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Energy
• Something that enables an object to work
is called energy.
• What are some different forms of energy?
– Potential
– Electrical
– Mechanical
– Kinetic
Potential Energy
• Potential Energy: energy that is stored
and held in readiness to do work.
• Any substance that can do work has
potential energy.
– Fossil Fuels
– Electric Batteries
– Food
Potential Energy
• Work is required to elevate objects
• This potential energy is called
– Gravitational Potential Energy (GPE)
• The amount of gravitational potential
energy is equal to the work done against
gravity by lifting it.
Energy
• Energy is the physical agent that allows work to be done.
• Energy has many forms:
– In motion (KE)
– In position (GPE)
– In the physical/chemical properties of materials (EPE)
• Energy, like work, is measured in Joules (J).
A
B
Which path will take the most work to get
the ball to the top of the tower?
Gravitational Potential Energy
• Work = Force x Distance
• Upward Force = weight
– Work = Weight x Distance
– Work = mg
x Distance
– GPE = mg x Distance (height)
• GPE = mgh
Which path will take the most work to get
the ball to the top of the tower?
GPE Example
A cannon fires a 10kg
cannon ball 150m into the
air.
What is the GPE at its
highest point?
h=150m
GPE  mgh
GPE  10kg (9.8 sm2 )(150m)
GPE  14, 700 J
GPE Example
A cannon fires a 10kg
cannon ball 150m into the
air.
What is the GPE at when
the cannonball lands back
on the ground?
GPE  mgh
GPE  10kg (9.8 m s2 )(0m)
GPE  0J
Kinetic Energy
Kinetic Energy: is the energy of motion
1 2
KE  mv
2
Note that KE quadruples when the velocity
doubles
How far would a car skid?
30km/hr
Skid 10m
KE 
60km/hr
120km/hr
1
m(30) 2
2
Skid 40m
KE 
1
m(60) 2
2
Skid 160m
KE 
1
m(120) 2
2
Sample Problem #1 (KE)
• A train (m = 340000kg) travels along a stretch of
track with a velocity of 16m/s.
Tunnel
KE  mv
1
2
KE 
1
2
2
 340000kg 16 s 
KE  43,520, 000 J
m
2
What is
the KE of
the train?
• A train (m = 340000kg) travels along a stretch of
track with a velocity of 16m/s. KE  43,520, 000 J
How much work is required to stop the train in 84.6s?
v2  v1  at
v22  v12  2ad
F  ma
W  Fd
v2  v1
m
a  .189s2
a
t
v22  v12
d
d  677.2m
2a
F  6430.6N
W  6430.6 N (677.2m)
W  43,520, 000 J
Work-Energy Theorem
• A train (m = 340000kg) travels along a stretch of track
with a velocity of 16m/s. How much work is required to
stop the train in 84.6s?
KE  43,520, 000 J
W  KE
W  0 J  43,520, 000 J
W  43,520, 000 J
Elastic Potential Energy
EPE: is the energy stored in a spring or
flexible object
1 2
EPE  kx
2
• k is the spring constant (Units N/m) (material dependent)
– Which spring would be harder to compress?
• x is the distance compressed or stretched
(Units m)
The Law of Conservation of Energy
• The Law of Conservation of Energy: in a
closed and isolated system, the total
energy remains constant.
– Energy can not be destroyed.
– Energy transforms, but the total amount
never changes.
Conservation of Energy
What energy transformation take place in this
example?
1
1 2 1
1
1 2 1
2
2
2
2
mv1  mgy1  kx1  I1  wother  mv2  mgy2  kx2  I 2
2
2
2
2
2
2
Conservation of Energy
What energy transformation take place in this
example?
1
1 2 1
1
1 2 1
2
2
2
2
mv1  mgy1  kx1  I1  wother  mv2  mgy2  kx2  I 2
2
2
2
2
2
2
Conservation of Energy
What energy transformation take place in this
example?
1
1 2 1
1
1 2 1
2
2
2
2
mv1  mgy1  kx1  I1  wother  mv2  mgy2  kx2  I 2
2
2
2
2
2
2

Conservation of Energy
PE
PE
PE + KE
PE + KE
KE
Simple Harmonic Motion
X1
mgh1
V=0
X2
mgh0
V=0
Hooke’s Law
Fs  kx
Understanding the MVE – Hoops, Anyone?
