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Grade: upto college level


A 57-kg woman runs up a flight of stairs having a rise of 4.5 m in 3.5 s. what average power must she supply?

5 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
471 Points
							As the woman runs up a flight of stairs, there are two forces will act on the woman. One is the gravitational force which is in the downward direction and other one is the normal force N of the floor which balances the weight of the body (gravitational force) and the direction is just opposite to the direction of gravitational force.
So, F = N
= weight of the body
= mg (Since, weight of the body = mg)
Here m is the mass of the body and g is the acceleration due to gravity.
To obtain the force F exerted by the woman to propel herself up the stairs, substitute 57 kg for m and 9.8 m/s2 for g in the equation F = mg,
F = mg
= (57 kg) (9.8 m/s2)
= (560 kg.m/s2) (1 N/1 kg.m/s2)
= 560 N
In accordance to the reference of woman the stairs are moving down. Thus the woman will exert a force in the down ward direction.
To obtain the work done W by the woman, substitute 560 N for F and 4.5 m for s in the equation W = Fs,
W = Fs
= (560 N) (4.5 m)
= (2500 N.m) (1 J/1 N.m)
= 2500 J
As the direction of force is in the direction of displacement, thus the work done is positive here.
To obtain the average power P supplied by the woman, substitute 2500 J for W and 3.5 s for t in the equation P = W/t,
P = W/t
= 2500 J/3.5 s
= (710 J/s) (1 W/ 1 J/s)
= 710 W
From the above observation we conclude that, the average power P supplied by the woman would be 710 W.

5 years ago
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