Question

A 600 kg elevator starts from rest. It moves upward for 4.50 s with constant acceleration until it reaches its cruising speed, 1.75 m/s. (a) What is the average power of the elevator motor during this period

Answer 1

Answer:

408.33watts

Explanation:

Power is expressed according to the formula

Power = work done/time

Power = force × distance/time

Power = Force× velocity

Find the force

F = mv/t

F = 600×1.75/4.5

F = 1050/4.5

F=233.33N

Get the power

Power = 233.33×1.75

Power = 408.33watts

Hence the average power of the elevator motor during this period is 408.33m/s