1. Chapter 6: Conservation of Energy1

2. Introduction Energy – ability to perform work.
Unit of energy (and of work) – Joules (J)
1 J = 1 N- m = 1 kg- m2 / s2
Forms of energy:
Light, chemical, nuclear, mechanical, electrical, sound, heat, kinetic, elastic, magnetic etc.
Each type of energy is calculated differently.
Maybe do example of conservation (like cookies) and counter example. Force is NOT conserved.

3. Law of Conservation of Energy
In any natural process, total energy is always “conserved”, i.e. energy can not be created nor destroyed.
Can be transformed from one form to another.
Can be transferred from one system to
another.
In science, any law of conservation is a very powerful tool in understanding the physical universe.
Maybe do example of conservation (like cookies) and counter example. Force is NOT conserved.

4. Work by Constant Force Work – transfer of energy as a result of force.
Constant force – its magnitude and direction unchanged.
The force acts on the object throughout the process.
Only component of force parallel to direction of motion performs work!

5. Work by Constant Force
Work done by force F in moving the object at constant speed through displacement Dx :
W = (F cos q) Dx
F cos q = component of F parallel to Dx.
Work is a scalar quantity.
Unit = Joule (J) J = 1 N-m F x  f N W

6. Work by Constant Force W = (F cos q) Dx 1. If Dx = 0, then W = 0
If you held a 50 kg bag on your head without moving for 3 hours, would you have done work?

7. W = (F cos q) Dx 2. If q = 0, cos 0 = 1 ie F and Dx are parallel
and W = F Dx x F

8. W = (F cos q) Dx
3. If q = 90o, cos 90o = 0, ie F and Dx are perpendicular to each other
and W = 0 F x  f N W
Work done by the normal force N is zero

9. W = (F cos q) Dx
4. If q = 180o, cos 180o = -1, ie F and Dx are antiperpendicular to each other
and W = negative. F x  f N W
Work done by the frictional force f is - F Dx

10. Example:
A 30 N chest was pulled 5 m across the floor at a constant speed by applying a force of 50 N at an angle of 30o. How much work was done by tension T?
W = F Dx cos q
= (50 N) (5 m) cos (30)
= 217 Joules T mg N f
30o
50 N

11. Example:
A 30 N chest was pulled 5 m across the floor at a constant speed by applying a force of 50 N at an angle of 30o. How much work was done by gravity? T mg N f
30o
50 N mg f mg
90o Dx
W = mg Dx cos q
= 30 x 5 cos(90) = 0

12. Example:
A 30 N chest was pulled 5 m across the floor at a constant speed by applying a force of 50 N at an angle of 30o. How much work was done by friction f?
50 N mg f
To find the magnitude of f:
Consider x-component of forces acting.
Since constant velocity: Fnet = 0
So Tcos30o – f = 0
Thus f = 43.3 N.
W = 43.3 x (cos180o) x 5 = -217 J T mg N f f Dx
180o

13. Work by Variable Force W = Fx Dx Work = area under F vs x plot
F = kx
Spring F = k x
Area = ½ k (x)2 = Wspring

14. Example
A dart gun with a spring constant k = N/m is compressed 8.0 cm. How much energy will it transfer to a dart when the spring is released?
W = ½ k(x)2

15. Example
As you ride in an elevator going upward at constant speed, work done on you by the normal force from the floor of the elevator
is negative.
is zero.
is positive.
is zero, negative or positive depending on the speed of the elevator.

16. Kinetic Energy
Suppose a constant force F is applied horizontally on a small rigid object of mass m.
Its velocity will change, say from vo to v as it moves through a distance x (without rotation). Work done on the object
W = Fx x [F parallel to x]
= m ax x [Recall: v2 = vo2 + 2a(x-xo)]
= ½ m (v2 – v02) = ½ mv2 - ½ mv02
The quantity ½ . mass . (velocity)2 is called kinetic energy of the object.

17. An object of mass m, moving with velocity v, has kinetic energy
K = ½ mv2
K  m: If m is doubled, K will double.
K  v2 If v is doubled, K will quadruple.
Can K ever be zero?
Can K ever be negative?
W = K Can W ever be negative?

