Work and Energy - PowerPoint Presentation

Slide1

Work and Energy

©2009 by Goodman & Zavorotniy Slide2

The most powerful concepts in science are called 
"conservation principles". These principles allow us to 
solve problems without worrying much about the details 
of a process.

We just have to take a snapshot of a system initially and 
finally; by comparing those two snapshots we can learn 
a lot.

Conservation PrinciplesSlide3

A good example is a jar of candy.

If you know that there are fifty pieces of candy at the 
beginning. And you know that no pieces have been 
taken out or added...you know that there must be 50 
pieces at the end.

Now, you could change the way you arrange 
them...move them around, whatever...but you still will 
have 50 pieces.

Conservation Principles

50 pieces

still

50

pieces

still 50 piecesSlide4

In that case we would say that the number of pieces of 
candy is conserved.

That is, we should always get the same amount, 
regardless of how they are arranged as long as we take 
into account whether any have been added or taken 
away.

Conservation Principles

50 pieces

still

50

pieces

still 50 piecesSlide5

We also have to be clear about the system that we're 
talking about. If we're talking about one candy jar...we 
can't suddenly start talking about a different one and 
expect to get the same answers.

We must define the system whenever we use a 
conservation principle.

Conservation Principles

50 pieces

still

50

pieces

still 50 piecesSlide6

Energy is a conserved property of nature. It is not 
created or destroyed, so in a closed system we will 
always have the same amount of energy.

The only way the energy of a system can change is if 
it is open to the outside...if energy has been added or 
taken away.

You could ask "what is energy?"

It turns out that energy is so fundamental, like space 
and time, that there is no good answer to that 
question.

Conservation of EnergySlide7

However, just like space and time, that doesn't stop 
us from doing very useful calculations with energy.

We may not be able to define energy, but because it 
is a conserved property of nature, it's a very useful 
idea.

Conservation of EnergySlide8

If we call the amount of energy that we start with as "Eo" and the amount we end up with as "Ef" then we would say 
that if no energy is added to or taken away from a system 
that

Eo = Ef

It turns out that there are only two ways to change the 
energy of a system. One is with heat, which we won't deal 
with here; the other is with "Work", "W".

If we define positive work as that work which increases the 
energy of a system our equation becomes:

Eo + W = Ef

Conservation of EnergySlide9

Work can only be done to a system by an external 
force; a force from something that is not a part of the 
system.

So if our system is a box sitting on a table...and I come 
along and push the box, I can increase the energy of 
the box...I am doing work to the box.

WorkSlide10

Previously, we learned that the amount of work done to a 
system, and therefore the amount of energy increase that 
the system experiences is given by:

Work = Force x Distanceparallel

OR

W = Fdparallel

This is still valid, but we have to bring a new 
interpretation to that equation based on what we know 
about vector components. First, let's review some more 
basic facts.

WorkSlide11

W = Fdparallel

If the object that is experiencing the force does not 
move (if dparallel = 0) then no work is done: W = 0. The energy of the system is unchanged.

If the object moves in the direction opposite the 
direction of the force (for instance if F is positive and 
dparallel is negative) then the work is negative: W < 0. The energy of the system is reduced.

If the object moves in the same direction as the 
direction of the force (for instance if F is positive and 
dparallel is also positive) then the work is positive: W > 0. The energy of the system is increased.

WorkSlide12

W = Fdparallel

This equation gives us the units of work. Since force is 
measured in Newtons (N) and distance is measured in 
meters (m) the unit of work is the Newton-meter (N-m).

And since N = kg-m/s2; a N-m also equals a kg-m2/s2.

However, in honor of James Joule, who made critical 
contributions in developing the idea of energy, the unit 
of energy is also know as a Joule (J).

1 Joule = 1 Newton-meter = 1 kilogram-meter2/second2

1 J = 1 N-m = 1 kg-m2/s2

Units of Work and EnergySlide13

Eo + W = Ef

Since work changed the energy of a system: the 
units of energy must be the same as the units of 
work.

