Question

A 63 kg gymnast climbs 15 m up a vertically hanging rope. He completed the distance in 70 seconds. How much power did his body output to achieve the climb?

Answer 1

The work required is equal to the potential energy when he's at the top. This is given by 

$E=mgh=63kg \times 9.8 \frac{m}{s^2} \times 15m=9261J$

The power is this energy divided by the time it took to do the work

$P= \frac{E}{t}= \frac{9261J}{70s} =132.3W$

Answer 2

Answer:b

Explanation:

hope it helps