The coefficient of friction between the two blocks displayed in the...

In the frame of the descending elevator, the effective acceleration due to gravity is $g_{eff} = g - a$.

The normal reaction between the block of mass $m$ and the block of mass $M$ is $N = m(g - a)$.

The maximum static friction force between the two blocks is f_\max = \mu N = \mu m(g - a).

Let $T$ be the tension in the string. For the system to remain in equilibrium, the net force on each block must be zero. With the impending motion of $m$ to the left, the friction force $f$ on $m$ acts to the right, and the friction force on $M$ acts to the left.

For block $M$ in the horizontal direction:
$T - f = 0 \Rightarrow T = f$

For block $m$ in the horizontal direction:
$F - T - f = 0 \Rightarrow F = 2f$

The maximum horizontal force F_\max without disrupting equilibrium corresponds to f = f_\max:
F_\max = 2f_\max = 2\mu m(g - a)

When the applied force is twice this maximum value, F = 2F_\max = 4\mu m(g - a). The system accelerates. Let $a'$ be the magnitude of the acceleration of the masses. Block $m$ accelerates to the left and block $M$ accelerates to the right with the same magnitude $a'$.

The equations of motion for the blocks are:
For block $M$: T - f_\max = M a'
For block $m$: F - T - f_\max = m a'

Adding these two equations yields:
F - 2f_\max = (m + M)a'

Substituting the values of $F$ and f_\max:
$4\mu m(g - a) - 2\mu m(g - a) = (m + M)a'$

$2\mu m(g - a) = (m + M)a'$

$a' = \dfrac{2\mu m(g - a)}{m + M}$

Answer: $2\mu m(g - a)$ and $\dfrac{2\mu m(g - a)}{m + M}$