The units of both work and energy are the Joule.

Units of Work and EnergySlide14

Force and Work

Previously, the force 
was either parallel, 
anti-parallel, or 
perpendicular.

Let's look at those 
three cases.

F

Δ

x

v Slide15

F

Δ

x

v

Force and Work

W = Fdparallel = 0

F

Δ

x

v

F

Δ

x

v

W = Fdparallel = F

Δ

x

W = Fdparallel = -F

Δ

xSlide16

v

F

How do we interpret:

W = Fdparallel

in this case.

Force and Work

Δ

xSlide17

After breaking F into 
components that are 


parallel

and

perpendicular

to the direction of motion, we 
can see that no Work 
is done by the 
perpendicular 
component, only by 
the

parallel

component.

FAPP

Fparallel

v

Fperpendicular

Force and Work

Δ

x

θSlide18

W = F

Δ

x

cos

θ

Th

e work done on an 
object by a force is 
the product of the 
magnitude of the 
force and the 
magnitude of the 
displacement times 
the cosine of the 
angle between them.

FAPP

Fparallel

v

Fperpendicular

Force and Work

Δ

x

θSlide19

This is a similar case. 
We just have to find the 
component of force that 
is parallel to the object's 
displacement.

Force and WorkSlide20

The interpretation is the 
same in this case, just 
determine the angle 
between the force and 
displacement and use:

W

= FΔ

x

cos

θ

Fparallel

Fperpendicular

θ

FAPP

Δ

x

Force and WorkSlide21

1

A force F is at an angle θ above the horizontal is 
used to pull a heavy suitcase of weight mg a 
distance d along a level floor at constant velocity. 
The coefficient of friction between the floor and the 
suitcase is μ. The work done by the force F is:

A

Fdcos

θ - μ

mgd

B

Fdcos

θ

C

-μ mgD

2Fdsin θ - μ mgd

EFdcos θ - 1

v

F

θSlide22

2

A force F is at an angle θ above the horizontal and 
is used to pull a heavy suitcase of weight mg a 
distance d along a level floor at constant velocity. 
The coefficient of friction between the floor and the 
suitcase is μ. The work done by the friction force 
is:

A

Fdcos

θ - μ

mgd

B

0

C

-μ

d(mg - Fsinθ)

D2Fdsin θ - μ mgd

EFdcos θ - 1

v

F

θSlide23

3

A force F is at an angle θ above the horizontal and 
is used to pull a heavy suitcase of weight mg a 
distance d along a level floor at constant velocity. 
The coefficient of friction between the floor and the 
suitcase is μ. The work done by the normal force 
is:

A

Fdcos

θ - μ

mgd

B

0

C

-μ

mgdD

2Fdsin θ - μ mgd

EFdcos θ - 1

v

F

θSlide24

4

A force F is at an angle θ above the horizontal and 
is used to pull a heavy suitcase of weight mg a 
distance d along a level floor at constant velocity. 
The coefficient of friction between the floor and the 
suitcase is μ. The work done by the gravitational 
force is:

A

Fdcos

θ - μ

mgd

B

0

C

-μ

mgdD

2Fdsin θ - μ mgd

EFdcos θ - 1

v

F

θSlide25

5

A force F is at an angle θ above the horizontal and 
is used to pull a heavy suitcase of weight mg a 
distance d along a level floor at constant velocity. 
The coefficient of friction between the floor and the 
suitcase is μ. The work done by the net force is:

A

Fdcos

θ - μ

mgd

B

0

C

-μ

mgdD

2Fdsin θ - μ mgd

EFdcos θ - 1

v

F

θSlide26

6

A 4 kg ball is attached to a 1.5 m long string and 
whirled in a horizontal circle at a constant speed of 
5 m/s. How much work is done on the ball during 
one period?

