On flat ground, a 70-kg person requires about 300 W of metabolic power to walk at a steady pace of 5.0 km/h (1.4 m/s). Using the same metabolic power output, that person can bicycle over the same ground at 15 km/h.

A 70-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?a. 300 W b. 315 W c. 350 WW d. 370 WW

C 350W

Explanation:

Given power output to walk on a flat ground to be 300W, h = 0.05x, v = 1.4m/s

m = 70kg and g =9.8m/s².

x = horizontal distance covered

Total energy used = potential energy used in climbing and the energy used in a walking the horizontal distance.

E = mgh + 300t

Where t is the time taken to cover the distance

x = vt and h = 0.05vt

So

E = mg×0.05×vt + 300t

Substituting respective values

E = 70×9.8×0.05×1.4t +300t = 348t

P = E/t = 348W ≈ 350W.

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