Let T = Tension on both sides of the string

a = Acceleration of the blocks

µ = 0.25 Consider the motion of block B,

W_{B }– T = ma ...(1)

10 – T =(10/2)a

Consider the motion of block A

T – µW_{A} = ma

T – 0.25 X 20 = (20/g)a ......(2)

Add equation (1) and (2)

10 – 5 = (30/g)a a = 1.63m/sec^{2} ........(3)

Now using the relation,

v^{2 }= u^{2} + 2as

v^{2} = 0 + 2X 1.63 X 1

v **= 1.81m/sec**