PUM Module
Work and Energy
Adapted from The Active Physics Learning guide
by A. Van Heuvelen and E. Etkina
In this chapter, students develop the notion of work and energy by observing different
ways that an object can gain the ability to smash a piece of chalk (Problems 3-7). The
students then learn to represent the transformation of energies in the work-energy
process in multiple ways, including verbally, in a work-energy bar chart and in
mathematical equations (Problems 8-13). By representing an equality in different ways,
the student will firm up their concept of equality and the ways that it is represented
(Practiced especially in Problems 1-2). The students derive and use mathematical
expressions for calculating gravitational potential energy and kinetic energy (Problems
14-22). Next the students derive and use expressions to calculate elastic potential energy
and universal gravitational potential energy (Problems 23-33). Finally, the students use
all they have learned in a series of problems (Problems 34-48).
# and Goal of Lesson
1. Using a bar chart to
represent transformations of a
conserved quantity
2. Recognizing work and
mechanical energy, practice
recognizing patterns
3. Introduce internal energy
and review material
4. Represent processes with
work-energy bar chart
5. Physical significance of
positive/negative gravitational
potential energy
6. Represent gravitational
potential energy mathematically
7. Represent kinetic energy
mathematically
8. Represent internal energy
mathematically
9. Work=area under force vs.
displacement graph
10. Lab: Use potential energy of
spring to measure its spring
constant
11. Understand physical
significance of negative
Class
work
1
Homework Assessment
2
3, 4, 5
6, 7
8, 9, 10
11, 12
13, 14
15, 16
17
18
19
20
21
22, 23
25
26, 27
28, 29, 30
31, 32
33, 34
35
1
24
NJCCS
universal gravitation and use
mathematical representation
12. Problem Solving
13. Problem Solving
14. Problem Solving
15. Lab: Applying knowledge of
work-energy principle to
measure spring constant
36, 37, 38
41, 42
44, 45
48,
39, 40
43
46, 47
49, 50
Physics Conceptual Goals
--Learn to identify a system and the initial and final states of a physical process.
--Understand the concept of work and how work is related to concepts of dynamics.
--Learn different kinds of energy and how to describe them mathematically.
--Understand energy transformation processes in a system and system energy changes
caused by external interactions (work).
Physics Procedural Goals
--Be able to describe mechanical processes with words, pictures, energy bar charts, and
mathematical representations.
--Be able to apply knowledge about work and energy to real-life situations.
Scientific Abilities Goals
Is able to extract the information from representation correctly
Is able to construct new representations from previous representations
Mathematics Conceptual Goals we need to add to these using what we actually do,
may be Eva could help here
--To represent an equality in words, graphs, and mathematical expressions.
--To manipulate and solve algebraic expressions.
--To model and solve contextual problems using various representations, such as graphs,
tables, and equations
Mathematics Procedural Goals
--Problem solving: solve problems that arise in mathematics and in other contexts.
--Communication: analyze and evaluate the mathematical thinking and strategies of
others.
--Representations: relate and use representations to organize, record, and communicate
mathematical ideas
--Representations: use representations to model and interpret physical, social, and
mathematical phenomena.
2
Lesson 1.: Bar Chart Practice
In problems 1 & 2, we want students to practice representing an equality graphically and
mathematically.
1. Represent: Use a bar chart to represent transformations of a conserved quantity
In this exercise, P represents the amount of money in your pocket. Catm represents the amount of money in
your ATM card, and Cgift represents the amount of money on a rechargeable Best Buy gift card.
The total amount of money you have remains the same before and after unless you earn some or spend
some -- right? Fill in the bar charts. Show that the two sides balance by writing a mathematical equation
below the bar chart using the given values.
Situation before and after
P = money in your pocket
Catm = money in your ATM card
Cgift = money in your gift card
a) You have no money in your pocket, $60 in your
before
after
earn
ATM account, and a gift card with $20 on it.
or
Pi+CATM,i+CGift,i+spend =Pf +CATM,f+CGift,f
You withdraw $20 cash from the ATM.
+$60
Represent this transaction graphically on the bar chart
and mathematically in the space below. Did you earn or
spend any money?
0
$60 + $20 = $20 + $40 + $20
-$60
$80 = $80
b) Next, you buy a snow shovel for $10 cash at Jones
Hardware,.
Represent this transaction graphically on the bar chart
and mathematically in the space below. Did you earn
or spend any money?
$20 + $40 + $20 - $10 = $10 +
$40 + $20
before
earn
or
Pi+CATM,i+CGift,i+spend =Pf
+$60
after
+CATM,f+CGift,f
0
$70 = $70
-$60
c) After returning from the hardware store, you spend
three hours shoveling snow for an old lady who gives
you $20 cash.
Represent this transaction graphically on the bar chart
and mathematically in the space below.
$10 + $40 + $20 + $20 = $30 + $40
+ $20
before
earn
or
Pi+CATM,i+CGift,i+spend =Pf
+$60
0
$90 = $90
-$60
3
after
+CATM,f+CGift,f
Situation before and after
P = money in your pocket
Catm = money in your ATM card
Cgift = money in your gift card
d) When you are finished shoveling, you spend $20
cash to put gas in your car so you can drive to the Best
Buy.
Represent this transaction graphically on the bar chart
and mathematically in the space below.
before
+$60
earn
or
after
Pi+CATM,i+CGift,i+spend=Pf +CATM,f+CGift,f
0
$30 + $40 + $20 - $20 = $10 + $40 +
$20
-$60
$70 = $70
e) At Best Buy, you purchase a "Cher's Greatest Hits"
DVD Box Set for $40. You empty out your gift card
and use your ATM card to pay for the rest.
Represent this transaction graphically on the bar chart
and mathematically in the space below.
before
earn
or
Pi+CATM,i+CGift,i+spend=Pf
+$60
after
+CATM,f+CGift,f
0
$10 + $40 + $20 - $40 = $10 + $20
$30 = $30
-$60
f) What happens next? What do you do with the Cher
DVDs? Continue the story and make a graph to match.
Below, represent the transaction mathematically.
For example, while I am at the store, I
take my $10 cash and put it onto my
Best Buy gift card.
$10 + $20 = $20 + $10
$30 = $30
before
+$60
earn
or
after
Pi+CATM,i+CGift,i+spend=Pf +CATM,f+CGift,f
0
-$60
g) What happens next? Continue the story and make a
graph to match.
For example, back at home, I shovel
more snow for the old lady and earn
$50 cash.
before
earn
or
Pi+CATM,i+CGift,i+spend=Pf
+$60
0
$20 + $10 + $50 = $50 + $20 + $10
-$60
4
after
+CATM,f+CGift,f
Situation before and after
P = money in your pocket
Catm = money in your ATM card
Cgift = money in your gift card
h) Does this graph show something that could
happen? If not, explain why not. If so, describe a
situation it could match.
No, the cash in my pocket cannot just
disappear. Even if it is lost, the loss of money
should be recorded in the shaded region of
the graph.
before
earn
or
Pi+CATM,i+CGift,i+spend =Pf
+$60
after
+CATM,f+CGift,f
0
-$60
$50 + $20 + $10 ≠ $20 + $20 + $10
$80 ≠ $50
i)
How can you tell if one of these graphs represents a situation that could really happen?
Explain.
The total amount of money on both sides of the vertical line should be equal.
5
Lesson 1-Homework:
Represent: Use a bar chart to represent transformations of a conserved quantity
2. This time the exercise concerns the food in your house, and we are going to make bar charts with
estimated quantities instead of exact numbers. Make sure the total mass of food remains the same unless
you buy some or throw some out. Fill in the bar charts. Show that the two sides balance by writing a
mathematical equation below the bar chart using values from the chart.
Situation before and after
FP = food in your pantry
FR = food in the refrigerator
FS = food on the stove
ΔUate = food "U" ate.
a) You have some food in your pantry and
before
after
buy or
refrigerator already.
FP,i +FR,i + FS,i+ discard =FP,f +FR,f +FS,f +∆Uate
You go shopping and come home with bags of
groceries. You put some away in the pantry and
+
some away in the refrigerator.
Students can pick their own units for
measuring the food.
2.5 lbs + 5 lbs + 5 lbs = 5 lbs + 7.5 lbs
12.5 lbs = 12.5 lbs
b) The situation starts from where we left off in the
previous question.
You take some ingredients from the pantry and
some from the refrigerator and make a meal on the
stove.
5 lbs + 7.5 lbs = 3.75 lbs + 6.25 lbs + 2.5
lbs
12.5 lbs = 12.5 lbs.
0
-
before
buy or
after
FP,i +FR,i + FS,i+ discard =FP,f +FR,f +FS,f +∆Uate
+
0
-
c) Next, you eat half of the food you cooked, and
store the rest in the refrigerator.
before
FP,i +FR,i +
3.75 lbs +6.25 lbs +2.5 lbs = 5 lbs +6.25
lbs + 1.25 lbs
+
12.5 lbs = 12.5 lbs
0
-
6
after
buy or
FS,i+ discard =FP,f +FR,f +FS,f +∆Uate
d) The next day you come home with a takeout
from White Castle. You eat the sack of White
Castles and throw out your leftovers from the day
before.
before
FP,i +FR,i +
after
buy or
FS,i+ discard =FP,f +FR,f +FS,f +∆Uate
+
5 lbs + 6.25 lbs – 1.25 lbs + 1.25 lbs =
3.75 lbs + 6.25 lbs + 1.25 lbs
0
11.25 lbs = 11.25 lbs
-
e) Why do we show ΔUate on the "after" side but not on the "before" side?
The food that is represented by ΔUate is stuck in my stomach! On the “before” side of the
table, we only track food that can be changed into other types of food. Therefore, we
don’t need to keep track of food in my stomach.
7
Lesson 2.
Observational Skills: Recognizing Work and Mechanical Energy
System and Process: Suppose you had a picture of some part of the world at one
particular time. It is usually too complicated to keep track of everything happening all of
the time. Usually, we are interested in only what is happening in one small part of that
world for some short time interval. We will circle that part of that world that is of interest
and call it the system—everything inside this line. Everything outside the system is called
its environment. As we gain experience, we will find that making an appropriate choice
of system will simplify our analysis. As life progresses, the system changes. We call a
process the change in the system from some initial time (from its initial state) to some
final time (the system’s final state). There can be interactions between objects in the
system and between one or more objects in the environment with one or more objects in
the system. Forces that objects inside the system exert on each other are called internal
forces; forces that objects outside the system exert on the objects inside the system, are
called external forces.
Students should recognize the two similarities amongst these different experiments. In
each, an object has an ability to smash chalk. You can change this ability by exerting a
force on the object over a given distance.
3. Observe and find a pattern We describe three experiments involving a well-defined
system and a process in which the system changes from an initial state to a final state. At
the end of this process, we find that the system has the potential to do something it
couldn’t do before—to smash a piece of chalk into many pieces. Draw arrows indicating
the direction of the external force that you (outside the system) exert on a system object
and the displacement of the object while you exert the force. Fill in the table that follows
to help you complete this activity.
(a) The system includes a 1-kg block with a flat bottom and a string attached to the top,
the Earth, and a piece of chalk. You (outside the system) pull up on the string so that the
1-kg block slowly rises 0.5 m above the piece of chalk. After this lifting process, you
release the block. It falls and breaks the chalk.
1 kg
Lift
Lift
1 kg
chalk
(b) The system includes a 1-kg dynamics cart that can roll on a low-friction horizontal
dynamics track and a piece of chalk that is taped to the fixed vertical end of the track.
You (outside the system) push the cart so that it rolls faster and faster toward the chalk at
the end of the dynamics track and breaks the chalk when it hits it.
8
wall
Push
wall
(c) The system includes a slingshot that holds a piece of chalk. You (outside the system)
slowly pull back on the sling. When you release the sling, the chalk shoots out at high
speed and hits the wall causing the chalk to break.
Pull
wall
wall
Experiment
Draw arrows indicating the
direction of the force you
exerted on an object in the
system ( FY on S).
(a)
(b)
(c)
Draw an arrow indicating the
displacement of the system
object while you were exerting
the force ( d ).
(d) Look for a pattern of what was done to the system to give it the chalk-smashing
potential. Then, devise a new physical quantity to describe this pattern. Be explicit.

