Work efficiency running up stairs

Question

A 70 kg physicist is running up the stairs of the physics building and makes it up 48 m vertically in 1 minute.

a) What is the physicist's power output while running up the stairs?

b) If the physicist's work efficiency is just 3% at what rate were they using metabolic energy?

c) If the physicist's metabolism can provide 5.6 x 10⁶ J of energy before they need a rest, how long could they continue running up the stairs?

Answer 1

a) When the physicist runs his kinetic energy is converted into potential energy at the end of the journey.

Let's call this potential energy, the output energy, E_out

But potential energy is given by

E_out = mgΔh

where m is the mass of physicist, g is acceleration due to gravity ( g=9.80m/s²), and Δh is change in height. Plugging in the values gives

E_out = (70kg)(9.81m/s²)(48m - 0) = 32961.6 J

The power output, P = output energy/time = E_out/t = 32961.6J/60s = 549.4 J/s = 549.4 W

b) At 30% efficiency, the physicist' metabolic rate is (32961.6J)*(100/3) = 1098720 J per minute.

In other words, the metabolic energy release is 1098720 J per minute.

c) If 1098720J of energy is released per minute, how long will it take 5.6 x 10⁶ J?

==> (5.6 x 10⁶ J)/(1098720J/minute) ≈ 5.1 minutes

Therefore, the physicist could continue running for 5.1 minutes = 306 seconds before coming to rest!