Work is transferred energy. Doing work is the act of transferring the energy. 6.1 Work

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1 Ch 6 Work & Energy

2 Work is transferred energy. Doing work is the act of transferring the energy. 6.1 Work

3 Work is a Scalar Quantity A force does positive work when it has a vector component in the same direction as the displacement.

4 Work is energy transferred to or from an object by means of a force acting on the object.

5 Energy transferred to the object is positive work, and energy transferred from the object is negative work.

6 The Forces that Count W F cos ( 60 ) F 1 s

7 Which force does more work?

8 Work = (F cos q) s

9 W = F s cos q

10 Assumptions: W = F s cos q Force is constant. If force is not constant, average force is assumed. The angle q is between force and displacement.

11 Is work done? Car does not move Car moves.

12 6.1 + and - Work

13 6.1 + and - Work

15 Loading a Van

16 Less Effort? Less Work? Ramp is frictionless.

17 Which involves less work? Method now involves ramp friction.

18 6. Work-Energy Theorem

19 6. Work-Energy Theorem W KE f KE o W 1 Mv 1 f Mv o

20 6. Work-Energy Theorem Positive work results in a gain in KE. Negative work results in a decrease in KE.

21 6. Work-Energy Theorem Positive work results in a gain in KE.

22 6.3 Work Done by Fg

23 Does path matter? Down slope vs. Height drop?

24 6.3 Work Done by Fg

25 Same work vertically speaking? For down the slope: W ( Mgsin( 5 deg )) s For vertically down (height drop): W Mgssinq

27 GPE

28 Is Work Done?

29 Path Dependency?

30 6.4 Conservative Forces Work done on moving object is independent of the path taken between two end points. Example: Gravitational Force.

31 6.4 Conservative Forces

32 6.4 Conservative Forces W gravity mg h o h f

33 6.4 Closed Path

34 6.4 Non-Conservative Forces Work done on object depends on path taken. Example: Friction

35 6.5 Cons. of Mech. Energy

37 Pendulum Lab Set-Up razor blade support rod sewing thread steel ball lab table floor carbon paper

38 6.5 Cons. of Mech. Energy

39 KE vs. Displacement

41 It s all about the height! KE PE 1 m v m g H

42 Prob 33 Cons. of Energy

43 Prob 33 Cons. of Energy KE PE 1 m v m gh

44 Prob 33 Mass cancels: v g ( r) v 9.8 m s ( 1.1m ) v 6.6 m s

45 Prob # *4 Cons. of Energy

46 Prob # *4 Cons. of Energy PE KE m g y 1 m v

47 Prob # *4 Mass Cancels: g y 1 v gy v

48 Prob # *4 Solve for v: v g y v 9.8 m s ( 4m 3m) v 4.43 m s

49 Prob # *44 Pool Slide h

53 Prob # *44 Pool Slide 1 GPE mg h mv 1 mg H mg h mv

54 Prob # *44 Cancel m: 1 gh gh v 1 gh v H g

55 Prob # *44 v =x/t 1 v H h g H 1 a t 1 x g t

56 Prob # *44 Compute: H m s ( 0.500s ) m 0.500s 9.8 m s H 6.33m

57 When Force Applied Varies

58 Does height affect Distance?

59 Does height affect Distance? PE W m g h FD

60 Does Mass affect Distance?

61 Does Mass affect Distance? PE W m g m g m g h h h F k D k F N D k m gd h k D

62 Does Dip affect Distance?

63 Does Dip affect Distance? It's all about the height difference!

64 Energy Transfer

65 Energy Transfer

66 Energy Transfer

67 v o 5g R v bottom 5gR

68 iii. From what height (H) along the track must the object be released so that it just barely holds on to the track at the top of the loop?

69 Since F N is just about zero, F c m g. Mv Mg F c R g v top R v top gr at the loop top

70 Conservation of Energy m g h o 1 m v top m g h f PE o KE f top PE f top

71 m g h o 1 m v top m g ( R) gh o 1 v top g( R) 1 gh o ( gr) g( R)

72 1 gh o ( gr) g( R) gh o gr 1 h o (.5) R

73 Alternatively, PE lost KE gained m g( h R) g( h R) g( h R) 1 m 1 v 1 ( gr) v

74 Then, h R 1 ( R) h 1 ( R) R h 1 R h (.5) R

75 iv. From what height (H) along th track must the object be released so that it just barely reaches the position Q shown in the diagram?

