U U U$ $ $ # $ $ $ ) .

Earth

Initial state Final state y

0

µ = 0

Ki Ugi Usi W Kf Ugf Usf 'Uint

0

1. Sketch the initial-final states. 2. Choose a system.

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.


For the process in the skills box, some of these terms are zero:

s i f g fU K U# $

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 Prediction Outcome

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?

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.

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.


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


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


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): i fU W U$ # .

2. The work done by an external force F!

oriented at an angle " relative to the displacement

d!

of the system object: cosW Fd "# .

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


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

R on BT!

exerted by the rope on the box is equal in magnitude to the gravitational force E on BF!

exerted by Earth on the box (see the force diagram in Fig. 6.7b). Since the magnitude of the

gravitational force is Bm g , we find that the magnitude of the tension force for this process

is R on B BT m 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 iy and moving upward at a negligible very slow speed v . The final state is when the box

is at position fy 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:

i i f fK U W K U$ $ # $ .

Since the box is traveling at onstant speed, the kinetic energy of the system does not change.

Therefore i fK K# and we can subtract them from both sides of the equation. This leaves us with

i fU W U$ # .

ALG

6.3.1


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 :

& ' & '0 0

R on B R on B cos cos 0 cos0f i f iW T d T y y mg y y"# # % # %

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:

& ' 0

cos0g i f i g fU mg y y U$ % #

We can now write an expression for the change in the gravitational potential energy of the

system: g f g i f iU U mgy mgy% # % . 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:

gU mgy# (6.3)

where m is the mass of the object, 9.8 N kgg # , 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 on CYF!

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.


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:

H on Ci fK W K$ #

or

H on C f iW K K# % .

We know that the work in this case equals 0

H on C H on Ccos0F d F d# . Thus,

H on C f iF d K K# % .

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:

C C H on C m a F# .

Rearrange a kinematics equation from Chapter 1 (2 2

f i 2x x x xv v a d# $ ) to get an expression for the

displacement of the cart in terms of its initial and final speeds and its acceleration:

2 2

f i

C2

v vd

a

%# .

Now, insert these expressions for force and displacement into the left side of the equation

H on C f iF d K K# % :

2 2 2 22 2f i f i

H on C C C C C f C i

C

1 1( ) ( )

2 2 2 2 2

v v v vF d m a m m v m v

a

* +%# # % # %, -

. /

We can now set the result equal to the right part of the same equation:


2 2

C f C i

1 1

2 2f im v m v K K% # %

It appears that 2

C

1

2m v is an expression for the kinetic energy of the cart.

Kinetic energy: The kinetic energy of an object is:

21

2K mv#

(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.

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:

0g i fU K$ #

2

2

10

2

1

2

2( )

i f

i f

f i

mgy mv

gy v

v gy

( $ #

( #

( #

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:

22( ) 2(9.8 m/s )(20 m) 20 m/sf iv gy# # #

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 iv , what expression can we use to

determine its maximum height above its launching position before it starts descending?

Answer: 2 / 2iv 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:

& ' & '

2

2

0

1cos

2

1cos 0

2

2( )

f

f

i f

f i

W K

Fd mv

mg y mv

v gy

"

$ #

( #

( 0 #

( #


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 cosW Fd "# 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.


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 Force F exerted by the

scale on the spring

Spring 2 distance x

stretched Force F exerted by the

scale on the spring

0.000 m 0.0 N 0.000 m 0.0 N

0.050 m 1.0 N 0.030 m 1.0 N

0.100 m 2.0 N 0.060 m 2.0 N

0.150 m 3.0 N 0.090 m 3.0 N

0.200 m 4.0 N 0.120 m 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


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:

S on O –xF 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 O on SF 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.


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 ( 0x # ) 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 H on S average( )F :

H on S average

0( )

2

kxF

$# .

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:

2

H on S average

1 1( ) ( )

2 2W F x kx x kx# # # .

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:

21

2sU kx# (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 ( S on O xF 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


spring constant = 900 N/mk . 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:

i fU W U$ #

0i s i f s fK U K U( $ $ # $

2 21 10 0

2 2kx mv( $ # $

2 21 1

2 2kx mv( #


If we multiply both sides of the equation by 2, divide by m , and take the square root, we have

kv x

m# .

