91Ó°ÊÓ

Mass of the particle = 1.0kg

$\mathbf{U}\left(x\right)\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{x}}{\mathbf{4}}}\mathbf{}\mathbf{J}$

We can use the relation$\mathit{E}\mathbf{=}\mathit{K}\mathbf{+}\mathit{U}$to solve the problem. The values of x for various conditions can be directly determined from the graph, after plotting it.

The energy at $x=5.0\mathrm{m}$ is

$$\begin{gathered} E=K+U=2.0\mathrm{J}-5.7\mathrm{J} \\ E=-3.7\mathrm{J} \end{gathered}$$

A plot of the potential energy curve (SI units understood) and the energy E (the horizontal line) is shown for $0≤x≤10m$.

The problem asks for a graphical determination of the turning points, which are the points on the curve corresponding to the total energy computed in part (a). The result for the smallest turning point (determined, to be honest, by more careful means) is x=1.3 m

And the result for the largest turning point is x = 9.1 m .

Since$K=E-U$, then maximizing$K$involves finding the minimum of$U$. A graphical

determination suggests that this occurs at$x=4.0\mathrm{m}$, which plugs into the expression

$E-U=-3.7-\left(-4x^{-x/4}\right)$to give $K=2.16\mathrm{J}≈2.2\mathrm{J}$. Alternatively, one can measure from the graph from the minimum of the U curve up to the level representing the total energy E and thereby obtain an estimate of $K$at that point.

As mentioned in the previous part, the minimum of the $U$curve occurs at$x=4.0\mathrm{m}$

The force (understood to be in newtons) follows from the potential energy, using Eq. 8-20 (and Appendix E if students are unfamiliar with such derivatives).

$F=\frac{dU}{dx}=\left(4-x\right)e^{-x/4}$

This revisits the considerations of parts (d) and (e) (since we are returning to the minimum of $U\left(x\right)$)— but now with the advantage of having the analytic result of part (g). We see that the location that produces$F=0$ is exactly $x=4.0\mathrm{m}$.