A conservative force F(x)acts on a 2.0 kgparticle that moves along an x axis. The potential energy $\mathit{U}\left(x\right)$associated with $\mathit{F}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is graphed in Fig. 8-63. When the particle is at $\mathit{x}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$, its velocity is $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{5}\frac{\mathbf{m}}{\mathbf{s}}$. What are the (a) magnitude and (b) direction of $\mathit{F}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$at this position? Between what positions on the (c) left and (d) right does the particle move? (e) What is the particle’s speed at x=7.0 m?

Mass of the particle m=2.0 kg

At x = 2.0 m,$v=-1.5\frac{\mathrm{m}}{\mathrm{s}}$

From graph,

$$\begin{gathered} Atx=2.0\mu U \\ =-8J \end{gathered}$$

$$\begin{gathered} Atx=7.0mU\left(x\right) \\ =-16J \end{gathered}$$

If we know the U(x) for a system, we can find the force acting on the system |F(x)| by differentiating the potential energy with respect to distance.
Mechanical energy of an isolated system is equal to the kinetic energy + potential energy

Formula:

$$\begin{gathered} F=\frac{-dU\left(x\right)}{dx} \\ ∆E_{mec}=∆k+∆U \end{gathered}$$

The force at $x=2.0m$is,

Take a slope of this graph between the region x= 1 m to x = 4 m

$$\begin{gathered} F=-\frac{dU\left(x\right)}{dx} \\ \Rightarrow F=\left[\frac{U\left(x=4m\right)-U\left(x=1m\right)}{4m-1m}\right] \\ \Rightarrow F=\left[\frac{-\left(1.75\right)-\left(-2.8\right)}{3m}\right] \\ \Rightarrow F=4.9N \end{gathered}$$

Since slope is negative, the force would be in positive x direction.

At x=2.0 m we can find potential energy, as

$$\begin{gathered} U\left(x=2.0m\right)=U\left(x=1.0m\right)+\left(-4.9\frac{J}{m}\right)\left(1.0\right) \\ =-7.7J \end{gathered}$$ (i)

$\Rightarrow \Delta k=\frac{1}{2}\times m{v}^2$

$\Rightarrow \Delta k=\frac{1}{2}\times (-1.5{)}^2\times 2$

$\Rightarrow k=2.25J$ (ii)

So the total mechanical energy is,

$\Delta E_{mec}=\Delta k+\Delta U$ (iii)

$$\begin{gathered} ∆E_{mec}=-7.7\mathrm{J}+2.25\mathrm{J} \\ =-5.45 \\ \approx -5.5\mathrm{J} \end{gathered}$$ (iv)

At 5.5 J, there are two points on the graph$x\approx 1.5m$and x=13.5 m. Therefore, the particle will confined in the region $1.5m<x<13.5m$.

The left boundary is at$x=1.5m$

From the above result, the right boundary is at$x=13.5m$

At x=7.0 m, we get$U\left(7\right)=-17.5J$. So the total energy is

$E_{mec}=K+U\left(7\right)$

$\Rightarrow E_{mec}=-5.5J$

$U\left(7\right)=-17.5\mathrm{J}$

Determine the kinetic energy as:

$$\begin{gathered} K=E_{mec}-U\left(7\right) \\ \Rightarrow K=-5.5-(-17.5) \\ \Rightarrow K=-5.5+17.5 \\ \Rightarrow K=12J \\ K=\frac{1}{2}\times {\mathrm{mv}}^2 \end{gathered}$$

Determine the velocity as:

$$\begin{gathered} \Rightarrow v\left(7\right)=\sqrt{\frac{2K}{m}} \\ \Rightarrow v\left(7\right)=\sqrt{\frac{2\times 12}{2}} \\ \Rightarrow v\left(7\right)=\sqrt{12} \\ \Rightarrow v\left(7\right)=3.5\hspace{0.33em}\frac{m}{s} \end{gathered}$$