Q: 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 in pulling the entire rope just off the ground such that the rope becomes vertical. (take g = 10 m/s^2)

(a) 5.5 W

(b) 6 W

(c) 10.5 W

(d) None of these

**Click to See Answer : **

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

W_{1} = 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