1. Chapter 11 Energy and Its Conservation
Physics

2. Section 11.1 The Many Forms of Energy
Essential Questions:
How is a system’s motion related to its kinetic energy?
What is gravitational potential energy?
What is elastic potential energy?
How are mass and energy related?

3. A Model of the Work-Energy Theorem
The work-energy theorem states that doing work on a system causes a change in the energy of that system.
Work is the process that transfers energy between a system and the external world.
When an agent performs work on a system, the system’s energy increases. When the agent does work on its surroundings, the system’s energy decreases.

4. Modeling Transformations
There are many types of energy.
Bar graphs are helpful in keeping track of the different types of energy in a system.

5. Throwing a Ball
Link

6. Catching a Ball
Link

7. Translational Kinetic Energy
Recall that translational kinetic energy (KE or KEtrans) is due to an object’s change in position.
KE = ½ mv2
m = mass
v = velocity
Kinetic energy is proportional to
the object’s mass. KE is also
proportional to the square of the
object’s speed.

8. KE = ½ mv2 KE = ½ (1600 kg)(12.5 m/s)2 KE = 125,000 J KE = 125 kJ
Ex: A 1600 kg car travels at a speed of 12.5 m/s. What is its kinetic energy?
KE = ½ mv2 KE = ½ (1600 kg)(12.5 m/s)2 KE = 125,000 J KE = 125 kJ

9. Rotational Kinetic Energy
Energy due to rotational motion is known as rotational kinetic energy (KErot).
KErot = ½Iω2
I = moment of inertia
ω = angular velocity

10. Potential Energy
Energy that is stored due to interactions between objects in a system is called potential energy.
Not all potential energy is due to gravity. A spring that is compressed is an example.
PE = mgh

11. PE = mgh PE = (60.0 kg)(9.80 m/s2)(3.5 m) PE = 2058 J PE = 2100 J
Ex: Deanna, with a mass of 60.0 kg, climbs 3.5 m up a gymnasium rope. How much energy does a system containing DeAnna and Earth gain from this climb?
PE = mgh PE = (60.0 kg)(9.80 m/s2)(3.5 m) PE = 2058 J PE = 2100 J

12. Gravitational Potential Energy
Energy stored due to gravity is called gravitational potential energy (GPE). The height to which the object has risen is determined by using a reference level.
Wg = mgh
m = mass
g = gravitational acceleration
h = height

13. EX: You lift a 7.30 kg bowling ball from the storage rack and hold it up to your shoulder. The storage rack is 0.61 m above the floor and your shoulder is 1.12 m above the rack. a) When the bowling ball is at your shoulder, what is the ball-Earth system’s gravitational potential energy relative to the floor?
m = 7.30 kg g = 9.8 m/s2 hr = 0.61 m hs = 1.12 m GPE(shoulder to floor)
GPE = mgh
GPE = (7.30 kg)(9.8 m/s2)(1.12 m)
GPE = J
GPE = 80.1 J

14. EX: You lift a 7.30 kg bowling ball from the storage rack and hold it up to your shoulder. The storage rack is 0.61 m above the floor and your shoulder is 1.12 m above the rack. B) When the bowling ball is at your shoulder, what is the ball-Earth system’s gravitational potential energy relative to the rack?
m = 7.30 kg g = 9.8 m/s2 hr = 0.61 m hs = 1.12 m GPE(shoulder to rack)
GPE = mgh
GPE = (7.30 kg)(9.8 m/s2)(0.51 m)
GPE = J
GPE = 36 J

15. EX: You lift a 7.30 kg bowling ball from the storage rack and hold it up to your shoulder. The storage rack is 0.61 m above the floor and your shoulder is 1.12 m above the rack. C) How much work was done by gravity as you lifted the ball from the rack to shoulder level?
m = 7.30 kg g = 9.8 m/s2 hr = 0.61 m hs = 1.12 m W = ?
W = Fd
W = -(7.30 kg)(9.80 m/s2)(0.51 m)
W = J
W = -36 J

16. Elastic Potential Energy
Elastic Potential Energy is stored energy due to an object’s change in shape. Systems that include springs, rubber bands, and trampolines often have elastic potential energy.

