Question

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

• A. $5N$
• B. $3N$
• C. There is no friction between stone and ice
• D. $4N$

Answer 1

The correct option is D

$4N$

Initial velocity of the stone = $20m{s}^{-1}$.
Final velocity of the stone, $v=0$(finally the stone comes to rest).
Distance covered by the stone = 50 m.
According to third equation of motion:
$v^2=u^2+2as$
0 × 0 = 20 × 20 + 2 × a × 50
$\Rightarrow$$a=-4m{s}^{-2}$.

The negative sign indicates that acceleration is in the opposite direction to the motion of the stone.
Mass of the stone, $m=1kg$.
From the newton's second law of motion:
Force = mass × acceleration. Therefore,
$F=1\times -4=-4N$.
Hence, the force of friction between the stone and the ice is 4N. The negative sign indicates that friction oppposes the relative motion.