Question

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?

Answer 1

Answer:

The required power is 471.5W

Explanation:

Let d be the horizontal distance traveled on trade mil

Vertical distance traveled=0.05d

Time takes to travel distance d=d/5 hr  

The energy for vertical motion is given as:

$E=mgh\\E=70kg(9.81m/s^{2} )0.05d\\E=34.3d$

And the Power is given as:

$Power=Energy/time\\Power=\frac{34.3d}{\frac{d}{5} }\\ Power=171.5W$

So the total power is:

$Power_{total} =300W+171.5W\\Power_{total} =471.5W$

The required power is 471.5W