Answer to Question #240857 in Molecular Physics | Thermodynamics for Kelani

Question

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.

__W

Answer 1

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

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

The power is

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