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

Initial velocity of the stone , u=20 m/s
Final velocity of the stone v=o
Distance covered by the stone S=50 m
Since `V^2-u^2`
2= 2as
`0-20^2=2a xx50`
`F=-4ms^(-2)`
Force of friction, F=ma=-4N