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

with the invention of the automobile, the woman of the house could drive to the store and get groceries. women were also free to socialize more. during world war ii women were allowed to work in factories to the war effort and could drive to work or ride mass transit buses to their jobs.