$FIGUREEX10.28$shows the potential energy of a $500g$particle as it moves along the x-axis. Suppose the particle’s mechanical energy is $12J$.

a. Where are the particle’s turning points?

b. What is the particle’s speed when it is at $x=4.0m$?

c. What is the particle’s maximum speed? At what position or positions does this occur?

d. Suppose the particle’s energy is lowered to $4.0J$. Can the particle ever be at $x=2.0m$? At$x=4.0m$?

It is the change in an object's location with regard to time.

Particle's mass $$\begin{gathered} m=500g \\ =0.5kg \end{gathered}$$,

Particle's mechanical energy $ME=12J$.

Formula to determine the kinetic energy of a particle,

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

$KE$= kinetic energy,

$m$= particle's mass,

$v$= particle's velocity,

$ME=PE+KE$

= particle's mechanical energy.

Consider the graph of potential energy against distance,

The turning points of the particle are the sites where the starting and final mechanical energies are equivalent. It is clear from the graph that the particle has a mechanical energy of $12J$at localid="1647865494698" $$\begin{gathered} x=1mand \\ x=8m \end{gathered}$$.

Particle's total energy at $x=1m$is,

$TE=12J$

Particle's potential energy is $PE=8J$.

Which is at $x=4m$. Thus, kinetic energy of the particle is,

$$\begin{gathered} KE=ME-PE \\ =12-8 \\ =4J \end{gathered}$$

Particle's velocity at $x=4m$,

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ 4=\frac{1}{2}\times 0.5\times v^2 \\ v=4m/s \end{gathered}$$

Hence, particle's velocity at$x=4m$is$4m/s$.

The particle's maximum velocity occurs near the bottom of the graph, where potential energy is $0$. The bottom of the graph is located at point$x=6m$in the diagram above.

Particle's potential energy is $PE=0J$.

Which is at $x=6m$. Thus, particle's kinetic energy is,

$$\begin{gathered} KE=ME-PE \\ =12-0 \\ =12J \end{gathered}$$

Particle's velocity at $x=6m$,

$$\begin{gathered} ∆KE=\frac{1}{2}m{v}^2 \\ 12=\frac{1}{2}\times 0.5\times v^2 \\ {v}_{max}=6.92m/s \end{gathered}$$

Particle's maximum velocity will occur at$x=6m$and is$6.92m/s$.

$4J$lowers the particle mechanical energy. As a result, the particle's new mechanical energy at $x=2m$,

$$\begin{gathered} ME=12-4 \\ =8J \end{gathered}$$

Kinetic energy of particle at $x=2m$,

$$\begin{gathered} KE=ME-PE \\ =8-4 \\ =4J\ne 0 \end{gathered}$$

Since the kinetic energy is larger than $0$, the particle can reach a point $x=2m$. If the mechanical energy of the particle is lowered by $4J$. Particle's kinetic energy at $x=4m$is,

$$\begin{gathered} KE=ME-PE \\ =8-8 \\ =0J \end{gathered}$$

Since the kinetic energy is equal to$0$, therefore, even if the particle's mechanical energy is lowered by$4J$, it can never reach at point$x=4m$.