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Question

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?

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Answer

Initial velocity of the stone, u = 20 m/s

Final velocity of the stone, v = 0 m/s

Distance covered by the stone, s = 50 m

According to the third equation of motion:

v2 = u2 + 2as

Where,

Acceleration, a

(0)2 = (20)2 + 2 × a × 50

a = – 4 m/s2

The negative sign indicates that acceleration is acting against the motion of the stone.

Mass of the stone, m = 1 kg

From Newton’s second law of motion:

Force,

F = Mass x Acceleration

F= ma

F= 1 × (– 4) = – 4 N

Hence, the force of friction between the stone and the ice is – 4 N.

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