91Ó°ÊÓ

Mass of the particle m=2.0 kg

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

From graph,

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

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

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

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

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)

$⇒\mathrm{Δ}k=\frac{1}{2}×m{v}^2$

$⇒\mathrm{Δ}k=\frac{1}{2}×(-1.5{)}^2×2$

$⇒k=2.25J$ (ii)

So the total mechanical energy is,

$\mathrm{Δ}{E}_{mec}=\mathrm{Δ}k+\mathrm{Δ}U$ (iii)

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

At 5.5 J, there are two points on the graph$x≈1.5m$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

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

$⇒E_{mec}=-5.5J$

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

Determine the kinetic energy as:

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

Determine the velocity as:

$$\begin{gathered} ⇒v\left(7\right)=\sqrt{\frac{2K}{m}} \\ ⇒v\left(7\right)=\sqrt{\frac{2×12}{2}} \\ ⇒v\left(7\right)=\sqrt{12} \\ ⇒v\left(7\right)=3.5 \frac{m}{s} \end{gathered}$$