Question

A single conservative force$\mathit{F}\left(x\right)$acts on a 1.0 kg particle that moves along an x axis. The potential energy $\mathit{U}\left(x\right)$ associated with role="math" localid="1661227819293" $\mathit{F}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is given by $\mathbf{U}\left(x\right)\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{x}}{\mathbf{4}}}\mathbf{}\mathbf{J}$, where xis in meters. Atthe particle has a kinetic energy of 2.0 J. (a) What is the mechanical energy of the system? (b) Make a plot of $\mathit{U}\left(x\right)$as a function of xfor $\mathbf{0}\mathbf{\le }\mathit{x}\mathbf{\le }\mathbf{10}\mathbf{}\mathit{m}$, and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of xthe particle can reach and (d) the greatest value of x the particle can reach. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of xat which it occurs. (g) Determine an expression in newtons and meters for $\mathit{F}\left(x\right)$as a function of x. (h) For what (finite) value of xdoes$\mathit{F}\left(x\right)\mathbf{=}\mathbf{0}$?

Short answer

1. $\mathrm{E}=-3.7\mathrm{J}$
   
2. The graph is drawn below
3. $\mathrm{x}=1.3\mathrm{m}$
   
4. $\mathrm{x}=9.1\mathrm{m}$
   
5. localid="1661228528777" $K=2.2\mathrm{J}$
   
6. localid="1661228532069" $x=4.0\mathrm{m}$
   
7. localid="1661228354484" $F=\frac{dU}{dx}=\left(4-x\right)e^{-x/4}$
   
8. $x=4.0\mathrm{m}$
   

Step-by-step solution

Given Data

Mass of the particle = 1.0kg

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

Understanding the concept

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.

Step 3(a): Calculate the mechanical energy of the system 

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

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

Step 4(b); Make a plot of U(x) as a function of x for 0≤x≤10 m

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

Step 5(c): Determine the least value of x the particle can reach

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

Step 6(d): Determine the greatest value of x the particle can reach

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

Step 7(e): Determine the maximum kinetic energy of the particle

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 $K=2.16\mathrm{J}\approx 2.2\mathrm{J}$. 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.

Step 8(f): Determine the value of x at which it occurs

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

Step 9(g): Determine the expression in newtons and meters for F(x) as a function of x

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).

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

Step 10(h): Calculate for what (finite) value of x does F(x)=0

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 $x=4.0\mathrm{m}$.