A particle can move along only an xaxis, where conservative forces act on it (Fig. 8-66 and the following table). The particle is released at x = 5.00 mwith a kinetic energy ofK = 14.0 Jand a potential energy of U = 0. If its motion is in the negative direction of the xaxis, what are its (a) K and (b) U at x 2.00 mand its (c)K and (d) U at x = 0? If its motion is in the positive direction of the xaxis, what are its (e) K and (f) U at x = 11.0 m, its (g) K and (h) U at x = 12.0 m, and its (i) K and (j) U at x = 13.0 m? (k) Plot U(x) versus x for the range x = 0to x = 13.0 m

Next, the particle is released from rest at x = 0. What are (l) its kinetic energy at xand (m) the maximum positive position${\mathit{x}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$it reaches? (n) What does the particle do after it reaches${\mathit{x}}_{\mathbf{max}}$?

1. Initial position of particle$x=5m$
2. Kinetic energy at$x_5\mathrm{is}{K}_5=14\mathrm{J}$.
3. Potential energy at $x_5\mathrm{is}{U}_5=0\mathrm{J}$.

The problem is based on the principle of conservation of mechanical energy. According to this principle the energy can neither be created nor be destroyed; it can only be internally converted from one form to another if the forces doing work on the system are conservative in nature. For the conservative system, mechanical energy of the system is conserved. Using this, we can find the potential energies at different locations.

Formula:

$E=K+U$

The work from$x=3.0\mathrm{m}$ to x = 2.0 m is

$$\begin{gathered} W=F_2∆x \\ =\left(5.00\right)\left(-1.00\right) \\ =-5.00\mathrm{J} \end{gathered}$$.

so the potential energy at$x=2.00\mathrm{m}\mathrm{is}{U}_2=+5.00\mathrm{J}$

Since mechanical energy is conserved, we have $E_{max}=14.0\mathrm{J}$, so the kinetic energy at$x=2.00\mathrm{m}$is

$$\begin{gathered} K_2=E_{max}-U_2 \\ =14.0-5.00 \\ =9.0\mathrm{J} \end{gathered}$$

The work from x = 2.00 m to x =0 m is

$$\begin{gathered} W=F_1∆x \\ =\left(3.00\right)\left(-2.00\right), \\ =-6.00\mathrm{J} \end{gathered}$$

so the potential energy at x = 0 m is

$$\begin{gathered} U_0=6.00J+U_2 \\ =\left(6.00+5.00\right)\mathrm{J} \\ =11.0\mathrm{J} \end{gathered}$$

So, the kinetic energy at$K_0=E_{max}-U_0=\left(14.0-11.0\right)\mathrm{J}=3.00\mathrm{J}$.

The work from x = 8.00 m to x = 11.0 m is

$$\begin{gathered} W=F_3∆x \\ =\left(-4.00\right)\left(3.00\right) \\ =-12.0\mathrm{J} \end{gathered}$$,

so the potential energy at x = 11.0 m is$U_1=12.0\mathrm{J}$ .

The kinetic energy at x = 11.0 m is

$$\begin{gathered} K_{11}=E_{max}-U_{11} \\ =\left(14.0-12.0\right)\mathrm{J} \\ =2.00\mathrm{J} \end{gathered}$$

Solve for the work as:

$$\begin{gathered} W=F_4∆x \\ =\left(-1.00\right)\left(1.00\right) \\ =-1.00\mathrm{J} \end{gathered}$$,

So the potential energy at x = 12.0 m is

$$\begin{gathered} U_{12}=1.00J+U_{11} \\ =\left(1.00+12.0\right)\mathrm{J} \\ =13.0\mathrm{J} \end{gathered}$$

Thus, the kinetic energy at x = 12.0 m is

$$\begin{gathered} K_{12}=E_{max}-U_{12} \\ =\left(14.0-13.0\right) \\ =1.00\mathrm{J} \end{gathered}$$

There is no work done in this interval from x = 12.0 m to x = 13.0 m. So, the potential energy at x = 13.0 m is $U_{13}=13.0\mathrm{J}$

There is no work done in this interval from x = 12.0 m to x = 13.0 m. So, the kinetic energy at x 13.0 m. is$K_{13}=1.0\mathrm{J}$

Plot U(x) versus xfor the range t x = 0 m to x 13 m.

We can say from the graph that the particle is “falling” down the 0 < x < 3 slopes of the well, gaining kinetic energy as it does so, and is certainly able to reach x = 5. SinceU = 0 at x = 5 , then its initial potential energy has completely converted to kinetic energy K = 11.0J

This is not sufficient to climb up and out of the well on the large x side (x > 8), but does allow it to reach a “height” of 11 at x = 10.8 m. This is a “turning point” of the motion.

When the particle is stopped at x = 10.8 m, it is accelerated to the left by the force it gains enough speed as a result that it is eventually able to return to x = 0, where it stops again.

Determine the work done on the particle from 0 m to 5 m as:

$$\begin{gathered} W=F_1{d}_1\cos \theta +F_2{d}_2\cos \theta \\ =3\times 2\times \cos 0°+5\times 1\times \cos 0° \\ =11\mathrm{J} \end{gathered}$$

Since, the force is conservative determine the negative work done on the particle as:

$$\begin{gathered} -W=U\left(5\right)-U\left(0\right) \\ -W=0-U\left(0\right) \\ U\left(0\right)=W=11\mathrm{J} \end{gathered}$$

Determine the total value of the kinetic energy as:

$$\begin{gathered} E_{Total}=K\left(0\right)+U\left(0\right) \\ =0+11 \\ =11\mathrm{J} \end{gathered}$$

Determine the kinetic energy at x = 5 m as:

$$\begin{gathered} E_{Total}=K\left(5\right)+U\left(5\right) \\ K\left(5\right)=11\mathrm{J} \end{gathered}$$

Consider the maximum positive position of the particle for it to reach the 0 kinetic energy is when the particle potential energy is equal to the total energy that is 11 J.

From the graph obtained in part (k)

The value is

$$\begin{gathered} U\left(x\right)=11\mathrm{J} \\ {X}_{max}=11\mathrm{m} \end{gathered}$$

Consider as the particle reaches $X_{max}$it starts travelling to the left by the force $F_3$and start gaining the kinetic energy in that direction.

Then, the particle returns back to $x_3=0\mathrm{m}$and at this position the kinetic energy of the particle becomes zero again and then it stops.