What average power is genereted by a `90.0 kg` mountain climbs who climbs a summit of height `600 m` in `90.0 min`?

Answer

100

Explanation

We assume the climber has negligible speed at , both the beginning and the end of the climb .Then `K_(f) = K_(p)` and the work done by the muscles is
`W_("ne") = 0 + (U_(f)- U_(i)) = mg(y_(f) - y_(i))`
`= (90.0 kg) (10.0m//s^(2))(600m)`
` = 5.40 xx 10^(5)J`
The evarege the power delivered is
`P= (W_(nc))/(Delta t)= (5.40 xx 10^(5)J)/((90 min)(60s//1min)) = 100W`