1. Energy Potential Energy (PE) Kinetic Energy (KE)
The Energy of Position
PE = Fd
PE = (Force)(distance)
SI Units = Joules “J”
Kinetic Energy (KE)
The Energy of Motion
KE = ½ mv2
KE = ½ (mass)(velocity)2
SI Units = Joules “J”
Gravitational PE
PEg= mgh
Elastic PE
PEe= ½ kΔ x2

2. Energy lost = Energy Gained
Energy is the capacity to do work and Work is the measure of transferred energy.
Heat is the transfer of energy due to a change in temperature, but WORK is not due to a temperature change..
Pushing the oars through the water.
The “pusher” looses energy.
The “water and surroundings” gain energy.
Energy lost = Energy Gained
Other Energy Types
Gravitational
Nuclear
Electromagnetic
Force of Gravity pulls
us toward Earth. This
is due to our position
in the gravitational
field
The nuclear force
maintains the integrity
of matter due to the
electron position
relative to the nucleus.
The force of attraction
due to oppositely
charged particles and
their position in the
electromagnetic field.

3. Work: Force along the direction of displacement.
Work & Energy
Work: Force along the direction of displacement.
Work is done only when components of a force are
parallel to a displacement.
1. Scalar quantity
2. SI Units “J” joule
3. W = Fd = mad
W = Fd(CosΘ),
Wnet = Fnetd(CosΘ)
The sign of work…
W = +; when displacement is in the same
direction as the force applied.
W = -; when displacement is in the opposite
If Wnet = +; the object speeds up
If Wnet = - ; the object slows down.
The weight lifter is doing work against gravity since gravity opposes positive vertical motion.

4. How is pulling this wagon different?
How much work is done on a suitcase that
is raised 3.00m if it has a mass of 20.0 kg?
W = Fd;
W = (20.0kg)(9.81 m/s2)(3.00m)
= 589 J
How is pulling this wagon different? FA
You will move this wagon horizontally, however, the force applied is at an angle relative to the horizontal.

5. How much work is done pulling the wagon 10
How much work is done pulling the wagon 10.0 m left if the wagon and load have a mass of 70.0 kg, the force applied is 125N and the angle of the handle is 40º with the horizontal.?
W = Fd Cos 
You will move this wagon horizontally, however, the force applied is at an angle relative to the horizontal.
FAx = FaCos 40
= 125N Cos 40
= 95.8 N
Solve using the force in the direction of
the motion.
W = Fd
= (95.8N)(10.0 m)
= 958 J
FA=125N
40º
Direction of
motion.

6. Potential Energy
Stored Energy
Gravitational Potential Energy
1.  Depends on height from ground zero!
2. Position relative to gravitational source.
PE = mgh
Δ PE = W (work)
= F Δ d
= ma Δ d
Δ PE = mg Δ h
= mg(hf – hi)
ΔPE = PEf – PEi
The potential energy of an object is relative. Since it is based on where the zero position, (base-level), is defined in each problem, the PE is calculated relative to that level.
Potential energy depends of the properties of an object and the interaction of the object with the environment.

7. Elastic Potential Energy
Stored energy in any compressed or stretched object.
Dependent upon how much the spring is compressed or stretched vs it’s relaxed length.
The resting line can be determined by examining the length of the spring when no external forces are acting on it…this is also called the relaxed length of the spring.
PEelastic = ½kx2
where k = spring constant
x = distance compressed or stretched
Spring Constant: the parameter that expresses how resistant a spring is to being compressed or stretched. (SI Units N/m)

8. Problem Solving = (1,590 J) . (9.81 m/s2)(2.45 m) = 66.2 kg PEg = mgh;
ΔX = 75 m
di = 104 m
Problem Solving
In 1993, Javier Sotomayor from Cuba set a record in the high jump by clearing a vertical distance of 2.45 m. If the PEg associated with his jump at the top point of his trajectory was 1,590 J, what was his mass?
= (1,590 J)
(9.81 m/s2)(2.45 m)
= 66.2 kg
PEg = mgh;
Peg = m gh
Situated 4080 m above sea level, La Paz, Bolivia, is the highest capital in the world. If a car with a mass of 905 kg is driven to La Paz from a location that is 1860 m above sea level, what is the increase in potential energy of the car?
ΔPE = mg(hf – hi)
= (905 kg)(9.81 m/s2)(4080 m – 1860 m)
= 1.97x107 J

9. PEe = ½ kΔd2 = ½ (32 N/m)(179 m – 104 m) 2 = 9.0 x 104 J
Suppose a 51 kg bungee jumper steps off the Royal Gorge Bridge, in Colorado. The bridge is situated 31 m above the Arkansas River. The bungee cord’s spring constant is 32 N/m, the cord’s relaxed length is 104 m, and its length is 179 m when the jumper stops falling. What is the total potential energy associated with the jumper at the end of his fall? Assume the bungee cord has negligible mass.
df = 179 m
k = 32 N/m
PEe = ½ kΔd2 = ½ (32 N/m)(179 m – 104 m) 2
= 9.0 x 104 J
 When a 2.00 kg mass is attached to a vertical spring, the spring is stretched 10.0 cm so that the mass is 50.0 cm above the table.
a)  What is the gravitational potential energy associated with this mass
relative to the table?
b)  What is the spring’s elastic potential energy if the spring constant is
400.0 N/m?
c)  What is the total potential energy of this system?
PEg = mgh = (2.00kg)(9.81m/s2)(0.500m) = 9.81 J
PEelastic= (1/2)(400.0N/m)(0.100m)2 = 2.00 J
PEt = PEg + PEelastic = 9.81J J = 11.81J

