As a particle moves along an x axis, a force in the positive direction of the axis acts on it. Figure 7-50shows the magnitude Fof the force versus position x of the particle. The curve is given by $\mathit{F}\mathbf{=}\mathit{a}\mathbf{/}{\mathit{x}}^{\mathbf{2}}$, with $\mathbf{a}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{.}{\mathbf{m}}^{\mathbf{2}}$. Find the work done on the particle by the force as the particle moves from$\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$to $\mathbf{x}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$by (a) estimating the work from the graph and (b) integrating the force function.

It is given that,

1. From the figure, 7-50 is the force (F) in Newton vs. displacement (x) in meters.
2. Force is$F=\frac{a}{x^2}$
3. $a=9.0\mathrm{N}.{\mathrm{m}}^2$
4. Displacements are given as $x_1=1.0\mathrm{m}$and$x_2=3.0\mathrm{m}$

The problem deals with the work done which is the fundamental concept of physics.Work is the displacement of an object when force is applied on it.Use the concept of work done. For the first part, find the work done by counting the squares and by approximation. For the second part, find the work done by integration for given force and limits ${\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$

Formula:

$W=Fd$

Where,

F is force, dis displacement and Wis the work done.

Now, find the work done from the graph by counting the squares. Roughly, 12 squares are betweenthegiven displacements${\mathrm{x}}_1=1.0\mathrm{m}$and${\mathrm{x}}_2=3.0\mathrm{m}$.

For each square, F = 1N and x = 0.5m.

Therefore,

The work done for one square is,

$$\begin{gathered} W_1=Fd \\ =1\times 0.5 \\ =0.5\mathrm{J} \end{gathered}$$

So, for 12 squares, work done is

$$\begin{gathered} W_1=12\times 0.5=6\mathrm{J} \\ {\mathrm{W}}_1=6\mathrm{J} \end{gathered}$$

Hence, work done from the graph is $W_1=6\mathrm{J}$.

The force and the displacement limits are given,

$$\begin{gathered} W_2={\int }_{1.0}^{3.0}Fdx \\ ={\int }_{1.0}^{3.0}\left(\frac{a}{x^2}\right)dx \\ ={\int }_{1.0}^{3.0}a\left(x^{-2}\right)dx \\ {W}_2=a{\left|-\frac{1}{x}\right|}_{1.0}^{3.0} \\ =-\left(\frac{a}{3}-\frac{a}{1}\right) \\ =\frac{2a}{3} \end{gathered}$$

It is known that $a=9.0{\mathrm{Nm}}^2$,

$$\begin{gathered} W_2=\frac{2\times 9.0}{3} \\ =6\mathrm{J} \end{gathered}$$

$W_2=6\mathrm{J}$

Hence, work done by integrating the force function is$W_2=6\mathrm{J}$.