The loaded cab of an elevator has a mass of $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{kg}$and moves 210 mup the shaft in 23 sat constant speed. At what average rate does the force from the cable do work on the cab?

The mass of elevator is, $\mathrm{m}=3.0\times {10}^3\mathrm{kg}$.

The distance is, $\mathrm{d}=210\mathrm{m}$.

The time is, $∆\mathrm{t}=23\mathrm{s}$.

The rate at which force does the work on the object is called as power due to the force. The work done on a particle by a constant force during its displacement is given as

$\mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}}$

Formula:

$$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}} \\ {\mathbf{P}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{W}}{\mathbf{∆}\mathbf{t}} \end{gathered}$$

The cab of the elevator moves with constant speed. So the cab is not accelerated; hence, the forces acting on the cab are balanced. The forces acting on the cab arethetension in the cable holding the cab and the gravitational force. Hence work done by the cable is

$\begin{array}{rcl}\mathrm{W}& =& \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{d}}\\ & =& \mathrm{Tension}\times \mathrm{displacement}\\ \mathrm{W}& =& \mathrm{mgd}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\mathrm{W}& =& 3.0\times {10}^3\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 210\mathrm{m}\\ & =& 6.2\times {10}^6\mathrm{J}\end{array}$

The power due to force is the rate at which force does the work on the object.

So,

${\mathrm{P}}_{\mathrm{avg}}=\frac{\mathrm{W}}{∆\mathrm{t}}$

Substitute all the value in the above equation.

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{avg}}& =& \frac{6.2\times {10}^6\mathrm{J}}{23\mathrm{s}}\\ & =& 2.7\times {10}^5\mathrm{W}\end{array}$

The rate of doing work on the cab by the force is calculated to be 270 kW.