In each case, a force exerted by on an object in the same direction as the object is moving
causing the chalk-smashing potential of the system to increase. I will call this new
physical quantity “chalk-smashing booster.”
9
4. Observe and find a pattern In the chalk smashing activities, you found that the
external force you exerted on an object in a system over a certain distance gave the
system the potential to smash a piece of chalk. The force you exerted on the object in the
system was always in the direction of the displacement of that object. Suppose that a
friend, outside the system, decides to save the chalk in the first two experiments by
exerting with her hand an opposing force on the block or on the cart after they are
released. In each case she pushes on the moving object opposite to the direction of its
velocity. Below, give the direction of the force your friend exerts on the moving object
relative to its displacement as she stops it, thus causing the system to lose its potential to
break the chalk.
(a) The system includes a 1-kg block with a flat bottom and a
string attached to the top, the Earth, and a piece of chalk. You
(outside the system) pull up on the string so that the 1-kg block
slowly rises 0.5 m above the piece of chalk. After this lifting
process, you release the block and it starts falling. Your friend
then starts pushing upward on the falling block, slowing it
down, and the block does not break the chalk.

(b) The system includes a 1-kg dynamics cart that can roll
on a low-friction horizontal dynamics track and a piece of
chalk that is taped to the fixed vertical end of the track.
You (outside the system) push the cart so that it rolls faster
and faster. You then stop pushing. Just before the cart
reaches the chalk, your friend pushes it in a direction
opposite to its direction of motion. This causes the cart to
slow down and stop so that it does not break the chalk.
d
Ff on b

Ff on b

d
(c) How could you modify the definition of the quantity you devised
in the previous
problem to account for the system’s loss of the chalk-breaking potential thanks to your
friend’s intervention?
I should change the name to “chalk-smashing potential changer.” By exerting a force on
the object in the same direction as the object is moving, I increase the potential for the
object to smash chalk. If I exert a force on the object in the opposite direction of the
moving object, then I decrease the potential for the object to smash chalk.
5. Observe and find a pattern Consider a
system that includes the Earth and a 1-kg block.
(a) You (outside the system) hold a string tied to
the block so that it stays about 1 cm above a
table. A piece of chalk is placed on the table
under the block. If you release the block and it
falls on the chalk, the chalk does not break (it’s
too close to the chalk).
10
You are holding
the string
Motion
Next you slowly walk about 2 m beside the table, continually keeping the block 1 cm
above the surface. After walking the 2 m, the block hangs over a second identical piece of
chalk. Draw the force exerted by the string on the block and the displacement of the
block as you walked the 2 m.
FY on b
d
(b) Discuss whether the vertical force the string exerted on the block while moving it
 caused the system to have a better chance of breaking the
horizontally above the tabletop

second piece of chalk than the first piece. Revise the quantity you devised in the last two
activities to account for this result. Your revision will involve the angle between the
external force exerted on the system and the system object’s displacement. We call this
quantity work.
The ability for the block to smash the chalk does not change during this process. If the
force exerted on the object and the displacement are at right angles to each other, then
the chalk-smashing ability does not change.
11
Lesson 3.
This problem sets up the introduction of another form of energy, internal energy, by
having students observe that work done on a system can cause the system to warm up.
6. Observe and find a pattern A system consists of a crate and a rough horizontal
surface on which it sits (see the illustration below). The surface is made of a special
material that changes color when it changes temperature (these materials are called
thermoplastics; they are made of regular plastic with added thermo chromatic materials).
You, outside the system, pull on a rope attached to the crate so that it moves slowly at
constant velocity. You do positive work by pulling the crate for about 10 m. Describe
how the system (block and surface) is different after you do the work than before the crate
started moving.
The crate and surface are warmer after the crate started moving.
7. Describe You do work on a system to change its potential to do something—for
example, to smash chalk or to make touching surfaces of objects in a system warm. In
Activities 3 through 6, the work done on the system by the external force caused different
types of changes in the system. Below, we describe each type of change in the system as a
result of the work done on it. Devise a name for that type of change.
(a) The external force caused the block to move higher above the Earth’s surface.
Falling potential
(b) The external force caused the cart to move faster and faster.
Speed potential
(c) The external force caused the slingshot to stretch.
Stretching potential
(d) The external force caused the surfaces of the touching objects to warm.
warming potential
12
Work
System
FW on S
System
FW on S
FW on S
ds
ds



A woman catches
a ball thrown at her
Since the force exerted by the woman on the
box is in the opposite direction of the
displacement,
W = - Fw on b*d
A woman 
pulls a box
 upwards
Since the force exerted by the woman on the
box is in the same direction as the
displacement,
W = Fw on b*d
System
FW on S
dFs W on S