18. Kinetic Energy F vo v Dx
If more than one force acts on the object, the change in kinetic energy is the total work done by all the forces acting on the object.
Total work Wtotal = DK
This is called the Work- Energy Principle

19. Total Work Total work Wtotal = DK = change in K.E.
Wtotal = First find work done by each force ,
then add them.
= first add all the forces and then then calculate work done by the net force
Q: What is the total work done on an object moving with constant velocity?
(A) Positive (B) Negative (C) Zero

20. WT = F Dx cos q = (50 N)(5 m) cos (30) = 217 J
Example 1:
A 30 N chest was pulled 5 m across the floor at a constant speed by applying a force of 50 N at an angle of 30o. How much work was done by all the forces acting on it?
WT = F Dx cos q = (50 N)(5 m) cos (30) = 217 J
WN = 0, Wmg = 0 Wf = -217J
Wtotal = (-217) = 0 T mg N f
30o
T = 50 N

21. Example 2:
A 30 N chest was pulled 5 m across the floor by applying a force of 50 N at an angle of 30o. How much work was done by all the forces acting on it if its speed changed from 2 m/s to 5 m/s in the process?

22. Example W FN V
You are towing a car up a hill with constant velocity. The work done on the car by the normal force is:
1. positive 2. negative 3. zero T
correct
Since the direction of the force is positive the value of work will be positive.
it's negative because it's trying to slow down the car. The normal force does no work because it acts in a direction perpendicular to the displacement of the car.
Work done by gravity?
Work done by tension?

23. Example: Block w/ friction
A block is sliding on a surface with an initial speed of 5 m/s. If the coefficient of kinetic friction between the block and table is 0.4, how far does the block travel before stopping? x y mg N f
Y direction: F=ma
N-mg = 0
N = mg
Work
WN = 0
Wmg = 0
Wf = f Dx cos(180)
= -mmg Dx
W = D K
-mmg Dx = ½ m (vf2 – v02)
-mg Dx = ½ (0 – v02)
mg Dx = ½ v02
Dx = ½ v02 / mg
= 3.1 meters
5 m/s

24. POTENTIAL ENERGY Potential Energy (U) = stored energy.
It is the energy an object possess due to its position or its configuration.
There are different types of potential energy:
Gravitational potential energy.
Elastic potential energy.
Electrical potential energy.

25. Gravitational Potential Energy
To move mass m from initial point to final point at constant velocity, an external force FEXT, equal but opposite to the force of gravity must be applied.
WEXT = F cos y = mgh
= (mg)(cos0)h
= mgh
Wg = mgcos(180o)(h)
= -mgh mg
Yo = 0
Yf = h
FEXT

26. Gravitational Potential Energy
The quantity mgh is called gravitational potential energy near the earth’s surface:
U = mgh
where U = 0 wherever we have chosen h = 0
More general way of writing gravitational potential energy is
U = mg h
FEXT
U = 0 mg

27. The gravitational potential energy of an object of mass m at a height h is
Ug = mgh
Ug = mgh and Wg = -mgh ie Ug = - Wg.
Ug  m
Ug  h
A reference level where h = 0 can be chosen to be at any convenient location.
Only change in Ug is important.
Ug = mgh = mg(hf – hi)

28. Gravitational Potential Energy
Example 3: A block slid down a plane
If a block is slid down a plane.
Distance moved down the plane = s
Vertical height moved = h
How much work will the force of gravity do? h  mg S
W = F cos x
Wg = mg . cos . S
But S = y /cos
Wg = (mg)(cos)(h/cos) = mgh

29. Elastic Potential Energy
Elastic (Spring) force F = -kx
Work done by spring force W = ½ kx2
Energy stored in the spring when compressed or stretched by distance x
Elastic Potential Energy U = ½ kx2

30. Example
A dart gun with a spring constant k = N/m is compressed 8.0 cm.
(a) How much energy will it transfer to a dart when the spring is released?
(b) With what speed will the dart (mass 200 g) leave the gun?

31. Example
Emil throws an orange straight up and catches it at the same point it was thrown.
How much work is done during the orange’s free fall?
If the orange was thrown upward from a 1.2 m height above the ground and falls to the ground, how much work is done by gravity? [Mass of orange = .25 kg]

32. Conservative and Non Conservative Forcesh  mg
Dropped
Sliding down a plane
Thrown
Work done by force of gravity, Wg = - mgh
Independent of path

33. Gravitational Potential Energy
To move mass m from initial point to final point at constant velocity, an external force FEXT, equal but opposite to the force of gravity must be applied.
WEXT = F cos y = mgh
= (mg)(cos0)h
= mgh
Wg = mgcos(180o)(h)
= -mgh mg
Yo = 0
Yf = h
FEXT

