Question

A uniform rope of linear mass density $\lambda $ and length $l$ is coiled on a smooth horizontal surface. One end is pulled up by an external agent with constant vertical velocity $v$.Choose the correct option(s).
(A). Power developed by external agent as a function of $x$is $P = \lambda xgv$
(B). Power developed by external agent as a function of $x$is $P = (\lambda {v^2} + \lambda xg)v$
(C). Energy lost during the complete lift of the rope is zero.
(D). Energy lost during the complete lift of the rope is $\dfrac{{\lambda l{v^2}}}{2}$

Answer 1

Hint: Here we apply the concept of average power which is defined as total work done per unit time.
Formula Used: ${P_{avg}} = \dfrac{{{W_{total}}}}{T}$

Complete step by step answer:
Average power is total work done per unit time.
${P_{avg}} = \dfrac{{{W_{total}}}}{T} = \dfrac{{{W_1} + {W_2}}}{T}................(1)$
For ${W_1}$,
$dw = Fdx = \lambda xgdx$
${W_1} = \int {dw = \int\limits_0^L {\lambda xgdx} } = \dfrac{{\lambda g{L^2}}}{2}$ $\left( {\because \lambda = \dfrac{m}{L}} \right)$
From work energy theorem
$W = \Delta K.E = \dfrac{1}{2}M{v^2} - 0 = \dfrac{1}{2}M{v^2}$
Put the value of ${W_1}$ and ${W_2}$in equation (1)
${P_{avg}} = \dfrac{{\lambda g{v^2} + \dfrac{1}{2}\lambda L{v^2}}}{{\dfrac{L}{v}}}$ $\left( {T = \dfrac{L}{v}} \right)$
$ = \dfrac{{\lambda vgL}}{2} + \dfrac{1}{2}\lambda {v^3}$

Hence option B is correct.

Note: In this question it is said that a uniform rope of a length and a linear mass density which is coiled in on the horizontal smooth surface and from one side of the rope is pulled up with a constant velocity and we have to choose among the average power applied by the external agent in pulling the entire rope and the two condition of energy.
Other method,
The mass of the rope $m = \lambda dx$
The force or weight of the rope $F = \lambda gdx$
The power acting on the rope $P = \lambda gdx \times v = \lambda gv\int\limits_0^l {dx} $
The average power $P = \lambda gv\left[ x \right]_0^l = \lambda gvl$