A woman carries a box while walking at a
constant pace
Since the force exerted by the woman on the
box is perpendicular to the displacement,
W=0
Tips
--Make sure to always identify a system for energy processes. It is especially important
gravitational potential energy. If the Earth is a part of the system and interactions between
the Earth and objects in the system are internal, then the system possesses gravitational
potential energy . Since we consider the force of the Earth on objects to be part of the
system, the earth does not do work on the system. If the Earth is not part of the system,
then the force that it exerts on system objects does work but the system does not possess
gravitational potential energy.
--When there are two surfaces rubbing against each other, and both surfaces are in the
system, then the internal energy of the system increases. There is no work done by friction
because the surfaces are both inside the system. Alternatively, you could exclude the
supporting surface and include the work done by friction that surface does on an object
moving across the surface.
13
Lesson 3-Homework
8. Pose a problem Describe a real-life situation in which an external force does the
following:
(a) Positive work on a system;
Two women push a broken car off the road. The car and road are in the system
(b) Positive work on a system but with a value that is less than in part a;
A UPS delivery man lifts a 20 lb box 3 feet above the ground.
(c) Negative work on a system;
An outfielder in a baseball game catches a flyball. The baseball and the earth are in the
system.
(d) Zero work even though an object in the system moves.
A man carries a bowling ball across a room.
Students use the work-energy concepts that they have developed in a different way,
strengthening their understanding of the concepts. Instead of indentifying the type of
energy in a given situation, the students invent a situation that illustrates a given workenergy process.
9. Design an experiment For each item below, describe one real-life experiment that is
consistent with the work-energy process. (Do not mention the experiments described
earlier.)
(a) Positive work causes an increase in the gravitational energy of the system.
An elevator lifts a family from the 1st floor to the 8th floor of a building.
(b) Positive work causes an increase in the kinetic energy of the system.
A person pushes a car down the road.
(c) Positive work causes an increase in the elastic potential energy of the system.
A person stretches a rubber band.
(d) Kinetic energy in the system is converted to gravitational potential energy.
A car rolls up a hill, gradually coming to a stop.
(e) Kinetic energy in the system is converted to elastic potential energy.
A car rolls into a large spring, compresses it, and comes to a stop.
(f) Gravitational potential energy in the system is converted to internal energy.
A roller coaster rolls from the top of a hill down to the bottom where brakes cause the
coaster to slow to a stop.
(g) Gravitational potential energy in the system is converted to elastic potential energy.
A bungee jumper jumps from the top of bridge. As she falls, the bungee cord stretches
causing the jumper to slow down and to stop.
14
In this problem, students practice identifying when work should be positive and when it
should be negative.
10. Regular Problem: As you are building the foundation for a house, you lift a 10 kg
cinder block from a pile of cinder blocks. You have to pick it up a distance of .80 m.
You then carry it for 10 m to the foundation that you are building and set it back down on
the ground by lowering it a distance of .80 m. How much work do you exert on the
block-earth system when you pick the block up, when you carry it to the house, and when
you set it back down? What is the total work that you exert on the system? List your
assumptions.
As you lift the block:
W=78.4 J
As you carry the block:
W=0 J
As you set the block down:
W=-78.4 J
Total work that you exert on the system:
W=0 J
15
Lesson 4.
Reasoning Skills: Constructing a work-energy bar chart
Qualitative work-energy bar charts provide a concrete way to represent work-energy
processes and to reason qualitatively about them. They are similar to the money and food
bar charts that we used at the beginning of the module. In a bar chart, a bar represents
each type of energy initially in the system, as are the final energies of the system. If an
external object does work on the system, then there is a nonzero work bar. The place
marker for the work bar is shaded to indicate that it is not a type of energy but is instead a
process involving an interaction between a system object and an object outside the
system.
To learn how to represent a situation with a bar chart follow the steps indicated in the
sketch. We illustrate how to draw a bar chart for 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. In the final situation, the spring is relaxed. Note that Ug = gravitational
potential energy, Us = elastic potential energy, and Uint = internal potential energy.
1. Make initial-final sketches.
2. Choose a
system.
Initial state
Final state
y
vi > 0
yf > 0
xf = 0, spring
relaxed
xi < 0, spring
µ=0
compressed
0
yi = 0
vi = 0
Earth
3. Use initial sketch to choose
bars for initial energies:
 Spring compressed
Ki Ugi Usi
W
Kf Ugf
Usf ∆Uint
4. Use final sketch to choose
bars for final energies:
 Car moving
 Car position elevated
0
5. Decide if work done
by external object
Now examine the height of the bars on the bar chart for the initial and final states of the
system. How can you explain that the length of the Usi bar is equal to the sum of the
lengths of Kf and Ugf bars?
16
11. Represent and reason Fill in the table that follows.
Experiment: Description Draw a sketch
Construct a qualitative work-energy bar
of system and process.
showing initial and
chart.
final states
A motor pulls a roller
coaster up the first hill of
before
after
the track via a chain.
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
Initial state: The roller
coaster is at rest at the
+
bottom of the hill.
Final state: The roller
0
coaster is moving at a
moderate speed at the top
of the hill.
System: Includes the
roller coaster, chain and
Earth but excludes the
motor that pulls the chain
up the hill.
12. Bar Chart Jeopardy In the table that follows, describe in words and then with a
sketch a process (the system, its initial and final situations, and any work done on the
system) that is consistent with the qualitative work-energy bar chart shown below. There
are many possible choices.
Bar chart for a process.
Describe in words one
Sketch the process
possible consistent
just described.
process.
A skier starts at rest at the
before
after
top of a ski jump. The
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
skier slides down the jump
and goes flying into the
+
air.
0
0
-
17
Lesson 5.
Representing Energies Mathematically
In this problem, students recognize that gravitational potential energy is measured
relative to a particular point in space.
13. Represent and Reason Complete the table that follows.
Word description
Sketch the initial and Complete the work-energy bar chart for this
of a process.
final state. Circle
process
the system.
before
after
(a) Hector lifts a
20 N brick 1 m off
Ki + Ug,i + Usi + W = Kf + Ug,f + Us,f +∆Uint
40 J
of the surface of a
table that is 1 m
20 J
high. He then
0
moves the brick
-20 J
horizontally over a
piece of chalk on
-40 J
the floor.
b) Eva lifts an
identical brick 2 m
from the floor to a
spot right next to
Hector’s brick. It
is hanging over a
piece of chalk on
the floor.
before
after
Ki + Ug,i + Usi + W = Kf + Ug,f + Us,f +∆Uint
40 J
20 J
0
-20 J
-40 J
When dropped, the bricks in parts a and b will both smash the chalk on the
floor the same amount. You may found different potential energies for each
system, though. How can this be?
The energy of each system measures its ability to smash a particular piece
of chalk. For Hector’s final state, the 20 J of energy measures the system’s
ability to smash the chalk on the table. For Eva’s system, the 40 J of energy
measures its ability to smash the chalk on the floor.
c) Redraw your energy-bar chart for Hector, but use it to represent the
system’s ability to smash the chalk on the floor. Is the final state identical to
Eva’s final state in part b?
18
d) Redraw your energy-bar chart for Eva, but use it to represent the
system’s ability to smash the chalk on the table. Is the final state identical to
Hector’s final state in part a?
14. You smash open walnuts on a picnic table by lifting a block from the
table to a height of 1 m above the nuts and then dropping the block on nuts.
Draw a work-energy bar chart representing the system’s ability to smash
these walnuts as you lift the block.
Your friend also has walnuts in his tree house 10 m above the table top.
Draw a work-energy bar chart representing the system’s ability to smash
these walnuts as you lift the block 1 m above the picnic table below.
What is the physical significance of a negative gravitational potential
energy?
19
Lesson 5-Homework
15. A motor has slowly pulled a roller coaster half way up a 50 m high hill.
Draw a work-energy bar chart for the coaster’s trip so far, using the ground
as the reference point.
Draw another work-energy bar chart for this trip, but use the top of the 50 m
high hill as the reference point.
Draw another work-energy bar chart, this time using the top of the loop-theloop as the reference level.
20
16. Reason Jeff and Jim are both demolition experts skilled in using a
wrecking ball to destroy old buildings. The motion of the wrecking ball is
shown below.
When asked to draw work-energy bar charts for the motion of the wrecking
ball, Jeff and Jim draw the bar charts below.
Jim’s
Jeff’s
before
after
before
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
+
0
0
-
-
When would Jeff’s be correct and when would Jim’s be correct?
Jeff’s would be correct if the gravitational potential energy measured the earth-ball’s
ability to smash something at the ball’s final height. Jim’s would be correct if the
gravitational potential energy measured the earth-ball’s ability to smash something at
the ball’s initial height.
How should Jim and Jeff label their work-energy bar charts to prevent any
more confusion?
For example:
Jim should label his: “Measuring the potential to smash something at the ball’s initial
height.”
21
Jeff should label his: “Measuring the potential to smash something at the ball’s final
height.”
Lesson 6.
17. Derive the relation: To develop a mathematical expression for gravitational
potential energy, we analyze the following situation. To build the foundation for a new
skyscraper, a construction company needs to drive metal poles into granite stone. To do
this, a crane lifts a massive block at a constant rate from a height yi above the pole to a
height yf above the pole. The crane then drops the massive block onto the top of the pole,
which we have set to y=0.
a) Here is a picture of the initial and final states of the
process.
- yf
- yi
-0
-0
b) Complete the work-energy bar chart for this process. Write a mathematical expression
representing this process.
before the block
is lifted
after the block is
lifted
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ug,i + W = Ug,f
(c) Draw a free-body diagram for the block. Use it to find an expression for the force that
the crane motor exerts on the cable-block system in terms of its mass and the
gravitational constant g.
F m on cb
Fm on cb FE on cb
m
Fm on cb mg
m

