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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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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: 

v 2 = u 2 + 2as 

Where, 

Acceleration, a 

(0)2 = (20)2 + 2 × a × 50 

a = – 4 m/s

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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