A $2$ kg block is pressed against a vertical wall by a horizontal force $F$. The coefficient of static friction between the block and the wall is $0.5$. Find the minimum value of $F$ that keeps the block from slipping down. (Take $g = 10$ m/s².)

**Setup.** Normal reaction from the wall: $N = F$. Friction acts vertically (upward) to support the block's weight. **Limiting condition.** The block is on the verge of sliding when static friction is maximal: $\mu_s N = mg$. **Solve.** $\mu_s F = mg \Rightarrow F = \dfrac{mg}{\mu_s} = \dfrac{2 \times 10}{0.5} = 40$ N. **Distractor logic.** Option A ($20$ N) drops the $\mu_s$ factor entirely (treating friction as if it were $1.0$). Option D ($50$ N) over-counts by including $\mu$ on both sides.