91Ó°ÊÓ

A tiny, isolated item to which physical or chemical qualities, such as volume, density, or mass, may be assigned.

Particle's mass $$\begin{gathered} m=20gm \\ =0.02kg \end{gathered}$$

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.

Consider the graph of potential energy against distance,

Since the particle has been released from its rest at $x=1m$, it will go in a path where its potential energy reduces, as shown in the diagram. Potential energy drops in the right direction of the graph. As a result, the particle will go in the correct direction.

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

The highest speed will occur near the bottom of the curve, when the potential energy difference is greatest. The bottom of the graph appears at $x=4m$in the graph above.

The difference in potential energy at localid="1647864003221" $$\begin{gathered} x=1mand \\ x=4m \end{gathered}$$is $3J$may be seen in the graph above. The potential energy differential is transformed to kinetic energy according to the rule of energy conservation. Hence,

localid="1647864043650" $$\begin{gathered} ∆PE=∆KE \\ =3J \end{gathered}$$

Maximum velocity of the particle can be calculated by,

$∆KE=\frac{1}{2}m{v}^2$

Substituting the values,

$$\begin{gathered} 3=\frac{1}{2}×0.02×v^2 \\ v=17.32m/s \end{gathered}$$

Hence, particle's maximum velocity is $17.32m/s$.

The energy held in an item as a result of its location relative to a zero point is known as potential energy.

The graph's turning points are the points in motion when the end potential energy equals the starting potential energy is equal at $$\begin{gathered} x=1mand \\ x=6m \end{gathered}$$, which is the case in the preceding graph. As a result, the motion's turning point are$$\begin{gathered} x=1mand \\ x=6m \end{gathered}$$.