A uniform rope of mass 1 kg and length 1m is lying on the ground. One end of the rope is pulled up by a worker with the constant velocity of 1m/s. The average power supplied by the worker in pulling the entire rope just off the ground such that the rope becomes vertical. $(t a k e g = 10 m / s^{2})$

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5.5 W

6 W

10.5 W

None of these

answer is A.

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Detailed Solution

Average power =<P>= $\frac{Total work done}{Total time}$

$= \frac{Total increase in energy}{time} = \frac{mg \frac{1}{2} + \frac{1}{2} {mv}^{2}}{\frac{l}{v_{}}} = 5.5 W$

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