1. Fan Cart

6. F N F NA F g

7. Fan Cart F N F NA F g

8. Fan Cart F N F NA F g

9. Fan Cart F N F NA F g =mg

10. Fan Cart F N =mg F NA F g =mg

11. Fan Cart F N =mg F NA =.12N F g =mg

12. Fan Cart F N =mg F NA =.12N F g =mg m=.550 kg

13. Fan Cart F N =mg F NA =.12N F g =mg Given: m=.550 kg Find a=?m/s/s FRS F=ma F = a m

14. Fan Cart F N =mg F NA =.12N F g =mg Given: m=.550 kg Find a=?m/s/s FRS F=ma F = a m

15. Fan Cart F N =mg F NA =.12N F g =mg Given: m=.550 kg Find a=?m/s/s FRS F=ma F = a m

16. Fan Cart F N =mg F NA =.12N F g =mg Given: m=.550 kg Find a=?m/s/s FRS F=ma F = a m

17. Fan Cart F N =mg F NA =.12N F g =mg Given: m=.550 kg Find a=?m/s/s FRS F=ma F = a m.12N =.55kg

18. Fan Cart F N =mg F NA =.12N F g =mg Given: m=.550 kg Find a=?m/s/s FRS F=ma F = a m.12N =.55kg

19. Fan Cart F N =mg F NA =.12N F g =mg Given: m=.550 kg Find a=?m/s/s FRS F=ma F = a m.12N =.22m/s/s.55kg

20. Fan Cart F N =mg F NA =.12N F g =mg Given: m=.550 kg Find a=?m/s/s FRS F=ma F = a m.12N =.15m/s/s.80kg

21. Cart / Falling Object

22. Cart / Falling Object

23. Cart / Falling Object

24. Cart / Falling Object

25. Cart / Falling Object

26. Cart / Falling Object FgFg

27. Cart / Falling Object F g =mg

28. Cart / Falling Object F g =mg FNFN

29. Cart / Falling Object F g =mg FNFN FTFT

30. Cart / Falling Object F g =mg FNFN FTFT

31. Cart / Falling Object F g =mg FNFN FTFT FTFT

32. Cart / Falling Object F g =mg FNFN FTFT FTFT Net F =ma F T =ma

33. Cart / Falling Object F g =mg FNFN FTFT FTFT Net F =m c a F T =m c a

34. Cart / Falling Object F g =mg FNFN FTFT FTFT Net F =m c a F T =m c a Net F =m f a F g -F T = m f a

35. Cart / Falling Object F g =mg FNFN FTFT FTFT Net F =m c a F T =m c a Net F =m f a F g -F T = m f a m f g-F T =m f a

36. Cart / Falling Object F g =mg FNFN FTFT FTFT Net F =m c a F T =m c a Net F =m f a F g -F T = m f a m f g-F T =m f a m f g-m c a=m f a

37. Cart / Falling Object F g =mg FNFN FTFT FTFT Net F =m c a F T =m c a Net F =m f a F g -F T = m f a m f g-F T =m f a m f g-m c a=m f a m f g=m f a+m c a m f g = a (m f +m c )

38. Cart / Falling Object F g =mg FNFN FTFT FTFT Net F =m c a F T =m c a Net F =m f a F g -F T = m f a m f g-F T =m f a m f g-m c a=m f a m f g=m f a+m c a m f g = a (m f +m c )

39. Cart / Falling Object F g =mg FNFN FTFT FTFT NetF X =m c a F T =m c a Net F =m f a F g -F T = m f a m f g-F T =m f a m f g-m c a=m f a m f g=m f a+m c a m f g = a (m f +m c ) Acceleration = Force due gravity on falling object divided by total mass

40. Cart Falling Object Lab F G – F T = m f a F T =m c a m f g-m c a=m f a m f g=m c a+m f a m f g = (m c +m f )a m f g = a Weight =a (m c +m f ) (combined mass)

41. Block / Falling Object

42. Block / Falling Object F N F T F fr F T F G F G =m b g

43. Block / Falling Object F N F T F fr F T F G F G =m b g

44. Block / Falling Object F N F T F fr F T F G F G =m b g

45. Block / Falling Object F N F T F fr F T F G F G =m b g

46. Block / Falling Object F N F T F fr F T F G =m f g F G =m b g

47. Block / Falling Object F N F T F fr F T F G =m f g F G =m b g

48. Block / Falling Object F N F T F fr F T Net F =m b a F T -F fr =m b a F G =m f g F G =m b g

49. Block / Falling Object F N F T F fr F T Net F =m b a F T -F fr =m b a F G =m f g F T -  F N =m b a F G =m b g

50. Block / Falling Object F N F T F fr F T Net F =m b a F T -F fr =m b a F G =m f g F T -  F N =m b a Net F =m f a F T -  m b g=m b a m f g-F T =m f a m f g-m f a=F T F G =m b g

