QuestionDetails: Two masses are connected by a rope (which does not stretch)woun

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QuestionDetails: Two masses are connected by a rope (which does not stretch)wound around 2 pulleys as shown (the pulleys ahve negliible massand spin wtih negligible resistance). When the system is released,block A begins to drop and block B begins to slide to the right.The coeffiecient of kinetic friction between block B and thehorizontal surface is k=0.2. Block B has a massof 5kg and block A has a mass of 4kg. At what rate is A accleratingdownward. (Hint neither block is in static equilibrium, and the tensionin the rope will be one of your unknowns. Applying Newton's secondlaw to each block will give two equations to obtain a third youwill need to consider the relationship between the verticalacceleration of A to the horizontal accleration of B that isimposed by the rope) QuestionDetails: Two masses are connected by a rope (which does not stretch)wound around 2 pulleys as shown (the pulleys ahve negliible massand spin wtih negligible resistance). When the system is released,block A begins to drop and block B begins to slide to the right.The coeffiecient of kinetic friction between block B and thehorizontal surface is k=0.2. Block B has a massof 5kg and block A has a mass of 4kg. At what rate is A accleratingdownward. (Hint neither block is in static equilibrium, and the tensionin the rope will be one of your unknowns. Applying Newton's secondlaw to each block will give two equations to obtain a third youwill need to consider the relationship between the verticalacceleration of A to the horizontal accleration of B that isimposed by the rope)

Explanation / Answer

2 T - MB g = MB a /2      where a is the acceleration ofblock A MA g - T = MAa       T = MAg - MA a 2 (MA g - MA a ) - MB g   = MB a /2    This can now be solved for a (MB / 2 + 2 MA) a = (2MA  - MB) g a = (8 - 1) g / (2.5 + 8) = .67 g 2 (MA g - MA a ) - MB g   = MB a /2    This can now be solved for a (MB / 2 + 2 MA) a = (2MA  - MB) g a = (8 - 1) g / (2.5 + 8) = .67 g

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