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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 the escalator is running so that the escalator steps move at a speed of `0.5m^-1`. 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? |
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Answer» Work done by the man in climbing the escalator, `W=mgh=90xx10xx20sin30^@=9000J` With the escalator stationary, man takes `10s` to climb up, therefore average power `P=W/t=9000/10=900W` Speed of the man when escalator is stationary, `v=20/10=2ms^-1` In the second case, the speed of the man relative to the moving escalator is also `2ms^-1`. His speed relative to the ground is `0.5+2=2.5ms^-1`. According, the time taken by the man in climbing is `20/2.5=8s` The power developed by the man as seen from the ground reference is then, on the average: `Power=W/l=9000/8=1125W` |
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