1. Work, Energy and Power AP style

2. Energy Energy: the currency of the universe.
Everything has to be “paid for” with energy.
Energy can’t be created or destroyed, but it can be transformed from one kind to another and it can be transferred from one object to another.

3. Work = Force x displacement
Doing WORK is one way to transfer energy from one object to another.
Work = Force x displacement
W = F∙d
Unit for work is Newton x meter. One Newton-meter is also called a Joule, J.

4. Work- the transfer of energy

5. Work = Force · displacement
Work is not done unless there is a displacement.
If you hold an object a long time, you may get tired, but NO work was done on the object.
If you push against a solid wall for hours, there is still NO work done on the wall.

6. For work to be done, the displacement of the object must be along the same direction as the applied force. They must be parallel.
If the force and the displacement are perpendicular to each other, NO work is done by the force.

7. But since they are in opposite directions, now it is NEGATIVE work. d
For example, in lifting a book, the force exerted by your hands is upward and the displacement is upward- work is done.
Similarly, in lowering a book, the force exerted by your hands is still upward, and the displacement is downward.
The force and the displacement are STILL parallel, so work is still done.
But since they are in opposite directions, now it is NEGATIVE work. F d F d

8. Therefore, NO work is done by your hands.
On the other hand, while carrying a book down the hallway, the force from your hands is vertical, and the displacement of the book is horizontal.
Therefore, NO work is done by your hands.
Since the book is obviously moving, what force IS doing work???
The static friction force between your hands and the book is acting parallel to the displacement and IS doing work! F d

9. Both Force and displacement are vectors
So, using vector multiplication,
Work = 𝑭 • 𝒅
This is a DOT product- only parallel components will yield a non-zero solution.
In many university texts, and sometimes the AP test, the displacement, which is the change in position, is represented by “s” and not “d”
W = 𝑭 • 𝒔
The solution for dot products are NOT vectors.
Work is not a vector.
Let’s look an example of using a dot product to calculate work…

10. An object is subject to a force given by
F = 6i – 8j as it moves from the position r = -4i + 3j to the position r = i + 7j (“r” is a position vector)
What work was done by this force?
First find the displacement, s
s = Dr =
rf – ro =
(i + 7j) - (-4i + 3j) =
5i + 4j
Then, do the dot product W = F · s
(6i – 8j) · (5i + 4j) =
30 – 32 = -2 J of work

11. Example
How much work is done to push a 5 kg cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity?
W = Force · displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp?
W = 25 N x 7 m
W = 175 J
7 m
F = 25 N
3 m

12. Example
How much work is done to carry a 5 kg cat to the top of a ramp that is meters long and 3 meters tall at a constant velocity?
W = Force · displacement
What force is required to carry the cat?
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height?
d = height NOT length
W = mg x h
W = 5 kg x 10 m/s2 x 3 m
W = 150 J
7 m
3 m

13. If all four ramps are the same height, which ramp would require the greatest amount of work in order to carry an object to the top of the ramp?
W = F∙ Dr
For lifting an object, the distance, d, is the height.
Because they all have the same height, the work for each is the same!

14. The force required is your weight!
And,….while carrying yourself when climbing stairs or walking up an incline at a constant velocity, only the height is used to calculate the work you do to get yourself to the top!
The force required is your weight!
Your Force
Vertical component of d
Horizontal component of d

15. How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long?
ZERO, because your Force is vertical, but the displacement is horizontal.

16. Example
Displacement = 20 m
A boy pushes a lawnmower 20 meters across the yard. If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees, how much work did he do?
F cos q q F
W = (F cos q )d
W = (200 N cos 50˚) 20 m
W = 2571 J

17. NOTE: If while pushing an object, it is moving at a constant velocity,
the NET force must be zero.
So….. Your applied force must be exactly equal to any resistant forces like friction.
S F = ma
FA – f = ma = 0