10. Kinetic Energy I love Physics KE = ½ mv2 KE  m KE  v2 Δ KE = work
Δ KE = Fd
Δ KE = mad
I love Physics
Δ KE = KEf - KEi
Δ KE = ½ mv2f - ½ mv2i
Δ KE = ½ m(v2f - v2i) = Fd

11. Δ KE = work
Δ KE = ½ m(v2f - v2i)
Imagine a snowboarder going down a steep hill. The hill makes a 50° angle with the horizontal and the total mass of the board and rider is 101 kg.
FA= F||= Fnet
F|| = mgSin 50
= 759 N
If his initial velocity is zero, what would his final velocity be at the end of the m run if there was no friction.
a = F|| / g
= 7.73 m/s2
d = vf2 – vi2 ; 2a
2ad = vf
=55.6 m/s
=124 mph!

12. Δ KE = work = F Δ d
Δ KE = ½ m(v2f - v2i)
What was the final Kinetic Energy of the snowboarder?
KE = ½ mv2f = (0.5)(101 kg)(55.6 m/s)2
= 1.56x105 J
What was the change in Kinetic Energy of the snowboarder?
m = 101 kg
a = 7.73 m/s2
Vi = 0
Vf = 55.6 m/s
Since the initial velocity was zero,
Δ KE = ½ mv2f = 1.56x105 J or
F d = 759N x 200.0m
1.52x105J

13. Δ KE = work = F Δ d
Δ KE = ½ m(v2f - v2i)
m = 101 kg
Vi = 0
F|| = 759N
What would the change in Kinetic Energy be if the coefficient of friction was 0.092?
FA – Ff = Fnet
F|| - FN = Fnet
F|| - mgCos 50 = Fnet
759N - (0.092)(101 kg)(9.81 m/s2)Cos 50 = Fnet
759N – 58.6 N = Fnet
700. N
a = 6.93 m/s2
Vf= 52.6 m/s
= 118 mph
Δ KE =1.40x105J
Fnet d = N x 200.0m
= 1.40x105J

15. Conservation of Energy
Pendulum
Relationships
PEg = max
KE = 0
PEg = KE
PEg = 0
KE = max
We know that energy cannot be lost or gained, but it can change forms.
KEi + PEi = KEf + PEf

16. PEg = max because it is in
the highest position
KE = 0 (Vi = 0)
The decrease in Potential Energy is equal to the increase in Kinetic Energy.
Half way indicates PEg = KE
PEg = 0 (ground level)
KE = max because it is at
its terminal velocity.
(highest)

17. The Law of Conservation of Energy states that energy can neither be created nor destroyed. Energy simply changes forms.
When the 5.0 kg ball falls from the roof of a building 16 meters high, we can use the conservation of energy to determine the final velocity of the ball by first looking at the potential energy of the system.
PE = mgh
PE = (5.0 kg)(9.81 m/s2)(16 m)
PE = 780 J
PE = KE
780 J = ½ mv2
v = [(2)(780J)/( 5.0 kg)]1/2
v = 18 m/s

18. In a system the conservation of energy can be verified through the equation KEi + PEi = KEf + PEf
An 8.0-kg flower pot falls from a window ledge 12.0 m above a sidewalk.
a)  What is the kinetic energy of the pot just as it reaches the side walk.
b)  Determine the speed of the pot just before it strikes the walk.
Solution
a) KEi + PEi = KEf + PEf
mgh= KEf
(8.0kg)(9.81 m/s2)(12.9 m) = 940 J
940 J = ½ mv2
[(2)(940 J/(8.0 kg)]1/2 = 15 m/s

19. Problem: A 10.0 g pebble is placed in a sling shot with a spring constant of N/m and is stretched back m.
What is the maximum velocity the pebble will acquire?
If shot straight up, what is the maximum height the pebble will reach?
In each situation energy is conserved if heat effects and air resistance are negligible.
At the beginning, when the slingshot is "cocked“
Etotal = KEi + PEg + Pee
Etotal = ½mv2 + mgh + ½kx2
Etotal = ½m(0)2 + mg(0) + ½(200.0)(0.500)2
Etotal = 25 J

20. The maximum velocity is obtained after the pebble is released but before it has achieved any altitude.
Etotal = KEi + PEg + Pee
Etotal = ½mv2 + mgh + ½kx2
25 J = ½( kg)(v)2 + mg(0) + ½ k(0)2 √
2(25 J) = √(v)2
( kg)
= 70.7 m/s
At its maximum height, the velocity of the pebble is zero. Etotal = KEi + PEg + Pee
Etotal = ½mv2 + mgh + ½kx2
25 J = ½m(0)2 + (0.010 kg)(9.8 m/s2)h + ½k(0)2
25J = h = 255 m
(0.010 kg)(9.8 N/kg)

21. The more time it takes to do the work, the less power you use.
Power = Work = W = Fd
Time t t
P = mad or mg d =
t t
F·v
SI Unit: J/s or Watt, W (This honors James Watt,
the developer of the steam
engine)
 The more time it takes to do the work, the less power you use.