A body of mass M is dropped from a height h o | vedclass

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(B) Let the body strike the sand floor with velocity $v$. By the law of conservation of energy,the potential energy at height $h$ is converted into ki

Vedclass · Accepted Answer

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

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(B) Let the body strike the sand floor with velocity $v$. By the law of conservation of energy,the potential energy at height $h$ is converted into kinetic energy just before impact: $Mgh = \frac{1}{2} Mv^2$.When the body enters the sand,it travels a distance $x$ before coming to rest. Let $F$ be the average resistive force of the sand.Applying the work-energy theorem: The net work done on the body is equal to the change in its kinetic energy.The forces acting on the body inside the sand are the weight $Mg$ (downwards) and the resistive force $F$ (upwards).Work done by net force = $(F - Mg)x = \Delta K = 0 - \frac{1}{2} Mv^2$.Since $\frac{1}{2} Mv^2 = Mgh$,we have $(F - Mg)x = Mgh$.$Fx - Mgx = Mgh \implies Fx = Mgh + Mgx$.$F = Mg \left( 1 + \frac{h}{x} \right)$.

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

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