A

9 J

B

4.5 J

C

Zero

D

2 J

E

8 JSlide27

A book of mass "m" is lifted vertically upwards a distance 
"h" by an outside force. How much work does that 
outside force do on the book?

Gravitational Potential Energy

W = Fdparallel

 Since a = 0, Fapp = mg

W = (mg) dparallel

 Since F and d are in the same direction 
 ...and dparallel = h

W = (mg) h

W = mgh

Fapp

mgSlide28

Gravitational Potential Energy

But we know that in general, Eo + W = Ef.

If our book had no energy to begin with, Eo = 0, then

W = Ef

 But we just showed that we did W = mgh to lift 
 the book... so

mgh = Ef

 Or

Ef = mgh

 The energy of a mass is increased by an   amount mgh when it is raised by a height "h".Slide29

Gravitational Potential Energy

The name for this form of energy is Gravitational Potential 
Energy (GPE).

GPE = mgh

One important thing to note is that changes in gravitational 
potential energy are important, but their absolute value is 
not.

You can define any height to be the zero for height...and 
therefore zero for GPE. But whichever height you choose 
to call zero, changes in heights will result in changes of 
GPE.Slide30

7

A 2 kg block is held at the top of an incline plane. 
What is the gravitational potential energy of the 
block?

A

80 J

B

60 J

C

50 J

D

40 J

E

30 JSlide31

8

A 2 kg block released from rest from the top of an 
incline plane. There is no friction between the 
block and the surface. How much work is done by 
the gravitational force on the block?

A

80 J

B

60 J

C

50 J

D

40 J

E

30 JSlide32

Kinetic Energy

Imagine an object of mass "m" at rest at a height "h".

If dropped, how fast will it be traveling just before striking the 
ground?

Use your kinematics equations to get a formula for v2.

v2 = vo2 + 2ad

 Since vo = 0, d = h, and a = g

v2 = 2gh

 And we can solve this for "gh" 

gh = v2 / 2

 We're going to use this result later.Slide33

Kinetic Energy

In this example, we dropped an object. While it was 
falling, its energy was constant...but changing forms.

It had only gravitational potential energy, GPE, at 
beginning, because it had height but no velocity.

It had only kinetic energy, KE, just before striking the 
ground, as it had velocity but no height.

In between, it had some of both.Slide34

Kinetic Energy

Now let's look at this from an energy perspective.

No external force acted on the system so its energy is constant. 
Its original energy was in the form of GPE, which is "mgh".

Eo + W = Ef 

 W = 0 and Eo = mgh

mgh = Ef

 Solving for gh yields

gh = Ef/m

 Now let's use our result from kinematics

 gh = v2 / 2v2 / 2 = Ef/mEf = 1/2 mv2 This is the energy an object has by  
 virtue of its motion: its kinetic energySlide35

Kinetic Energy

The energy an object has by virtue of its motion is called 
its kinetic energy.

The symbol we will be using for kinetic energy is KE.

Like all forms of energy, it is measured in Joules (J).

The amount of KE an object has is given by:

KE = 1/2 mv2Slide36

9

A stone is dropped from the edge of a cliff. Which 
of the following graphs best represents the stone's 
kinetic energy KE as a function of time t?

A

B

C

D

ESlide37

10

A ball swings from point 1 to point 3. Assuming 
the ball is in SHM and point 3 is 2 m above the 
lowest point 2. Answer the following questions. 
What happens to the kinetic energy of the ball 
when it moves from point 1 to point 2?

A

Increases

B

Decreases

C

Remains the same

D

Zero

E

More information is requiredSlide38

11

A ball swings from point 1 to point 3. Assuming 
the ball is in SHM and point 3 is 2 m above the 
lowest point 2. Answer the following questions. 
What is the velocity of the ball at the lowest point 
2?

A

2.2 m/s

B

3.5 m/s

C

4.7 m/s

D

5.1 m/s

E

6.3 m/sSlide39

12

A 2 kg block released from rest from the top of an 
incline plane. There is no friction between the 
block and the surface. What is the speed of the 
block when it reaches the horizontal surface?