18. ●Does the gravitational force do any work?
A 5.0 kg box is pulled 6 m across a rough horizontal floor (m = 0.4) with a force of 80 N at an angle of 35 degrees above the horizontal. What is the work done by EACH force exerted on it? What is the NET work done?
●Does the gravitational force do any work?
NO! It is perpendicular to the displacement.
● Does the Normal force do any work?
No! It is perpendicular to the displacement.
● Does the applied Force do any work?
Yes, but ONLY its horizontal component!
WF = FAcosq x d = 80cos 35˚ x 6 m = J
● Does friction do any work?
Yes, but first, what is the normal force? It’s NOT mg!
Normal = mg – FAsinq
Wf = -f x d = -mN∙d = -m(mg – FAsinq)∙d = J
● What is the NET work done?
J – 7.47 J = J
Normal FA f q mg

19. W = F • d (this is a DOT product!!)
For work to be done, the displacement of the object must be in the same direction, as the applied force. They must be parallel.
If the force and the displacement are perpendicular to each other, NO work is done by the force.
So, using vector multiplication,
W = F • d (this is a DOT product!!)
(In many university texts, as well as the AP test, the displacement is given by d = Dr, where r is the position vector: displacement = change in position
W = F • Dr
If the motion is in only one direction,
W = F • Dx Or
W = F • Dy

20. An object is subject to a force given by
F = 6i – 8j as it moves from the position r = -4i + 3j to the position r = i + 7j (“r” is a position vector)
What work was done by this force?
First find the displacement, Dr
rf – ro =
(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then, complete the dot product W = F · Dr
(6i – 8j) · (5i + 4j) =
30 – 32 = -2J of work

21. Varying Forces The rule is… “If the Force varies, you must integrate!”
If the force varies with displacement-
in other words,
Force is a function of displacement,
you must integrate to find the work done.

22. Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N. What is the work done by the force?

23. Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 – x.
What work is done to stretch the spring 2 meters beyond its equilibrium position?

24. Graphing Force vs. postion
If you graph the applied force vs. the position, you can find how much work was done by the force.
Work = F·d = “area under the curve”.
Total Work = 2 N x 2 m + 3N x 4m = 16 J
Area UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force, N
Position, m F
Net work = 16 J – 2 J = 14 J d

25. The Integral = the area under the curve!
If y = f(x), then the area under the f(x) curve from x = a to x = b is equal to

26. Therefore, given a graph of Force vs
Therefore, given a graph of Force vs. position, the work done is the area under the curve. 𝑊= 𝐹∙𝑑𝑟

28. Work and Energy

29. Work = Force x distance = transfer of energy
Work and Energy
Often, some force must do work to give an object potential or kinetic energy.
You push a wagon and it starts moving. You do work to stretch a spring and you transform your work energy into spring potential energy.
Or, you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy.
Work = Force x distance = transfer of energy

30. Kinetic Energy
the energy of motion
K = ½ mv2

31. Where does K = ½ mv2 come from??
Did your teacher just amazingly, miraculously make that equation up??
Hmmm…

32. The “Work- Kinetic Energy Theorem”
S Work = DK
S F • d = DK = ½ mvf2 - ½ mvo2
College Board AP Objective: You are supposed to be able to derive the work-kinetic energy theorem….. (this does not mean to take a derivative- it means to start with basic principles- S S F= ma, S W = S F •d, and kinematics equations and develop the equation for kinetic energy)
So…. do it!! Then you’ll see where the equation we call “kinetic energy” comes from.
(Hint: start with S F = ma and use the kinematics equation that doesn’t involve time.)

33. NET Work by all forces = D Kinetic Energy SF·d = D½ mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)?
The work done by the forces stopping the car = the change in the kinetic energy
SF·d = D½ mv2
With TWICE the speed, the car has
FOUR times the kinetic energy.
Therefore it takes FOUR times the stopping distance.