F E on cb


0
0
Fm on cb  mg
22
(d) Write an expression for the work the cable does on the block during its displacement
yf – yi. Put this into the energy expression from part b.
W=mg(yf – yi)
Ug,i + mg(yf – yi) = Ug,f
e) Examine the expression that you derived in part d. Do you see that the change in the
block’s gravitational potential energy is equal to the change in the quantity mgy? Discuss
how this expression can be used to write an expression for determining the gravitational
potential energy of the block-Earth system.
Ug = mgy
23
Lesson 6-Homework
18. Test the relation You are the head engineer for the construction company discussed
in the last problem. Before you build the machine to drive the poles into the ground, you
need to test your expression for gravitational potential energy.
Describe experiments that you will run to test whether the gravitational potential energy
of a system depends on the mass and positions of the objects in it. Include a sketch.
What does the relationship predict will happen in each experiment? What are your
assumptions? Record your observations and discuss your output.
In the first experiment, we will drop a block from two different heights onto a piece of
Styrofoam. The expression for gravitational potential energy predicts that the block
dropped over the greater distance will dent the Styrofoam more than the block dropped
over the smaller distance.
As expected, the block dropped from higher up made a larger dent than the block
dropped over the shorter distance. This result supports the idea that the potential energy
of a system increases as the height of the object increases.
In the second experiment, we will drop a small peddle and a large brick from the same
height onto a piece of Styrofoam. The expression for gravitational potential energy
predicts that the more massive brick will dent the Styrofoam more than the small stone.
As expected, the brick made a larger dent than the smaller stone. This result supports the
idea that the potential energy of a system increases as the mass of the object increases.
24
Lesson 7.
In this problem, students derive a mathematical expression for kinetic energy.
19. Derive the relation In a car crash-test facility, engineers evaluate the reaction of a
car to an impact on its side. To create such an impact, a hydraulic piston pushes a 1000
kg block on wheels over a distance d. This causes the block to accelerate from an initial
to a final velocity. To measure the car smashing potential of this block, let’s determine
the change in its kinetic energy after the piston pushes it a distance d.
Here is a picture of the initial and final state of the
process.
vi
vf
(a) Draw a free-body diagram for the block. Use it to find
an expression for the force that the piston exerts on the
block in terms of its mass m and acceleration a.
Fs on b
d
Fp on b
Fs on b
ab=Fp on b/m
Fp on b = abm
(b) Use a kinematics equation to convert the acceleration a in the equation from part a
into an expression involving the block’s speeds vi and vf. Substitute this into the
expression for force from part a.
ab=(vf2-vi2)/2d
Fp on b =((vf2-vi2)/2d) m
(c) Substitute the expression for force from part b) into the expression for Work: W=Fd
and simplify.
Wp on b=((vf2-vi2)/2d)md
Wp on b=((vf2-vi2)/2)m
Wp on b=1/2 m vf2-1/2 m vi2
d) Use a using a work energy bar chart, develop a mathematical representation of this
process in terms of Work, initial kinetic energy and final kinetic energy. Compare this
expression with the one from part c).
before the block
is lifted
after the block is
lifted
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
25
Ki + W = K f
W=Kf - Ki
(d) Do you see that the work that the piston did on the block equals the change in a
quantity: (1/2) m vf 2 – (1/2) m vi2? Discuss how these expressions can be used to write an
expression for the kinetic energy of the block.
K=1/2 m v2
In this process, the only type of energy that changes is kinetic energy. The kinetic energy
of a certain state only depends on the speed of the object. Looking at the expression
above, then, we see that K = ½ m v2
26
Lesson 7-Homework
In this activity, students practice using the expression for kinetic energy that they
developed in the last activity.
20. Regular Problem A crane lifts a 50-kg crate so that the crate’s speed increases from
0 m/s to 5.0 m/s in a vertical distance of 10.0-m. What is the force that the crane exerts on
the crate? Specify the system, its initial and final states, and any assumptions you made.
Explain how the assumptions affect your answer.
-10 m
-0 m
earth
before the block
is lifted
earth
after the block is
lifted
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
W=Kf + Ug,f
Fcrane on crated =Kf + Ug,f
Fcrane on crate = (Kf + Ug,f)/d
Fcrane on crate = (1/2 m v2 + mgh)/d
Fcrane on crate = (625 J + 5000 J)/10m
Fcrane on crate = 562.5 N
I assume that the crate accelerates at a constant rate upwards. This means that the force
exerted by the crane on the crate is constant. If the acceleration is not constant, then the
force exerted by the crane on the crate is constant.
27
Lesson 8.
21. Find the relation Determine an expression for the change in internal energy due to
friction in a system that consists of a crate and a rough horizontal surface on which it
slides. You, outside the system, pull on a rope attached to the crate so that it moves slowly
at constant velocity. At the end of the process, the bottom of the block and the surface on
which was moving have became warmer.
System
(a) Write an expression for the work done on the system by the external force of the rope
on the crate as the rope pulls the block a distance s across the surface.
Wr on c=Fr on c s
(b) Choose the crate alone as the system (a different system
than in the sketch) and draw a free-body diagram for the
crate. Apply Newton’s second law for the horizontal x-axis.
How are FR on C (rope on crate) and Fs on C (surface friction
on crate) related?
Fs perpendicular on C
Fs parallel on C
FR on C
FE on C
(c) Now, combine (a) and (b) to write an expression for the work done by the force exerted
by the surface through friction on the crate. Is it positive or negative?
Ws on c=-Fs on cs. This work is negative.
before
after
(d) Represent the process with a bar
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
chart. The system is the crate, the
surfaces, and the Earth. The rope is
+
outside. In the initial state, the crate is
moving but the surfaces are cool; in
0
the final state, the crate is still moving
with the same velocity but the
surfaces are warmer. Think of what
force does work on the system and what happens to the
internal energy of the system as the result of this
process.
(e) Examine the bar chart. Write an expression for the change in internal energy and
decide whether it increases or decreases.
28
Ki + W = Kf + ∆Uint
W = ∆Uint
Internal energy increases
Tip
It is important to understand that the bottom surface of the crate is hotter as is the rough
surface on which it moves. Also, there may be parts of the surface rubbed off—a form of
chemical internal energy change (such as the skid marks caused by a car coming to an
abrupt stop). The system’s internal energy increase due to friction is ∆Uint = +fS on O s.
Lesson 8.-Homework
Here the students practice using the expression for kinetic energy and internal energy.
22. Regular Problem When a normal car slows down, the force of friction exerted by the
brakes on the wheels slows the car down, converting the car’s kinetic energy into useless
internal energy in the form of thermal energy. When a hybrid car slows down, an
electrical generator helps to brake the wheels, converting 20% of the car’s kinetic energy
into reuseable electrical energy. The car’s electrical motor will use this electrical energy
later on to propel the car. If a hybrid car slows down from 60 mph to 0 mph, what is the
amount of kinetic energy that the electrical generator converts into electrical energy?
The car has a mass of 1200 kg.
Soon afterwards, the hybrid car is traveling at 50 mph. How far can the car’s
electrical motor propel the car using this electrical energy? The force exerted on the car
by air and moving parts through friction at this speed is 350 N.
before
after
Ki + Ug,i + Us,I+ Uele + W = Kf + Ug,f + Us,f + Uele + ∆Uint
Ki = Uele + ∆Uint
½ m v2 = Uele + ∆Uint
.2 (½ m v2)= Uele
Uele=86,000 J
before
after
Ki + Ug,i + Us,I+ Uele + W = Kf + Ug,f + Us,f + Uele + ∆Uint
29
Uele = Ffriction d
d = Uele / Fair on car
d = 246 m
23. Reason Consider all of the Activities that you have done so far. What is the
relationship between the initial energy of the system, the external work done on the
system, and the final energy of the system? Express this relationship mathematically.
initial energy of system + work = final energy of system
Generalized Work-Energy Principle The initial energy in the system Ui plus any work
W done on objects in the system by objects outside the system equals the final energy Uf
in the system.
Ui + W = Uf
(Ki + Ug i + Us i + Uint i + …) + W = (Kf + Ug f + Us f + Uint f +…)
or
(Ki + Ug i + Us i + …) + W = (Kf + Ug f + Us f + Uint +…)
The energy can take many different forms: kinetic K, gravitational potential Ug, elastic
potential Us, internal energy change ∆Uint, and others introduced in later chapters. The
unit of energy is the joule (J), where 1 J = 1 N•m.
24. Reason An 1 kg ball is thrown upwards with an initial speed of 20 m/s. Neglect air
resistance. Fill out the table below.
Describe the motion Find the maximum
Find the maximum
Find the maximum
of the object in
height that the
height that the
height that the
words and sketch
object reaches using object reaches using object reaches using
the situation.
kinematics.
Newton’s laws and
your knowledge of
kinematics.
energy.
The object
2ad = vf2 - vi2
a=FE on ball/m
Ki = Ug,f
2
2
2
2
accelerates
d = (vf - vi )/2a
2ad = vf - vi
½ mvi2= mgh
downwards as it
d=20m
d = (vf2 - vi2)/2a
h=1/2 vi2/g
moves. The object
d=20 m
h=20 m
moves upwards until
30
it comes to a stop
and then moves
downwards.
Discuss whether the height of the ball is the same for all three methods. Explain.
Yes, all three methods predict that the ball will travel to a maximum height of 20 m. The
three methods are based on two explanations of motion. In one, a force exerted on an
object by another causes an object to accelerate. In the other explanation, an object’s
potential to perform a task is conserved during a process.
31
Lesson 9.
Calculating the potential energy in a spring and universal gravitational
potential energy.
Optional: To prepare for deriving potential energy stored in a spring and universal
gravitational potential energy, students learn to calculate the work done when the force
exerted on the system changes. To do this, they sum together the work done over
segments with constant force. They are also introduced to finding work by calculating
the area underneath the force vs. distance graph.
25. Regular problem: You are the coach of the two-man U.S. Olympic
Bobsled Team. At the beginning of a race, one of the team members pushes
the bobsled and its driver for 50 meters along the level track. For the first 20
meters, the athlete exerts a 400 N force in the horizontal direction on the sled
and driver. For the next 20 meters, the member exerts a force of 350 N on
the two. For the final 10 meters, he exerts a force of 300 N on the two. The
total mass of the bobsled and driver is 330 kg. The frictional force exerted
by the surface of the ice on the sled is 300 N. Let’s calculate how fast the
bobsled and driver are moving after these first 50 meters.
a) Below are sketches of the initial and final states. Circle the objects to be
included in the system.
vi=0
vf
Let’s determine the work exerted by the teammate on the system. Since we
can only calculate work done on a system when the force exerted on it is
constant, we must calculate separately the work for each segment of constant
force.
b) Over the first 20 meters, how much work did the teammate exert on the
system?
W=400 N * 20 m
32
W=8000 J
c) Over the next 20 meters, how much work did the teammate exert on the
system?
W=350 N * 20 m
W=7000 J
d) Over the last 10 meters, how much work did the teammate exert on the
system?
W=300 N * 10 m
W=3000 J
e) What is the total amount of work that the teammate does on the system?
W=8000 J + 7000 J + 3000 J
W=18,000 J
f) You can also find the total work the teammate does on the system from a
graph! Graph the force exerted by the teammate on the system vs. the
position of the bobsled. Find the total area of the space between the graph
and the x axis. This is the work done on the system. Does it match your
answer from part e?
F (N)
400350300-
0
20
40 50
x (m)
W= 8000 J + 7000 J + 3000 J
W= 18000 J
33
g) Now, let’s consider the energy changes for the entire 50-meter push.
Draw a work-energy bar chart and write a mathematical expression
representing this process.
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
W 
1
2
mv 2  Eint
h) How fast are the bobsled and teammate moving after the first 50 m?
W
1
2
mv 2  Fi sled d
v  2(W  Fmi  sledd )
v=4.2 m/s
34
Lesson 9-Homework
26. Represent and reason Complete the table that follows for three processes. The goal
is to devise a graphical method to determine the work done by an externally generated
force on a system object. Note: P = person and O = object.
Word description
of a process.
Sketch the process.
Draw FP on O x -versusyo (for vertical motion)
or xo (for horizontal
motion) graphs.
(a) Rona lifts a
backpack from the
floor to the desk,
exerting a constant
upward force. The
backpack and the
Earth (not Rona)
are the system.
(b) Kruti catches a
medicine ball in
the gym. The ball
and the Earth are
the system but not
Kruti. Her hands
move back toward
her body while
stopping the ball.
(c) Carlos stretches
a horizontal rubber
cord (it behaves
like a spring) with
a spring constant k.
The spring and the
Earth are the
system but not
Carlos.
Describe how to
use the graph to
find the work
done by the force.
The work done by
Rona on the
system is equal to
the area beneath
the graph.
FP on O
x
y
ytable
FP on O
x
ytable y
The work done by
Carlos on the
system is equal to
the area beneath
the graph.
FP on O
x
y
ytable
35
The work done by
Kruti on the
system is equal to
the area beneath
the graph.
To prepare students to derive an expression for energy stored in a spring, students derive
Hooke’s Law from experimental data in this problem.
27. Derive the relation As you stretch a spring, the force exerted by the spring on you
increases. Let’s determine how this force increases as the spring is stretched. To do this,
we hang two springs with different stiffnesses from a rod and gradually add objects of
different masses to the other ends of the springs. We label the distance each spring has
stretched from the unstretched length as “x.”
a) Sketch one spring before and after you added objects of different masses to its end. Be
certain to label the amount the spring has stretched as “x.”
x
b) Draw a free body diagram for an object hanging at the end of the spring. Consider this
object to be the system. What is the direction of the force exerted by the spring on the
object?
Fs on obj
FE on obj
c) Measure the stretch of the spring for objects with different masses hanging on its end.
Mass of object (kg)
0
.5
1
1.5
2
2.5
3
3.5
Spring 1 stretch (cm)
0
1.2
2.6
3.7
5.1
6.2
7.5
8.8
Spring 2 stretch (cm)
0
2.4
5.1
9.8
9.9
12.7
15.1
17.3
d) Create a graph showing Fs on o versus ( x ). What is the meaning of the slope of this
function?


