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Sloan [31]
3 years ago
14

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 accomplis

h this feat?
___ W
Mathematics
1 answer:
Komok [63]3 years ago
8 0

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.

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