34. A car going 20 km/h will skid to a stop over a distance of 7 meters
A car going 20 km/h will skid to a stop over a distance of 7 meters. If the same car was moving at 50 km/h, how many meters would be required for it to come to a stop? The velocity changed by a factor of 2.5, therefore the stopping distance is 2.52 times the original distance: 7 meters x 6.25 = meters

35. Example, Wnet = SF∙d = DK
A 500 kg car moving at 15 m/s skids 20 m to a stop.
How much kinetic energy did the car lose?
DK = ½ mvf2
DK = -½ (500 kg)(15 m/s) 2
DK = J
What is the magnitude of the net force applied to stop the car?
SF·d = DK
SF = DK / d
SF = J / 20 m
F = N

36. SF∙d = D ½ mv2
A 0.02 kg bullet moving at 90 m/s strikes a block of wood. If the bullet comes to a stop at a depth of 2.5 cm inside the wood, how much force did the wood exert on the bullet? F = 3240 N

37. Example Wnet = SF∙d = DK
A 500 kg car moving on a flat road at 15 m/s skids to a stop.
How much kinetic energy did the car lose?
DK = D½ mvf2
DK = -½ (500 kg)(15 m/s)2
DK = J
How far did the car skid if the effective coefficient of friction was m = 0.6?
Stopping force = friction = mN = mmg
SF·d = DK
-(mmg)·d = DK
d = DK / (-mmg) *be careful to group in the denominator!
d = J / (0.6 · 500 kg · 9.8 m/s2) = m

39. First, a review of Conservative Forces and Potential Energy from the notes…
What we have covered so far- the College Board Objectives state that the student should know:
The two definitions of conservative forces and why they are equivalent.
Two examples each of conservative and non-conservative forces.
The general relationship between a conservative force and potential energy.
Calculate the potential energy if given the force.
Calculate the force if given the potential energy.

40. Conservation of Mechanical Energy

41. Mechanical Energy E = ½ mv2 + U
Mechanical Energy = Kinetic Energy + Potential Energy
E = ½ mv U

42. Potential Energy Stored energy
It is called potential energy because it has the potential to do work.

43. Example 1: Spring potential energy in the stretched string of a bow or spring or rubber band. Us= ½ kx2
Example 2: Chemical potential energy in fuels- gasoline, propane, batteries, food!
Example 3: Gravitational potential energy- stored in an object due to its position from a chosen reference point.

44. Gravitational potential energy
Ug = weight x height
Ug = mgh

45. The Ug may be negative. For example, if your reference point is the top of a cliff and the object is at its base, its “height” would be negative, so mgh would also be negative.
The Ug only depends on the weight and the height, not on the path that it took to get to that height.

46. “Conservative” forces - mechanical energy is conserved if these are the only forces acting on an object.
The two main conservative forces are: Gravity, spring forces
“Non-conservative” forces - mechanical energy is NOT conserved if these forces are acting on an object.
Forces like kinetic friction, air resistance (which is really friction!)

47. Conservation of Mechanical Energy
If there is no kinetic friction or air resistance, then the total mechanical energy of an object remains the same.
If the object loses kinetic energy, it gains potential energy.
If it loses potential energy, it gains kinetic energy.
For example: tossing a ball upward

48. Conservation of Mechanical Energy
The ball starts with kinetic energy…
Which changes to potential energy….
Which changes back to kinetic energy
PE = mgh
What about the energy when it is not at the top or bottom?
E = ½ mv2 + mgh
Energybottom = Energytop
½ mvb2 = mght
K = ½ mv2
K = ½ mv2

49. Examples dropping an object box sliding down an incline
tossing a ball upwards
a pendulum swinging back and forth
A block attached to a spring oscillating back and forth
First, let’s look at examples where there is NO friction and NO air resistance…..

50. Conservation of Mechanical Energy
If there is no friction or air resistance, set the mechanical energies at each location equal. Remember, there may be BOTH kinds of energy at any location. And there may be more than one form of potential energy, U!
E = E2
U1 + ½mv12 = U2 + ½ mv22

51. Some EASY examples with kinetic and gravitational potential energy…

52. A 3. 0 kg Egg sits on top of a 12 m tall wall
A 3.0 kg Egg sits on top of a 12 m tall wall. What is its potential energy? U = mgh U = 3 kg x 10 m/s2 x 12 m = 360 J If the Egg falls, what is its kinetic energy just before it hits the ground? Potential energy = Kinetic energy = 360 J What is its speed just before it strikes the ground? 360 J = K = ½ mv2 2(360 J) / 3 kg = v2 v = m/s

53. Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive. If the window was 80 m high, how fast was the brush moving just before it hit the ground? (g = 10 m/s2) mgh1 + ½ mv12 = mgh2 + ½ mv22 mgh = ½ mv2 gh = ½ v2 2gh = v2 Don’t forget to take the square root!

