The friction coefficient between the two blocks shown in the figure...

Let $T$ be the tension in the string and $f$ be the frictional force between the two blocks.

For the system to remain in equilibrium, both blocks must be at rest.

Considering the forces on the upper block $m$ in the horizontal direction, the applied force $F$ acts to the left, while the tension $T$ and static friction $f$ act to the right.

$F = T + f$

Considering the forces on the lower block $M$ in the horizontal direction, the tension $T$ acts to the right, and the static friction $f$ exerted by $m$ acts to the left.

$T = f$

Substituting $T = f$ into the first equation, we get:

$F = 2f$

The maximum value of static friction is $f = \mu mg$.

Therefore, the maximum horizontal force $F$ that can be applied without disturbing equilibrium is:

F_{\text\max} = 2\mu mg

When the applied force is doubled, the new force is $F' = 4\mu mg$. The blocks will now accelerate. Let $a$ be the acceleration of block $m$ to the left. Due to the string constraint, block $M$ will accelerate to the right with the same magnitude $a$.

The kinetic friction between the blocks is $f = \mu mg$.

Writing the equations of motion for both blocks:

For block $m$ moving to the left:

$F' - T - f = ma$

$4\mu mg - T - \mu mg = ma$

$3\mu mg - T = ma$

For block $M$ moving to the right:

$T - f = Ma$

$T - \mu mg = Ma$

Adding the two equations of motion:

$(3\mu mg - T) + (T - \mu mg) = ma + Ma$

$2\mu mg = (M + m)a$

Solving for the acceleration $a$:

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