The acceleration of the blocks (A) and (B) respectively in situation shown in the flgure is: (pulleys and strings are massless)

The F.B.D. of block A and B arefrom constraint, the acceleration of A and B are ‘2a’ and ‘a’ respectively. Applying Newton's second law to blocks A and B, we get4g-T = 4(2a)---------(i)2T-5g = 5(a) ---------(ii)Solving we get acceleration of A and B as 2g7