A block of mass $m=5\text{ kg}$ is placed on a rough horizontal surface with coefficient of static friction ${\mu }_s=0.4$. A horizontal force of $F=15\text{ N}$ is applied to the block. What is the frictional force $f$ acting on the block? (Take $g=10{\text{ m/s}}^2$)

Solution

1. Given: mass $m=5\text{ kg}$, coefficient of static friction ${\mu }_s=0.4$, applied force $F=15\text{ N}$, and $g=10{\text{ m/s}}^2$.
2. Calculate the maximum static friction: $$\begin{array}{rl}{f}_{\text{max}}& ={\mu }_{s}mg\\ & =0.4\times 5\times 10\\ & =20\text{ N}\end{array}$$.
3. Compare the applied force to the maximum static friction: since $$\begin{array}{rl}F& =15\text{ N}<f_{\text{max}}\\ & =20\text{ N}\end{array}$$, the block does not move.
4. Because the block is at rest, the static frictional force $f$ must equal the applied force: $$\begin{array}{rl}f& =F\\ & =15\text{ N}\end{array}$$.
5. Hence the answer is (A).