91Ó°ÊÓ

The work energy theorem tells that the work done on the body results in a change in its kinetic energy.

$$\begin{gathered} {\mathbf{W}}_{\mathbf{total}}\mathbf{=}{\mathbf{K}}_{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{1}}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{∆}\mathbf{K} \end{gathered}$$

The kinetic energy of the system is given by,

$K=\frac{1}{2}m{v}^2$

Here,m is the mass of the system, andv is velocity of the system.

It is given that the elevator is hoisted by its cables at constant speed, then there will not be any change in their kinetic energies because the kinetic energy depends on the velocity of the system.

As the mass of the system is same and the velocity is constant, so no factor is responsible to bring any change in the kinetic energy of the system. Therefore, the final kinetic energy will be equal to the initial kinetic energy of the system.

$K_2=K_1$

The work done will be as follows.

$$\begin{gathered} W_{tot}=K_2-K_1 \\ =0 \end{gathered}$$

Therefore, the total work done on the elevator will be zero.