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Energy
Energy = the potential to do work
And, in reverse…
work done = change in energy
Units: Joules
W=∆E
Energy
Energy = the potential to do work
And, in reverse…
work done = change in energy
Units: Joules
W=∆E
Two major types of energy
1) Potential energy – energy that is stored due to an
object’s position.
Examples: gravitational potential energy
elastic potential energy
2) Kinetic energy - energy of motion
Gravitational Potential Energy
Stored energy due to an object’s height.
The higher the object, the more gravitational potential
energy.**
PEg = Fgd = mgh

PEg = mgh
Elastic Potential Energy
Energy stored due the compression or stretching of elastic
material, such as a spring.
Fspring = kx (Hooke’s Law)
where k is the spring constant and x is the distance the spring
has been stretched or compressed.
Pespring = ½ kx2
Potential Energy and Work
We said earlier that W = Δ E
and we know that potential energy depends only on the
position of an object…
THIS IMPLIES THAT WHEN DOING WORK, THE PATH
DOESN’T MATTER --- ONLY THE END POSITION.
Work doesn’t depend on path.
Imagine a mass m lifted a vertical distance h following two different
paths.
1. straight up:
–– the work done against gravitational
force is:
F2
h
d
0
W
=
F
h
cos
0
g
θ
W = mg h
mg
W depends on
VERTICAL
distance moved,
not on the path
2. along a ramp a distance d (no friction)
– minimum force needed is F2 = mg sinθ.
– the work done against gravitational
force is:
W = F2 d cos 00 = mg sinθ d
W = mg h
Work doesn’t depend on path.
Imagine a mass m lifted a vertical distance h following two different
paths.
1. straight up:
–– the work done against gravitational
force is:
F2
h
d
0
W
=
F
h
cos
0
g
θ
W = mg h
mg
Also notice:
W = the
potential energy
of the block!
2. along a ramp a distance d (no friction)
– minimum force needed is F2 = mg sinθ.
– the work done against gravitational
force is:
W = F2 d cos 00 = mg sinθ d
W = mg h
A ramp reduces the force needed – makes life easier
h
d
θ
h
1. W = F h = mg  h
2. W = F  d = mg sinθ  h/sinθ = mg  h
W = F h
or
F
d (along the ramp)
Work and PE – You do
a) A cart filled with bricks is
pulled at constant speed up
a ramp. The cart has a
mass of 3.0kg and the height
of the ramp is 0.45 m.
What is the PEg of the cart at
the top of the ramp?
b) It required 14.7 N to pull the cart up the ramp,
and the ramp is 0.9 m long. How much work was
needed to pull the cart up the ramp?
Work and PE – You do
a) A cart filled with bricks is
pulled at constant speed up
a ramp. The cart has a
mass of 3.0kg and the height
of the ramp is 0.45 m.
What is the PEg of the cart at
the top of the ramp?
b) It required 14.7 N to pull the cart up the ramp,
and the ramp is 0.9 m long. How much work was
needed to pull the cart up the ramp?
Both answers are 13.2 J!
Kinetic energy
Kinetic energy is the energy of motion.
KE = ½ m v2
where v = velocity
This means that a moving object possesses the capacity to
do work.
Example: A hammer by virtue of its motion can be used to do
work in driving a nail into a piece of wood.
(1)
A father pushes his child on a sled on level ice, a distance 5 m
from rest, giving a final speed of 2 m/s. If the mass of the child
and sled is 30 kg, how much work did he do?
W = ∆ KE = ½ m v2 – 0 = ½ (30 kg)(2)2 = 60 J
(2)
What is the average force he exerted on the child?
W = Fd = 60 J, and d = 5 m, so F = 60/5 = 12 N
(3) A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the
road does work on the car through a backward-directed friction force.
How much work must this friction force do in order to stop the car?
vi = 45 km/h
vf = 0
Ffr
d
(work done
by friction
force)
(3) A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the
road does work on the car through a backward-directed friction force.
How much work must this friction force do in order to stop the car?
W = ∆ KE = 0 – ½ m u2 = – ½ (1000 kg) (45 x1000 m/3600 s)2
W = – 78125 J = – 78 kJ
(the – sign just means the work leads to a decrease in KE)
vi = 45 km/hr
vf = 0
Ffr
d
(work done
by friction
force)
(4) Instead of slamming on the brakes the work required to stop the car is
provided by a tree!!!
What average force is required to stop a 1000-kg car going at 45 km/h if the
car collapses one foot (0.3 m) upon impact?
W = – 78125 J
W = – F d = – ½ m u2
F=
-W
78125
=
d
0.3
F = 260 x 103 N
The net force acting on the car is F
corresponds to 29 tons hitting you OUCH
Do you see why the cars should not be rigid. Smaller collapse distance,
gretaer force, greater acceleration. For half the distance force would
double!!!!
OUCH, OUCH
a = (v2 - u2 )/2d = - 520 m/s2 HUGE!!!!
KE and Work – You do
 Determine the KE of a 625 kg roller coaster
that is moving with a speed of 18.3 m/s.
 What would the KE be if the roller coaster
doubled its speed?
KE and Work – You do
 Determine the KE of a 625 kg roller coaster
that is moving with a speed of 18.3 m/s.
1.05 X 105 J
 What would the KE be if the roller coaster
doubled its speed?
4.19 X 105 J