JEE MainPhysicsToughWork Power & Energy

A uniform rope of linear mass density $\lambda$ and length 1 is
coiled on a smooth horizontal surface. One end is pulled up with
constant velocity v. Then the average power applied by the external
agent in pulling the entire rope just off the ground is

Show answer and solution

Initial K.E. = 0
Initial P.E. = 0
When the rope is just pulled off the table.
Final K.E. =

\frac{1}{2}(\lambda l)v^{2}

Final P.E

= \frac{\left( \lambda\mathcal{l} \right)g\mathcal{l}}{2}

Time taken

t = 1/v

Average power =

\frac{\text{ net change in energy }}{\text{ time }}

= \frac{\frac{1}{2}\lambda\mathcal{l}v^{2} + \lambda\mathcal{l}g\mathcal{l/}2}{\mathcal{l/}v} = \frac{1}{2}\lambda v^{3} + \frac{\lambda\mathcal{l}vg}{2}

Source: Grass-G / P-06 Physics paper 14 October.docx

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