A

3.2 m/s

B

4.3 m/s

C

5.8 m/s

D

7.7 m/s

E

6.6 m/sSlide40

13

A satellite with a mass m revolves around Earth in 
a circular orbit with a constant radius R. What is 
the kinetic energy of the satellite if Earth’s mass is 
M?

A

½ mv

B

mgh

C

½GMm/R2

D

½ GMm/R

E

2Mm/RSlide41

14

An apple of mass m is thrown in horizontal from 
the edge of a cliff with a height of H. What is the 
total mechanical energy of the apple with respect 
to the ground when it is at the edge of the cliff?

A

1/2mv02

B

mgH

C

½ mv02- mgH

D

mgH - ½ mv02

E

mgH + ½ mv02Slide42

15

An apple of mass m is thrown in horizontal from 
the edge of a cliff with a height of H. What is the 
kinetic energy of the apple just before it hits the 
ground?

A

½ mv02 + mgH

B

½ mv02 - mgH

C

mgH

D

½ mv02

E

mgh - 1/2 mv02Slide43

16

A 500 kg roller coaster car starts from rest at point 
A and moves down the curved track. Ignore any 
energy loss due to friction. Find the speed of the 
car at the lowest point B.

A

10 m/s

B

20 m/s

C

30 m/s

D

40 m/s

E

50 m/sSlide44

17

A 500 kg roller coaster car starts from rest at point A 
and moves down the curved track. Ignore any 
energy loss due to friction. Find the speed of the car 
when it reaches point C.

A

10 m/s

B

20 m/s

C

30 m/s

D

40 m/s

E

50 m/sSlide45

18

Two projectiles A and B are launched from the 
ground with velocities of 50 m/s at 60 ̊ and 50 
m/s at 30 ̊ with respect to the horizontal. 
Assuming there is no air resistance involved, 
which projectile has greater kinetic energy when 
it reaches the highest point?

A

Projectile A

B

Projectile B

C

They both have the same none zero kinetic 
energy

D

They both have zero kinetic energy at the 
highest point

E

More information is requiredSlide46

19

An object with a mass of 2 kg is initially at rest 
at a position x = 0. A non-constant force F is 
applied to the object over 6 meters.

What is the total work done on the object?

A

200 J

B

150 J

C

170 J

D

190 JE

180 JSlide47

20

An object with a mass of 2 kg is initially at rest at 
a position x=0. A non-constant force F is applied 
to the object over 6 meters.

What is the velocity of the object at 6 meters?

A

150 m/s

B

25 m/s

C

300 m/s

D

12.25 m/sE

Not enough informationSlide48

21

A metal ball is held stationary at a height h0 above the floor and then thrown upward. Assuming the 
collision with the floor is elastic, which graph best 
shows the relationship between the total energy E of 
the metal ball and its height h with respect to the 
floor?

A

B

C

D

ESlide49

22

A toy car travels with speed vo at point x. Point Y 
is a height H below point x. Assuming there is no 
frictional losses and no work is done by a motor, 
what is the speed at point Y?

A

(2gH+1/2vo2)1/2

B

vo-2gH

C

(2gH + vo2)1/2

D

2gH+(1/2vo2)1/2

E

vo+2gHSlide50

Elastic Potential Energy

Energy can be stored in a spring, this energy is called 
Elastic Potential Energy.

Hooke observed the relationship between the force 
necessary to compress a spring and how much the spring 
was compressed.

Fspring = -kx

k represents the spring constant and is measured in N/m.

x represents how much the spring is compressed.

The - sign tells us that this is a restorative force. Slide51

Elastic Potential Energy

The work needed to compress a spring is equal to the area 
under its force vs. distance curve.

