1. Chapter 5 Work and Energy

2. Review  x = v i  t + ½ a  t 2  x = ½ (v i + v f )  t v f = v i + a  t v f 2 = v i 2 + 2a  x

3. F = ma Kinetic friction is a resistive force exerted on a moving body by its environment.

4. Section 5-1 Work

5. Work The product of the component of a force along the direction of displacement and the displacement.

6. Work = Force x Displacement Instead of using  x, use ‘d’. W = Fd Note: It’s the F net doing the work.

7. Work is done only when the components of a force are parallel to the displacement.

8. No Movement, No Work

9. When the net force is at an angle to the direction of the displacement.

11. W = Fd(cos  If  is 0, then cos 0 = 1; therefore, the equation is just W = Fd

13. Units of Work From the equation, we get Newton-meters (N-m). Another name for a N-m is a Joule (J) 1 J = 1 N-m

14. Ex 1: Andy is out pulling his wagon again, if he pulls the wagon with a force of 60 N at an angle of 35 degrees to the ground. How much work is done, if he pulls it 15 m?

15. G: F = 60 N, d = 15 m,  =35 o U: W = ? E: W = Fdcos  S: W =

16. Ex 2: Greg applies a net force of 550 N, at an angle of 33 degrees to the ground, to help move a car. If he does 2500 J of work, how far does he move the car?

17. G: F =550 N, W=2500 J,  =33 o U: d = ? E: W = Fd(cos  or d =W/(Fcos  ) S: d =

18. Work is a scalar It’s (+) when the force is in the same direction as the displacement. It’s (-) when the force is in the direction opposite of the displacement.

20. (+) Net Work 1. Object speeds up 2. Work is done on the object.

21. (-) Net Work 1. Object slows down 2. Work is done by the object.

22. Section 5-2 Energy

23. Energy The ability to do work.

24. Kinetic Energy(K.E.) Energy of an object due to its motion.

25. K.E. depends upon speed and mass. ½ mv 2 is called ‘Kinetic Energy’ K.E. = ½ mv 2

26. Ex 3: A 7 kg shot put moves at a speed of 3.00 m/s. (A) What is the K.E. of the shot put have? (B) How fast must a 1.5 kg basketball move to have the same K.E.?

27. (A) G: m shot =7 kg, v shot =3m/s U: K.E. shot = ? E: K.E. shot =½ m shot v shot 2 S: K.E. shot =

28. (B) G: K.E. bb = 31.5 J, m =1.5 kg U: v bb = ?

29. (B) G: K.E. bb = 31.5 J, m =1.5 kg U: v bb = ? E:

30. S:

31. S: v bb =

32. Work–Kinetic Energy Theorem The net work done on an object is equal to the change in the kinetic Energy of the object.

33. Net Work = change in K.E. W net =  K.E. = K.E. f - K.E. i F net dcos  ½mv f 2 - ½mv i 2 F net dcos  W net =  K.E. = K.E. f - K.E. i = ½mv f 2 - ½mv i 2 They are all equal to each other.

34. Ex 4: On a frozen pond, Joe kicks a 10 kg sled, giving it an initial speed of 2.2 m/s. How far does the sled move if the coefficient of friction between the ice and the sled is 0.10?

35. G: m = 10kg, v i = 2.2 m/s, v f = 0 m/s,  k =0.10 U: d = ? E: W net =  K.E. F net dcos  = ½mv f 2 - ½mv i 2 The net force is only due to friction, so F net =  k mg. Because it is slowing down.

36.  k mgdcos  = - ½mv i 2 The force acts in the opposite direction of motion, so  = 180. d = (- ½mv i 2 )/ (  k mgcos 

37. S: d =

38. Potential Energy (P.E.) Energy associated with an object due to the position of the object.

39. Examples: Wound clock spring, compressed spring, or an object above the floor or ground.

40. Gravitational Potential Energy (P.E. g ) The P.E. associated with an object due to the position of the object relative to the Earth or some other gravitational source.

41. Potential Energy Equation Derivation We know energy is work P.E. = W P.E. = Fd

42. The force acting on the body is its weight and the distance it falls is the height. P.E. g = F g h P.E. g = mgh This is called Gravitational P.E.

