Q. An iron nail is dropped from a height $h$ from the level of a sand bed. If it penetrates through a distance $x$ in the sand before coming to rest, then average force exerted by the sand on nail is
A) $mg\left( \frac{h}{x} + 1 \right)$
B) $mg\left( \frac{x}{h} + 1 \right)$
C) $mg\left( \frac{h}{x} - 1 \right)$
D) $mg\left( \frac{x}{h} - 1 \right)$
Explanation
(a) The nail has fallen through a total vertical distance of $(h + x).$ Hence loss in tis potential energy = $mg(h + x).$ If average retarding force exerted by sand on nail is $f$, then work done $W = - Fx$
For equilibrium,
$mg(h + x) = Fx$ or $F = mg\frac{h + x}{x} = mg\left( \frac{h}{x} + 1 \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.