Solve and Evaluate

& '900 N m0.40 m 40 m s

0.090 kg

kv x

m# # #

.

This is reasonable for the speed of an arrow fired from a bow. Let’s also check the units of k

xm

to make sure they really are equivalent to m

s:

2

N kg m m(m) = (m) = .

m kg s m kg s

�

� � �

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 2kx x kx1 # ?

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

on the car, which slows the car to a stop. Figure 6.14(a)


The friction force does work on the car:

0

friction cos180 –k kW f d f d# # .

The negative work done by the friction force causes the car’s kinetic energy to decrease to zero

(Fig 6.14b). Mathematically,

i (– ) 0kK f d$ #

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 ( internal 0U) 2 ). 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 internal0 U# ) --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 internal0 U# ) where internalU) is greater than zero. How can we resolve this?

ALG

6.3.3


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 R on BT


must equal the magnitude of the friction force that carpet exerts on the box C on Bkf . 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:

0

R on B cos 0W T d#

Substituting R on B C on BkT f# and 0cos 0 1.0# into the above, we get:

C on BkW f d# $

The only system energy change is its internal energy internalU) . The work-energy

equation for pulling the box across the surface process is:

internalW U# ) .

After inserting the expression for the work done on the box and rearranging, we get:

internal kU f 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

internal kU f d) # $ , (6.7)

where kf 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 internal kU f 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

friction – kW f 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.


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:

int

2

R on C

2

R on C

0

1

2

1( )

2

i f

i k

i k

K U

mv f d

mv N d3

$ #

( #

( #

The magnitude of the upward normal force R on CN that the road exerts on the car equals the

magnitude of the downward gravitational force E on C CF m g# that Earth exerts on the car. Thus

the above becomes:


2

C C

1

2i km v m gd3#

Solve and Evaluate Rearranging the above and canceling the car mass, we get;

& '& '& ' 1 mph2 2 0.70 9.8 N kg 24 m (18.1 m/s) 40 mph

0.45m si kv gd3

* +# # # #, -

. /

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

energy is represented mathematically as an equation we call the generalized work-energy

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:

i fU W U$ # (6.8)

where intg sU K U U U# $ $ $ $"

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.

ALG

6.3.4-

6.3.5


Testing Experiment Table 6.6 Testing the generalized work–energy equation.

Testing experiment Prediction based on the generalized

work-energy equation

Outcome

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?

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.

gi f= U K

2

i f

1

2mgy mv# .

For each experiment f

v is the same if

iy h# is the same.

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.

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.

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 ( C on ElT )

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.

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.

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.

! 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.

ALG

6.4.1-

6.4.13


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.

i fU W U$ #

2 0

C on El

1( ) ( – 0) cos 0 2

i f fmv T y mgy$ #

2

C on El

1( ) 2

i f f

mv T y mgy$ #

Solve and Evaluate

! Use the mathematical description of

the process to solve for the desired

unknown quantity.

! Evaluate the result. Does it have the

correct units? Is its magnitude

reasonable? Do the limiting cases

make sense?

2

C on El –2 f

mvT mg

y#

2(1000 kg)(4.0 m/s)

(1000 kg)(9.8 N/kg) – = 8500 N2(6.0 m)

# .

! The result has the correct units. The force that the cable exerts is

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


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:

i fU W U$ #

, ,0s i f g fU K U( $ # $

2 21 1

2 2i f fkx mv mgy( # $

Solve and Evaluate Dividing both parts of the equation by (1/2)x2 we get:


2 0

2

12( sin 37 )

2f i

i

mv mgx

kx

$#

2 o

2

12 (60 kg)(15 m/s) (60 kg)(9.8 N/kg)(3.0 m)sin 37

21740 N m 1700 N m

(3.0 m)

4 5$6 78 9# # :.

