CBSE Notes, Lectures

CBSE - Physics - Force and Laws of Motion

Force and Laws of Motion

NCERT Exercise

A stone of 1 kg is thrown with a velocity of 20 m s−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?

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, v2 - u2
2 = 2as,
Or, 0 - 202 = 2a × 50,
Or, a = – 4 ms-2
Force of friction, F = ma = – 4N

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