36
Spring 1 Stretch vs. Force Exerted on Object
10
9
Stretch of Spring (cm)
y = -0.2575x - 0.2533
8
2
R = 0.9908
7
6
5
4
3
2
1
0
-40
-35
-30
-25
-20
-15
-10
-5
-1 0
Force Exerted by Object on Spring (N)
-1/slope = spring constant = 3.88 N/cm
e) Repeat the experiment for the second spring. How are the two spring different from
eachother?
Spring 2 Stretch vs. Force Exerted on Object
20
18
R2 = 0.9994
16
Stretch of Spring (cm)
y = -0.4994x + 0.0208
14
12
10
8
6
4
2
0
-40
-35
-30
-25
-20
-15
-10
-5
Force Exerted by Object on Spring (N)
-1/slope = spring constant = 2.00 N/cm
The first spring is stiffer than the second spring!
37
0
Lesson 10.-Lab
If time does not permit you to work through this problem, have students
together state that Uspring depends on k and x from their intuition. Then
give them Uspring = ½ kx2.
28. Find the relationship In this problem, we will determine an expression for the
elastic potential energy of a stretched spring. The spring is the system and you stretch it
from its equilibrium position to some final position x. You are outside the system. The
spring is very easy to stretch at first but gets more and more difficult as you stretch it
farther. (Remember that FY on S = -k x – the force that “you” exert on the “spring”)
(a) Draw a picture of the situation showing the system and the external object exerting a
force on it. Place the origin of the x-axis at the end of the un-stretched spring.
∆x
(b) Look at the graph [FY on S versus-x] and think
how you can find the work that you did while
stretching the spring from 0 to x.
W=Area under curve – WE CAN SPLIT THE
STRETCH INTO SMALL SEGMENTS FOR
WHICH THE FORCE IS ALMOST CONSTANT,
THEN THE WORK WILL BE THE SUM OF
AREAS UNDER THE CURVE OR REPLACE
THE FORCE WITH THE AVERAGE FORCE –
HALF THE MAX VALUE
W=1/2 * x * kx
W=1/2 kx2
Fy on s
x
x

(c) Examine the expression that you derived from part a. Do you see that the work that
1
you did on the spring equals kx2 ? How does this expression relate to the elastic
2
potential energy of the spring that you have stretched?
The work that I have done on the spring is stored in the spring as elastic potential energy.
 is ½ kx2.
The value of this energy
38
Students will practice designing and running an application experiment. They will use
the expression for potential energy stored in a spring, use work-energy bar charts, decide
where the zero point for gravitational potential energy should be, and perform
mathematical manipulations to determine the height from which an object should be
released.
29. Application Experiment You have a Bouncy Buddy, a toy figure that hangs from a
spring. (a) Design two experiments to determine the spring constant of this spring.
Describe your procedure, record your data, and provide the results below. Decide
whether the results of the two methods agree with each other.
There are multiple ways to determine the spring constant of Bouncy Buddy. In one way,
students lie the spring horizontally on a tabletop and exert various forces on it with a
spring scale. The students then measure the amount by which the spring stretches and
use this to determine the spring constant. Encourage students to plot data points in
Microsoft Excel and use the plot fitting tool to calculate the spring constant.
(b) Use your knowledge of energy and systems to predict the maximum distance the
spring will stretch if the Bouncy Buddy is released at rest with the spring in its relaxed
(not stretched) position. Adjust the horizontal rod holding the loop so the feet of the
Bouncy Bunny will just touch the table when released (but the Buddy’s body does not
touch the table).
Students should draw a labeled sketch of the situation, being certain to label the zero
point for gravitational potential energy, draw a work-energy bar chart, and
mathematically determine the height from which the bouncy buddy should be released.
The students may set any point to be the origin, but the zero point for gravitational
potential energy shown on the sketch should be consistent with the values shown in the
bar chart.
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ug,i = Us,f
mgh = ½ k h2
½ kh2-mgh = 0
h=0
½ kh-mg=0
h=2mg/k
39
Students practice drawing a sketch, identifying the system, selecting a zero point for
gravitational potential energy, creating a work-energy bar chart, and mathematically
manipulating equations to solve for the desired value.
30. Regular Problem The Bouncy Buddy used in the earlier Bungee application
experiment can be used for another experiment, an inverse Bungee jumping application.
Place the Buddy’s wooden ring on a horizontal rod that can move up and down on a ring
stand. Predict the initial position of the rod so that when the Buddy starts seated on the
tabletop with the spring stretched above and is then released, the Buddy just makes it to
the rod but does not hurt his head.
-h
- ∆x
earth
before
-0
earth
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Us,i = Ug,f
½ k ∆x2 = mgh
h=∆x+l (l is the relaxed length of the spring)
½ k ∆x2 = mg(∆x+l)
½ k ∆x2 - mg∆x – mgl = 0
solve for ∆x
40
Lesson 10-Homework
In this problem, the student selects a system, uses a work-energy bar chart, and
mathematically solves for the compression of the spring.
31. Regular Problem Instead of traditional brakes, a spring slows down a new type of
roller coaster. The coaster has 5.0 x 105 J of kinetic energy before it compresses the
spring and stops. The spring constant of the spring is 2.0 x 104 N/m. How far is the spring
compressed? What are the system and the initial and final states that you chose for the
situation?
earth
earth
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ki = Us,f
Ki = ½ k ∆x2
x  2 K i / k
∆x=7 m
41
In this problem, the student selects a system, uses a work-energy bar chart, and
mathematically solves for the compression of the spring.
32. Regular Problem A model airplane launcher uses an elastic cord to accelerate a
small wood and paper airplane to flight speed. The .005 kg plane must be moving at 1
m/s to fly. If the elastic band has a spring constant of 120-N/m, how far should you
stretch the elastic band so that the plane will accelerate to its flight speed?
v=0 m/s
v=1 m/s
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Us,i = Kf
1/2 k x2 = ½ m v2
x = sqrt(m/k) v
x = 6.45 mm
42
Lesson 11.
33. Derive a relation: To launch satellites into space quickly and
inexpensively, NASA wants to design an “elevator” spacecraft to pull new
satellites from the ground into space along a special tether. The new satellite
will move at a constant velocity upwards along the tether. As a scientist
working for NASA, you need to determine the amount of work the elevator
spacecraft will exert on the new satellite to move it into space.
a) Below are sketches of the initial and final states of this process.
Elevator spacecraft
Elevator spacecraft
v
Rs
v
New satellite
New satellite
Ri
Rf
Earth
Earth
b) Draw a free body diagram for the new satellite at the initial and final
states. Remember, the force exerted on the new satellite by the earth
decreases as the new satellite moves away from the earth. At any distance r
from the center of the earth, the force exerted by the earth on the new
satellite is:
F

E on S
GM m
e s
2
r
43
c) Just like the previous problem, the force exerted by one object on
another is changing. To calculate the total work exerted on the new
satellite, we have to calculate a separate work each time the force
exerted on the satellite changes. In this case, though, the force is
constantly changing, so we have to calculate the work done over
infinitely many tiny segments.
W 2 
W1 


GM e m s
r22
GM e m s
r12
* (r)
Very small change in
distance
* (r)
WT  W1  W 2  W 3  ...W  

44
GM e m s
ri

GM e m s
rf
FE on satellite vs. Distance of Satellite from Center of the Earth
0
0
r500
i
0
Distance of Satellite from Center of the Earth
1000 1500 2000 2500 3000 rf 3500
0
0
0
0
0
0
500

F E on -satellite
1000
1500
2000
2500
The graph above shows how the force exerted by the earth on the new
satellite changes as the satellite moves away from the earth. Why is the
force negative?
By finding the area of the shaded region, we can determine the work done by
the earth on the satellite. Doing, this we get the same answer as before:
WT  
GM e m s
ri

