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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? - Physics Cbse class 11
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 (finally the stone comes to rest)

Distance covered by the stone, s = 50 m

According to the third equation of motion:

v2 = u2 + 2as

where, a = acceleration

(0)2 = (20)2 + 2 × a × 50

0 = 400 + 100 a

-400 = 100 a

Therefore, a = - 4 m/s2

Concept Note:- 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 × 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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