final_eaxm_solutions_F05

final_eaxm_solutions_F05 - Physics 40A - Final Exam...

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Physics 40A - Final Exam Solutions - Fall 2005 Problem 1 [25 points] A box of mass m 1 = 10 kg rests on a surface inclined at 30° to the horizontal. It is connected by a massless cord, which passes over a massless and frictionless pulley, to a second box of mass m 2 , as shown in the figure. (a) If the coefficient of static friction is μ s = 0.4, what is the maximum value that m 2 can have if the boxes are to remain at rest? (b) If the coefficient of kinetic friction is k = 0.3, and m 2 = 10 kg, determine the acceleration of the system. Solution: (a) We apply Newton's second law to m 2 T ­ m 2 g = m 2 0 ⇒ T = m 2 g 1 If m 2 is very large m 1 would tend to move up the plane so the frictional force is directed down the plane. Thus, Newton's second law applied to m 1 gives N ­ m 1 g cos = m 1 0 N = m 1 g cos (perpendicular to plane) T ­ m 1 g sin ­ f s = m 1 0 (up the plane) Combining with Eq. (1) gives m 2 g = m 1 g sin  f s Since the maximum value of f s is f s, max = s N s m 1 g cos , the maximum value of m 2 is given by m 2 g = m 1 g sin  s m 1 g cos Therefore, m 2 = m 1 [ sin s cos ] m 2 = 10 kg [ sin 30 °  0.4 cos30 ° ] = 8.46 kg (b) Since m 2 is now greater than 8.46 kg, m 1 will move up the plane. Newton's second law applied to m 1 and m 2 gives m 1 a = T ­ m 1 g sin f k m 2 ­ a = T ­ m 2 g T = m 2 g ­ a Setting f k k N k m 1 g cos we get m 1 a = m 2 g ­ a ­ m 1 g sin ­ k m 1 g cos Solving for a gives a = m 2 g ­ m 1 g sin k m 1 g cos m 1 m 2 a = 9.8 m/s 2  10 kg 1 ­ sin 30 ° ­ 0.3cos30 ° 20 kg = 1.18 m/s 2 1
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Physics 40A - Final Exam Solutions - Fall 2005 Problem 2 [20 points] A child stands 1.2m from the center (axis of rotation) of a playground merry-go-round. When the merry-go-round rotates at a rate just exceeding 0.345 rev/s, the child slips off. What is the coefficient of static friction between the child and the merry-go-round? Solution: In the radial direction, outward from the center of the merry-go-round, the only force acting is the frictional force. It acts toward the center of the merry-go-round. Newton's second law applied to the child gives F r = ma r f s = m v 2 r where v is the velocity of the child and r
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This note was uploaded on 04/16/2008 for the course PHYS 40 taught by Professor Ellison during the Winter '08 term at UC Riverside.

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final_eaxm_solutions_F05 - Physics 40A - Final Exam...

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