Unit 8
Work and Energy
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Work
 Work – the result of applying a CONSTANT FORCE on a body
and moving it through a displacement d.


W  FParallel d  Fd

F

d
8-1
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Kinetic Energy & Work-Energy Principle
 Work-Energy Principle - the net work done on a body is equal to
the change in its kinetic energy.
Kinetic Energy
KE  mv
1
2
Work Energy Theorem WTotal
2
 KE2  KE1  KE
m

Units of W & KE W   kg 2  m  N  m  J 
 s 
Pronounced as a Joule
8-2
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Work Done by a Constant Force
 Consider the bulldozer below. It has an initial velocity v1 at time t1
and velocity v2 at time t2.
 How can we determine the net work done on the dozer by the
constant force?

t1 ,v1

t1 ,v1
WTotal  KE2  KE1  KE
v1  v2

t2 , v2
v1  v2

t2 , v2
 What kind of work do we have in the second case?
8-3
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Negative Work

When a shuttle is launched, the force acts in
the direction of the displacement and
produces a positive work.
 When the chute stops the shuttle, the force
acts in the opposite direction of the
displacement and produces a negative work.
W 0

F

F

x
W 0
W  Fx

x
8-4
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
A Closer Look at Work
w  fd cos
 Observe the animation below.
 Pay close attention to the angle of the Tension relative
to the direction of motion
8-5
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
A Closer Look at Work
 What is the angle between
the direction of motion and
the Tension in the picture
below?
 Look at WS 35 #4.
w  fd cos
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
8-6
A Closer Look at Work
 WS 35 4b.
w  fd cos
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
8-7
A Closer Look at Work
 WS 35 4c.
 How does the work done in this figure compare with
that done on 4b?
w  fd cos
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
8-8
A Closer Look at Work
 WS 35 4d.
 How does the work done in this figure compare with
that done on 4c?
w  fd cos
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
8-9
A Closer Look at Work
 WS 35 4e.
 What is the angle between the direction of motion and
the Tension in the picture below?
w  fd cos
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
8-10
Gravitational Potential Energy
 The energy a body has due to its position from the
“ground.”
 The “ground” can be any surface that represents
the origin: a table top, the planet’s surface, the
floor of an airplane,….
 The Lowest Point is ALWAYS equal to zero.
 The equation for GPE is as follows:
GPE  m gy
GPE  mg y2  y1   GPE
y2

