Question

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

• A. $1.5m/s$
• B. $2.0m/s$
• C. $1.7m/s$
• D. $1.9m/s$

Answer 1

The correct option is D $1.9m/s$

Friction will oppose the motion
$$\begin{array}{cc}& \text{Net force}=2000\times g+4000=24000N\\ & \text{Power of lift}=60\text{HP}\\ & \text{Power = Force}\times \text{Velocity}\\ & v=\frac{P}{F}=\frac{60\times 746}{24000}\\ & v=1.86\approx 1.9m/s\end{array}$$