76 h o R from the ground

77 v. If the object is released from position P as shown, what two forces make up the centripetal force at the top of the loop?

78 F c m g F N

79 vi. If the object is released from position P as shown, what would be the speed at the top of the loop

80 m g Mv F N F c R m g F N R v m at the loop top

81 m gh 1 m v top m g h o h f g h o h f 1 m v top 1 v top v top g h o h f

82 Prob # **83 Cons. of Energy

84 What s the minimum r for just barely making the loop? Since F N is just about zero, F c Mv top Mg F c v R top gr m g. Prob # **83 at the loop top

85 Conservation of Energy Prob # **83 1 m v o 1 m v top m g h KE bottom KE top PE top

86 Prob # **83 1 m v o 1 m v top m g ( r) 1 v o 1 v top g( r) 1 v o 1 ( gr) g( r)

87 1 v o 1 ( gr) g( r) 1 v o gr 1 Prob # **83 r v o (.5) g

88 Prob # **83 r r v o (.5) g 4.0 m s (.5) 9.8 m s 0.37m

89 Problem #35

90 Pythagorize to get initial vertical speed: 14 m s V i y 13.0 m s V i y 14 m s 13.0 m s V i y m s

91 Now use the 3 rd equation to get apex height: ay V f y V i y y m s 9.8 m s y 1.38m

92 Alternatively, 1 m V i y 1 V i y gh m g H

93 1 V i y m s H H g m s H 1.38m

94 third method of solution v o 14.0 m s 1 m v o 1 m v f m g H KE before KE after PE after 1 v o 1 v f gh

95 H 1 v o g 1 v f H v o g v f H 1.38m

96 Problem # *43

97 Skier starts from rest. Crest of second hill is circular with radius of r 36 m. Neglect friction and air resistance What is the height (h) of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

98 Solution: PE 1st KE nd There is no PE nd gained from the h reference given. Ov erarching Concept:The PE lost form the first hill is completely converted to the KE gained at the top of the second hill.

99 PE converts to KE: gh 1 v nd v nd gr since F 0 and m g Nnd m v nd r

100 gh 1 ( gr) Since h = h, the diameter of the circle do es not have to be taken into acco unt when determining h. h h 36.0m 18m

101 Problem # **46

102 Problem # **46 Girl starts from rest at top of frictionless, circular, waterslide. F N is zero when girl leaves slide surface. At what angle q does she leave the surface?

103 Solution: PE 1st PE nd KE nd There is some PE nd left overfrom the h 1 = r reference given. The new h is h r cos q.

104 Overarching Concept: The PE lost form position is converted to the KE gained and new PE at position.

105 r q f r r cos q f

106 F N mg cos q q q q mg

107 gh 1 gh 1 v nd There is no initial KE at the top of slide.

108 g( r) g rcos q 1 v nd g( r) g rcos q 1 g cos q r 1 cos q 1 cos q 1 cos q 1 1

109 cos q cos q 1 ( 1.5) cos q q acos ( 0.667) q 48.deg

110 Energy Transfer

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112

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114 A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown. How much work is done by the force as the block moves from the origin to x = 8.0 m?

115

116

117

118 Pendulum Lab Set-Up razor blade support rod sewing thread steel ball lab table floor carbon paper

119 razor blade longer, thicker string

120 razor blade

121

122 Conservation of Mechanical Energy An Inelastic Collision PE = KE m m g h = 1/ m v 1 m g h = 1/v PE KE v = g h Finding velocity of ball just after hitting floor

= 1 2 kx2 dw =! F! d! r = Fdr cosθ. T.E. initial. = T.E. Final. = P.E. final. + K.E. initial. + P.E. initial. K.E. initial =

Practice Template K.E. = 1 2 mv2 P.E. height = mgh P.E. spring = 1 2 kx2 dw =! F! d! r = Fdr cosθ Energy Conservation T.E. initial = T.E. Final (1) Isolated system P.E. initial (2) Energy added E added