For exercise, a 50 kg athlete climbs stairs for 10 s. The athlete’s metabolic

Question

For exercise, a 50 kg athlete climbs stairs for 10 s. The athlete’s metabolic power is 1200 W. Assume the efficiency of the human body is 25%.
a. In applying the energy equation ( ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W ) to the system consisting of the

earth and the athlete, which terms are positive? Which are negative? (If any terms are zero, ignore them.) €

b. Determine the metabolic energy expended by the athlete, in food calories.

c. Determine the increase in potential energy of the system.

d. How high did she climb?

Explanation / Answer

efficeincy = 25%

so work used = 1200*10 * 25/100 = 3000 J

= 717.703 calories

increase in potential enrgy = 3000 J

height =h

50*9.8*h = 3000

h = 6.122m

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