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A block of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended.The coefficient of Kinetic friction between block and the table is given.When the block A is sliding on the table,the tension in the string is given by _____________. (AIPMT-2015)
Dear Student ,here in this question the m1 rests on the table and the mass m2 is connected by a light string .Let the tension on the string is T .So , when the block A starts to slide then ,T - F' = m1a ....(1)and m2g -T = m2a ....(2)Now , F' = frictional force = Mu * m * gkm1gNow solving equations 1 and 2 we get a = m2g – mu * m1* g / m1 + m2 put this value in 2 You will get the required tensuion
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