Determine the velocity of each block when system is started from rest & block B gets displacement by 2m. Take $µk=0.2$ between block A & horizontal surface.

Consider length of string w.r.t. block A as XA &

Length of string w.r.t block B as XB

So, applying C.S.L.M. we get

$2XA+ XB= C $

But as B moves down, XB increases & XA decreases

So, rewriting eqn

$-2XA+ XB= C \text {______(i) }$

Differentiating w.r.t. t

$-2VA+VB=0............2VA=VB\text {_____(ii) }$

Differentiating once more w.r.t. t

$2aA=aB \text {____(iii) }$

Applying NSL to block A

$\sum Fx= max \space\space \&\space\space \sum Fy=0 \\ T - 0.2(N)=10aA \space\space N=10g \\ T - 0.2(10g)=10aA \text {_____(iv) }$

Applying NSL to block A

$\sum Fy= may \\ 5g - T= maB \text {_____(v) } $

Solving eqns (iii)(iv)&(v),we get

$T=34.335 N \\ aA= 1.4715 m/s^2 \\ aB= 2.934 m/s^2 $

Now, after displacement of 2m, let velocity of B be VB

$(VB)2 = 0+ 2aB(2).... VB = 3.43 m/s $

Putting in eqn (ii), $VA=1.715 m/s $

Velocity if A is $1.715 m/s$ & Velocity of B is $3.43 m/s $