Question

A 1.0x10^3 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of 4.0x10^3 N hinders the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s?

Answer 1

The power of the motor is 64.9 kW

Explanation:

We start by analying the forces acting on the elevator-load system. We have:

- The weight of the system,

$W=Mg$ (downward)

where

$M=1.0\cdot 10^3 + 800 = 1800 kg$ is the total mass of the elevator + load

$g=9.8 m/s^2$ is the acceleration of gravity

- The frictional force,

$F_f = 4.0 \cdot 10^3 N = 4000 N$

Also acting downward

- The force applied by the motor to push the elevator upward, F, acting upward

So the equation of motion is

$F-mg-F_f = ma$

where a is the acceleration. However, we know that the elevator is moving at constant speed, so the acceleration is zero:

a = 0

And the equation becomes

$F-mg-F_f = 0$

which means that we can find the force applied by the motor:

$F=mg+F_f = (1800)(9.8)+4000 = 21640 N$

Now we can calculate the power exerted by the motor, given by

$P=Fv$

where

F = 21,640 N is the force applied

v = 3.00 m/s is the speed

Substituting,

$P=(21640)(3.00)=64,920 W = 64.9 kW$

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