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Energy
Today we’ll discuss….
1.
2.
3.
4.
5.
Recap “Work”
Define Energy
Types of Energy & their Formulas
Work-Energy Theorem
Conservation of Energy (COE!)
Energy (E)
• Recall “Work”  W = F*d (Joules)
• Energy - The ability to do work (“work waiting to happen”).
Measured in JOULES (J)
• Types:
• Kinetic (KE) – Energy of Motion
• Gravitational Potential (UG) – Energy given when you
lift an object up
• Elastic or Spring Potential (US) – stored in a spring or
rubber band when you stretch it
• Heat (Q)
• Nuclear, Light, Sound
Kinetic Energy (KE)
• Energy of an object resulting from motion
1 2
KE  mv
2
What is the kinetic energy of a 2500kg car that
is moving 10m/s?
• KE = ½ mv2
• = ½ (2500 kg)(10 m/s)2 = 125,000 J
Example
• An 875 kg compact car speeds up from
22.0 to 44.0 m/s while passing another
car. What were its initial and final
energies?
Kinetic Energy (KE)
• Ex: If you triple an object’s velocity, how much
does its kinetic energy increase?
1 2
KE  mv
2
1
2
xKE  m 3v 
2
x 9
Kinetic Energy (KE)
• Is it possible for two objects to have the same p
(little p stands for _____?) but different KE?
Gravitational Potential Energy (UG)
U g  mgh
m=mass, 9.8 m/s2, h=height
Ex: You lift a 2.00 kg textbook from the floor to
a shelf 2.1 m above the floor. What is the
book’s gravitational potential energy?
• Ug = mgh = (2kg)(9.8 m/s2)(2.1m) = 41.2 J
What is the potential energy once it hits the
floor?
• Ug = mgh = (2kg)(9.8 m/s2)(0m) = 0 J
Energy “Loss”
• Energy is never really gone, it just
transfers into a different form
• This “loss” is typically due to:
• Friction or
• Heat
Heat Energy – i.e. “Friction” (Q)
Qlevel  mg    d
Qincline  mg cos  d
On a level surface
On an incline
Friction
• A 50 kg skier goes 100 meters down a 45
degree hill. How much work was done by
friction if μ = 0.1?
• The skiier then hits a flat slope, and coasts
for another 20 meters. How much work
was done by friction (assume μ = 0.1)?
Work-Energy
Theorem
• Work-Energy Theorem: Work = Energy. One
can be changed into the other.
• When work is done on an object, there is a
change in energy.
• It took work to lift my book off the desk. When I hold it
in the air, it has UG. The work I put into lifting the
book equals the UG.
• If I drop the book, it has the potential (hence “potential
energy”) to do WORK on an egg below it!
• ΔE = W
Work-Energy Theorem
• A 105 g hockey puck is sliding across ice. A
A
player exerts a constant 4.5 N force over a
distance of 0.15 m.
How much work does the player do?
W = F*d
W= (4.5N)(0.15m)= 0.68J
What is the change in the hockey puck’s energy?
ΔE = W
ΔE= 0.68J
Conservation of Energy
(“COE”)
Recall…
1. Types of Energy & their Formulas
1.
2.
3.
4.
KE = ?
UG = ?
Qlevel = ?
Qincline = ?
2. Work-Energy Theorem
1. W = E
Law of Conservation of Energy -
COE
• Energy can neither be created nor
destroyed –
• It can only be changed from one form to
another.
• E “start” = E “end”
• The energy at the “start” must be equal to
the energy at the “end”
• The “start” and the “end” of a problem are
arbitrary points that you get to choose
depending on what the problem is asking
COE in Action
• Let’s think about lifting the book again
The book now has
UG because it has
height off the ground.
I did WORK to lift
the book.
I then drop the book
Halfway down, the book
has UG and KE!
(has height and is moving)
Just before hitting the
ground, all of the book’s
energy has been transferred
into KE (no more height, but
is still moving)
COE: The Format!
Energystart with = Energyend with + “friction (if any)”
This is how you will solve all COE problems!
Your “friction” is heat loss due to friction (in other words,
Qlevel and Qincline)
Sample Problems!
Sample #1 – worked out
• I do 600 Joules of work to lift a 10 kg bucket.
• A) How much UG does it have?
600J because W = E
• B) How high does it go?
E = UG = mgh
600J = (10kg)(9.8)(h)  h = 6.12 m
• C) How fast will it be going when it hits the ground?
W = UG (at the top) = KE (at the bottom)
600J = (1/2)mv2  v = 11 m/s
• D) How fast is it going halfway down?
UG (top) = UG (halfway) + KE (halfway)
mgh
=
mgh
+ (1/2)mv2  m’s cancel
(9.8 * 6.12 m) = (9.8*3.06 m) + ½ (v2)
v = 7.74 m/s
2. Tarzan, standing on a 20-meter high cliff,
swings on a vine to the jungle floor below.
How fast is he traveling when he reaches
the jungle floor?
“START”
“END”
20 m
20 m
2. Tarzan, standing on a 20-meter high cliff,
swings on a vine to the jungle floor below.
How fast is he traveling when he reaches
the jungle floor?
Estart = Eend
“START”
UG = KE
mgh = (1/2)mv2 
(note: m’s cancel !)
“END”
20 m
20 m
(9.8*20) = (1/2)(v2)
v = 19.8 m/s
• #3. Tarzan, in an attempt to escape from Cheetah
who has gone “mad,” is running on the jungle floor
at 10 m/s. He swings on a vine toward a 10-meter
tall cliff. Will Tarzan reach the top of the cliff
safely?
“END”
“START”
10 m
10 m
• #3. Tarzan, in an attempt to escape from Cheetah
who has gone “mad,” is running on the jungle floor
at 10 m/s. He swings on a vine toward a 10-meter
tall cliff. Will Tarzan reach the top of the cliff
safely?
Estart = Eend
KE = UG
“START”
“END”
(1/2)mv2 = mgh
(m’s cancel)
(1/2)(102) = (9.8*h)
h = 5.1 meters – NO 
10 m
10 m
Conservation of Energy
E before = E after + Q’s
Examples: Dropping a book, pendulums, etc
Friction
• Suppose you push a 4 kg box to give it a
speed of 8 m/s and it slides to a stop 5
meters away.
• Find the friction force.
• Find the coefficient of friction.