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A stone of `1kg` is thrown with a velocity of `20 ms^(-1)` across the frozen surface of lake and comes to rest after travelling a distance of `50m`. What is the force of friction between the stone and the ice?

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Correct Answer - `-4 N`
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0
Distance covered by the stone, s = 50 m
Since, `v^(2)-u^(2)=2as`
Or, `0-20^(2)=2axx50`,
Or , `a=-4 ms^(-2)`
Force of friction, F=ma= -4N

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