17. Mass
Albert Einstein recognized yet another form of potential energy that is proportional to the object’s mass. He demonstrated that mass represent a form of energy. This energy is called the rest energy (E0) and can be calculated using the following formula:
E0 = mc2

18. Other Forms of Energy Chemical and Nuclear Energy Thermal Energy
Fossil fuels, chemical bonds, bonds inside an atom’s nucleus.
Thermal Energy
Heat transfer
Electrical Energy
Power plants
Radiant Energy
Light energy

19. Section 11.1 The Many Forms of Energy
Did We Answer Our Essential Questions?
How is a system’s motion related to its kinetic energy?
What is gravitational potential energy?
What is elastic potential energy?
How are mass and energy related?

20. Section 11.2 Conservation of Energy
Essential Questions:
Under what conditions is energy conserved?
What is mechanical energy, and when is it conserved?
How are momentum and kinetic energy conserved or changed in a collision?

21. The Law of Conservation of Energy
The law of conservation of energy states that in a closed, isolated system, energy can neither be created nor destroyed; rather, energy is conserved.

22. Mechanical Energy
The sum of the kinetic energy and potential energy of the objects in a system is the system’s mechanical energy (ME).
ME = KE + PE

23. Conservation of Mechanical Energy
When mechanical energy is conserved, the sum of the system’s kinetic energy and potential energy before an event is equal to the sum of the system’s kinetic energy and potential energy after that event.
KEi + PEi = KEf + PEf

24. Conservation and Other Forms of Energy
Frictional Forces
Thermal Energy
Air Resistance
Sound Energy

25. Ex: A 22. 0 kg tree limb is 13. 3 m above the ground
Ex: A 22.0 kg tree limb is 13.3 m above the ground. During a hurricane, it falls on a roof that is 6.0 m above the ground. a) Find the kinetic energy of the limb when it reaches the roof. Assume that the air does no work on the tree limb.
KEi + PEi = KEf + PEf
½mv2 + mgh = KEf+ mgh
½(22.0 kg)(0 m/s)2 + (22.0 kg)(9.80 m/s2)(13.2 m) = KE + (22.0 kg)(9.80 m/s2)(6.0 m)
J = KE J
KEf = J
KEf = 1600 J
m = 22.0 kg hlimb = 13.3 m hroof = 6.0 m vi = 0.0 m/s g = 9.80 m/s2 KEf = ?

26. v = 12 m/s KEf = ½mv2 1600 J = ½(22.0 kg)v2 v = 12.06045378 m/s
Ex: A 22.0 kg tree limb is 13.3 m above the ground. During a hurricane, it falls on a roof that is 6.0 m above the ground. B) What is the limb’s speed when it reaches the roof?
m = 22.0 kg hlimb = 13.3 m hroof = 6.0 m vi = 0.0 m/s g = 9.80 m/s2 KEf = ?
KEf = ½mv2
1600 J = ½(22.0 kg)v2
v = m/s
v = 12 m/s

27. Ex: A skier starts from rest at the top of a 45
Ex: A skier starts from rest at the top of a 45.0 m high hill, skis down a 30° incline into a valley, and continues up a 40.0 m high hill. The heights of both hills are measured from the valley floor. Assume that friction is negligible and ignore the effect of the ski poles. A) How fast is the skier moving at the bottom of the valley?
45.0 m
40.0 m
30°

28. KEi + PEi = KEf + PEf vf = 29.7 m/s ½mvi2 + mgh = ½mvf2 + mgh
Ex: A skier starts from rest at the top of a 45.0 m high hill, skis down a 30° incline into a valley, and continues up a 40.0 m high hill. The heights of both hills are measured from the valley floor. Assume that friction is negligible and ignore the effect of the ski poles. A) How fast is the skier moving at the bottom of the valley?
h1 = 45.0 m h2 = 40.0 m vi = 0.0 m/s g = 9.80 m/s2 θ = 30°
KEi + PEi = KEf + PEf
½mvi2 + mgh = ½mvf2 + mgh
½vi2 + gh = ½vf2 + gh
½(0)2 + (9.8)(45) = ½vf2 + (9.8)(0)
(9.8)(45) = ½vf2
vf = 29.7 m/s

