1. WORK, POWER, & ENERGY

2. A General Definition of Work
Work is a scalar quantity equal to the product of the displacement x and the component of the force Fx in the direction of the displacement.
Work = Force component X displacement
Work = (F cos ) x

3. In order for work to be done, three things are necessary:
There must be an applied force.
The force must act through a certain distance, called the displacement.
The force must have a component along the displacement.
d = W = 0
W = Fd
W = Fd cosθ

4. A teacher applies a force to a wall and becomes exhausted.
Read the following statements and determine whether or not they represent examples of work.
A teacher applies a force to a wall and becomes exhausted.
NO, displacement doesn’t occur
A book falls off a table and free falls to the ground.
Yes, displacement in the direction of force

5. You are carrying groceries to the car.
NO, the displacement and the applied force are NOT in the same direction

6. A truck carries a box in it’s bed 100 m.
NO, This is not an example of work. The force is upward on the box but the displacement is along the ground.

7. W = F d W = (Newtons )(meters) W = Nm W = Joule (J)
The units of work are;
W = F d
W = (Newtons )(meters)
W = Nm
W = Joule (J)
In customary; W = foot pounds

8. You do twice as much work
You do twice as much work. If work is equal to force times distance, twice the distance with the same force is twice the work.
If the distance is the same but the force doubles you do twice as much work since work equals force times distance.

9. Positive Work F x Force F contributes to displacement x.
Example: If F = 40 N and x = 4 m, then
Work = (40 N)(4 m) = 160 Nm
Work = 160 J
1 Nm = 1 Joule (J)

10. Negative Work x f The friction force f opposes the displacement.
Example: If f = -10 N and x = 4 m, then
Work = (-10 N)(4 m) = - 40 J
Work = - 40 J

11. Work of a Force at an Angle
x = 12 m
F = 70 N
60o Fy
Work = Fx x
Work = (F cos ) x Fx
Work = (70 N) Cos 600 (12 m) = 420 J
Only the x-component of the force does work!
Work = 420 J

12. Example What work is done by a 60 N force in dragging the bag a distance of 50 m when the force is transmitted by a handle making an angle of 30 with the horizontal?
W= FcosΘd
W= (60 N)(cos 30º)(50m)
W= J

13. Note: Work is positive since Fx and x are in the same direction.
Example: A lawn mower is pushed a horizontal distance of 20 m by a force of 200 N directed at an angle of 300 with the ground. What is the work of this force?
300
x = 20 m
F = 200 N F
Work = (F cos q ) x
Note: Work is positive since Fx and x are in the same direction.
Work = (200 N)(20 m) Cos 300
Work = J

14. Graph of Force vs. Displacement
Assume that a constant force F acts through a parallel displacement Dx.
Force, F
The area under the curve is equal to the work done. F x1 x2
Area
Work = F(x2 - x1)
Displacement, x

15. Example for Constant Force
What work is done by a constant force of 40 N moving a block from x = 1 m to x = 4 m?
40 N
Force, F
Displacement, x
1 m
4 m
Area
Work = F(x2 - x1)
Work = (40 N)(4 m - 1 m)
Work = 120 J

16. POWER is the rate at which work is done.
Power = (work)
(time)
P = W t
P = Joules
sec
P = J/s = Watts = W
In customary; Power = horsepower= hp
746 W = 1 hp

17. P = Fv v = d/t P = W t = Fcosθ d t = F cosθ v Nm/s m/s N m/s2 = 3730 N
Example A 0.25 hp motor is used to lift a load at the rate of 5 cm/s. How great a load can it lift at this constant speed? (1 hp = 746 W)
P = 0.25 hp (746 W/hp) = W
v = 0.05 m/s
v = d/t
P = W t
= Fcosθ d t
= F cosθ v
P = Fv
Nm/s
m/s
= 3730 N N
m/s2
= kg

19. Energy is the ability to do work or that which can be converted into work..
When something has energy, it is able to perform work or, in a general sense, to change some aspect of the physical world

20. In mechanics we are concerned with two kinds of energy:
KINETIC ENERGY: K, energy possessed by a body by virtue of its motion.
K = ½ mv2
Units: Joules (J)
POTENTIAL ENERGY: U, energy possessed by a system by virtue of position or condition.
U = m g h
Units: Joules (J)

21. An object that possesses kinetic energy must be moving since kinetic energy is the energy of motion.
When an object is raise above the ground it gains a certain amount of potential energy. If the same object is raised twice as high, it gains (a) four times as much potential energy (b) twice as much potential energy (c) no additional potential energy.
The raised object now has twice as much potential energy. The mass and the acceleration due to gravity did not change so the only quantity that changed is the height which doubled. The doubling of height increases PE by a factor of two.

