For AB as a whole: from the equilibrium condition,
kx-(m+M)gsinθ-μ(m+M)gcosθ=0，
So the spring is in compression at this time. So, A is wrong.
When B, A and B are just separated, the elastic force between AB is zero. According to Newton's second law,
Along the inclined plane, mgsinθ+μmgcosθ=ma,
Get a=gsinθ+μgcosθ,
According to Newton's second law, the accelerations of A and B are the same, so B is correct.
C, in the process of A and B from release to maximum velocity V, for AB as a whole, according to the kinetic energy theorem, we get
-(m+M)gLsinθ-μ(m+M)gcosθ？ L+W bullet = 12(m+M)v2
Work done by the spring on A W =12 (m+m) V2+(m+m) GLS in θ+μ (m+m) GCOS θ? L, so c is wrong.
D, in the process of releasing from A and B to the maximum speed V, for B, according to the kinetic energy theorem, it is obtained as follows.
The work done by the resultant force on b is W =△Ek= 12mv2, so d is correct.
So: BD.