If there is a net nonzero force on a moving object, can the total work done on the object be zero? Explain, using an example.

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

$$\begin{gathered} {\mathit{W}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}\mathbf{=}{\mathit{K}}_{\mathbf{2}}\mathbf{-}{\mathit{K}}_{\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{∆}\mathit{K} \end{gathered}$$

The expression for the work is given by,

$W=\overrightarrow{F}.\overrightarrow{s}$

Here, F is the force and is the displacement.

When an object is moving in circular path then work done is zero, since in this situation displacement of the object is zero.

The work done is given by,

$$\begin{gathered} W=\overrightarrow{F}.\overrightarrow{s} \\ =Fs\cos \theta \end{gathered}$$

Here,$\theta$ is the angle between the force and displacement.

For the one complete revolution of circle, displacement of the object is zero.

Substitute 0 for s in the above equation.

$$\begin{gathered} W=F\left(0\right)\cos \theta \\ =0 \end{gathered}$$

The kinetic energy of an object moving in a circular path is zero because the kinetic energy of the object does not change.

Therefore, the total work done on the object will be zero if there is a net non-zero force on a moving object.