Question

A stone of mass 'm' falls from a height 'h' into sand pit and moves moves down into the sand through a distance 's' before it comes to the state of rest. Calculate the resistance force offered by the sand on the stone.

• A. $mg(1+\frac{h}{s})$
• B. $mg\frac{h}{s}$
• C. mg(s+h)
• D. $mg(1-\frac{s}{h})$

Answer 1

The correct option is A $mg(1+\frac{h}{s})$

Using sign convention, Upward direction= Positive sign, Downward direction= Negative sign,

Mass of stone = $m$, $u=0$, acceleration= $-g$, distance= $-h$

Let velocity of stone is $v_1$ when it falls on the sand.

Using $v^2=u^2+2as$ $\Rightarrow v_1^2=2\times (-g)\times (-h)$ $\Rightarrow v_1^2=2gh$

Let acceleration is $a$ when stone is passing through sand.

Final Velocity of stone is $v_2$=0, distance=$-s$,

Using $v^2=u^2+2as$ $\Rightarrow v_2^2=v_1^2-2as$ $\Rightarrow 0=2gh-2as$

$\Rightarrow a=g\frac{h}{s}$

Stone is going downward and stops so resistance force retard the stone so acceleration is acting upward as shown in figure.

Let resistance force is $F$.

From the FBD of the stone ,

$F-mg=ma$ $\Rightarrow F=m(g+a)$ $\Rightarrow F=mg(1+\frac{h}{s})$