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Types of Energy

How many types/forms of energy can you list?
Gravitational Potential Energy (PE)
The work done to lift (or lower) something is equal to
its change in gravitational potential energy
PE = Fd = (mg)Δh
= mg (h2 – h1)
m
h
PE=mgh
PE: in Joules
h=0 (reference point)
m: in kg
h: in m
Example:
Find the potential energy of a 60.0 kg student sitting on a
80.0 cm high stool…
a) relative to the floor in the room
b) relative to the floor in the room 1 floor below (3.00 m lower)
c) relative to the ceiling 2.50 m above the student
Example:
Find the potential energy of a 60.0 kg student sitting on a
80.0 cm high stool…
a) relative to the floor in the room.
b) relative to the floor in the room 1 floor below (3.0 m lower)
c) relative to the ceiling 2.50 m above the student
a) PE = mgh
= 60.0 x 9.8 x 0.800
= 470 Joules
b) PE = mgh
= 60.0 x 9.8 x 3.8
= 2200 Joules
c) PE = mgh
= 60 x 9.8 x (-2.5)
= -1470 Joules
Kinetic Energy (KE)
The energy of motion. Moving things have
kinetic energy.
 Proportional to mass of the object and the square
of its velocity.

KE = ½
KE: Kinetic Energy (J)
m: mass (kg)
v: velocity (m/s)
2
mv
Example 1
Which has more kinetic energy: a 1200
kg car moving at 20 km/hr or a
transport truck (m=60 t) moving at 3
km/hr?
Example
1.
2.
How much energy does a 4500 kg car have if it is
traveling at 35 km/hr?
If that same car had 1 MJ of energy, how fast is
it moving?
1.
2.
How much energy does a 4500 kg car have if it is
traveling at 35 km/hr?
KE = ½ mv2
= 0.5 * 4500 * (35/3.6)2
= 212673 J = 2.1 x 105 J
If that same car had 1 MJ of energy, how fast is
it moving?
KE = ½ mv2
1,000,000 = ½ * 4500 * v2
v2 = 444 m2/s2
v = 21.08 m/s = 76 km/hr
To Do…
 Read
pgs. 217-224
 Do Ch. 11 P.P. (1-8)
Skateboard Push

