Question

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

Answer 1

Here, $m=1 \mathrm{~kg}, u=20 \mathrm{~m} \mathrm{~s}^{-1}$

$v=0, \quad s=50 \mathrm{~m}, \quad F=?$

From, $v^{2}-u^{2}=2 a s$,

$0-(20)^{2}=2 a \times 50=100 a$

or $a=\frac{-400}{100}=-4 \mathrm{~m} \mathrm{~s}^{-2}$

$\therefore \quad F=m a$,

$F=1(-4)=-4 \mathrm{~N}$

Negative sign indicates that force of friction is opposing the motion of the ball.