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

in Alaska, a paratrooper survived a 370-m fall into a snow bank when his parachute failed to

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

AP

5.5-5.7


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 work-

energy equation:

2 0

S on P

1cos180 0

2i imv mgy F d$ $ #

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:

2

i

S on P

1

2mv mgd

Fd

$#

2

5

1(90 kg)(54 m/s) (90 kg)(9.8 N/kg)(1.07 m)

2 1.2 x 10 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 380 cm 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


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 380-cm 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


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:


g i s fU U#

21

2imgy kx( #

Solve and Evaluate Solving the above for the spring constant gives:

2

2 imgyk

x#

The length of the cord is = 40 mL and the cord stretches 35 mix # . Thus, the distance of the

person’s initial position iy above the final 0fy # position is:

40 m 35 m 75 mi fy L x# $ # $ # .

This means that the spring constant has a value:

& '& '& '

& '22

2 70 kg 9.8 N kg 75 m284 N m

35 m

imgyk

x# # #

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


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 Analysis

(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:

1 1

1

2 2

2

1.0 kg; 10 m/s

–6.0 m/s

4.0 kg; 0 m/s

4.0 m/s

ix

fx

ix

fx

m v

v

m v

v

# #

#

# #

#

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

(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


that hit the cart is flattened in the

collision.

1 1

1

2 2

2

1.0 kg; =10 m/s

= - 4.2 m/s

4.0 kg; 0 m/s

3.55 m/s

ix

fx

ix

fx

m v

v

m v

v

#

# #

#

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

(3) In this experiment the 1.0-kg sand-

filled 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.

1 1

1

2 2

2

1.0 kg; 10 m/s

2.0 m/s

4.0 kg; 0 m/s

2.0 m/s

ix

fx

ix

fx

m v

v

m v

v

# #

#

# #

#

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

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 impulse-

momentum. 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).


Table 6.8 Types of collisions.

Elastic collisions Inelastic collisions Totally inelastic 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.

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.

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


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 b i b 0x ip mv$ # ; the final component is


& 'bB bB ix ip m M v# $ . For part II, the initial kinetic energy of the system transforms into the

system’s gravitational potential energy.

Part I: & 'b b B bB i imv m M v# $

Part II: & ' & '2

b B bB b B bB

1

2i fm M v m M gy$ # $

where bm is the mass of the bullet, BM is the mass of the block, b iv is the initial speed of the

bullet, bB iv is the speed of the bullet + block immediately after the collision, and bB fy is the y-

coordinate of the bullet + block at its highest point.

We wish to determine b iv . From the part I equation, we get:

b B Bb bB bB

b b

1i i i

m M Mv v v

m m

* +$# # $, -

. / .

We don’t know bB iv . We can get this speed from the part II equation:

& ' & '2

bB bB

bB bB

1

2

2

i f

i f

m M v m M gy

v gy

$ # $

( #

Now we combine these two equations to eliminate bB iv and solve for b iv :

b bB bB 1 1 2i i f

M Mv v gy

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:

& '& 'b

1.0 kg1 2 9.8 N kg 0.20 m 200 m s

0.01 kgiv

* +# $ #, -. /

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: int200 J; 2 J; and 198 Ji gf fK U U# # ) # .

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.


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:

PowerU

Pt

)# #

) (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:

PowerW

Pt

# #)

(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 .


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

cos cosW Fd mgd

Pt t t

" "# # #) ) )

Solve and Evaluate

& '& '& '13.6 kg 9.8 N kg 1.0 m cos 0cos170 W

0.80 s

mgdP

t

" 0# # #

)

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 .


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 ( Air + Rolling on CF!

). 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 (0

E on C C sin 4xF m g# ) 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)


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),

Air + Rolling on C( ) cos180

(680 N) (680 N)F xW x

P vt t t

) 0 )# # # % #) ) )

.

The magnitude of the power for the part (b) process when the car is going up the hill is:

& '0

Air + Rolling on C( sin 4 ) cos180(680 N + 957 N) 1637 N

F mg xW xP v

t t t

$ ) 0 )# # # % #) ) )

Solve and Evaluate Substitute in the above to determine the power in each case. For part (a):

& '& ' 4(680 N) 680 N 27 m s 1.8 10 W 24 hpP v# # # ; #

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:

& '& ' 4(680 N + 957 N) ( 1637 N) 1637 N 27 m s 4.4 10 W 59 hpx

P vt

)# % # % # # ; #

).

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: 4 = 8.0 10 W or 107 hpP ; .

