MCAT Question A Day - 3/11/14 - Answer!
What is the average power output of a 70-kg woman who runs up an incline to a height of 2 meters in 2 seconds?
A. 343 W
B. 686 W
C. 1372 W
D. 1715 W
Pav = (70 kg)(9.8 m/s2)(2m) / (2s) = 686 kg m2/s3 = 686 J/s = 686 W
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The correct answer is (B). Pav = w/t, but the work done (w) is mgh; therefore,
Pav = (70 kg)(9.8 m/s2)(2m) / (2s) = 686 kg m2/s3 = 686 J/s = 686 W