22. Example Find the kinetic energy of a 3200 N automobile traveling at 75 km/h?
Fg = 3200 N
v = 20.8 m/s
m = W/g = kg
K = ½ mv 2
= ½ (326.5kg) (20.8m/s)2
= 7.09 x104 J

23. U = mgh = 0.25kg (9.8m/s2) (0.2m) = 0.49 J U = mgh
Example A 250 g object is held 200 mm above a workbench that is 1 m above the floor. Find the potential energy relative to
a. the bench top
U = mgh
= 0.25kg (9.8m/s2) (0.2m)
= 0.49 J
m = 0.25 kg
h = 0.2 m
b. the floor
h = 1.2 m
U = mgh
= 0.25kg (9.8m/s2) (1.2m)
= 2.94 J

24. LAW OF CONSERVATION OF ENERGY
The law of conservation of energy states that:
"Energy is neither created nor destroyed."
Energy can be transformed from one kind to another,
but the total amount remains constant.

25. For mechanical systems involving conservative forces, the total mechanical energy equals the sum of the kinetic and potential energies of the objects that make up the system.

26. The ball has a maximum U at point --------
The ball has a maximum K at point
The ball has a maximum speed at point

27. U = mgh = (5kg)(9.8m/s2)(3m) = 147J Etotal = U+ K
Example
A 5 kg rock is lifted to a height of 6.0m.
a) What is the work done to lift the rock?
W = FcosΘd
W = 5kg(9.8m/s2)(cos 0)(6.0m)
W= 294 J
b) What is the potential energy of the rock at 6.0m? The Kinetic Energy?
U = mgh
U = 5kg(9.8m/s2)(6.0m)
U = 294 J K = 0J
c) What is the U and K when the rock is at 3m?
U = mgh = (5kg)(9.8m/s2)(3m) = 147J
Etotal = U+ K
294 J = 147 J + K K = 147J

28. d) What is the K and U just before the rock hits the ground?
U = mgh = (5kg)(9.8m/s2)(0m) = 0J
Etotal = U + K
294 J = 0 J + K K = 294J
e) What is the velocity of the rock just before it hits the ground?
Utop + Ktop = Ubottom + Kbottom
mgh = ½ mv2
(9.8m/s2)(6m)
V = m/s

29. #9 of homework

30. Einitial = Efinal Ki + Ui = Kf + Uf mgh = ½ mv2 = 5.6 m/s
Example A 40 kg pendulum ball is pulled to one side until it is 1.6 m above its lowest point. What will its velocity be as it passes through its lowest point?
m = 40 kg
h = 1.6 m
h=0
Einitial = Efinal
Ki + Ui = Kf + Uf
mgh = ½ mv2
= 5.6 m/s

31. WORK, POWER, & ENERGY Day 3

32. CONSERVATIVE AND NON-CONSERVATIVE FORCES
The work done by a conservative force depends only on the initial and final position of the object acted upon. An example of a conservative force is gravity. The work done equals the change in potential energy and depends only on the initial and final positions above the ground and NOT on the path taken.

33. Friction is a non-conservative force and the work done in moving an object against a non-conservative force depends on the path. For example, the work done in sliding a box of books against friction from one end of a room to the other depends on the path taken.

34. W =  K = Kf –Ko WORK-ENERGY PRINCIPLE:
The work of a resultant external force on a body is equal to the change in kinetic energy of the body.
W =  K = Kf –Ko

35. W = ΔK = ½ mvf2 - ½ mvo2 Fd = ½ mvf2 – ½ mvo2 F = - ½ mvo2 d
Example 1 What average force F is necessary to stop a 16 g bullet traveling at 260 m/s as it penetrates into wood at a distance of 12 cm?
W = ΔK = ½ mvf2 - ½ mvo2
m = kg
d = 0.12 m
vo = 260 m/s
vf = 0 m/s
Fd =
½ mvf2 – ½ mvo2
F = - ½ mvo2 d
F = N

