A body of mass m is dropped from a height h on a sand floor....

A body of mass $m$ is dropped from a height $h$ on a sand floor. If the body penetrates $x \mathrm{~m}$ into the sand, the average resistance offered by the sand to the body is (A) $m g\left(\frac{h}{x}\right)$ (B) $m g\left(1+\frac{h}{x}\right)$ (C) $m g h+m g x$ (D) $m g\left(1-\frac{h}{x}\right)$

A body of mass $m$ is dropped from a height $h$ on a sand floor. If the body penetrates $x \mathrm{~m}$ into the sand, the average resistance offered by the sand to the body is
(A) $m g\left(\frac{h}{x}\right)$
(B) $m g\left(1+\frac{h}{x}\right)$
(C) $m g h+m g x$
(D) $m g\left(1-\frac{h}{x}\right)$

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Key Concepts

Average Resistive Force

This key concept involves calculating a constant force that, over the penetration distance, performs the same work as any variable resistance might. By equating the work done by this average force (the force multiplied by the displacement x) to the gravitational energy lost (which includes the drop height and the penetration distance), one can derive the expression for the average resistance offered by the sand.

Gravitational Potential Energy

This concept refers to the energy an object possesses because of its position in a gravitational field. In the context of the problem, the mass m held at height h has gravitational potential energy that is converted into kinetic energy as it falls, and eventually into work done against the sand's resistance when it penetrates the sand.

Work-Energy Principle

The work-energy principle states that the total work done on an object is equal to its change in kinetic energy. In this scenario, the kinetic energy acquired by the mass during the fall, together with the additional work done as it continues into the sand, is fully dissipated by the work done against the average resistive force of the sand over the distance x.

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