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 -  μW_A = 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