Question

In the rope climb, a 75 kg athlete climbs a vertical distance of 5.0 m in 9.0 s. What minimum power output was used to accomplish this feat?
___ W

Answer 1

Answer:

The minimum power output used to accomplish this feat is 408.625 watts.

Step-by-step explanation:

The minimum power is that needed to overcome potential gravitational energy at constant velocity. From Principle of Energy Conservation, Work-Energy Theorem and definition of power we obtain the following relationship:

$\dot W = m\cdot g \cdot \dot y$ (Eq. 1)

Where:

$m$ - Mass of the athlete, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$\dot y$ - Climbing rate, measured in meters per second.

$\dot W$- Power, measured in watts.

By the consideration of constant velocity, we get that the climbing rate is represented by:

$\dot y = \frac{s}{t}$ (Eq. 2)

Where:

$s$ - Travelled distance, measured in meters.

$t$ - Time, measured in seconds.

And by substituting on (Eq. 1), the following expression is found:

$\dot W = \frac{m\cdot g\cdot s}{t}$

If we know that $m = 75\,kg$, $g = 9.807\,\frac{m}{s^{2}}$, $s = 5\,m$ and $t = 9\,s$, then the minimum power output is:

$\dot W = \frac{(75\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}{9\,s}$

$\dot W = 408.625\,W$

The minimum power output used to accomplish this feat is 408.625 watts.