W = 1/2 (base) (height)

W = 1/2 (x) (F)

W = 1/2 (x) (kx)

W = 1/2 kx2Slide52

Elastic Potential Energy

The energy imparted to the spring by this work must be 
stored in the Elastic Potential Energy (EPE) of the spring:

EPE = 1/2 kx2

Like all forms of energy, it is measured in Joules (J).Slide53

23

A force of 20 N compresses a spring with spring 
constant 50 N/m. How much energy is stored in 
the spring?

A

2 J

B

5 J

C

4 J

D

6 J

E

8 JSlide54

24

A block with a mass of m slides at a constant 
velocity V0 on a horizontal frictionless surface. The block collides with a spring and comes to rest 
when the spring is compressed to the maximum 
value. If the spring constant is K, what is the 
maximum compression in the spring?

A

V0 (m/k)1/2

B

KmV0

C

V0K/m

D

m V0/K

E

V0 (K/m)1/2

F(V0m/K)1/2Slide55

25

A block of mass m is placed on a frictionless 
inclined plane with an incline angle

θ.

The block is just in contact with a free end of an unstretched 
spring with a spring constant k. If the block is 
released from rest, what is the maximum 
compression in the spring?

A

kmgsin

θ

B

kmgcos

θ

C

(2mgsinθ)/

kD

(mg)/kEkmgSlide56

Recalling conservation of energy, we can now solve more 
complicated problems if energy is conserved.

A roller coaster is at the top of a track that is 80 m high. How fast 
will it be going at the bottom of the hill?

Eo + W = Ef 

 W = 0

Eo = Ef

 E0 = GPE, Ef = KEGPE = KE Substitute GPE and KE equations

mgh = 1/2 m v2

 Solving for v yieldsv2 = 2gh

 Substitute the values and solve

v2 = 2 (9.8) 80 This method works with many energy v =39.6 m/s problemsSlide57

26

A student uses a spring (with a spring constant of 
180 N/m) to launch a marble vertically into the air. 
The mass of the marble is 0.004 kg and the spring 
is compressed 0.03 m. How high will the marble 
go?Slide58

27

A student uses a spring gun (with a spring 
constant of 120 N/m) to launch a marble vertically 
into the air. The mass of the marble is 0.002 kg 
and the spring is compressed 0.04 m.

How fast will it be going when it leaves the gun?Slide59

28

A student uses the lab apparatus shown below. A 5 
kg block compresses a spring 6 cm. The spring 
constant is 300 N/m. What will the block's velocity 
be when released?Slide60

29

A mass is attached to an ideal spring on a 
smooth horizontal surface. It is displaced an 
amount Δxo and released. Which of the following is true?

A

I only

B

II only

C

III only

D

I and II only

E

I and III only

I. The KE is largest when the mass passes 
through the equilibrium point.

II. The PE is largest when the mass has a 
displacement of ±Δxo.III. The PE and KE will never 
be equalSlide61

GPE and Escape Velocity

The expression GPE = mgh only works near the surface 
of a planet. As an object goes to a great height a more 
general expression is needed based on Universal 
Gravitational Force: FGravity = GMm/r2 .

Deriving this accurately requires Calculus, but you can 
remember it by thinking of it this way

W = FΔxcosθ

 FGravity = GMm/r2; cosθ = -1; d = r

W = -(GMm/r2)r

UG = -GMm/rSlide62

GPE and Escape Velocity

To escape a planet, an object needs to have sufficient 
kinetic energy to overcome its negative gravitaional 
potential energy.

Eo + W = Ef

UG + KE + W = 0

-GMm/r + 1/2 mv2 + 0 = 0

1/2 mv2 = GMm/r

v2 = 2GM/r

v = (2GM/r)1/2 

Note that escape 
velocity is 
independent of the 
mass of the 
escaping object. It 
only depends on the 
mass and radius of 
the object being 
escaped from.Slide63

30

A rocket of mass 5,000 kg will have _______ escape 
velocity as a 10,000 rocket?

