A stone weighing 1 kg and sliding on ice with a velocity of 2 m/s is stopped by friction in 10 sec. The force of friction (assuming it to be constant) will be

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-20 N

-0.2 N

0.2 N

20 N

answer is B.

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Detailed Solution

u = 2m/s, v = 0, t =10 seca=v−ut=0−210=−210=−15Friction force = ma = 1 x (-0.2) = -0.2 N

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