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:
[tex]\dot W = m\cdot g \cdot \dot y[/tex] (Eq. 1)
Where:
[tex]m[/tex] - Mass of the athlete, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]\dot y[/tex] - Climbing rate, measured in meters per second.
[tex]\dot W[/tex]- Power, measured in watts.
By the consideration of constant velocity, we get that the climbing rate is represented by:
[tex]\dot y = \frac{s}{t}[/tex] (Eq. 2)
Where:
[tex]s[/tex] - Travelled distance, measured in meters.
[tex]t[/tex] - Time, measured in seconds.
And by substituting on (Eq. 1), the following expression is found:
[tex]\dot W = \frac{m\cdot g\cdot s}{t}[/tex]
If we know that [tex]m = 75\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]s = 5\,m[/tex] and [tex]t = 9\,s[/tex], then the minimum power output is:
[tex]\dot W = \frac{(75\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}{9\,s}[/tex]
[tex]\dot W = 408.625\,W[/tex]
The minimum power output used to accomplish this feat is 408.625 watts.