Figure shows a plot of potential energy Uversus position x.of a 0.200 kg particle that can travel only along an xaxis under the influence of a conservative force. The graph has these values: , ${\mathit{U}}_{\mathbf{A}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{J}\mathbf{,}\mathbf{}{\mathit{U}}_{\mathbf{C}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{J}$, and ${\mathit{U}}_{\mathbf{D}}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{J}$. The particle is released at the point where Uforms a “potential hill” of “height” ${\mathit{U}}_{\mathbf{B}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{J}$, with kinetic energy 4.00 J. What is the speed of the particle at: (a) x = 3.5 m (b) x= 6.5 m? What is the position of the turning point on (c) the right side (d) the left side?

$$\begin{gathered} 1.Themassofparticleism=0.200\mathrm{kg} \\ 2.ThekineticenergyofaparticleatBis{K}_B=4.00\mathrm{J} \\ 3.U_A=9.00\mathrm{J} \\ 4.U_B=12.00\mathrm{J} \\ 5.U_C=20.00\mathrm{J} \\ 6.U_D=24.00\mathrm{J} \end{gathered}$$

The problem deals with the law of conservation of energy. According to the law of energy conservation, energy can neither be created, nor be destroyed. By using the principle of conservation of energy, calculate the kinetic energy at any point along the path of the particle using conservation of energy. Then, using the equation of kinetic energy, calculate the speed at the required position. To find the position of the turning point, consider the slope of the line having the point of potential energy greater than total energy and less than total energy.

Formulae are as follow:

$$\begin{gathered} U_1+K_1=U_2+K_2 \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$

where, K is kinetic energy, Uis potential energy, m is mass and vis velocity.

The position x = 3.5 has potential energy, $U_A=9.00\mathrm{J}$

The equation for conservation of mechanical energy is,

$U_1+K_1=U_2+K_2$

Applying this equation for the potential $U_A\mathrm{and}{U}_B$

localid="1661400948788" $$\begin{gathered} U_B+K_B=U_A+K_A \\ \mathrm{Solving}\mathrm{this}\mathrm{equation}\mathrm{for}{K}_A, \\ {K}_A=U_B+K_B-U_A=12.00+4.00-9.00=7.00\mathrm{J} \end{gathered}$$

Now,

$K=\frac{1}{2}m{v}^2$

Rearranging it for velocity v,

$$\begin{gathered} v=\sqrt{\frac{2\times K}{m}}=\sqrt{\frac{2\times 7.00}{0.200}} \\ v=8.37\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the particle at x = 3.5 m is, v = 8.37 m/s

At the position, x = 6.5, the potential energy is zero. So, the kinetic energy at this position will be equal to the total energy of a particle. So,

$$\begin{gathered} K=U_B+U_B \\ K=12.00+4.00=16.00J \\ \mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{equation}\mathrm{for}\mathrm{velocity}\mathrm{relating}\mathrm{kinetic}\mathrm{energy}, \\ v=\sqrt{\frac{2\times K}{m}}=\sqrt{\frac{2\times 16.00}{0.200}} \\ v=12.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the particle at x =6.5 m is, v = 12.6 m/s.

At the turning point, kinetic energy of a particle is zero. This can happen when its potential energy is equal to the total mechanical energy.

Let us consider the motion of a particle from x = 7 m to x = 8m.

As the potential energy at x= 7m is 0J and at x =8m is $U_D=24.00\mathrm{J}$. The turning point must lie in this region as the total energy is 16.00 J.

The slope of line AC can be expressed as,

$$\begin{gathered} \frac{16.00-0}{x_R-7.0}=\frac{24.00-0}{8.0-7.0} \\ \mathrm{By}\mathrm{solving}\mathrm{this}\mathrm{equation}\mathrm{of}{x}_R, \\ {x}_R=7.67\mathrm{m} \end{gathered}$$

Hence, the position of the turning point on the right side is, $x_R=7.67\mathrm{m}$.

Applying the same logic as part c for the left side, let us consider the motion of a particle

from x = 1m to x= 3 m.

The slope of line PR can be expressed as,

Hence,the position of the turning point on the left side, $x_L=1.73\mathrm{m}$.

Therefore,the speed of a particle at any position can be found using the equation of conservation of energy. To calculate the turning point, the slope of the line can be used