Work and energy

Force pulling B towards ground is Gravitational (which i would assume to be 10)
and pulling A is solely tension as table is frictionless

As A and B is connected by a piece of string thus when B is pulled towards ground A is pulled towards the pulley

Now consider work done and energy changes

WD by gravitational forces on the system acts solely on B, becoz A moves horizontally so by WD=FS, where S is zero, WD must be Zero too

for B, also apply WD=FS, where F = mg = 20, S = 1.5m
(please note that here WD = PE (mgh) as well)

so that WD by G on the system is 30

but as while B drops A is pulled, so this energy is changed to KE of both blocks

thus u can use 30 = 0.5(mava^2)+0.5(mbvb^2), where Va and Vb is equal, putting all the numbers given u should get 30 = V^2(0.5(3)+0.5(2))
V^2 = 30/2.5 = 12
V = 3.46ms-1

the key to this kind of mechanic questions is that u have to identify all kind of movements and how energy changed, the rest is simply equation applying