A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000N, the speed of the elevator at full load is close to:  $\left(1HP=746W,g=10m{s}^{-2}\right)$

$\begin{array}{l} Both mg and friction are acting downwards.\\ If elevator is moving up with uniform velocity,\\ then\\ F_{fr}V +F_{mg}V= P\\ \Rightarrow 4000\times V+(2000\times 10)\times V=(60\times 746) \\ \Rightarrow V = \frac{60\times 746}{4000+20000} \\ \Rightarrow V=1.865m/s\approx 1.9m/s\end{array}$