Question

In climbing up a rope, a 62 kg athlete climbs a vertical distance of 5.0 m in 9.0 s. What minimum power output was used to accomplish this feat?

Short answer

The minimum power used by the athlete to accomplish the feat is 337.56 W.

Step-by-step solution

Definition of power

Power is directly proportional to the work done on an object. It is defined as the rate of doing work or the work done per unit time. Mathematically, power is represented as\(P = \frac{W}{t}\), where W is the work done and t is the time taken.

Identification of given data

The mass of the athlete is \(m = 62\;{\rm{kg}}\).

The vertical distance (or height) covered by the athlete is \(h = 5\;{\rm{m}}\).

The time taken by the athlete is \(t = 9\;{\rm{s}}\).

Determination of the work done on the athlete

The work done on the athlete is the change in its potential energy, which is given by:

\(\begin{aligned}W &= PE\\ &= mgh\\ &= \left( {62\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {5\;{\rm{m}}} \right)\\ &= 3038\;{\rm{J}}\end{aligned}\)

Determination of the minimum power used by the athlete

The power used by the athlete to accomplish the task is the work done per unit time. It is given as:

\(\begin{aligned}P &= \frac{W}{t}\\ &= \frac{{3038\;{\rm{J}}}}{{9\;{\rm{s}}}}\\ &= 337.56\;{\rm{W}}\end{aligned}\)

Thus, the minimum power used is 337.56 W.