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A 60.0-kg woman runs up a 3.00 m high flight of stairs in 3.50 s, starting from rest and reaching a final speed of 2.00 m/s. Taking g = 9.80 m/s^2, what is her power output?
A504 W
B34.3 W
C538 W
D1884 W
Answer & Solution
Correct answer: C. 538 W
1. The work going into mechanical energy is W = KEf + PEg = one half m vf squared plus mgh.
2. The kinetic energy term is one half (60.0 kg)(2.00 m/s)^2 = 120 J, and the potential energy term is (60.0 kg)(9.80 m/s^2)(3.00 m) = 1764 J.
3. Adding these gives a total work of 1884 J.
4. Dividing by the time, P = 1884 J divided by 3.50 s, gives P = 538 W.
5. 1884 W is the total work left undivided by the time, and 504 W and 34.3 W come from omitting one of the two energy terms before dividing by time.
_Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.7 Power_
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