Question

A 679-kg elevator starts from rest and moves upward for 3.30 s with constant acceleration until it reaches its cruising speed, 1.63 m/s. (a) What is the average power of the elevator motor during this period? (b) How does this amount of power compare with its power during an upright trip with constant speed?

Answer 1

Power = Work/time

Work = force x distance

in the first case, the force exerted = F =ma

if the elevator started from rest and got to a speed of 1.63 m/s in 3.3 secs, then its accel was 1.63m/s / 3.3s = 0.494 m/s^2

then the force acting on the elevator is found from newton's second law:

sum of forces = ma
sum of forces = tension pulling up elevator - mg = ma
tension force on elevator = m(g+a) = 679kg(9.8+0.494)m/s/s
= 6989.62N

the distance the elevator travels in 3 seconds is:

dist = 1/2 at^2 = 0.5(0.494)(3.3)^2 = 2.69 m

the power = force x distance/ 3.3 s = 5697.6 Watts

in the second case, the power is NOT zero (wouldnt it be a wonderful world if we could run machines at constant speed and not do work)

Power = Work/time = Force x distance /time; but notice that we have a distance/time term, which we know to be velocity

in the case of constant motion, we have that Power = Force x velocity

here the force pulling up the elevator is equal to the weight of the elevator = mg = 6654.2N

so power = 6654.2N x 1.63 m/s = 10,846.34 W

the power output is greater at constant speed, since the elevator is traveling faster during cruising compared to the acceleration phase