36. W = ΔK = ½ (1200kg)(25m/s)2= 3.75x105 J = 47 kW
Example 2 An advertisement claims that a certain 1200-kg car can accelerate from rest to a speed of 25 m/s in a time of 8.0 s. What power must the motor produce to cause this acceleration?
m = 1200 kg
vo = 0 m/s vf = 25 m/s
t = 8 s
W = ΔK
= ½ (1200kg)(25m/s)2= 3.75x105 J
= 47 kW

37. W = U W = mg(hf-ho) WORK-ENERGY PRINCIPLE Part II:
The work of a resultant external force on a body is equal to the change in gravitational
potential energy of the body.
W = U
W = mg(hf-ho)

38. For mechanical systems involving non-conservative forces, the total mechanical energy equals the sum of the kinetic and potential energies – the work done against friction of the objects that make up the system.

39. Some more examples of non-conservation

40. Summary: Energy Gains or Losses:
Gravitational Potential Energy
U = mgh
Spring Potential Energy
Kinetic Energy
Work Against Friction
Work = fx

41. Summary: Conservation of Energy
The basic rule for conservation of energy:
mgho
½kxo2
½mvo2 =
mghf
½kxf2
½mvf2 +
Work against friction: fk x
Remember to use the absolute (+) value of the work of friction. (Loss of energy)

42. ΔU = ΔK ΔU = m2gh – m1gh = 0.7 ( 9.8) (1.2) - 0.8 (9.8) (1.2)
6.8 In an Atwood machine the two masses are 800 g and 700 g. The system is released from rest. How fast is the 800-g mass moving after it has fallen 120 cm?
ΔU = ΔK
ΔU = m2gh – m1gh
= 0.7 ( 9.8) (1.2) (9.8) (1.2)
= J
m1 = 0.8 kg
m2 = 0.7 kg
h = 1.2 m
loss of U = gain in K
= 1.17 J
= 1.25 m/s

43. At point A Energy: UA + KA At point B UA + KA = KB = 4.4 m/s
Example If friction forces are negligible and the bead has a speed of 2 m/s at point A, a. What will be its speed at point B? A
vA = 2 m/s
hA = 0.8 m
hB = 0 m C B
At point A
Energy: UA + KA
At point B
UA + KA = KB
= 4.4 m/s

44. At point C UA + KA = UC + KC = 3.14 m/s
b. What will be its speed at point C?
hC = 0.5 m A
At point C
UA + KA = UC + KC C B
= 3.14 m/s

45. GENERAL CASE
In real life applications, some of the mechanical energy is lost due to friction. The work due to non-conservative forces is given by:
  WNC = Δ K + Δ U

46. WFf = Δ K + Δ U = Ff d = 0.015(9.8)(0.5-0.8) = - 0.04 J
Example Suppose the bead in Prob. 6.9 has a mass of 15 g and a speed of 2 m/s at A, and it stops as it reaches point C. The length of the wire from A to C is 250 cm. How large is the average friction force that opposes the motion of the bead?
m = kg d = 2.5 m
vA = 2 m/s vC = 0 m/s
hA = 0.8 m hC = 0.5 m
WFf = Δ K + Δ U = Ff d
= 0.015(9.8)( ) = J
= 0 - ½ (0.015)(2)2 = J
= x N

47. Example A 64 N block rests initially at the top of a 30 m plane inclined at an angle of 30. If k = 0.1. Find the final velocity at the bottom of the plane from energy considerations.
Fg = 64 N
m = 64/9.8 = 6.5 kg
l = 30 m
μ = 0.1 l
h= 30 sin 30˚= 15 m h
At the TOP
U = mgh
= 64(15)
= 960 J

48. At the BOTTOM U = K + WFf Ff = μ FN = μ Fg cos 30˚ = 0.1 (64 cos 30˚ )
Fg = 64 N
m = 64/9.8 = 6.5 kg
l = d = 30 m
μ = h= 15 m
At the BOTTOM
U = K + WFf
Ff = μ FN = μ Fg cos 30˚
= 0.1 (64 cos 30˚ )
= 5.54 N
WFf = Ff d
= 5.54 (30)
= J
K = U - WFf =
= J
K = ½ mv2
= m/s

50. The correct ranking is C>H>E>D>G>B>F>A
We are interested in one half of the mass of the load as well as the speed of the car/load squared.

52. The ranking would be F>A=B>C>D>E
The ranking would be F>A=B>C>D>E. We need to concern ourselves with the mass and change in height.