An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.20 ${\mathrm{ms}}^{-1}$. What would be the minimum horse power of the motor to be used?   [CGPMT 2015, UK PMT 2015)

Given, mass of elevator = 500 kg
Velocity = 0.20 ${\mathrm{ms}}^{-1}$
Weight of elevator = 500 x 9.8 = F
Now, power, P= Fv = 500 x 9.8 x 0.20 = 980 W
Therefore, hp-rating of motor  $=\frac{1}{746}\times 980=1.31\mathrm{hp}$