Homework Help: physics

Posted by Anonymous on Thursday, June 28, 2012 at 8:19am.

a 70-kg man climbs a mountain 1200 m high (measured from the base) in 4 h and uses 9.8 kcal/min. calculate his power consumption in watts. what is his power output in useful work? what was the efficiency of this man during the climb?

physics - Elena, Friday, June 29, 2012 at 11:20am
9.8 kcal min = 41 030.64 J/min
41 030.64 J/min•60 • 4 = 9847353.6 J.
Useful work
mgh = 70•9.8•1200 = 823200 J.
mgh/t=823200/4•60 = 3430 J/min

η =(3430/41030.64) •100% = =0.086%

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