Question

A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x\, cm$ into the sand, the average resistance offered by the sand to the body is

• A. $Mg\left(\frac{h}{x}\right)$
• B. $Mg\left(1+\frac{h}{x}\right)$
• C. $Mgh + Mgx$
• D. $Mg\left(1-\frac{h}{x}\right)$

Answer 1

Answer: (B)

$Mg\left(1+\frac{h}{x}\right)$

Answer 2

the body strikes the sand floor with a velocity $v$, then Potential energy = Kinetic energy $Mgh=\frac{1}{2}M\upsilon^{2}$ With this velocity $v$, when body passes through the sand floor it comes to rest after travelling a distance $x$. Let $F$ be the resisting force acting on the body. Net force in downward direction $= Mg - F$ Work done by all the forces is equal to change in KE $\left(Mg-F\right)x=0-\frac{1}{2}M\upsilon^{2}$ $\left(Mg-F\right)x=-Mgh$ or $Fx=Mgh+Mgx$ or $F=Mg\left(1+\frac{h}{x}\right)$