1. Work and Energy
Dr. Robert MacKay
Clark College

2. Introduction What is Energy?
What are some of the different forms of energy?
Energy = $$$

3. Overview W K U Work (W) Kinetic Energy (K) Potential Energy (U)
All Are measured in Units of Joules (J)
1.0 Joule = 1.0 N m W K U

4. Overview W K U Work Kinetic Energy Potential Energy Heat Loss

5. Work and Energy Dr =4.0 m W= F Dr = 6.0 N (4.0m) = 24.0 J F= 6.0 N
Work = Force x displacement
W = F d
Actually
Work = Force x displacement parallel to force
Dr =4.0 m
W= F Dr
= 6.0 N (4.0m)
= 24.0 J
F= 6.0 N

6. Work and Energy Dr = 8.0 m F= - 6.0 N W= F Dr = -6.0 N (8.0m) =-48 J
Work = Force x Displacement parallel to force
Dr = 8.0 m
F= N
W= F Dr
= -6.0 N (8.0m)
=-48 J

7. Work and Energy Dr = 6.0 m F= ? N W= 60 J
Work = Force x Displacement parallel to force
Dr = 6.0 m
F= ? N
W= 60 J

8. Work and Energy Dr = ? m F= - 50.0 N W= 200 J
Work = Force x Displacement parallel to force
Dr = ? m
F= N
W= 200 J

9. Work and Energy Dr = 8.0 m F= + 6.0 N W= 0
Work = Force x Displacement parallel to force
Dr = 8.0 m
F= N
W= 0
(since F and d are perpendicular

10. Figure 7.2  If an object undergoes a displacement ∆r under the action of a constant force F, the work done by the force is F∆r cos .
Fig. 7.2, p.184

11. Work and Energy Dr = 8.0 m F= + 6.0 N 40° Work = F Dr
Dot product W = F Dr cos (40)
Dr = 8.0 m
F= N
40°
W= (6.0N) [8.0m cos(40) ]= 36.8 J

12. Work and Energy Dr = 8.0 m F= + 6.0 N 40° Work = F Dr
Dot product W = F Dr cos(40)
Dr = 8.0 m
F= N
40°
W= (6.0N cos(40) ) [8.0m]= 36.8 J

13. Total Area ~S (DA) Work ~ S (FxDx) ~ ~ Fig. 7.7a, p.189
Figure 7.7  (a) The work done by the force component Fx for the small displacement ∆x is Fx ∆x, which equals the area of the shaded rectangle. The total work done for the displacement from xi to xf is approximately equal to the sum of the areas of all the rectangles.
Fig. 7.7a, p.189

14. Figure 7.7  (b) The work done by the component Fx of the varying force as the particle moves from xi to xf is exactly equal to the area under this curve.
Fig. 7.7b, p.189

15. Work Variable Force (q=0)
Work = F Dr cosq = Favg Dx

16. Springs 101
Spring Constant k, stiffness = 50 N/m

17. Work Variable Force (q=0)
Work = F D cosq = F D

18. Work Variable Force (q=0)
Work = “Area” units of N m (Joules)
W= 0.5*(100N)(4m) - 0.5(50N)2m)
= +200 J -50 J = 150 J

19. Potential Energy, U Gravitational Potential Energy Springs Chemical
Pressure
Mass (Nuclear)
Measured in Joules

20. Potential Energy, U The energy required to put
something in its place (state)
Gravitational Potential Energy
Springs
Chemical
Pressure
Mass (Nuclear)

21. Potential Energy U=(mg) h
Gravitational Potential Energy = weight x height
U=(mg) h
m = 2.0 kg
4.0 m

22. Potential Energy
U=(mg) h
U=80 J
m = 2.0 kg
4.0 m
K=?

23. Potential Energy to Kinetic Energy
U=(mg) h
m = 2.0 kg
K E= 0 J
PE=40 J
2.0 m
1.0 m
KE=?

24. Potential Energy of a springx 1 U=
kx2 2

25. Potential Energy of a springx 1 U=
kx2 2
For a spring with stiffness k= 80 N/m,
what is its potential energy when stretched 0.1m?
How about 0.2 m?

26. Potential Energy of a springx 1 1 U=
kx2 =
80 N/m (0.1m)2 2 2
= 0.40 J
For a spring with stiffness k= 80 N/m,
what is its potential energy when stretched 0.1m?
How about 0.2 m?

27. Potential Energy of a springx 1 1 U=
kx2 =
80 N/m (0.2m)2 2 2
= 1.60 J
For a spring with stiffness k= 80 N/m,
what is its potential energy when stretched 0.1m?
How about 0.2 m?

