Question

An elevator m=800kg has a maximum load of 600kg the elevator goes up 30m at a constant speed of 4ms_¹ what is the average power output of the elevator motor if she elevator is fully loaded with it's maximum m weight (neglect friction)

Answer 1

m = 800 + 600 = 1400 kg
gain of potential energy in (30/4) seconds = m g h = 1400 * 9.81 * 30 Joules
power = energy/time =1400 * 9.81 * 30 / (30/4) = 1400 * 9.81 * 4 Watts
You could also say power = force * speed = 1400 * 9.81 * 4

Answer 2

thank you