Question

A uniform rope of linear mass density $\lambda$ and length l 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:

• A. $\frac{1}{2}\lambda l{v}^2+\frac{\lambda l^{2}g}{2}$
• B. $\lambda lgv$
• C. $\frac{1}{2}\lambda v^3+\frac{\lambda lvg}{2}$
• D. $\lambda lgv+\frac{1}{2}\lambda v^3$

Answer 1

Answer: (C)

$\frac{1}{2}\lambda v^3+\frac{\lambda lvg}{2}$

Average power is total work done per unit time.

$P_{avg}=\frac{W_{total}}{T}$

$=\frac{W_1+W_2}{T}.....................\left(1\right)$

For W₁,

$dw=Fdx$

$=\lambda xgdx$

$W_1=\int dw=\underset{0}{\overset{L}{\int }}\lambda xgdx(\because \lambda =\frac{m}{L})$

$W_1=\frac{\lambda g{L}^2}{2}$

From work energy theorem

$W=\Delta K.E$

$=\frac{1}{2}M{v}^2-0$

$W_2=\frac{1}{2}M{v}^2$

Put the values of W₁ and W₂ in equation (1)

$P_{avg}=\frac{\lambda g{v}^2+\frac{1}{2}\lambda L{v}^2}{\frac{L}{v}}(T=\frac{L}{V})$

$=\frac{\lambda vgL}{2}+\frac{1}{2}\lambda v^3$