example of power

Problem 3.1:
An elevator must lift 1000 kg a distance of 100 m at a velocity of 4 m/s. What is the average power the elevator exerts during this trip?


Solution for Problem 3.1
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.