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?

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

answer; how far out of place which the object finished relative to the starting location;