A uniform rope of mass 1 kg and length 1 m is lying on the ground. One end the rope is pulled up by a worker with the constant velocity of 1 m/s. The average power supplied by the worker…

Ans: (a)

Sol: $\displaystyle Average \; Power = \frac{Total \; Work \;done}{Total \; time}$

$\displaystyle P_{avg} = \frac{W_1 + W_2}{t}$

W₁ = Gain in P.E

$\displaystyle W_1 = m g \frac{L}{2} $ (Since Center of mass lies at mid of rope of length L)

$\displaystyle W_1 = 1 \times 10 \times (1/2) = 5 J $ …(i)

$\displaystyle W_2 = \Delta K.E = \frac{1}{2} m v^2 $

$\displaystyle W_2 = \frac{1}{2} \times 1 \times 1^2 = 0.5 J $ …(ii)

$\displaystyle t = \frac{L}{v} = \frac{1}{1} = 1 sec $

$\displaystyle P_{avg} = \frac{W_1 + W_2}{t}$

$\displaystyle P_{avg} = \frac{5 + 0.5}{1}$

= 5.5 W