Question

A stone of 1 kg is thrown with a velocity of $20m{s}^{-1}$ 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

Given, initial velocity of the stone, $u=20m{s}^{-1}$
final velocity of the stone, $v=0$, mass,
massof stone, $m=1kg$ and
distance covered by the stone, $s=50m$.
Let the acceleration be $a$.
By using third equation of motion, $v^2-u^2$ = 2as, we get,
$0-{20}^2=2a\times 50$
$\Rightarrow a=-4m{s}^{-2}$
Since, no external force is acting on a body therefore, force of friction, $F=ma=1\times –4=-4N$.