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