91Ó°ÊÓ

The given force function is,

$\mathrm{F}={\mathrm{F}}_0\left(\frac{\mathrm{x}}{{\mathrm{x}}_0}-1\right)$

The particle moves from $\mathrm{x}=0\mathrm{to}\mathrm{x}=2{\mathrm{x}}_0$

We can use the equation of work fortheintegration method. For the graph method, the work done is equal to the area under the curve in the graph.

Formula:

$$\begin{gathered} \mathbf{W}\mathbf{=}{\mathbf{∫}}_{{\mathbf{x}}_{\mathbf{i}}}^{{\mathbf{x}}_{\mathbf{f}}}\mathbf{F}\left(x\right)\mathbf{d}\mathbf{x} \\ \mathbf{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{×}\mathbf{base}\mathbf{×}\mathbf{height} \end{gathered}$$

The given function of force is,

$\mathrm{F}={\mathrm{F}}_0\left(\frac{\mathrm{x}}{{\mathrm{x}}_0}-1\right)$

So,

We can plot the graph of F(x) using these points as

The work done, from the graph, is simply the area under the curve.

In the above graph,we see that the area under the curve from $\mathrm{x}=0\mathrm{to}\mathrm{x}=2{\mathrm{x}}_0$is equal to the area under the curve for $\mathrm{x}=0\mathrm{to}\mathrm{x}=2{\mathrm{x}}_0$and both are opposite to each other

So, the work done from the graph is $\mathrm{W}=0\mathrm{J}$

Using the integration method, we have

$$\begin{gathered} \mathrm{W}={∫}_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{F}\left(\mathrm{x}\right)d\mathrm{x} \\ \mathrm{W}={∫}_0^{2{\mathrm{x}}_0}{\mathrm{F}}_0\left(\frac{\mathrm{x}}{{\mathrm{x}}_0}-1\right)d\mathrm{x} \\ \mathrm{W}={\mathrm{F}}_0{\left(\frac{{\mathrm{x}}^2}{2{\mathrm{x}}_0}-\mathrm{x}\right)}_0^{2{\mathrm{x}}_0} \\ \mathrm{W}=2{\mathrm{x}}_0-2{\mathrm{x}}_0 \\ \mathrm{W}=0\mathrm{J} \end{gathered}$$

Therefore, work done is 0 J.