A 90 kg man runs up an escalator while it is not in operation in 10s. What is the average power developed by the man. Suppose next that the escalator is running so that the escalator steps move at a speed of 0.6 m/s. What is then the power developed by the man as seen by the ground reference if he moves at the same speed relative to the escalator steps as he did when the escalator is not in operation ?

A 90 kg man runs up an escalator while it is not in operation in 10s.

1.	A 90 kg man runs up an escalator while it is not in operation in 10s. What is the average power developed by the man. Suppose next that the escalator is running so that the escalator steps move at a speed of 0.6 m/s. What is then the power developed by the man as seen by the ground reference if he moves at the same speed relative to the escalator steps as he did when the escalator is not in operation ?
Answer» Solution :Work done by the man in CLIMBING the escalator W = mgh `= 90 xx 9.8 xx 17 sin 45^(@)` = 10611 J With the escalator stationary, man takes 10s to climb up THEREFORE, average power `P = (W)/(t) = (10611)/(10) = 1061.1` W Speed of the man when escalator is stationary `v = (17)/(10) = 1.7 m//s` In the SECOND CASE the speed of the man relative to the moving escalator is also 1.70 m/s. His speed relative to ground = 0.6 + 1.7 = 2.3 m/s. Accordingly the time taken by the man for climbing is = `(17)/(2.3) = 7.4` s The power developed by the man as seen from the ground reference is then, on the average : `rArr "Power" = (W)/(t) = (10611)/(7.4) = 1434` W.