A

larger

B

smaller

C

equalSlide64

31

Four objects are thrown with identical speeds in different 
directions from the top of a building. Which will be moving 
fastest when it strikes the ground?

A

B

C

D

h

E

All will be the sameSlide65

32

Four objects are thrown with identical speeds in different 
directions from the top of a building. Which will strike the 
ground soonest?

A

B

C

D

h

E

All will be the sameSlide66

33

Four objects are thrown with identical speeds in different 
directions from the top of a building. Which will go the 
highest?

A

B

C

D

h

E

All will be the sameSlide67

34

Four objects are thrown with identical speeds in different 
directions from the top of a building. Which will land 
farthest from the base of the building?

A

B

C

D

h

E

All will be the sameSlide68

35

Four objects are thrown with identical speeds in different 
directions from the top of a building. Which will have the 
have the greatest horizontal component of velocity at its 
maximum height?

A

B

C

D

h

E

All will be the sameSlide69

36

Four objects are thrown with identical speeds in different 
directions from the top of a building. Which will have the 
have the greatest kinetic energy at its maximum height?

A

B

C

h

D

All will be the sameSlide70

37

A rocket is launched from the surface of a planet 
with mass M and radius R. What is the minimum 
velocity the rocket must be given to completely 
escape from the planet's gravitational field?

A

2GM/16R2

B

(2GM/R)1/2

C

(GM/R)1/2

D

2GM/RSlide71

Power

It is often important to know not only if there is enough energy 
available to perform a task but also how much time will be 
required.

Power is defined as the rate that work is done:

P = W / t

Since work is measured in Joules (J) and time is measured in 
seconds (s) the unit of power is Joules per second (J/s).

However, in honor of James Watt, who made critical 
contributions in developing efficient steam engines, the unit 
of power is also know as a Watt (W).Slide72

P = W / t Since W = Fd parallel

P = (Fd parallel) / t 

 Regrouping this becomesP = F(d parallel / t)

 Since v = d/t

P = Fv parallel

So power can be defined as the product of the force applied 
and the velocity of the object parallel to that force.

PowerSlide73

A third useful expression for power can be derived from our 
original statement of the conservation of energy principle.

P = W / t

 Since W = Ef - E0 P = (Ef - E0) / t 

So the power absorbed by a system can be thought of as the 
rate at which the energy in the system is changing.

PowerSlide74

38

A driver in a 2000 kg Porsche wishes to pass a 
slow moving school bus on a 4 lane road. What is 
the average power in watts required to accelerate 
the sports car from 30 m/s to 60 m/s in 9 seconds?

A

1,800 W

B

5,000 W

C

10,000 W

D

100,000 W

E

300,000 WSlide75

39

A force F is applied in horizontal to a 10 kg block. 
The block moves at a constant speed 2 m/s across 
a horizontal surface. The coefficient of kinetic 
friction between the block and the surface is 0.5. 
The work done by the force F in 1.5 minutes is:

A

9000 J

B

5000 J

C

3000 J

D

2000 J

E

1000 JSlide76

40

A crane lifts a 300 kg load at a constant speed to 
the top of a building 60 m high in 15 s. The 
average power expended by the crane to 
overcome gravity is:

A

10,000 W

B

12,000 W

C

15,000 W

D

30,000 W

E

60,000 WSlide77

41

In a physics lab a student uses three light 
frictionless PASCO lab carts. Each cart is loaded 
with some blocks, each having the same mass. The 
same force F is applied to each cart and they move 
equal distances d. In which one of these three 
cases is more work done by force F?

A

cart I

B

cart II

C

cart III

D

the same work is done 
on each

E

more information is requiredSlide78

42

In a physics lab a student uses three light 
frictionless PASCO lab carts. Each cart is loaded 
with some blocks, each having the same mass. The 
same force F is applied to each cart and they move 
equal distances d. Which cart will have more kinetic 
energy at the end of distance d?

