A force $\overrightarrow{\mathbf{F}}$in the positive direction of an x axis acts on an object moving along the axis. If the magnitude of the force is $\mathbf{F}\mathbf{=}\mathbf{10}{\mathbf{e}}^{\mathbf{-}\mathbf{x}\mathbf{/}\mathbf{2}\mathbf{.}\mathbf{0}}\mathbf{}\mathbf{N}$, with x in meters, find the work done by $\overrightarrow{\mathbf{F}}$as the object moves from $\mathbf{x}\mathbf{=}\mathbf{0}$to $\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$by (a) plotting $\mathbf{F}\left(x\right)$and estimating the area under the curve and (b) integrating to find the work analytically.

It is given that,

Force,$\overrightarrow{F}=10{\mathrm{e}}^{-\mathrm{x}/2}$

Distance, $x=0$to$x=2.0\mathrm{m}$

By plotting F(x)vs Xfor different x, estimate the area under the curve, and from this curve, find the work done as the object moves from x=0to x=2.0 m. Using the formula of integration, find the work done by$\overrightarrow{\mathbf{F}}$as the object moves from x=0to x=2.0 m.

Formula is as follow:

$W=\int Fdx$

where, F is force, dx is small displacement and wis the work done.

Now,

$\overrightarrow{F}=10{\mathrm{e}}^{-\frac{\mathrm{x}}{2}}$

The plot of $F\left(x\right)vsx$is as below. The force is plotted on y axis and displacement is plotted on x axis.

Hence, area under the curve can be found by finding the area of each rectangle under the curve. This would be in the range of 12 J to 14 J ,

Now,

$W={\int }_0^n{x}^{n}dx$

As $x^n=10e^{-x/2},n=2,$

$W={\int }_0^{2}10e^{-x/2}dx$

Putting,$-\frac{x}{2}=t,$

$$\begin{gathered} W=20{\int }_0^{-1}{e}^{-t}dt \\ =-20\left(\frac{1}{e}-1\right) \\ =-20\left(\frac{1-e}{e}\right) \\ =12.6\mathrm{J} \\ \approx 13\mathrm{J} \end{gathered}$$

Therefore, the work done by $\overrightarrow{F}$as the object moves from x=0 to x=2.0 m by integrating to find the work analytically is 13 J.