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18 September, 15:28

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

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  1. 18 September, 15:50
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    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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