Question

An elevator lifts 2400 kg a distance of 55 m in 18.9 s. How much power does the

elevator demonstrate?

Answer 1

The work done by the elevator over the 55 meters is easily calculable: W = mgh = (2400)(9.8)(55) = 1293600Joules. The total time of the trip can be calculated from the velocity of the elevator: t = x/v=55/18.9 = 2.91. Thus the average power is given by: P =W/t=1293600/2.91=444536 Watts