GM e m s
rf
d) Consider a system that contains the earth and the new satellite. Fill in
the work-energy bar chart for this system as it moves from the initial

to final state.
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
45
Using the work-energy bar chart, write a mathematical expression that
represents the energies involved in this process.
Ki +W = Kf + Ug,f
e) Rearrange the terms in the equation above so that the terms
representing the potential energies are on one side of the equation and
the terms representing the work are on the other side of the equation.
Substitute in the expression for work derived in part c.
Ki + W = Kf + Ug,f
WT  
GM e m s
ri

GM e m s
rf
f) In the above equation, if you measured gravitational potential energy
the potential at the surface of the earth, then you can
with respect to
found that Ug,f = . If you did not set the initial potential energy to zero,
then you found that Ug = We can match Ug,i with the term 
Ug,f with the term

GM m
e s
r
f
GM m
e s and
r
i
.
In fact, Ug = 
GM m
e s.
r
Notice that the gravitational potential energy is negative. Explain why.
The expression measures the ability of the object-earth system to do work on something
infinitely far away from the earth. According to the expression, no matter how far apart
the earth and object are, they would still have a negative ability to do work on a second
object. The only object this system could never do positive work on is infinitely far away.
Therefore, this expression determines the work that the earth-object system could do on
an object infinitely far away from the earth.
46
Students practice algebraic manipulation while reconciling the new expression for
gravitational potential energy with the older one they learned.
34. Reason The two expressions for gravitational potential energy look very different.
The first one (Ug = mgy) was developed for processes with elevation changes on or near
the Earth’s surface. Does the new expression produce a similar result for such a change?
Suppose you lift a pile driver of mass m from position y to higher position y + ∆y, where
∆y is a relatively small elevation change.
(a) Use the first expression for gravitational potential energy to write an expression for
the gravitational potential energy change.
∆Ug= mg∆y
(b) Now, does the new expression for gravitational potential energy produce the same
result? The pile driver starts at distance r from the Earth’s center and ends at distance
r + ∆y from the center. You will have expressions for the initial and the final
energies. Find a common denominator and combine the two expressions. Note that g
= GM/r2. Can you get this expression to be the same as the expression in part (a)?
Ug,f - Ug,I = -GMEm/rf + GMEm/ri
Ug,f - Ug,I = -GMEm/r+ ∆y + GMEm/r
Ug,f - Ug,I = GMEm(-r/(r+ ∆y)*r + r+∆y /(r+ ∆y)*r)
Ug,f - Ug,I = GMEm(∆y /(r+ ∆y)*r)
Ug,f - Ug,I = GMEm/r2(∆y /(1+ ∆y/r))
Ug,f - Ug,I = GMEm/r2(∆y ) = mg∆y
47
Lesson 11-Homework
Students practice measuring gravitational potential energy with respect to a point very
far away.
35. Reason. Instead of moving the satellite into space using the space
elevator, NASA could also fire it from a cannon on the ground. To move the
satellite from the surface of the earth (6.3*106 m from the center of the earth)
to an altitude of 3.58*107 m from the center of the earth, how fast would the
cannon have to fire the new satellite? At its final height, the satellite should
not be moving relative to the earth.
a) Sketch the initial and final states of the system.
v=0
v
New satellite
New satellite
Rf
Ri
Earth
Earth
b) Represent this process using a work-energy bar chart.
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
c) Represent this process mathematically.
Ki + Ug,i = Ug,f
48
d) Determine the initial velocity of the satellite. (G=6.67*10-11
Nm2/kg2, ME=5.97*1024 kg)
Ki + Ug,i = Ug,f
Ki = Ug,f - Ug,i
½ m v2 = GMem(1/Ri – 1/Rf)
v = sqrt(2MeG(1/Ri – 1/Rf))
v = 1.02*104 m/s
e) Should NASA move the new satellite into space using the
elevator method from the last problem or the cannon method
from this problem? Why?
The force that the cannon would have to exert on the satellite to accelerate it to this
speed would be so large it would probably destroy the satellite.
49
Lesson 12.
Quantitative Reasoning: Problem-Solving Strategy: Work-Energy
Problems
Picture and Translate:
 Sketch the physical process described in the problem. For work-energy processes, the
sketch should include an initial state and a final state. Include a vertical y-axis used to
calculate gravitational potential energy.
 Decide your system choice. Objects such as the Earth, springs, and surfaces of
interacting objects are usually included in the system. Objects that belong to the system
do no work on each other but do possess different types of energy. External objects can
do work on the system objects, thus causing the system’s energy to change.
Simplify:
• Decide what internal or external interactions you can ignore.
Represent physically: Construct a work-energy bar chart. Use the bars to represent the
initial energies in the system, the work done on the system by any external objects, and
the final energies in the system. Consider whether the following change:
 A system object’s elevation above the Earth (gravitational potential energy);
 A system object’s speed (kinetic energy);
 An elastic system object (like a spring) stretches or compresses (elastic potential
energy);
 The surface temperatures of system objects increase as they rub against each other
while one moves relative to the other—there is (internal thermal energy change);
 A system object(s)’s shape during a collision changes (internal potential energy).
Represent Mathematically:
--Apply the generalized work-energy principle;
--Convert the bars in the bar chart into a mathematical description of the process (one
term for each bar in the bar chart).
Solve and Evaluate:
--Use the mathematical description of the process to determine the unknown. Evaluate
the results—units, magnitude, and limiting cases--to make sure they make intuitive sense.
Tip Be sure to identify a system before solving a problem. The system choice is very
important—it affects these descriptions. Also, decide what are the initial and final states
of the system before drawing bar charts.
50
36. Represent and reason In this activity we illustrate the steps of the problem solving
strategy while solving the following problem:
You decide to build a ski rope tow for the hill behind your dormitory. The cable is to pull
a 100-kg skier up a hill that is 18 m high. The skier starts at rest at the bottom and is to
move at speed 6.0 m/s at the top. The hill is 50 m long, and a 150-N friction force
opposes the skier’s motion. You want to buy a motor that provides the average tension
force that the cable needs to exert on the skier to pull her. How large is this force?
Picture and translate: The situation is represented in the Figure below.
We choose the skier, the hill, and the earth as the system. The motor and part of the cable
are outside the system—the cable does work on the skier. The initial state is when the
skier is at rest at the bottom of the hill; the final state is when the skier is moving near the
top of the hill. We need to find to find the force that the cable exerts on the skier.
Initial state
Final state
System
motor
yf > 0
vf > 0
>0
y
yi = 0
vi = 0
>0
0
Earth
Simplify: We assume a steady increase in the skier’s speed (constant acceleration) and
that the skier is a particle with a steady friction force opposing his motion.
Represent physically: The skier starts at the bottom of the hill and all energies are zero (vi
= 0 and yi = 0). Three forms of energy change as the skier moves up the hill. The skier’s
kinetic energy increases, the skier’s gravitational potential energy increases, and the
internal energy of the skis and snow increases because of friction. A bar chart
representing the process is shown below.
Ki
Ugi Usi
W
Kf Ugf
Usf ∆Uint Earth
internal
energy change
0
Represent mathematically
We apply the generalized
work-energy equation
—one term for each bar:
gravitational
energy
kinetic energy
W =
Kf
+ Ugf + ∆Uint
TR on S ∆s = (1/2)m vf2 + m g yf + fk ∆s
where y = 18 m and ∆s = 50 m, the actual distance traveled
along the hill.
51
Solve and evaluate: Substituting the known values into the above equation, we find:
T (50 m) = (1/2)(100 kg)(6.0 m/s)2 + (100 kg)(9.8 m/s2)(18 m) + (150 N)(50 m)
Notice that the work done by the cable causes three forms of energy to increase—the
terms on the right. Rearranging to solve for T, we find that T = 539 N, or 121 lb. Could
you hold on to the rope? You may need a less steep hill, less friction, a good weight loss
program, or a chair lift.
37. Represent and reason Word descriptions and pictorial representations of two workenergy processes are provided in the table that follows. Complete the last column of the
table by describing the processes with a bar chart and with a mathematical expression. Do
not solve for anything.
Word
description
(a) A stunt car has
an ejector seat that
rests on a vertical
spring compressed
a distance xi. When
the spring is
released, the seat
with its passenger is
launched out of the
car and reaches a
maximum height hf
above its starting
position.
(b) An elevator
while moving
down at speed vi
approaches the
ground floor and
slows to a stop in a
distance s.
Picture description
Construct a bar chart for the process and apply
in symbols the generalized work-energy
principle.
System
y
ti,
yi = 0,
vi = 0,
xi < 0.
0
before
tf,
yf = h,
vf = 0.
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
xf = 0
-
Earth
Us,i = Ug,f
System
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
y
+
ti, yi = s,
vi > 0
tf, yf = 0,
vf = 0
0
Earth
52
0
-
Ki + Us,i = Uint
38. Bar Chart Jeopardy Each of the two bar charts below describes real processes. Fill
in the table that follows to describe the possible processes. Do not solve for anything.
Bar chart
Draw an initialDescribe the
Use the bar
final sketch of a
process in words. chart to apply
process that might
in symbols the
be described by the
generalized
bar chart.
work-energy
equation to the
process.
(a)
For example: A
Ki + Ug,i = Kf +
car is rolling up a Ug,f
hill. At the
midway point, it
has a certain
velocity. When it
finally reaches
the top, the car is
moving at a
slower speed.
For example: A
W=Kf + Ug,f
cannon, which is
outside of the
system, fires a
stuntman across a
circus tent. The
graph shows the
energies of the
stuntman-earth
system when the
stuntman is at the
highest point in
his flight.
KiUgi Usi W Kf Ugf Usf ∆Uint
0
(b)
KiUgiUsi W
Kf UgfUsf∆Uint
0
53
Lesson 12-Homework
39. Regular Problem As a traffic cop investigating a car accident, you want to
determine how fast a car was moving before its driver began to brake. While braking, the
car left skid marks that are 70 m long. According to your reference book, the mass of the
car is 1600 kg and the coefficient of kinetic friction between the tires and road is .2. Was
the car traveling faster than the 40 mph speed limit? Fill in the table below to determine
the car’s initial speed.
Picture and
Represent using a
Represent
Solve and evaluate
simplify
work-energy bar chart mathematically
Ki = ∆Uint
v=sqrt(.4gd)
v=16.7 m/s
KiUgi Usi W Kf Ugf Usf ∆Uint
½ mv2=.2mgd
v=37 mph
No, the car was
not breaking the
speed limit.
40. Regular problem A new popular spring hockey game that Jay got for his birthday
uses springs to move a 0.0030-kg puck. It works like a regular table hockey game but
instead of hitting the puck, the players use small springs. Each spring has a 120-N/m
spring constant and can be compressed up to 0.020 m. How fast does the puck move
when it departs a spring that was initially compressed this distance? Fill the table below
Picture and
Represent using a
Represent
Solve and evaluate
simplify
work-energy bar chart mathematically
v=x sqrt(k/m)
v=4 m/s
KiUgi Usi W Kf Ugf Usf ∆Uint
Us,i=Kf
½ k x2 = ½ mv2
54
Lesson 13.
41. Regular problem As part of your new job as a car safety engineer, you have been
asked to predict the average force exerted on a crash test dummy during a simulated car
crash. The car accelerates to a speed of 20 m/s and then collides with a piston that stops
the car. The crash test dummy moves a total of 1.7 m as the car comes to a stop. The
dummy has a mass of 70-kg. Determine the average force that seat belt exerts on the
dummy. What assumptions did you make? Is the force exerted by the seat belt on the
dummy a safe amount?
vi
before
vf
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ki - W = 0
½ m v2 -Fbelt on dummyd = 0
Fbelt on dummy = ½ mv2/d
Fbelt on dummy = 8235 N
I assume that the dummy slows down at a constant acceleration. The force exerted on the
dummy is roughly 2050 lbs, which is large enough to seriously hurt a human being.
42. Regular problem Jim wants to get in shape for his new job as a test driver. He sets
his stationary bike on a high 80-N friction-like resistive force. He cycles for 30 min at a
speed of 18 mph. (a) Determine the internal energy change of the bicycle. (b) If Jim’s
body is 10 percent efficient at converting his metabolic energy into work in pedaling the
bicycle, determine Jim’s internal energy change (equal in magnitude to the metabolic
energy used). (c) How long must he exercise to produce 3.0 x 105 J of internal energy
while staying at this speed (not 30 minutes)? This amount of energy equals the energy
released by the body after eating a slice of bread.
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
a)
+
0
-
55
W = ∆Uint
∆Uint =FJ on b d
∆Uint =80 N * 14,400 m
∆Uint =1,200,000 J
b) .1 ∆Uint, Jim = ∆Uint
∆Uint, Jim = 12,000,000 J
c) 3.0*105 J (30 min / 1.2*107 J) = .75 min = 45 s.
43. Regular problem You are designing a new Bungee-jumping system for beginners.
An 80-kg cart (including its passenger) is to start at rest near the top of a 30o incline. The
uphill side of the cart is attached to a spring. The other end of the spring is attached
securely to a post farther up the hill. The spring is initially relaxed. After you are secure
in the cart, it is released and you coast 40 m down the hill before stopping. For every 1 m
that you coast down the hill, the height of the cart above the ground decreases by .5 m.
What is the spring constant of the spring that you should buy for this invention? Follow
the problem-solving strategy.