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,

E on OF mg#

where m is the mass of the object and 29.8 N kg 9.8 m sg # # 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 gU 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


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:

g i g f

g f g i

U W U

W U U

$ #

( # %

At this point it would be tempting to say that the work done by the elevator cable on the system is

cosW Fd "# , 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:

E S E S

E ISS E

M m M mW G G

R h R

* + * +# % % %, - , -$ . /. /


where EM is the mass of Earth, ER is the radius of Earth, Sm is the mass of the supplies, and

ISSh 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 , ,g f g iW U U# % , 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

E O

E O

g

M mU G

r <

# % (6.11)

where EM is the mass of Earth (245.98 10 kg; ), Om is the mass of the object, E Or < is the

distance from the center of Earth to the center of the object, and 11 2 26.67 10 N m kgG %# ; 1 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.

E S E S

E ISS E

E S

E ISS E

1 1

M m M mW G G

R h R

GM mR h R

* + * +# % % %, - , -$ . /. /

* +# % %, -$. /


& '& '& '11 2 2 24

6 5 6

6.67 10 N m kg 5.98 10 kg 1000 kg

1 1

6.37 10 m 3.50 10 m 6.37 10 m

%# % ; 1 ;

* +%, -; $ ; ;. /

93.26 10 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.

0s f s i s fW m gy m gy m gy# % # %

& '& '& '5 91000 kg 9.8 N kg 3.50 10 m 3.43 10 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 gU 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 gU 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.

21

2mv mgy#

& '& '2 2 9.8 N kg 2.5 m 7.0 m sv gy( # # #

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 1.6 N kgmg # . Using the same energy ideas, we can

estimate how high the jumper could jump on the Moon.

21

2mmv mg y#

& '& '

22 7.0 m s15.3 m

2 2 1.6 N kgm

vy

g( # # #


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:

i fU W U$ #

0i g i f g fK U K U( $ $ # $


2 P JJ

P

10 0 0

2

m mm v G

r

* +( $ % $ # $, -

. /

where Pm is the mass of the planet, Jm is the mass of the jumper, Pr is the radius of the planet,

and 11 2 26.67 10 N m kgG %# ; 1 is the gravitational constant.

Solve and Evaluate Solving for the speed of the jumper

P

P

2Gmv

r#

For the Moon, the escape speed is

& '& '11 2 2 22

6

2 6.67 10 N m kg 7.35 10 kg2370 m s

1.74 10 mv

%; 1 ;# #

;

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:

2Gmv

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

2 2 24

E

6

E

2 2 (6.67 Nm / kg )5.97 10 kg11200 m/s 11.2km s

6.40 10 km

Gmv

r

; ;# # # #

;

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


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

(83.00 10 m sc # ; )? 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.

2 p

p

Gmv c

r# # (6.13)

or

& '& '& '

11 2 2 24

22 8

2 6.67 10 N m kg 5.98 10 kg28.86 mm

3.00 10 m s

p

p

Gmr

c

%; 1 ;( # # #

;

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 Pierre-

Simon 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 301.99 10 kg; . All we need to do is use Eq.

(6.14) to find the radius of this dark star:

SunSun 2

2 .

Gmr

c#

Solve and Evaluate


& '& '& '

11 2 2 30

3

Sun 28

2 6.67 10 N m kg 1.99 10 kg 2.95 10 m 3 km

3.00 10 m sr

%; 1 ;# # ; :

;

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?


Summary

Words Pictorial and physical

representations

Mathematical 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)

cosW Fd "#

Kinetic energy (K) The energy of an

object of mass m moving at speed v. It is

a scalar quantity.

21

2K mv#

Gravitational potential energy ( gU ) 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.

gU mgy# (near Earth’s

surface)

A Bg

AB

m mU G

r# %

(general

expression)

Elastic potential energy ( sU ) The energy

of a stretched or compressed elastic object

(e.g., coils of a spring or stretched bow

string).

21

2sU kx#

Internal energy ( intU ) 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.

int kU f d) # (change

caused by friction)

Total energy (U) The sum of all the

energies of the system.

intg sU K U U U# $ $ $ $"

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.

i fU W U$ #

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.

All collisions:

i fp p> # >! !

Elastic collisions only:

i fK K> # >

For other collisions,

int 0U) 2

Chapter 6: Work and Energy -...

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