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A block of mass 2 kg is placed on a rough horizontal surface, coefficient of friction between the block and the surface is 0.4. A horizontal force of 10N is applied on the block. Then rate of work done by the force at t = 1 sec is (The 10 N force is applied at t = 0).

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10 W

22 N

8 W

16 W

answer is D.

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Detailed Solution

$a = \frac{10 - 0.4 \times 2 \times 10}{2} m / s^{2} = 1 m / s^{2}$

After 1 sec, v = u + at = 0 + 1 x 1 $m / s^{2}$ = 1 m/s
$∴$ Power P = F. $ν$ = 10 x 1 Watt = 10 Watt

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