Question

[IAPT final key is A; We gave A or D. The answer depends on the unstated relation between m and M.]

[Key Concept: When the block slides on the plate, kinetic friction μmg is the only horizontal force on the plate.]

$$F = 2\mu mg$$

Suppose together:

$$a = \left( \frac{F}{m + M} \right) = \frac{2\mu mg}{m + M}$$

$$\begin{matrix} & f = M \cdot a = \frac{2Mm\mu\text{ }g}{\text{ }m + M} \\ & f = \mu mg\left\lbrack \frac{2M}{\text{ }m + M} \right\rbrack \\ & \left( f_{s} \right)_{\max} = \mu mg \end{matrix}$$

If m > M :

then $f \leq \left( f_{s} \right)_{\text{max~}}$

block move together

$$a = \frac{2\mu mg}{m + M}$$

If $m < M$ :

then $f > \left( f_{s} \right)_{\text{max~}}$

our assumption is wrong.

$$a = \frac{\mu mg}{M}$$