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3 September, 13:05

A 70-kg person walks at a steady pace of 4.4 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?

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  1. 3 September, 13:11
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    P = 94 W

    Explanation:

    First we find the horizontal power. Since, the work done is in the form of kinetic energy of the person, while walking. Therefore:

    P₁ = K. E/t = mv²/2t

    where,

    P₁ = Power required for walking on flat surface = ?

    m = mass = 70 kg

    v = speed = (4.4 km/h) (1 h/3600 s) (1000 m/1 km) = 1.22 m/s

    t = time = 1 sec (For power calculation per second)

    Therefore,

    P₁ = (70 kg) (1.22 m/s) ²/2 (1 s)

    P₁ = 52.1 W

    Now, for the power required for vertical climb, we use the formula:

    P₂ = P. E/t = mgh/t

    P₂ = mgv

    where,

    P₂ = Power required for vertical climb = ?

    m = mass = 70 kg

    g = 9.8 m/s²

    v = vertical speed = 5% of 1.22 m/s (for 5% grade) = (0.05) (1.22 m/s)

    Therefore,

    P₂ = (70 kg) (9.8 m/s²) (0.061 m/s)

    P₂ = 41.9 W

    Therefore, the total metabolic power required, will be equal to:

    P = P₁ + P₂

    P = 52.1 W + 41.9 W

    P = 94 W
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