Question

A 68 kg hiker walks at 5.0 km/h up a 3.0 ∘slope.

What is the necessary metabolic power? Hint: You can model her power needs as the sum of the power to walk on level ground plus the power needed to raise her body by the appropriate amount.

Express your answer to two significant figures and include the appropriate units.

Answer 1

Power = Energy ÷ time

As the hiker walks up the slope, the hiker’s potential energy increases.
∆PE = 68 * 9.8 * ∆h = 666.4 * ∆h
Since the hiker is moving up a 3˚ slope, ∆h = sin 3˚ * distance up the slope.

Convert from km/h = m/s
1 km = 1000 m, 1 h 3600s
1 km/h = 1000/3600 = 10/36 = 5/18 m/s
5 km/h = 25/18 m/s ≈1.3889 m/s
Each second, the hiker moves approximately 1.3889 meters up the slope.

∆h = sin 3˚ * 25/18
∆PE = 666.4 * sin 3˚ * 25/18
This is approximately = 112.7968 J
Each second, the hiker’s potential energy increases this amount.
Power = Energy ÷ time, time = 1 second