51. Block / Falling Object F N F T F fr F T Net F =m b a F T -F fr =m b a F G =m f g F T -  F N =m b a Net F =m f a F T -  m b g=m b a m f g-F T =m f a m f g-m f a=F T F G =m b g

52. Block / Falling Object F N F T F fr F T Net F =m b a F T -F fr =m b a F G =m f g F T -  F N =m b a Net F =m f a F T -  m b g=m b a m f g-F T =m f a m f g-m f a=F T F G =m b g

53. Block / Falling Object F N F T F fr F T Net F =m b a F T -F fr =m b a F G =m f g F T -  F N =m b a Net F =m f a F T -  m b g=m b a m f g-F T =m f a m f g-m f a-  m b g=m b a m f g-m f a=F T F G =m b g

54. Block / Falling Object F N F T F fr F T Net F =m b a F T -F fr =m b a F G =m f g F T -  F N =m b a Net F =m f a F T -  m b g=m b a m f g-F T =m f a m f g-m f a-  m b g=m b a m f g-m f a=F T m f g-  m b g=m b a+m f a F G =m b g

55. Block / Falling Object F N F T F fr F T Net F =m b a F T -F fr =m b a F G =m f g F T -  F N =m b a Net F =m f a F T -  m b g=m b a m f g-F T =m f a m f g-m f a-  m b g=m b a m f g-m f a=F T m f g-  m b g=m b a+m f a m f g-  m b g = a (m f g-  m b g)=a (m f +m c ) F G =m b g

56. Block Falling Object F G – F T = m f a F T -F fr =m b a m f g-m b a-  m b g=m f a F T = m b a+F fr F T = m b a +  m b g m f g-  m b g=m f a + m b a m f g –  m b g = (m b +m f )a m f g-  m b g = a Weight -Friction =a (m b +m f ) (combined mass)

57. Lab Frictionless incline

58. Lab Frictionless incline

59. Lab Frictionless incline F N F g

60. Lab Frictionless incline F N  F gpara F gperp

61. Lab Frictionless incline F N = mgcos   F gll =mgsin  F gper = mgcos 

62. Lab 4 Frictionless incline F gpara = mgsin  ma F N = F gperp = mgcos  gsin  a The acceleration of a cart on a frictionless incline is dependent on the acceleration due to gravity and the sin of the incline angle. It is independent of the mass of the object.

63. Lab 5 Block on incline involving friction

64. Lab 5 Block on incline involving friction

65. Lab 5 Block on incline involving friction

66. Lab 5 Block on incline involving friction F g =mg

67. Lab 5 Block on incline involving friction F g =mg

68. Lab 5 Block on incline involving friction F g =mg FNFN

69. Lab 5 Block on incline involving friction F g =mg F gy FNFN

70. Lab 5 Block on incline involving friction F g =mg F gper FNFN

71. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gpara

72. Lab 5 Block on incline involving friction F g =mg F gper FNFN F gpara

73. Lab 5 Block on incline involving friction F g =mg F gperp FNFN F gpara F fr

74. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular

75. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel

76. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel mgcos 

77. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel mgcos  mgsin 

78. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel mgcos  mgsin  mgcos 

79. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel mgcos  mgsin  mgcos  F fr =  F N

80. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel mgcos  mgsin  mgcos  F fr =  F N  mgcos 

81. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel mgcos  mgsin  mgcos  F fr =  F N  mgcos  mgcos  =mgcos 

82. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel mgcos  mgsin  mgcos   f F gx = Static Friction mgcos  F fr =  F N  mgcos  mgcos  =mgcos 

83. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel mgcos  mgsin  mgcos   f F gx = Static Friction mgcos  F fr =  F N  mgcos  mgcos  =mgcos  If F gx > Static Friction mgsin  -  mgcos  =ma

84. Lab 5 Block on incline involving friction F g =mg F gy FNFN F gx F fr Perpendicular Parallel mgcos  mgsin  mgcos   f F gx = Static Friction mgcos  F fr =  F N  mgcos  mgcos  =mgcos  If F gx > Static Friction mgsin  -  mgcos  =ma

85. Lab 5 Block on incline involving friction Net F = Perpendicular to incline

86. Lab 5 Block on incline involving friction Net F = 0 Perpendicular to incline

87. Lab 5 Block on incline involving friction Net F = 0 Perpendicular to incline F gy =F N = mgcos 

88. Lab 5 Block on incline involving friction  F y = 0 Perpendicular to incline F perp =FN = mgcos 

89. Lab 5 Block on incline involving friction  F y = 0 Perpendicular to incline F perp =FN = mgcos 

90. Lab 5 Block on incline involving friction Fnet = 0 Perpendicular to incline F perp =F N = mgcos   F x = ____ if Force due to gravity parallel to incline equals static friction.

91. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F Net = 0 if Force due to gravity parallel to incline equals static friction.

92. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F Net = 0 if Force due to gravity parallel to incline equals static friction. F parallel =

93. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F Net = 0 if Force due to gravity parallel to incline equals static friction. F parallel = F fr

94. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F Net = 0 if Force due to gravity parallel to incline equals static friction. F parallel = F fr mgsin  ==  mgcos  sin  =  cos  sin  =  tan  cos 

95. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F Net = 0 if Force due to gravity parallel to incline equals static friction. F parallel = F fr mgsin  ==  mgcos  sin  =  cos  sin  =  tan  cos 

96. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F Net = 0 if Force due to gravity parallel to incline equals static friction. F parallel = F fr mgsin  ==  mgcos  sin  =  cos  sin  =  tan  cos 

97. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F Net = 0 if Force due to gravity parallel to incline equals static friction. F parallel = F fr mgsin  ==  mgcos  sin  =  cos  sin  =  tan  cos 

98. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F Net = 0 if Force due to gravity parallel to incline equals static friction. F parallel = F fr mgsin  ==  mgcos  sin  =  cos  sin  =  tan  cos 

99. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

100. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

101. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

102. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

103. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = ma if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

104. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = ma if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

105. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = ma if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

106. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = ma if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

107. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = ma if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

108. Lab 5 Block on incline involving friction F net = 0 Perpendicular to incline F perp =F N = mgcos  F net = ma if Force due to gravity parallel greater than incline friction F gparallel - F fr = ma mgsin  –  mgcos  ma gsin  –  cos  a

109. Lab 5 incline involving friction Force down – Friction = Net force mgsin  mgcos  ma gsin  - -  gcos  = a The acceleration of a cart incline is dependent on the acceleration due to gravity and angle of the incline and the coeffiecient of friction but not the mass of the object.

110. Lab 5 incline involving friction Force down – Friction = Net force mgsin  mgcos  ma gsin  - -  gcos  = a The acceleration of a cart incline is dependent on the acceleration due to gravity and angle of the incline and the coeffiecient of friction but not the mass of the object.

112. F net = F g heavy - F glight (m H +m L )a = m H g-m L g

113. F net = F g heavy - F glight (m H +m L ) a = m H g-m L g a = (m H g-m L g) (m H +m L )

114. Atwoods Pulley F net = F g heavy - F glight (m H +m L ) a = m H g-m L g a = (m H g-m L g) (m H +m L ) a = (Weight H – Weight L) ( total mass )

115. Atwoods Pulley F net = F g heavy - F glight (m H +m L ) a = m H g-m L g a = (m H g-m L g) (m H +m L ) a = (Weight H – Weight L) ( total mass )

116. Block on Incline Falling Mass

118. mg

119. Block on Incline Falling Mass mg mgsinf

120. Block on Incline Falling Mass mg mgsin 

121. Block on Incline Falling Mass mg mgsin   mgcos 

122. Block on Incline Falling Mass mg mgsin   mgcos  FTFT

123. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mg

124. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT

125. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT mfgmfg

126. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT m f g - mgsin 

127. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT m f g – mgsin  mgcos 

128. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT m f g – mgsin  mgcos  m  m f  a

129. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT m f g – mgsin  mgcos  m  m f  a Weight of Falling

130. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT m f g – mgsin  mgcos  m  m f  a Weight of Falling - Force parallel

131. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT m f g – mgsin  mgcos  m  m f  a Weight of Falling - Force parallel – Friction

132. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT m f g – mgsin  mgcos  m  m f  a Weight of Falling - Force parallel – Friction = Net Force

133. Block on Incline Falling Mass mg mgsin   mgcos  FTFT mfgmfg FTFT m f g – mgsin  mgcos  m  m f  a m f g – mgsin  mgcos  a  m  m f  Weight of Falling - Force parallel – Friction = Net Force