Question

A service elevator lifts 300 kg of cargo up to 10 stories (40m). The electric motor that supplies the mechanical energy for the lift uses 3000 kilojoules of electrical energy when performing this task within 20 seconds. How much potential energy does the elevator's cargo have at the end of this short ride (J)? How much work (change in mechanical energy) does this elevator system output by doing this task (J)? What is the energy % efficiency of the system? What is the mechanical power output of the elevator system during this task (Watts)?

Answer 1

a)

m = mass of cargo = 300 kg

h = height gained = 40 m

Potential energy of cargo is given as

PE = mgh

PE = 300 x 9.8 x 40

PE = 117600 J

b)

Change in mechanical energy = PE of cargo

Change in mechanical energy = 117600 J

c)

efficiency is given as

efficiency = Output x 100/Input = 117600 x 100/3000000 = 3.92

d)

Power output = PE/ t = 117600/20 = 5880 Watt