Work and Power

The work done by the elevator over the 100 meters is easily calculable: W = mgh = (1000)(9.8)(100) = 9.8×10⁵ Joules. The total time of the trip can be calculated from the velocity of the elevator: t = = = 25 s . Thus the average power is given by: P = = = 3.9×10⁴ Watts, or 39 kW.

To solve this problem we will use the equation P = Fv cosθ , Instead of the usual power equation, as we are given the velocity of the object. We merely need to calculate the force exerted on the object by the air resistance, and the angle between the force and the velocity of the object. Since the object has reached a constant speed, the net force on it must be zero. Since there are only two forces acting on the object, gravity and air resistance, the air resistance must be equal in magnitude and opposite in direction as the force of gravity. Thus F _a = - F _G = mg = 98 N, pointing upwards. Thus the force applied by air resistance is antiparallel to the velocity of the object. Thus:

Our first step must be to generate an expression for work. We have already seen that the work done by gravity after a distance h of free fall is equivalent to mgh . Can we take a time derivative of this expression? Of course: since h is a measure of displacement, its derivative will simply give us the velocity of the object: = = mgv . Thus, at any time during an objectís free fall, the power exerted by gravity is given by mgv . Recall that P = Fv . If we check our derived answer against this equation we find that we are correct.