Question

A stone of mass $2kg$ 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 $100m$. Calculate the frictional force between the stone and the ice.

• A. $3N$
• B. $2N$
• C. $1N$
• D. $4N$

Answer 1

The correct option is D $4N$

Given, mass of the stone, $m=2kg$
Initial velocity of stone, $u=20m{s}^{-1}$
Final velocity of stone, $v=0$ (stone comes to rest)
Distance covered by the stone, $s=100m$
Let acceleration of the stone be 'a', and force acting on the stone due to friction be, 'F'.
From the third equation of motion, we have, $v^2$ – $u^2=2as$
$\Rightarrow 0^2–{20}^2=2a\times 100$
$\Rightarrow –400=200a$
$\Rightarrow a=-2{ms}^{-2}$
We know, $F=m\times a=2\times -2=-4N$
Therefore, the magnitude of frictional force is $4N$. The negative sign shows that the direction of friction is opposite to direction of motion.