Question

A 65 kg mountain climber climbs through a vertical rope at a constant acceleration from rest to a

1 of 1.25 m/s in 10.0 s and feels no appreciable air resistance. How much power must the

speed m/s in 10.0s and alr must

climber imparts on the vertical rope?

Answer 1

The acceleration of the climber is:

$$a = \dfrac{v_f}{t} = \dfrac{1.25m/s}{10s} = 0.125m/s^2$$

where $v_f = 1.25m/s$ is its final speed, and $\Delta t = 10s$ is the time. The velocity is then depends on time as follows:

$$v(t) = at$$

According to the second Newton's law, the net force on the climber is:

$$F = ma$$

where $m = 65kg$ is his mass.

The power is then:

$$P(t) = F\cdot v(t) = ma\cdot at = ma^2t$$

This expression depends on time. I assume it is asked to find the average power during 10s. The average power is:

$$P_{av} = \dfrac{1}{\Delta t}\int_0^{\Delta t} P(t)dt = \dfrac{1}{\Delta t}\int_0^{\Delta t} ma^2tdt=ma^2\dfrac{\Delta t}{2}\\ P_{av} = \dfrac{65kg\cdot (0.125m/s^2)^2\cdot 10s}{2} \approx 5.1W$$

Answer. 5.1W.