54. Now… do one on your own
An apple falls from a tree that is 1.8 m tall. How fast is it moving just before it hits the ground? (g = 10 m/s2) mgh1 + ½ mv12 = mgh2 + ½ mv22 mgh = ½ mv2

55. And another one…
A woman throws a ball straight up with an initial velocity of 12 m/s. How high above the release point will the ball rise? g = 10 m/s2 mgh1 + ½ mv12 = mgh2 + ½ mv22 ½ mv2 = mgh h = ½ v2 / g

56. And another one…
Mario, the pizza man, tosses the dough upward at 8 m/s. How high above the release point will the dough rise? g = 10 m/s2 mgh1 + ½ mv12 = mgh2 + ½ mv22

57. Conservation of Mechanical Energy- another look
A skater has a kinetic energy of 57 J at position 1, the bottom of the ramp (and NO potential energy)
At his very highest position, 3, he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)
Mechanical energy = K + U
What is his kinetic energy at position 2, if his potential energy at position 2 is 25.7 J?
E = 57 J
U = 25.7 J
K = ??
E = 57 J

58. Conservation of Mechanical Energy… more difficult
A stork, at a height of 80 m flying at 18 m/s, releases his “package”. How fast will the baby be moving just before he hits the ground? Energyoriginal = Energyfinal mgh + ½ mvo2 = ½ mvf2
Vf = 43.5 m/s

59. Now you do one …
The car on a roller coaster starts from rest at the top of a hill that is 60 m high. How fast will the car be moving at a height of 10 m? (use g = 9.8 m/s2) mgh1 + ½ mv12 = mgh2 + ½ mv22 mgh1 = mgh2 + ½ mv22

60. Conservation of Energy with Springs
Example One: A 6 x 105 kg subway train is brought to a stop from a speed of 0.5 m/s in 0.4 m by a large spring bumper at the end of its track. What is the force constant k of the spring? Assume a linear spring unless told otherwise! Solution: All the initial kinetic energy of the subway car is converted into elastic potential energy. ½ mv2 = ½ kx2 Here x is the distance the spring bumper is compressed, x = 0.4 m.

61. K = ½ (0.25 kg) v2 = 2.5 J = Us = Ws v2 = 20 m2 / s2 v = 4.47 m/s
A spring with spring constant k = 500 N/m is compressed a distance x = 0.10 m. A block of mass m = kg is placed next to the spring, sitting on a frictionless, horizontal plane. When the spring and block are released, how fast is the block moving as it leaves the spring?
When the spring is compressed, work is required and the spring gains potential energy,
Us = ½ k x2
Us = ½ (500 N/m) (0.10 m)2
Us = 2.5 J
As the spring and mass are released, this amount of work done to the mass to change its kinetic energy from zero to a final value of K = ½ m v2
K = ½ (0.25 kg) v2 = 2.5 J = Us = Ws
v2 = 20 m2 / s2
v = 4.47 m/s

62. How fast is the block moving when the spring is compressed 0.05 m?
E = 2.5 J
2.5 J = ½ mv2 + ½ kx2

63. mgh1 + ½ mv12 + ½ kx12 = mgh2 + ½ mv22 + ½ kx22
E1 = E2
mgh1 + ½ mv12 + ½ kx12 = mgh2 + ½ mv22 + ½ kx22
Be careful about measuring height!