34. Negative, zero or positive depending on the speed of the elevator
1. As you ride an elevator going upward at constant speed, the work done on you by the normal force from the floor of the elevator is
negative
zero
positive
Negative, zero or positive depending on the speed of the elevator 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65

35. 2. What is the total work done on an object moving with constant velocity?
Positive
Negative
Zero 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65

36. 3. A passenger whose mass is 95 kg is seated in a jet airliner flying 1,200 m above the ground at a speed of 253 m/s. What is the kinetic energy of the passenger?
931 J
6.08 x 106 J
3.04 x 106 J
1.12 x 106 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65

37. 4. A passenger whose mass is 95 kg is seated in a jet airliner flying 1,200 m above the ground at a speed of 253 m/s. What is the potential energy of the passenger?
931 J
6.08 x 106 J
3.04 x 106 J
1.12 x 106 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65

38. 5. A spring has an elastic constant of 920 N/m
5. A spring has an elastic constant of 920 N/m. By how much should it be stretched in order to store 120 J of energy?
0.36 m
0.51 m
0.26 m
7.7 m
1.96 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65

39. Conservative Forces Forces which perform same amount of work
independent of the path taken are called
conservative forces. Eg: gravity, elastic force, electric force.
Work done by a conservative force depends only on the initial and final position and not on the path taken to get there.
Potential energy only goes with conservative forces. There is potential energy associated with gravity, elastic force, electric force.
WC = - U [ Ug = mgh and Uspring = ½ kx2]

40. Non Conservative Forces
Amount of work done by them depend on the path taken. Eg: friction, tension, push/pull.
There is no potential energy associated with non conservative forces.

41. Work - Energy w/ Conservative Forces
Total work = Work done by all types of forces.
= Work by conservative forces + work
by non-conservative forces
WTotal = WC + WNC
But WTotal = K and WC = - U
So, K = - U + WNC ie, WNC = K + U
WNC = K + U

42. Work - Energy w/ Conservative Forces
WNC = K + U
When only conservative forces are acting; WNC = 0
0 = K + U OR = (Kf – Ki) + (Uf – Ui)
OR Kf + Uf = Ki + Ui
The quantity K + U is called mechanical energy E
Thus Ef = Ei (Conservation of mechanical energy)

43. Skiing Example (no friction)
A skier goes down a 78 meter high hill with a variety of slopes. What is the maximum speed she can obtain if she starts from rest at the top? [Assume friction is negligible]
Conservation of energy:
Ki + Ui = Kf + Uf
½ m vi2 + m g yi = ½ m vf2 + m g yf
0 + g yi = ½ vf2 + g yf
vf2 = 2 g (yf-yi)
vf = sqrt( 2 g (yf-yi))
vf = sqrt( 2 x 9.8 x 78) = 39 m/s

44. Two similar objects are released from rest at the same time to slide down two frictionless slopes A and B of different inclines as shown in the figure below. Which of these statements is true about the motion of the two balls? They will reach the bottom with the same
Acceleration (B) Speed (C) time duration
(D) Same acceleration, speed, time duration A B h h
30o

45. Example
Suppose the initial kinetic and potential energies of a system are 75J and 250J respectively, and that the final kinetic and potential energies of the same system are 300J and -25J respectively. How much work was done on the system by non-conservative forces?
1. 0J J J J J
Initial = 75 J J = 325 J.
Final = 300 J - 25 J = 275 J.
Final -Initial = = -50 J. No energy goes into or comes out of the system overall, so some non-conservative force has to be accounted for.

46. Example
A dart gun with a spring constant k = N/m is compressed 8.0 cm.
How much energy will it transfer to a dart of mass 200 g when the spring is released?
What would be the escape speed of the dart?
If the dart was aimed upward, how high would it travel?
Uspring = ½ k(x)2 K = ½ mv2, Ug = mgh

47. Power Average Power P = rate at which work is being done.
= rate at which energy is being transformed.
P = DW / Dt Units: J/s = Watt
DW/ Dt = [F Dx cos(q)]/ Dt
= F (v Dt) cos(q)
i.e., P = F v cos(q)

48. Example:
How much electrical energy will a 75-watt light bulb use if left lit for 5 hours? [1 hour = 3,600 s]

49. Example:
A hot plate used 225,000 J of electrical energy in 5 minutes. What is the power (wattage) of the hot plate?