A 500 kg elevator starts from rest. It moves upward for 4.00 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?

(b) How does this power compare with the motor power when the elevator moves at its cruising speed?

1 answer

work:
acceleration = velocity/time = 1.75/4 = .4375
distance = .5*acceleration*t^2 = .5*.4375*16 = 3.5
energy = mass*gravity*distance + (mass/2)*v^2
=500*9.8*3.5+(500/2)*(1.75)^2
=17150.0+765.625 = 17915.625
power = work/time = 17915.625/4 = 4478.9063
answer for part a) 4478.9063

do not know how to do part b) could not figure out for myself