Question

An elevator's electric hoist is rated at 8.0 kW and has an efficiency of 90.0%. The hoist lifts the elevator car, mass 1225 kg, a distance of 9.00 m. (a) How much time is required? (b) How much electrical energy is used to perform this task?

Answer 1

Part a)

Power rated on the elevator is given as

`P = 8 kW`

`P = 8 * 10^3 W`

now the mass that is lifted above is given as

`m = 1225 kg`

height of the elevator lifted is

`h = 9 m`

now the potential energy is given as

`U = mgh`

`U = 1225 (9.8)(9) = 108045 J`

now power is defined as rate of energy

`P = (W)/(t)`

`8 * 10^3 = (108045)/(t)`

`t = 13.5 s`

so it will take 13.5 s to lift up

Part b)

Electrical energy used

`efficiency = 90`

so electrical energy used in this process will be 120050 J