The force of friction on block $B$ ( $F_{fr} = F_N \mu_s = F_N \mu_s \cdot 0.5$ ) must equal the weight of block $B$ ( $4 \mbox{ kg} \cdot g$ ):
$\begin{align*}F_N \cdot 0.5 = 4 \mbox{ kg} \cdot g \Rightarrow F_N = 8 \mbox{ kg} \cdot g\end{align*}$
Therefore, the acceleration of block $B$ is
$\begin{align*}a = \frac{F_N}{4 \mbox{ kg}} = 2 g\end{align*}$
The acceleration of both blocks must be the same. Therefore,
$\begin{align*}F = \left(16 \mbox{ kg} + 4 \mbox{ kg} \right) \cdot a = \left(16 \mbox{ kg} + 4 \mbox{ kg} \right) \cdot 2 g =...$
Therefore, answer (D) is correct.

Solution to 2001 Problem 73