1. Energy
3U/3UB Physics

2. Energy Energy is a scalar quantity measured in Joules (J).
1 J = 1 kg m2/s2 or 1 Nm
Energy is a measure of the ability of an object to do work.
If work is done ON an object it gains energy and is a positive value.
If work is done BY an object it loses energy and is a negative value.

3. Energy Types
Work – the energy pathway taken to change an amount of one type of energy and/or to change the type of energy.
W = Fd
The Work-Energy Theorem states that the work done is equal to the change in energy (type or amount).
W = ∆E
Kinetic energy is the energy of motion, an object of mass with velocity.
Ek = ½ mv2

4. Energy Types
Potential energy refers to the ability to do work – usually a stored energy. There are many types:
Gravitational Potential Energy – energy due to position above a gravitational body: Eg = mgh
Elastic Potential Energy – energy stored in stretched materials that have the ability to restore back to their original shape
Chemical Potential Energy – stored in molecular bonds and released when broken or combined with other elements (combustion, food)
Nuclear Potential Energy – energy released when atoms undergo fission or fusion

5. Energy Types
Mechanical Energy – the sum of the kinetic and gravitational potential energy.
Law of Conservation of Mechanical energy states the energy before an event is equal to the energy afterwards (ignoring dissipative forces).

6. Energy changes/conservation of energy
Example 1) An object is lifted up
You do work to lift the object up. If you used 20.0 J to do this, you lost J while the object gained 20.0 J of Gravitational Potential energy.
Work done by you = J, Work done on object = 20.0 J
Example 2) You apply a net force to a mass causing it to speed up
You do work by applying a net force over a displacement. The object gains kinetic energy. The loss in energy by you is gained by the object.
Note: We are ignoring energy lost due to friction (heat) or sound.

7. Energy changes/conservation of energy
Q1) You pull your sibling with 20.0 N [E] for 10.0 m [E]. What is the work done by you? W = Fs = 20.0 N (10.0 m) = 200. J The work done by you is – 200. J

8. Energy changes/conservation of energy
Q2) If for the previous problem, you applied the force at 30.0o N of E and N of friction exists; what is the work done on the system?
W = Fnet(s)
= (Fax – Ff)(s)
= [20.0N(cos 30.0o) – 6.00 N](10.0 m)
= 113 J
Note: the Work done by you is – 173 J (does not include friction!)

9. Energy changes/conservation of energy
Q3) An 8.0 kg bowling ball is thrown by Fred Flinstone from rest to m/s.
What’s the change in kinetic energy of the bowling ball?
∆Ek = ½ mvf2 – ½ mvi2
= ½ (8.0 kg)(30.0 m/s)2 - 0
= 3.6 x 103 J
b) What’s the work done by Fred?
W = ∆Ek
= -3.6x103 J

10. Energy changes/conservation of energy
Q4a) What’s the potential energy of a 35 kg girl at the top of a waterslide if she is 50.0 m off the ground?
Eg = mgh
= 35 kg (9.81 m/s2) (50.0 m)
= 1.7 x 104 J
Q4b) What’s the girl’s speed at the bottom if no energy is lost?
Total Mechanical Energy is conserved so (Eg)top + (Ek)top = (Eg)bottom + (Ek)bottom
1.7 x 104 J + 0 = 0 + ½ (35 kg) v2
v = 31 m/s

11. Energy changes/conservation of energy
Q5) A kg rollercoaster rolls down a track starting from rest at a height of m.
a) What is its speed at the bottom?
44.3 m/s
b) What is the total mechanical energy anywhere?
1.96 x 105 J
c) How fast is it at a height of 75.0 m?
22.1 m/s

12. Therefore, he utilized – 2.00 W of power
Power is the rate of energy use: P = ∆E/ ∆t = W/ ∆t
Ex1) A student pushes an object using 2.00 N over 60.0 m. If this was done in 1.00 minute, what power did he develop?
P = W/t
= Fd/t
= 2.00 N(60.0 m)/[1.00 min *60s/min]
= 2.00 W
Therefore, he utilized – 2.00 W of power

13. power
Ex2) A waterfall has water falling at 10.0 kg/s from a height of m. What power is developed in a turbine at the bottom, assuming the turbine is 100% efficient?
P = E/t
= mgh/t
= 10.0 kg (9.81 m/s2) (1000 m)/1 s
= 9.81 x 104 W
Note: if the turbine was 33% efficient:
Pout = e(Pin)
= 0.33(9.81 x 104 W)
= 3.24 x 104 W