43. Work is done only when a body is allowed to move or fall from a higher level to a lower level.

44. Both blocks acquire the same gravitational potential energy, mgh.

45. The same work is done on each block. What matters is the final elevation, not the path followed.

46. Difference between the P.E. of a body at a higher level and a lower level is the energy released to do work.

47. Zero Level Is the selected level an object is allowed to fall to, usually the floor or the ground.

49. EX 5: What is the P.E. of an Elevator having a mass of 500 kg, when its 25 m off the ground?

50. G: m = 500 kg, g = 10 m/s 2, h = 25m U: P.E. g = ? E: P.E. g = mgh S: P.E. g =

51. Ex 6: A student weighing 500 N walks up a flight of stairs to a height 3.5 m higher than started from. How much P. E. did he gain?

52. G:F g =mg=500 N, h=3.5 m U: P.E. g = ? E: P.E. g = mgh S: P.E. g =

53. Elastic Potential Energy (P.E. elastic ) The P.E. in a stretched or compressed elastic object.

54. When a force compresses or stretches a spring, elastic P.E. is stored in the spring.

55. P.E. elastic = ½ kx 2 Where: k = spring constant (N/m) x = distance compressed or stretched

56. The length of a spring with no external force on it is called the relaxed length.

58. The amount of P.E. stored depends upon how far the spring is stretched or compressed.

59. Section 5-3 Conservation of Energy

60. Mechanical Energy - Energy associated with the motion/positon of an object. - The sum of K. E. and all forms of P. E.

61. M.E. = K.E. + P.E.

62. In the absence of friction, the total M.E. remains the same at any point. This is called the Conservation of Mechanical Energy

63. Energy can not be created nor destroyed, it can just change its form of energy.

65. Although the amount of M.E. is constant, the form of energy can change. M.E. i = M.E. f K.E. i + P.E. i = K.E. f + P.E. f

66. ½ mv 2 i + mgh i = ½ mv 2 f + mgh f

70. Ex 8: Starting from rest, Ashley zooms down a frictionless slide from an initial height of 3.0 m. What is her speed at the bottom of the slide, if her mass is 55 kg?

71. The slide is frictionless, so M.E. is conserved. Also, the bottom of the slide is the zero level. Therefore, there is no P.E. at the bottom. Since Ashley starts from rest, v i = 0  she has no K.E i.

72. G: m = 55 kg, h = 3.0 m g = 10 m/s 2 U: v f = ? E: M.E. i = M.E. f K.E. i + P.E. i = K.E. f + P.E. f 0 + mgh i = ½ mv 2 f + 0

74. S: v f =

75. If you start from rest and end at the bottom (Zero Level) KE I = 0 J and PE f = 0 J. K.E. i + P.E. i = K.E. f + P.E. f P.E. i = K.E. f mgh i = ½ mv 2 gh i = ½ v 2

76. If you start from rest and end at the bottom (Zero Level) KE I = 0 J and PE f = 0 J. K.E. i + P.E. i = K.E. f + P.E. f P.E. i = K.E. f mgh i = ½ mv 2 gh i = ½ v 2

78. Energy conservation occurs even when acceleration varies.

80. In the example with the slide, we don’t know the shape of the slide; therefore, acceleration may not be constant. If acceleration is not constant, we can’t use our equations. Since the the slide is frictionless, ME is conserved and the shape plays no part in the ME.

81. ME is not conserved in the presence of friction. Kinetic Friction causes the KE to be converted into a non- mechanical form of energy (Heat). This and other forms of energy are difficult to account for and are considered to be ‘lost’.

82. Section 5-4 Rate of Energy Transfer

83. Power The rate at which energy is transferred. The rate at which work is done.

84. Power tells us ‘how fast work is done.’

85. Work Power = --------- time

86. W P = ----- t

87. Alternative equation Use the work equation and make a substitution. Fd P = ------ t

88. From Ch 2, the definition of speed: distance moved per unit of time.

89. Power Units From the equation we get: N-m J P = -------- = ------ s s

90. We call this a watt (W). One joule per second equals one watt.

91. 1000 watts = 1 kilowatt A non-metric unit of power commonly used in commerce and industry is the horsepower. 1 Hp = 746 W

92. Example Problem 9: Rebecca is pushing a lawnmower with a force of 40 N, for 12 m in 10 seconds. (a) what is the work done? (b) what is the power exerted?

93. (a) Work done: G:F = 40 N d = 12 m t = 10 s U: W= ?

94. E: W = Fd S: W=(40 N)x(12 m) S: W = 480 J

95. B) E: W P = ----- t

96. S: P =

97. S: P =

98. Example Problem 10: How much work can a 250 W motor do in 12 seconds?

99. G: P = 250 W t = 12 s U: W = ?

100. E: W P = ----- t W = Pt

101. S:W=