A70 kg person walks at a steady pace of 5.0 km/h on a treadmill at a 5.0% grade. (that is, the vertical distance covered is 5.0% of the horizontal distance covered.) if we assume the metabolic power required is equal to that required for walking on a flat surface plus the rate of doing work for the vertical climb, how much power is required? 370 w 350 w 315 w 300 w

Power, P = 350 watts

Explanation:

It is given that,

Mass of the person, m = 70 kg

Speed of the person, v = 5 km/h = 1.388 m/s

It is assumed that the metabolic power required is equal to that required for walking on a flat surface plus the rate of doing work for the vertical climb. Let P is the power required. It can be calculated as :

Since,

Here,

P = 47.60 watts

It can be assumed that, the required power to walk is 300 W. So,

P = 300 W + 47.60 W

P = 347.6 W

or

P = 350 W

So, the power required is 350 watts. Hence, this is the required solution.

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