An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ${\mathrm{ms}}^{-1}$. The frictional force opposing the motion is 4000 N. What is minimum power delivered by the motor to the elevator?

Here, m = 1800 kg

Frictional force, f = 4000 N

Uniform speed, v = 2 ${\mathrm{ms}}^{-1}$

Downward force on elevator is

$\mathrm{F} = \mathrm{mg} + \mathrm{f}$

     = $(1800 \mathrm{kg} \times 10 {\mathrm{ms}}^{-2})+4000 \mathrm{N} = 22000 \mathrm{N}$

The motor must supply enough power to balance this force. Hence,

$\mathrm{P} = \mathrm{Fv} = (22000 \mathrm{N})(2 {\mathrm{ms}}^{-1})$

  = $44000 \mathrm{W} = 44\times {10}^3 \mathrm{W} = 44 \mathrm{kW}$