1
1 2 1
1
1 2 1
2
2
2
2
mv1  mgy1  kx1  I1  wother  mv2  mgy2  kx2  I 2
2
2
2
2
2
2
Compressing
1
1 2
1
1 2
2
2
mv1  mgy1  kx1  wother  mv2  mgy2  kx2
2
2
2
2
The MVE
• A statement of the conservation of energy that
includes most of the forms of mechanical energy.
Energy Before
Energy After
1
1
1
1
1
1
mv12  mgy1  kx12  I 12  wother  mv2 2  mgy2  kx2 2  I 2 2
2
2
2
2
2
2
KE
GPE
EPE
RKE
External Work (Wo)
• Wo is any work done by the system or on the system
during an energy transition.
• In cases where energy is added to the system, Wo is
positive(+). Examples include motors and muscles.
• In cases where energy is lost by the system, Wo is
negative(-). Examples include friction and air resistance.
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?
Conservation of Energy Example
• A ball, initially traveling at a velocity of 14m/s is
rolled up a frictionless hill until it stops..
• How high up the hill did it go?
1
1 2
1
1 2
2
2
mv1  kx1  mgy1  wother  mv2  kx2  mgy2
2
2
2
2
2
1
v
2
h

2
mv1  mgh2
2
g
2
h2  10m
Understanding the MVE – Free Fall
1
1 2
1
1 2
2
2
mv1  mgy1  kx1  wother  mv2  mgy2  kx2
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.
MVE – Free Fall WS 11a #3
A beach ball is .82m above a picnic table which is .45m tall.
The table is on a 2.9m platform.
1
1
1
1
mv12  mgy1  kx12  wother  mv2 2  mgy2  kx2 2
2
2
2
2
• Find GPE of the ball at the
surface of the table
• Find GPE of the ball at the top
of the platform
Conservation of Energy
What are the kinetic and potential
energies at the following points?
Explain why.
W 0
TME  PE  KE
GPE  mgy
2
1
KE  2 mv
y  Almost Zer o
A
y
B
y
2
C
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  mgy
2
1
KE  2 mv
C
B
h
3h/4
h/4
Understanding the MVE - Launch
 The projectile will be launched up from the ground.
1
1
1
1
mv12  mgy1  kx12  wother  mv2 2  mgy2  kx2 2
2
2
2
2
Key Factor
What did the explosion do to the
ball?
Work-Kinetic Energy Principle?
The net work done on a body is
equal to the change in its
kinetic energy.
Understanding the MVE - Archery
Key Factor
What was the GPE of the arrow just as it struck the target?
Why?
1
1 2 1
1
1 2 1
2
2
2
2
mv1  mgy1  kx1  I1  wother  mv2  mgy2  kx2  I 2
2
2
2
2
2
2
8-25
Understanding the MVE – More Archery
• Let’s get a bulls eye hit this time!
1
1
1
1
mv12  mgy1  kx12  wother  mv2 2  mgy2  kx2 2
2
2
2
2
What was the GPE of the arrow at the beginning and the end of the
arrow’s flight?
Understanding the MVE – In the Factory
1
1 2 1
1
1 2 1
2
2
2
2
mv1  mgy1  kx1  I1  wother  mv2  mgy2  kx2  I 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).
MVE – Archery WS Intro #3
• An arrow (m=.15kg) is drawn back in a bow (k=1120N/m) a
distance of .35m.
• What is the EPE?
• What is the GPE
EPE  kx
1
2
1
1 2
1
1 2
2
2
mv1  mgy1  kx1  wother  mv2  mgy2  kx2
2
2
2
2
2
EPE  12 (1120 Nm )(.35m)2
EPE  68.6J
GPE  mgh
GPE  (.15kg )(9.8 sm2 )(1.3m)
GPE  1.9J
Conservation of Energy Examples
Simple Harmonic Motion
Simple Harmonic Motion:
Motion caused by a linear
restoring force that has a
period independent of
amplitude.
Period: The time required to
repeat one complete cycle
Amplitude: Maximum
displacement from
equilibrium.