29. KEi + PEi = KEf + PEf vf = 9.90 m/s ½mvi2 + mgh = ½mvf2 + mgh
Ex: A skier starts from rest at the top of a 45.0 m high hill, skis down a 30° incline into a valley, and continues up a 40.0 m high hill. The heights of both hills are measured from the valley floor. Assume that friction is negligible and ignore the effect of the ski poles. B) What is the skier’s speed at the top of the second hill?
h1 = 45.0 m h2 = 40.0 m vi = 0.0 m/s g = 9.80 m/s2 θ = 30°
KEi + PEi = KEf + PEf
½mvi2 + mgh = ½mvf2 + mgh
½vi2 + gh = ½vf2 + gh
½(0)2 + (9.8)(45) = ½vf2 + (9.8)(40.0)
49 = ½vf2
vf = 9.90 m/s

30. Ex: A skier starts from rest at the top of a 45
Ex: A skier starts from rest at the top of a 45.0 m high hill, skis down a 30° incline into a valley, and continues up a 40.0 m high hill. The heights of both hills are measured from the valley floor. Assume that friction is negligible and ignore the effect of the ski poles. C) Do the angles of the hills affect your answers?
NO!
45.0 m
40.0 m
30°

31. Analyzing Collisions
If the system is closed and isolate, then momentum and energy are conserved. However, the potential energy or thermal energy in the system might decrease, stay the same, or increase. Therefore, you cannot predict whether kinetic energy is conserved. We are going to look at four types of collisions.

32. Elastic and Inelastic Collisions
A collision in which the kinetic energy does not change is called an elastic collision. Collisions between hard objects, such as those made of steel, glass, or hard plastic, often are called nearly elastic collisions.
Elastic Collision
KEf = KEi

33. Elastic and Inelastic Collisions
A collision in which the kinetic energy does not change is called an elastic collision. Collisions between hard objects, such as those made of steel, glass, or hard plastic, often are called nearly elastic collisions.
Superelastic Collision
KEf > KEi

34. Elastic and Inelastic Collisions
A collision in which kinetic energy decreases is called an inelastic collision. Objects make of soft, sticky materials act in this way.
Perfectly Inelastic Collision
Inelastic Collision

36. Ex. In an accident on a slippery road, a compact car with a mass of 1150 kg moving at 15.0 m/s smashes into the rear end of a car with a mass of 1575 kg moving at 5.0 m/s in the same direction. A) What is the final velocity if the wrecked cars lock together?
m1 = 1150 kg v1 = 15.0 m/s m2 = 1575 kg v2 = 5.0 m/s
Sticky Collison
p1 + p2 = p'
m1v1 + m2v2 = (m1 + m2)v'
(1150)(15) + (1575)(5) = ( )v'
25125 = 2725v
v' = 9.2 m/s

37. Ex. In an accident on a slippery road, a compact car with a mass of 1150 kg moving at 15.0 m/s smashes into the rear end of a car with a mass of 1575 kg moving at 5.0 m/s in the same direction. B) How much kinetic energy decreased in the collision?
m1 = 1150 kg v1 = 15.0 m/s m2 = 1575 kg v2 = 5.0 m/s v' = 9.2 m/s
ΔKE = KEf – KEi
ΔKE = ½(m1 + m2)v'2 – [½m1v12 + ½m2v22]
ΔKE = ½(2725)(9.2)2 – [½(1150)(15)2 + ½(1575)(5.0)2]
ΔKE = – [ ]
ΔKE = J
ΔKE = J

38. Section 11.2 Conservation of Energy
Did We Answer Our Essential Questions?
Under what conditions is energy conserved?
What is mechanical energy, and when is it conserved?
How are momentum and kinetic energy conserved or changed in a collision?