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

$P = 8 kW$

$P = 8 \times 10^3 W$

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

$P = \frac{W}{t}$

$8 \times 10^3 = \frac{108045}{t}$

$t = 13.5 s$

so it will take 13.5 s to lift up

Part b)

Electrical energy used

$efficiency = 90$

$0.90 = \frac{Output}{Input}$

$0.90 = \frac{108045}{Input}$

$Input = 120050 J$

so electrical energy used in this process will be 120050 J