Question

The force on a particle is directed along an x axis and given by$\mathbf{F}\mathbf{=}{\mathbf{F}}_{\mathbf{0}}\left(\frac{x}{x_0}-1\right)$. Find the work done by the force in moving the particle from $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{to}\mathbf{}\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{0}}$ by (a) plotting $\mathbf{F}\left(x\right)$ and measuring the work from the graph and (b) integrating$\mathbf{F}\left(x\right)$.

Short answer

1. The net work done by the given force using graph is 0
2. The net work done by the given force using integration is 0

Step-by-step solution

Given

The given force function is,

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

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

Understanding concept of calculating work done

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:

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

(a) Plot  and calculate the work done from it  

The given function of force is,

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

So,

For x=0	F = -F0
For x=x	F = 0
For x=x	F = F0

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 $\mathrm{W}=0\mathrm{J}$

(b) Calculate the work done by using integration method  

Using the integration method, we have

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

Therefore, work done is 0 J.