-20 m
-0 m
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ug,i = Us,f
mgh = ½ k x2
k = 2mgh/x2
k = 20 N/m
56
40 m
Lesson 14.
In the first collision, students must recognize that a momentum and not an energy
approach should be used to model the first collision. Students then practice using
energies to solve the remaining part of the problem.
44. Regular problem An 80-kg skier comes off a slope traveling at speed 15 m/s and
moves onto a level snow-covered surface. The skier wearing a Velcro-covered vest runs
into a padded 20-kg cart, also covered with Velcro. The skier and cart, now stuck
together, compress a 1600-N/m spring on the other side of the cart (the spring’s other end
is mounted securely to a wall). The skier vibrates back and forth with the cart at the end
of the spring. What maximum distance did the skier and cart compress the spring? Follow
the problem-solving strategy.
before
Ps,i + Pc,i
after
+I =
Ps,f + Pc,f
+
0
-
Ps,i = Ps,f + Pc,f
msvi = msvf + mcvf
vf = ms/(ms+mc) vi
vf = 12 m/s
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ki = Us,f
½ mv2 = ½ kx2
x= v*sqrt(m/k)
57
x= 3 m
58
45. Equation Jeopardy The first column in the table that follows applies the generalized
work-energy equation to two different processes (in fact, there are many possible
processes described by each equation). For each mathematical description, construct a
sketch, a word description, and a bar chart that is consistent with the equation.
Generalized work-energy
equation applied to a
process.
(a)
(1/2) k xi2
= (1/2) m v2
+ mgy
Sketch a process that
might be described by
the equation. Describe
the process in words.
Ex: You shoot a water
balloon with a water
balloon slingshot across
a parking lot at your
friends. The second
state is when the balloon
is at the highest point in
its flight.
Construct a bar chart.
before
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ex: A girl starts at rest
at the top of a mountain
and skiis to the bottom.
(b)
mgd
= (1/2) k x2
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
59
after
Lesson 14: Homework
46. Evaluate the solution
Problem: A 40.000-N/m spring initially compressed 0.50 m is released and launches you
and your cart (100 kg total) up a 30o incline. What distance along the incline do you
travel before stopping? (On a 30o incline, every time you roll 2 m along the incline, you
move 1 m higher from the ground.)
2m
1m
Solution: (1/2)(40,000 N/m)(0.50 m) = (100 kg)(9.8 m/s2)y or y = 10.2 m.
(a) Identify any errors in the solution.
Your friend forgot to square the compression of the spring in ½ k x2. Also, answer gives
the height above the ground that the cart moves to when it stops, but the question asks for
the distance along the incline that the cart moves before it stops. Your friend also forgot
to sketch the process, to identify the system, and to create a bar chart.
(b) Provide a corrected solution if you find errors.
d
-y
-0
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
(1/2)(40,000 N/m)(0.50 m)2 = (100 kg)(9.8 m/s2)y or y = 10.2 m.
10.2 m / d = 1/2
d=20.4 m
47. Evaluate the solution
60
Problem: You are traveling in your 2000-kg Chevy at 20 m/s up a hill when you see a
goose crossing the road 24 m in front of you. You know from previous experience that
when you hit the brakes, a 16,000-N friction force opposes your motion. Will you hit the
goose? For every 20 m that you roll up the slope of the hill, the car’s height above sea
level increases by 1 m.
Solution: (1/2)(2000 kg)(20 m/s)2 = (16,000 N)x or x = 25 m. Oops!
(a) Identify any errors in the solution.
The student forgot to sketch the process, identify the system, represent with a bar chart,
and forgot to include the change in gravitational potential energy in the system.
(b) Provide a corrected solution if you find errors.
d
-y
-0
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ki = Ug,f +∆Uint
½ m v2 = mgy + Fb on cartd
½ m v2 = mg(d/20) + Fb on cartd
d= ½ mv2/(mg/20 + Fb on car)
d=23.5 m
you don’t hit the goose!
61
Lesson 15: Lab
In this lab, students will apply their knowledge of forces and energies to solve a practical
problem. They will also decide how assumptions affect one’s judgment about the
outcome of the experiment I.
48. Application experiment
Part I-The energy stored in the Hot Wheels launcher: The Hot Wheels car launcher
has a plastic block that can be pulled back to latch at four different positions. As it is
pulled back, it stretches a rubber band—a greater stretch for each of the four latching
positions. Your task is to use the generalized work-energy principle to determine the
elastic potential energy stored in the launcher in each of these launching positions.
Available equipment: Hot Wheels car, Hot Wheels track, Hot Wheels car launcher, meter
stick, two-meter stick, ruler, masking tape, timer, scale to measure mass, spring scale.
Write the following in your lab report:
a) Start by making a rough plan for how you will solve the problem. Make sure that
you use two methods to determine the energy. Write a brief outline of your
procedure including a labeled sketch.
b) In the outline of your procedure, identify the physical quantities you will measure
and describe how you will measure each quantity.
c) Construct free body diagrams, and energy and/or momentum bar charts wherever
appropriate.
d) Devise the mathematical procedure you will need in order to solve the problem.
Decide what your assumptions are and how they might affect the outcome.
e) Perform the experiment and record the data in an appropriate manner. Determine
the energies.
f) Use your knowledge of experimental uncertainties to estimate the range within
which you know the value of each energy.
Part II. Getting the Hot Wheels car to successfully make a loop-the-loop
Your task now is to determine the least energy launching position so that the car will
make it around the loop without loosing contact with the loop—on the FIRST TRY (do
not use a trial and error method). If you use the next lower energy setting, the car should
not make it around the loop. You may use the results you obtained from the previous
experiment.
Available equipment: Hot Wheels car, Hot Wheels track, Hot Wheels car launcher, meter
stick, two-meter stick, ruler, masking tape, timer, scale to measure mass, Hot Wheels
loop-the-loop and spin-out.
Start by connecting the car launcher to a level track, which is attached to a loop-the-loop.
Place a spin-out at the end of the track (to keep the car from leaving the table).
a) Start by making a rough plan for how you will solve the problem. Identify which
physics ideas you need for each part.
b) To help draw a free body diagram and/or an energy bar chart for each part of the
problem. Think of what conditions are necessary for the car to make the loop
successfully.
62
c) Use the free-body diagram and/or bar chart to devise the mathematical procedure
to solve each part of the problem. Combine the parts to complete the solution. In
the outline of your procedure, identify the physical quantities you will measure,
and describe how you will measure each quantity.
d) List the assumptions about objects, interactions and processes you are making in
the procedure. Discuss how the assumptions could affect your result.
e) Think of the experimental uncertainties. How will they affect your prediction?
f) Use the quantities that you measured and the mathematical procedure that you
devised to make a prediction. When you are ready with the prediction, call your
instructor so you can launch the car.
g) Perform the experiment and record your outcome. Did it match the prediction?
Explain if it did not. Were your assumptions invalid?
h) In which case should you use a random uncertainty instead of an instrumental
uncertainty?
i) When can you claim that your assumptions are valid?
63
Lesson 15: Homework
49. Regular problem At what height should you
release the cart in the loop-the-loop apparatus so
it just barely makes it through the loop without
falling. The mass of the cart is unknown. The
radius of the loop is R.
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ug,i = Ki + Ug,f
mgy = ½ m v2 + mg2R
a = Fnet/m
v2/R = mg/m
v2 = gR
mgy = ½ m gR+ mg2R
y=1/2 R + 2 R
y=5/2 R
50. Pose a problem Pose a problem that can be
solved with the generalized work-energy
principle using the situation depicted in the
illustration. Make a list of necessary givens and
assumptions to solve the problem.
You have designed a matchbox car course so that the cart illustrated to the right is
moving at 4 m/s when it reaches the top of the loop-the-loop. The radius of the loop is 1
m. Will the car stay on the track at the top of the loop-the-loop? How high from the
table surface should the car be released in order to be moving this fast at the top of the
loop?
64
Summary: Definitions and principles
Work W: Work is the product of the magnitude of the average force FEx on O that an
external environmental object exerts on a system object, the magnitude of the system
object’s displacement d, and the cosine of the angle between FEx on O and d.
W = (FEx on O cos ) d
The system gains energy if the work done on it is positive and loses energy if the work is
negative.
Kinetic energy K of a system object is one-half times the product of its mass m and the
square of its speed v:
K = (1/2) m v2
Gravitational potential energy Ug depends on the relative separation of an object of
mass m and the Earth’s center. When near the Earth’s surface, we calculate the Earthobject’s gravitational potential energy using
Ug = m g y
where y is the object’s elevation relative to a chosen zero reference level. When far from
the Earth, we use the expression
Ug = –(GmMEarth) /r
where r is the object’s distance from the center of the Earth.
Elastic potential energy Us is the energy stored in a stretched or compressed object and
depends on the force constant k (stiffness) of the elastic object and the distance x that the
elastic object is displaced from its equilibrium position:
Us = (1/2) k x2.
Increase in internal energy due to friction ∆Uinternal: When an object moves across a
surface with friction, the contacting surfaces warm slightly. If the surfaces are included in
the system, the increase in the system energy due to this friction is the product of the
magnitude of the average kinetic friction force Fk between the object and the surface and
the distance d that the object moves relative to a surface:
∆Uinternal = +Fk d.
If the surface is not in the system, then the work done by the external friction force is:
Wfriction = –Fk d.
Generalized work-energy equation: The total energy of a system is the sum of all the
energies. In the initial state Ui = Ki + Ugi + Usi , and in the final state U = Kf + Ugf + Usf
+∆Uint. If work is done on the system objects, then the energy can change, expressed
quantitatively as the generalized work-energy equation:
Ui + W = Uf.
Conservation of energy principle: If during a process, the net sum of the work done on
the objects in a system is zero, then the total energy in the system is conserved (the same
at the beginning as at the end). However, the types of energy in the system can change.
Ui = Uf.
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Test Questions
1) Bar Chart Jeopardy Each of the two bar charts below describes real processes. Fill
in the table that follows to describe the possible processes. Do not solve for anything.
Bar chart
Draw an initialDescribe the
Use the bar chart
final sketch of a
process in words.
to apply in
process that
symbols the
might be
generalized
described by the
work-energy
bar chart.
equation to the
process.
(a)
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
(b)
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
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Unless otherwise asked, include a sketch of the initial and final state of the process,
pick objects to be in the system, create a work-energy bar chart, and a write a
mathematical representation of the process.
2) In Metropolis, the comic book hero Superman uses his incredible strength to save the
lives of countless ordinary citizens. One day, the braking system on a passenger bus fails
when the bus is 50 m away from a busy intersection. The bus is moving towards the
intersection at 5 m/s. If Superman exerts a force of 300 N on the front of the bus, which
has a mass of 1000 kg, will he be able to stop the bus before it reaches the busy
intersection?
Answer:
d
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ki + W = 0
½ mv2 –FS on bus d = 0
d = ½ mv2 /FS on bus
d = 42 m
Superman is able to stop the bus before it reaches the busy intersection!
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3) When going down steep hills on a highway, the brakes on trucks may fail,
endangering the lives of the truck driver and passengers in nearby cars. To stop these
trucks before an accident occurs, highway engineers often add emergency pull-off ramps
at the bottom of steep hills. The upward incline of these ramps and the gravel surface
stop the run-away trucks. You are in charge of designing one of these ramps. At the
bottom of this hill, run-away trucks may reach a speed of 100 mph. For every meter that
a truck travels along the emergency ramp, the truck will move ½ meter higher. The force
exerted by the gravel on the wheels of the truck is 100 N and the force exerted by air on
the truck is roughly 400 N. How long should the ramp be to stop the truck? The
maximum mass of a truck is 15,400 kg.
Answer:
Students may also consider the air and road surface to be outside the system, in which
case the sketches and work-energy bar chart would change.
d
-y
-0
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Ki = Ug,f +∆Uint
½ m v2 = mgy + Fgravel on truck*d + Fair on truck*d
also on ramp,
d/y = 1 m /.5 m = 2
y=½d
½ m v2 = mg 1/2 d + Fgravel on truck*d + Fair on truck*d
d = ½ m v2 / (1/2 mg + Fgravel on truck + Fair on truck)
d = 199 m
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4) NASA engineers spend a great deal of time designing satellites to orbit the earth.
Assume for this problem that the orbit of a satellite around the earth is circular. Does the
earth do work on the satellite? In order to maintain its speed, will the satellite need to fire
rocket engines? Include a sketch, free body diagram, and work energy bar chart in your
answer.
Answer
Time A
Time B
v
Fearth on satellite

