Question

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

• A. $22kW$
• B. $44kW$
• C. $66kW$
• D. $88kW$

Answer 1

The correct option is B $44kW$

Here, $m=1800kg$
Frictional force, $f=4000N$
Uniform speed, $v=2m{s}^{-1}$
Downward force on elevator is
$F=mg+f$

$=(1800kg\times 10m{s}^{-2})+4000N$

$=22000N$
The motor must supply enough power to balance this force. Hence,
$P=Fv=\left(20000N\right)\left(2m{s}^{-1}\right)$
$=44000W=44\times {10}^3$

$=44kW$.