A block of mass m is placed upon another block of mass M , and the...

Since both blocks move together, they can be considered as a single system of total mass $(M+m)$.

The normal force exerted by the horizontal surface on the system is $N = (M+m)g$.

The kinetic friction force acting on the lower block from the surface is $f_k = \mu_k N = \mu_k(M+m)g$.

Applying Newton's second law to the system in the horizontal direction:

$F - f_k = (M+m)a$

Substituting the given acceleration $a = \dfrac{F}{2(M+m)}$ into the equation:

$F - \mu_k(M+m)g = (M+m) \left( \dfrac{F}{2(M+m)} \right)$

$F - \mu_k(M+m)g = \dfrac{F}{2}$

Rearranging the terms to solve for $\mu_k$:

$\mu_k(M+m)g = F - \dfrac{F}{2} = \dfrac{F}{2}$

$\mu_k = \dfrac{F}{2(M+m)g}$