Question #240857

What average mechanical power (in W) must a 61.5 kg mountain climber generate to climb to the summit of a hill of height 345 m in 49.0 min? Note: Due to inefficiencies in converting chemical energy to mechanical energy, the amount calculated here is only a fraction of the power that must be produced by the climber's body.


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1
Expert's answer
2021-09-24T09:26:41-0400

Let us consider the total energy needed to climb the mountain:

E=mgΔh=61.5kg9.81N/kg345m=2.08105J.E = mg\Delta h = 61.5\,\mathrm{kg}\cdot 9.81\,\mathrm{N/kg}\cdot 345\,\mathrm{m} = 2.08\cdot10^5\,\mathrm{J}.

The power is

P=EΔt=2.08105J49.060s=70.8W.P = \dfrac{E}{\Delta t} = \dfrac{2.08\cdot10^5\,\mathrm{J}}{49.0\cdot60^s} = 70.8\,\mathrm{W}.


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