64. What about objects suspended from a spring?
There are 3 types of energy involved: Kinetic, gravitational and spring potential However, since potential energy is determined using a reference point, you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached), rather than the position where the spring is unstretched. By doing so, the total potential energy, including both Us and Ug can be shown to be Us + Ug = U = ½ ky2 where y is the displacement measured from the equilibrium point.
equilibrium y
E = K + U

65. Mechanical energy will be lost in the form of heat energy.
If there is kinetic friction or air resistance, mechanical energy will not be conserved.
Mechanical energy will be lost in the form of heat energy.
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat.
Final energy – original energy = energy loss

66. Let’s try one…
A 2 kg cannonball is shot straight up from the ground at 18 m/s. It reaches a highest point of 14 m. How much mechanical energy was lost due to the air resistance? g = 10 m/s2 Final energy – original energy = Energy loss mgh – ½ mv2 = Heat loss 2 kg(10 m/s2)(14 m) – ½ (2 kg)(18 m/s)2 = ??

67. And one more…
A 1 kg flying squirrel drops from the top of a tree 5 m tall. Just before he hits the ground, his speed is 9 m/s. How much mechanical energy was lost due to the air resistance? g = 10 m/s2 Final energy – original energy = Energy loss

68. Don’t even think about it…
Sometimes, mechanical energy is actually INCREASED!
For example: A bomb sitting on the floor explodes.
Initially:
Kinetic energy = 0
Gravitational Potential energy = 0
Mechanical Energy = 0
After the explosion, there’s lots of kinetic and gravitational potential energy!!
Did we break the laws of the universe and create energy???
Of course not! NO ONE, NO ONE, NO ONE can break the laws!
The mechanical energy that now appears came from the chemical potential energy stored within the bomb itself!
Don’t even think about it…

69. Law of Conservation of Energy energy cannot be created or destroyed.
According to the
Law of Conservation of Energy
energy cannot be created or destroyed.
But one form of energy may be transformed into another form as conditions change.

71. Power

72. Power The rate at which work is done. 1. Power = Work ÷ time
Unit for power = J ÷ s
= Watt, W
What is a Watt in “fundamental units”?

73. Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 0.5 seconds. How powerful was he?
P = W / t
W = Fd
W = mg x h
W = 80 kg x 10 m/s2 x 2 m = 1600 J
P = 1600 J / 0.5 s
P = 3200 W

74. Since work is also the energy transferred or transformed, “power” is the rate at which energy is transferred or transformed.
Power = Energy ÷ time
This energy could be in ANY form: heat, light, potential, chemical, nuclear
Example: How long does it take a 60 W light bulb to give off 1000 J of energy?
Time = Energy ÷ Power

75. Since NET work = D K,
P = DK ÷ t
Example: How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 m/s in 5.0 s?
Power = ½ mv2 ÷ t

76. And yet another approach:
P = W ÷ t = (F·d) ÷ t = F · (d ÷ t)
P = F · v This is a DOT product!!!
Example: What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts

77. Calculus Applications
A particle of mass m moves along the y-axis as y(t)=at4-bt2+2t, where a and b are constants, due to an applied force, Fy(t) What is the power, P(t) delivered by the force?
P = F·v
We need to find both v and F!

78. Varying Forces
The rule is…
“If the Force varies, you must integrate!”

79. Work = F·d Power = F·v
If the force varies with displacement , in other words,
force is a function of displacement, F(x)
you must integrate to find the work done.
If the force is a function of velocity, F(v), you must integrate to find the power output.
(AND to determine velocity as a function of time, v(t) - all that natural log stuff we did for air resistance)
If the force varies with time, F(t)… well, that’s coming up next unit.

80. Examples:
If F(x) = 5x3, what work is done by the force as the object moves from x = 2 to x = 5?
𝑊= 𝑥 3 𝑑𝑥
If F(v) = 4v2, what power was developed as the velocity changed from 3 m/s to 7 m/s?
𝑃= 𝑣 2 𝑑𝑣

81. Another tricky example using definite integration …
A 2.88 kg particle starts from rest at x = 0 and moves under the influence of a single force, F = 6 + 4x -3x2 where F is in Newtons and x is in meters. Find the power delivered to the particle when it is at x = 3 m.
P = F·v, and finding the force at x = 3 is easy, but what is the velocity at x = 3 m? F = ma and integrating a(t) would yield velocity, but the variable here is NOT t - it’s x!
Hmmm….