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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 sec

$a = \frac{v - u}{t} = \frac{0 - 2}{10} = - \frac{2}{10} = - \frac{1}{5}$

Friction force =ma=1x(-0.2)=-0.2N

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