How do you get a person to go from zero
velocity to some non-zero velocity?
 You
do some work!
Energy & Work
Consider an object that is accelerated from initial velocity, vo, to
velocity, v, by a force, F (net force), over a distance, d.
What is the work done?
F = ma
F·d = ma·d
F·d = m(a·d)
2ad = v2 – vo2
F·d = m(v2 - vo2)
2
W = ½mv2 - ½mvo2
Kinetic Energy (energy of motion)
KE = ½mv2
W = KE - KEo
W = ΔKE (Work – Energy Theorem)
Conservative (Internal) vs. NonConservative (External) Forces
Conservative Forces
Total Energy will be conserved
Non-Conservative Forces
Total Energy will not be
conserved (amount of total
energy will change)
FGravity
FSpring Force
Fapplied
Fair
Ffriction
Work – Energy Theorem
W = ΔKE
The work done on an object by the net force equals
the change in kinetic energy of that object.
Example:
A 1000 kg car going 20 m/s brakes and comes to a stop.
How much work is done? What happens to the energy?
KE = ½mvo2 = ½(1000)(20)2 = 200000 J
W = ΔKE = 0 – 200000 J = –200000 J (lost energy)
Energy is dissipated as heat.
Conservation of Mechanical
Energy
Total Mechanical Energy (TE): the sum of an object’s Potential
Energy (PE) and Kinetic Energy (KE).
TE = PE + KE
TE = mgh + ½mv2
Law of Conservation of Energy:
Within a closed (no external forces), isolated system, the
total energy of an object is conserved.
E1 = E2
Einitial = Efinal
Example:
1. A pink soccer ball of mass 0.50 kg is kicked with an initial
v
velocity of 15 m/s. Find its…
a) velocity when it is 4.0 m high.
h
b) height when it is going 5.0 m/s.
15m/s
a)
Einital = Efinal
mgh1 + ½mv12 = mgh2 + ½mv22
0 + ½(0.5)(15)2 = (0.5)(9.8)(4.0) + ½(0.5)v2
56.3 = 19.6 + 0.25v2
v = 12 m/s
b)
56.3 = (0.5)(9.8)h + ½(0.5)(5.0)2
h = 10 m
2. Tarzan, a 75 kg Ape-man, swings from a branch 3.0 m
above the ground with an initial speed of 5.0 m/s.
a) Find his velocity when he swings past ground level.
b) Find the maximum height he swings to.
c) In a separate jump, what should his initial speed be if he is to
just reach a branch 4.0 m high?
b) & c)
3.0 m
a)
h=0
Note: in the physics jungle there is no air
resistance and the vines have no mass
and are unstretchable!
Note: The vine does NO WORK because it is always applying
its force perpendicular to the direction of motion
a)
mgho + ½mvo2 = mgh + ½mv2
gho + ½vo2 = 0 + ½v2
m cancels out; h = 0
v = √(vo2 + 2gho)
v = √((5.0)2 + 2(9.8)(3.0))
v = 9.2 m/s horizontal
b)
gho + ½vo2 = 0 + gh
h = vo2 + ho
2g
v=0
h = 4.3 m
c)
gho + ½vo2 = 0 + gh
vo = √(2gh - 2gho)
vo = 4.4 m/s
v=0
2. Tarzan’s less famous brother, George, swings from the
same 3.0 m-high branch with an initial speed of 3.5 m/s.
Unfortunately, he hits a tree 1.5 m above the ground. Find
his speed just before impact and the work done by the tree.
His mass is 90.0 kg.
3.0 m
1.5 m
mgho + ½mvo2 = mgh + ½mv2
m cancels out
gho + ½vo2 = gh + ½v2
v = √(vo2 + 2gho – 2gh)
v = √((3.5)2 + 2(9.8)(3.0 – 1.5))
v = 6.5 m/s
W
W
W
W
= ΔKE = KE - KEo
= ½mv2 – ½mvo2
= 0 – ½(90.0)(6.5)2
= -1900 J
Work – Energy Theorem
W = ΔTotal Energy
The work done on an object by a net force equals
the change in total energy of that object.
W = ΔKE + ΔPE
W = (KEf – KEo) + (PEf – PEo)
Question:
How much heat energy is produced when a “speed” skier
loses 300.0 m in elevation if he starts from rest and finishes
up going 40.0 m/s? His mass (including gear) is 90.0 kg.
Heat = Work done by friction
0m/s
Work = ΔPE + ΔKE
W = (PEf – PEo) + (KEf – KEo)
W = (mgh – mgho) + (½mv2 – 0)
= (0 – (90)(9.8)(300)) + (½(90)(40)2 – 0)
300m
W = -1.93 x 105 J
40m/s
h=0
193000 J of heat are produced
EXAMPLE: A 3.0 kg block slides down a
frictionless 58o incline with a slant length of 4.0 m.
a) Assuming no friction, what is the velocity at
the bottom of the plane?
b) Suppose the co-efficient of friction is 0.25.
What is the velocity of the block at the bottom?

EXAMPLE: A 3.0 kg block slides down a frictionless 58o incline with a
slant length of 4.0 m.
a) Assuming no friction, what is the velocity at the bottom of the
plane?
b) Suppose the co-efficient of friction is 0.25. What is the velocity of
the block at the bottom?

h = 4.0*sin58o
h = 3.39 m
PEtop = KEbottom
mgh = ½ mv2
(3.0)(9.8)(3.39) = 0.5(3.0)v2
v = 8.1 m/s
b) Work done by friction=FFd
FF = μFN = μ mgcos58o
=0.25(3.0)(9.8)cos58o
=4.07 N
W = (4.07)(-4.0) = -16.3 J (opposite direction
of motion
W = (KEf – KEo) + (PEf – PEo)
-16.3 = 0.5(3.0)v2 – (3.0)(9.8)(3.39)
v = 7.4 m/s
Homework