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Two blocks shown in fig, 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 the block B falls through a vertical distance of 1m, what is the velocity acquired by it? Neglect the friction in the pulley and the extension of the string.

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Let T = Tension on both sides of the string

a = Acceleration of the blocks

µ = 0.25 Consider the motion of block B, 

WB – T = ma ...(1)

10 – T =(10/2)a

Consider the motion of block A

T – µWA = ma 

T – 0.25 X 20 = (20/g)a ......(2) 

Add equation (1) and (2) 

10 – 5 = (30/g)a a = 1.63m/sec2 ........(3) 

Now using the relation, 

v2 = u2 + 2as 

v2 = 0 + 2X 1.63 X 1

 v = 1.81m/sec

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