Calculate the acceleration of block A and B in cases (a),(b), and (c ).

Explanation

(a) For block `2M, 2T=2Ma` …(i)
For block `M, Mg-t=M2a` …(ii)
Solving equation (i) and (ii), we get `a=g//3`
(b) For block 2M,
`2Mg sin theta-T'=2Ma`...(i)

For movable pulley,
`T'-2T=0xxa implies T'=2T` ...(ii)
For block `M,T-Mg=M2a` ...(iii)
Solving eq, (i), (ii) , and (iii), we get `a=(g[sin theta-1])/(3)`
This is negative, hence direction of accelerations will be opposite to those shown in fig.
Note: Here we assume that the itblock `2M` is moving down the plane. You may assume that it is moving up the plane. The magnitude of the acceleration found will remain the same.
(c) For block 2M,
`2Mg sin theta-T=2M(2a)` ...(i)

For block `M, 2T-Mg=Ma` ...(ii)
Solving Eqs (i) and (ii), we get `a=(g[4 sin theta-1])/(9)`.