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Velocity acquired by block - system released from rest:
Two blocks shown in the figure given below, have masses A = 20N and B = 10N and the coefficient of friction between the block A and the horizontal plane, μ = 0.25. If the system is released from rest, and block B falls through vertical distance of 1m, what is velocity acquired by it? Neglect friction in pulley and extension of string.
Sol: Let T = Tension on both sides of the string. a = Acceleration of the blocks μ= 0.25 Consider the motion of block B,
Consider motion of block A, T - μWA = ma
T - 0.25 X 20 = (20/g)a ...(2)
By adding equation (1) and (2)
10 - 5 = (30/g)a
a = 1.63 m/sec2 ...(3)
By using the relation, v2 = u2 + 2as
v2 = 0 + 2X 1.63 X 1
v = 1.81m/sec .......ANS
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