y
y1
8-11
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Elastic Potential Energy
 When a spring is compressed or stretched from its neutral position,
elastic potential energy is stored within the spring.
 k is a proportionality constant know as the spring constant or the
force constant.
 We use this constant in Hooke’s Law in order to determine the force
required to stretch or compress a spring.
Fspring  k x2  x1   kx
The work done to compress or stretch a spring is given
by
EPE  kx
1
2
2
8-12
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Compressing
 When a spring is compressed from its neutral position, elastic
potential energy is stored within the spring.
 According to Hooke’s law, the force needed to compress the spring
is
Fspring  k x2  0  kx2
The energy required to compress a spring is given by
EPE  12 kx2
2
Note: x is the displacement. x is the value of the number on
the number line
8-13
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Stretching
 When a spring is stretched from its neutral position, elastic potential
energy is stored within the spring.
 According to Hooke’s law, the force needed to stretch the spring is
Fspring  k x2  0  kx2
The energy required to stretch a spring is given by
EPE  kx2
1
2
2
8-14
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Law of Conservation of Energy
 The total energy is neither created nor destroyed in any process.
 Energy can be transferred from one form to another, and transferred
from one body to another, but the total energy amount remains
constant.
 Alternatively, energy can neither be created nor destroyed only
changed in form or transferred to another body.
8-15
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Conservation of Energy
 If there is no work being done on a system, then the total
mechanical energy of the system (the sum of its KE & PE)
remains constant.
 The following equations apply to the conservation of energy.
W 0
TME  PE  KE  a constant
GPE  m gy
2
1
KE  2 mv
“Energy before equals energy after.”
8-16
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Conservation of Energy
 What are the kinetic and potential
energies at the following points?
Explain why.
A
W 0
TME  PE  KE
GPE  m gy
2
1
KE  2 mv
y  Almost Zero
y
B
y
2
C
8-17
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Conservation of Energy Example
 A car’s engine (mcar = 1500 kg) puts
10,000 J of energy into getting the car
to the top of a hill.
 Calculate the GPE & KE of the car at
the three points below.
A
W 0
TME  PE  KE
GPE  m gy
2
1
KE  2 mv
C
h
B
3h/4
h/4
8-18
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
The MVE
 A statement of the conservation of energy that includes most of the
forms of mechanical energy.
Energy Before
Energy After
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
KE
GPE
EPE
RKE
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
8-19
The MVE
Energy before equals
1
2
mv  I  kx  mgy1  Wother 
2
1
2
1
1
2
2
1
1
2
energy after.
1
2
mv2  I2  kx2  mgy2
2
1
2
2
1
2
2
8-20
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Recognizing the Elements of the MVE
 In order to be successful in applying the MVE (conservation of
energy), we must first be able to recognize the individual elements
(energy types) found in the equation.
 In the next several slides, you will see different transitions from one
energy type to another.
 Attempt to understand why each selected term in the MVE is
important relative to the physical scenario observed.
 Let’s review again the MVE and its individual elements.
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
8-21
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Understanding the MVE – Free Fall
 For our first energy transition, we will exam the energies associated
with dropping a ball from a deck.
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
Key Factor
What is the ball’s final
height?
What is the ball’s GPE?
The height is always equal to
zero, and the GPE is
always equal to zero at the
Lowest Point.
8-22
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Understanding the MVE - Launch
 The next transition is similar to the previous except for the fact that
the projectile will be launched up from the ground.
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
Key Factor
What did the explosion do to the ball?
Do you remember the Work-Kinetic
Energy Principle?
The net work done on a body is equal
to the change in its kinetic energy.
8-23
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Understanding the MVE – Hoops, Anyone?
 Basketball demonstrates many wonderful energy transitions.
 Here we will analyze the scenario starting just as the ball leaves the player’s hand.
 However, just as with the mortar problem, the player exerted work on the ball
causing it to fly.
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
Key Factor
RKE changes very little during
the flight of a projectile.
However, you must still be able
to recognize it when it exists
because it does have a
significant impact on many
problems.
8-24
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Understanding the MVE - Archery
 Archery Problems are excellent examples of energy transitions.
 What energy type is associated with the pulling back of the bow?
 The releasing of the bow exerts work on the arrow in the same way
that the basketball player’s muscles exerted work on the basketball
and in the same way as the exploding powder exerted work on the
mortar.
1
1 2 1
1
1 2 1
2
2
2
2
Key Factor
What was the
GPE of the
arrow just as
it struck the
target?
Why?
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
8-25
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Understanding the MVE – More Archery
 Let’s get a bulls eye hit this time!
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
Key Factor
What was the GPE of the arrow at the beginning and the end of the arrow’s
flight?
We could treat it as zero since both points have the same GPE and are also the
lowest points in the problem.
8-26
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Understanding the MVE – In the Factory
 In this problem we will look at the work done by a conveyor belt.
 This work is done over several seconds; however, for the sake of analysis,
we will assume that the work was done instantaneously on the crate.
 We will also more closely examine the impact that friction has in MVE
problems.
1
1 2 1
1
1 2 1
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
2
Key Factor
What role did friction play in
this problem?
Friction resulted in the apparent
loss of energy to the system.
However, the energy is still
accounted for as work other
(WO).
8-27
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Understanding the MVE – On the Gym Floor
 In this problem we will look at the work done by a weight lifter lifting
weights.
 Again, this work is done over several seconds; however, for the sake of
analysis, we will assume that the work was done instantaneously on the
weight.
1
1 2 1
1
1 2 1
2
2
2
2
m
v

m
gy

kx

I


w

m
v

m
gy

kx

I

1
1
1
1
other
2
2
2
2
Key Factor
2
2
2
2
2
2
How much work is the
weight lifter doing
while holding the
weights in the air.
None!
No motion, no work!
Recall: W = f x d
8-28
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Understanding the MVE – On the Road
 Imagine a motorcycle is driving quickly into view from the left.
 What type of energy does it have?
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
Key Factor
When the rider squeezed his brakes, what element of
the MVE did he introduce?
Work Other (WO) – friction opposed the bike’s
motion bringing it to a stop.
Could you ignore the RKE in this problem?
NO!
It was overcome by the work too.
8-29
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Energy Transition – WS 38 # 1
 Although not required, a FBD may assist you in correctly
answering energy questions especially when looking at work.
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
8-30
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Energy Transition – WS 38 #2
 Again, read and understand the entire problem before
you attempt to solve any of the problem.
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2

8-31
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
How far does it fall?
 If the block slides a distance d down the plane, then
how far does it fall at the same time?
8-32
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Wrap it Up! WS 39 #1
 Try to understand what is happening in the
entire problem before you try to solve any
of the problem.
 Here is a good example.
8-33
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Wrap it up! WS 39 #1
 You have to love geometry!
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
8-34
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Wrap it up! WS 39 #2
 Again, read and understand the entire problem before
you attempt to solve any of the problem.
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
8-35
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
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