A

cart I

B

cart II

C

cart III

D

all three will have the 
same kinetic energy

E

more information is requiredSlide79

43

In a physics lab a student uses three light 
frictionless PASCO lab carts. Each cart is loaded 
with some blocks, each having the same mass.

The same force F is applied to each cart and they move 
equal distances d. Which cart will move faster at 
the end of distance d?

A

cart I

B

cart II

C

cart III

D

all three will move 
with the same speedE

more information is requiredSlide80

44

A box of mass M begins at rest with point 1 at a 
height of 6R, where 2R is the radius of the circular 
part of the track. The box slides down the 
frictionless track and around the loop. What is the 
ratio between the normal force on the box at point 2 
to the box's weight?

A

1

B

2

C

3

D

4

E

5Slide81

45

A ball of mass m is fastened to a string. The ball 
swings in a vertical circle of radius r with the other 
end of the string held fixed. Neglecting air 
resistance, the difference between the string's 
tension at the bottom of the circle and at the top of 
the circle is:

A

mg

B

2mg

C

3mg

D

6mg

E

9mgSlide82

Work and Energy

C

alculus BasedSlide83

Work

We learned before that work can be defined as:

This equation comes from the scalar product of the force 
vector and displacement vector.

If the force however is changing the previous equation for 
work does not apply. To account for the varying force like 
for a varying velocity when you are trying to find 
displacement requires the use of Calculus. Slide84

Work

F

x

By breaking the varying force into small segments we can 
precisely calculate the work done by a varying force.Slide85

Gravitational Potential Energy

Work is defined as the change in energy, so to find the potential 
energy due to gravity we integrate the gravitational force. Slide86

Power

The average power can be represented as:

To find the instantaneous power we use a limit expression 
in which the interval is brought to essentially zero

For a constant force the power equation can be represented in 
several waysSlide87

Work along a curved path

As you move from point 1 to point 2 along a curved 
path, the segments of the curve can be divided into an 
infinite amount of vector displacements denoted by d

l

.

ΦSlide88

Force and Potential Energy

We already have learned that work is the change in energy.

The force is dependent on the components of the potential so the 
force can be described as the negative gradient of the potential. Slide89

Energy Diagrams

U

A

-A

0

KE

A

-A

0

An energy diagram shows the relationship between the 
objects position and the amount of potential energy.

This energy diagram shows the change of the potential for a 
spring. At a maximum displacement (A or -A) the object on the 
spring has maximum potential and at zero the object has 
maximum kinetic energy.Slide90

Energy Diagrams

Some energy diagrams are far more complex than that of a 
spring. To find the magnitude of the force along the curve we 
have to take the derivative because

unstable equilibrium

stable

equilibrium

U

xSlide91

Energy Diagrams

In energy diagrams, stable equilibrium is anywhere the curve has 
a minimum and unstable equilibrium occurs when the curve has 
a maximum. To find the critical points take the derivative of the 
equation and set it equal to zero. The minimum is where the 
slope changes from negative to positive, and the maximum is 
where the slope changes from positive to negative.

Energy Diagrams to Force Diagrams

U

x

F

x

x

x

F

USlide92

46

Given the equation U(x) = x5 - 3x2 + 8. Find F(x).

A

5x4 + 6x

B

-5x4 + 6x - 8

C

-5x4 + 6x

D

-5x4 - 5x - 8

E

5x4 - 5x Slide93

47

Given F(x) = -24x3 + 3x2 + 8x - 7. Find U(x) from x = 0 to x = 1.

A

0 J

B

2 J

C

4 J

D

6 J

E

8 J Slide94

A

B

C

U(J)

(m)

0

A particle moves along the x-axis while acted on by 
a force parallel to the x-axis. The force corresponds 
to the potential energy function graphed.

c) What is the force on the particle 
when it is at point C?

a) What is the direction of the force on 
the particle when it is at point A?

b) What is the direction of the force 
on the particle when it is at point B?

Negative

Positive

Zero