Solution to 2001 Problem 73

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.

return to the 2001 problem list

return to homepage

Please send questions or comments to where X = physgre.