Question

A stone of $1$kg is thrown with a velocity of $20$m $s^{-1}$ across the frozen surface of a lake and comes to rest after travelling a distance of $50$m. What is the force of friction between the stone and the ice?

Answer 1

Before we proceed we have to make a free body diagram of the stone-ice system. In the figure, we are seeing that frictional force is acting in leftward while the stone in moving rightward direction. One thing we have to fix in our mind that frictional force will always act opposite to the direction of motion of the body.

Now on resolving the forces,

$f-ma$ or $f=ma$ ------(1)

also form the equation of motion $v^2=u^2+2as$, here final velocity is zero,so $v=0$

$u^2=-2as$

$a=\frac{-u^2}{2s}=-\frac{{20}^2}{2\times 50}=-4\frac{m}{se{c}^2}$

on substituting the given data in equation(1)

$f=1\times (-4)=-4N$