A stone of 1 kg is thrown with a velocity of `20 ms^-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?

Explanation

Here `m=1kg ,u=20 ms^-1`
`v=0, s=50m, F=?`
From `v^2-u^2=2as`
`0-(20)^2=2a times 50=100a`
or `a=(-400)/100=-4ms^-2`
`therefore F=ma`
`F=1(-4)=-4N`
Negative sign indicates that force of friction is opposing the motion of the ball.