The block of mass `M` and are arrenged as the situation in fig is shown.The coefficient of friction between two block is `mu`and that between the bigger block and the ground is `mu` find the acceleration of the block .

Explanation

For the motion of block `m`:

Along horizontal direction , `N_(1) = ma`
Along vertical direction `mg - (mu_(1) N_(1) + T) = m(2a)`
After substituting value of `N_(1)` in fig (i) , we have
`Mg - (mu_(1) ma + T) = m (2a)`
For the motion of block `M`:
Along verticle direction `sum F_(v) = 0`
or `N_(2) = T + mu_(1) N_(1) + Mg`
Along horizontal direction
`2T - (N_(1) + mu_(2)N_(2)) = Ma`
From equation (i) and (iii) we get
`2T - [N_(1) + mu_(2) (T + mu_(1) N_(1) + Mg)]= Ma`
or `2T - [ma + mu_(2) u_(2)(ma) + mu_(2) Mg]= Ma`...(v)
Now solving equation(ii) and (iv) , we get
`a = (2m - mu_(2)(M + m)g)/(M + m ( 5 + 2(mu_(1) - mu_(2))]`