28. Potential Energy of a springx 1 U=
kx2 2

29. Kinetic Energy K=1/2mv2
Table 7.1, p.194

30. Kinetic Energy, K
K =1/2 m v2
m=2.0 kg and v= 5 m/s
K= ?

31. Kinetic Energy K =1/2 m v2 m=2.0 kg and v= 5 m/s K= 25 J

32. Kinetic Energy K =1/2 m v2 if m doubles KE doubles
if v doubles KE quadruples
if v triples KE increases 9x
if v quadruples KE increases ____ x

33. Work Energy Theorm
K =1/2 m v2
F = m a

34. Work Energy Theorm
K =1/2 m v2
F = m a
F d = m a d

35. Work Energy Theorm K =1/2 m v2 F = m a F d =m a d
F d = m (v/t) [(v/2)t]

36. Work Energy Theorm K =1/2 m v2 F = m a F d = m a d
F d = m (v/t) [(v/2)t]
W = 1/2 m v2

37. Work Energy Theorm KE =1/2 m v2 F = m a F d = m a d
F d = m (v/t) [(v/2)t]
W = 1/2 m v2
W = ∆ KE

38. Work Energy W = ∆K
How much work is required to stop a 2000 kg car traveling at 20 m/s (45 mph)?

39. Work Energy W = ∆K W= ∆K =-1/2 m v2 =-1/2(2000 kg)(20 m/s)2
How much work is required to stop a 2000 kg car traveling at 20 m/s (45 mph)?
W= ∆K =-1/2 m v2
=-1/2(2000 kg)(20 m/s)2
= kg (400 m 2 /s 2)
= - 400,000 Joules

40. Work Energy W = ∆K W= ∆K = - 400,000 Joules F=weight=mg=-20,000 N
How much work is required to stop a 2000 kg car traveling at 20 m/s? If the friction force equals its weight, how far will it skid?
W= ∆K = - 400,000 Joules
F=weight=mg=-20,000 N
W=F d d=W/F=-400,000 J/-20,000N
= 20.0 m

41. Work Energy W = ∆K v = 20 m/s d=? m Same Friction Force v = 10 m/s

42. Conservation of Energy
Energy can neither be created nor destroyed only
transformed from one form to another
Total Mechanical Energy, E = U +K
In the absence of friction or other non-conservative forces
the total mechanical energy of a system does not change
E f=Eo

43. Conservation of Energy
U=100 J
K = 0 J
m = 1.02 kg
(mg = 10.0 N)
Constant E
{E = K + U}
Ef = Eo
U = 75 J
K = 25 J
10.0 m
U = 50 J
K = 50 J
U = 25 J
K= ?
No friction
No Air resistance
U = 0 J
K = ?

44. Conservation of Energy
U=100 J
m = 2.0 kg
K=0 J
Constant E
{E = K + U}
Constant E
{E = K + U}
Ef=Eo
5.0 m
No friction
U = 0 J
K = ?

45. Conservation of Energy
U =100 J
m = 2.0 kg
K = 0 J
Constant E
{E = K + U}
Constant E
{E = K + U}
Ef=Eo
5.0 m
No friction
v = ?
K = 100 J

46. Conservation of Energy
Constant E
{E = K + U}
Ef=Eo
Ef=Eo+Wother
Kf +Uf=Ko+Uo +Wother

49. Example 8.8

55. L cos(25°)

57. Power Work = Power x time 1 Watt= 1 J/s 1 J = 1 Watt x 1 sec
1 kilowatt - hr = 1000 (J/s) 3600 s = 3,600,000 J
Energy = $$$$$$
1 kW-hr = $0.04 = 4 cents

58. Power Work = Power x time W=P t [ J=(J/s) s= Watt * sec ] work = ?
when 2000 watts of power are delivered for 4.0 sec.

59. Power Energy = Power x time E =P t [ kW-hr=(kW) hr] or
[ J=(J/s) s= Watt * sec ]

60. Power Energy = Power x time
How much energy is consumed by a 100 Watt lightbulb when left on for 24 hours?
What units should we use?
J,W, & s or
kW-hr, kW, hr

61. Power Energy = Power x time
What is the power output of a duck who does 3000 J of work in 0.5 sec?
What units should we use?
J,W, & s or
kW-hr, kW, hr

62. Power Energy = Power x time E =P t [ kW-hr=(kW) hr] Energy = ?
when 2000 watts (2 kW) of power are delivered for 6.0 hr.
Cost at 4 cent per kW-hr?

63. Machines D d
Levers f F
Work in = Work out
f D = F d

64. Machines
d = 1 m
D =8 m
Levers
f=10 N
F=?
Work in = Work out
f D = F d

65. Machines d = 1 m D =8 m Levers f=10 N F=? Work in = Work out
10N 8m = F 1m
F = 80 N

66. Machines
Pulleys f
Work in = Work out
f D = F d D d F

67. Machines Pulleys f D d F Work in = Work out f D = F d
D/d = 4 so F/f = 4
If F=200 N
f=? f D d F

68. Machines Pulleys f D d F Work in = Work out f D = F d
D/d = 4 so F/f = 4
If F=200 N
f = 200 N/ 4 = 50 N f D d F

69. Machines F f Hydraulic machine d D Work in = Work out f D = F d
if D=20 cm , d =1 cm,
and F= 800 N, what is
the minimum force f?

70. Machines F f Hydraulic machine d D f = 40 N Work in = Work out
f D = F d
f 20 cm = N (1 cm)
f = 40 N
if D=20 cm , d =1 cm,
and F= 800 N, what is
the minimum force f?

71. Efficiency
Eout
Ein
Eloss

72. Efficiency
Eout= 150 J
Ein = 200 J
Eloss= ?

73. Machines F f Hydraulic machine d D f = 40 N Work in = Work out
f D = F d
f 20 cm = N (1 cm)
f = 40 N
if D=20 cm , d =1 cm,
and F= 800 N, what is
the minimum force f?

74. Machines F f Hydraulic machine d D if D=20 cm , d =1 cm, F= 800 N,
and f=60 N,
what is the effeciency?
Work in = 60 N (20 cm)=1200 N-cm
Work out = 800 N (1 cm)= 800 N-cm