Time A
Time B
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
Since the force exerted by the earth on the satellite is perpendicular to the motion of the
satellite, the earth does not do any work on the satellite while it is in orbit. In fact, no
object does work on the satellite while it is in orbit, so the satellite does not need to fire
its rocket engines in order to maintain its speed.
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5) Evaluate The Solutions:
Alan and Hector answered the following problem below. Identify any mistakes that they
made, explain why it is a mistake, and correct their answer below it.
On January 3rd, 2004, the Mars Exploration Rover Spirit made its descent to the red
planet. For most of the descent, a parachute slowed down the lander from its initial
velocity of 17,000 km/hr. To prevent the parachute from covering the lander on the
surface of Mars, though, the lander separated itself from the parachute when it was 10 m
above the surface. By this time, the parachute and lander’s rocket engines had slowed its
descent to 0 m/s. Over the last 10 m, the lander fell freely. To cushion its landing, large
airbags inflated around the lander. Imagine that the airbags compress 1 m at impact.
What is the average force exerted by the Martian surface on the lander? The lander has a
mass of 544 kg and the gravitational acceleration on Mars is 3.71 m/s2.
Alan’s Answer:
y
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
- 10m
0
-0m
-
System includes: lander, Mars, and surface of Mars
Ug,i + W = 0
mgy – Fsurf on systemd = 0
Fsurf on system=mgy/d
Fsurf on system=2*104 N
Corrections to Alan’s Answer:
Teacher Additions Answer: Since the surface of Mars is included in the system, the force
exerted by the surface on the lander is an internal force. Therefore, the surface of Mars
does not do work on the system, but rather creates a change in the internal energy of the
system.
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
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Hector’s Answer:
y
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
- 10m
+
0
-0m
-
System includes: only the lander
Ug,i + W = 0
mgy – Fsurf on systemd = 0
Fsurf on system=mgy/d
Fsurf on system=2*104 N
Corrections to Hector’s Answer:
Teacher addition answer: Since Hector did not consider the planet Mars to be part of the
system, the force exerted by Mars on the lander is an external force. The system, then,
never has any gravitational potential energy. Instead, Mars does work on the system.
before
after
Ki + Ug,i + Us,i + W = Kf + Ug,f + Us,f +∆Uint
+
0
-
WMars on sys + Wsurf on sys = 0
FMars on sys * y – Fsurf on sys * d = 0
Fsurf on sys = (FMars on sys * y) /d
Fsurf on system=2*104 N
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6) You are directly below your friend in a treehouse. You want to use a water balloon
slingshot to fire a water balloon at him. Your friend is 5 m above you, your water
balloon launcher has a spring constant of 100 N/m, and your water balloon has a mass of
2 kg. What is the minimum amount that you must stretch the slingshot in order to hit
your friend?
7) A meteor is moving at 20 m/s when it is 10,000 km away from the center of the moon.
When the meteor hits the moon, all of its kinetic energy is converted into internal energy.
Draw a sketch of the initial and final states of this process and draw a work-energy bar
chart. Write a mathematical expression describing this process in terms of the mass of
the moon and meteor, and the initial and final distances of the meteor from the center of
the moon. You do not need to solve for anything!
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