Find the acceleration of the block of mass M in the situation of figure. The coefficient of friction between the two blocks is `mu_1` and that between the bigger block and the ground is `mu_2`.

Answer

A, B

Explanation

Let the acceleration of the block M is a towards rilght. So the block m must to down with an acceleration 2a.
M it will also have acceleration a towards right. So it will experience two nertial forces as shown the free body diagram. From free body digram........1

`R_1-ma=0`
`rarr R_1=ma`..........i
Again, `2ma+T-Mg+mu_1R_1=0`
`rarr T=Mg-(2+mu_1)ma`.........ii
From free body diagram -2,
`T+mu_1R_1+Mg-R_2=0`
Putting the value of `R_1` from i
`R_2=T+mu_1 ma+mg`
Putting teh value of T from ii,
`R_2=(Mg-2ma-mu_1ma)=mu_1ma+Mg+mg`
R_2=Mg+mg-2ma`.....iii
Again fromn the free body diagram 2,
`T+T-R-Ma-mu_2R_2=0`
`rarr 2T-Ma-ma-mu_2(Mg+mg-2ma)=0`
Putting the values of `R_1 and R_2` from i and iii
`2T=(M+m)a+mu_2(Mg+mg-2ma)`....iv
From equation ii and iv we have
`2T=2mg-2(2+mu_1)ma`
`=(M+m)a+mu_2(Mg+mg-2ma)`
`rarr 2mg-mu_2(M+m)g`
`=a[M+m-2mu_2m+4m+2mu_1m]`
`rarr a=([2m-m_2(M+m)]g)/(M+m[5+2(mu_1-mu_2)])`