Question

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? 370 W 350 W 315 W 300 W

Answer 1

Power, P = 350 watts

Step-by-step explanation:

It is given that,

Mass of the person, m = 70 kg

Speed of the person, v = 5 km/h = 1.388 m/s

It is assumed that 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. Let P is the power required. It can be calculated as :

`P=(W)/(t)=(F.d)/(t)`

Since,
`(d)/(t)=v`

Here,
`v=5\% \ of\ 1.388=0.0694\ m/s`

`P=F* v`

`P=m* g* v`

`P=70* 9.8* 0.0694`

P = 47.60 watts

It can be assumed that, the required power to walk is 300 W. So,

P = 300 W + 47.60 W

P = 347.6 W

or

P = 350 W

So, the power required is 350 watts. Hence, this is the required solution.