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Question

# A stone of 1 kg is thrown with a velocity of 20 ms−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?

A

5N

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B

3N

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C
There is no friction between stone and ice
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D

4N

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Solution

## The correct option is D 4NInitial velocity of the stone = 20 ms−1 . Final velocity of the stone, v = 0 (finally the stone comes to rest). Distance covered by the stone = 50 m. According to third equation of motion: v2=u2+2as 0=20×20+2×a×50 a=−4 ms−2. 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 = mass × acceleration. Therefore, F = 1 × (-4) = - 4 N Hence, the force of friction between the stone and the ice is 4 N. The negative sign above indicates that the direction of friction is opposite to the direction of motion.

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