$FIGUREEX10.25$is the potential-energy diagram for a $500g$particle that is released from rest at A. What are the particle’s speeds at B, C, and D?

Velocity is a vector measurement of motion's direction and rate in physics. To put it another way, an object's velocity may be defined as the rate at which its location changes in respect to a frame of reference and time.

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

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,

The change in potential energy of an item is equal to the change in kinetic energy, according to the law of energy conservation. Hence,

$∆PE=∆KE$

Calculate the potential energy change from point A to point B,

$$\begin{gathered} ∆PE=5-2 \\ =3J \end{gathered}$$

Calculate the particle's velocity at point B,

localid="1647864344029" $$\begin{gathered} ∆PE=∆KE \\ =\frac{1}{2}m{v}_B^2 \end{gathered}$$

Substituting the values,

$$\begin{gathered} 3=\frac{1}{2}\times 0.5\times v_B^2 \\ {v}_B=3.464m/s \end{gathered}$$

Calculate the potential energy change from point A to point C,

$$\begin{gathered} ∆PE=5-3 \\ =2J \end{gathered}$$

Calculate the particle's velocity at point C,

localid="1647864364046" $$\begin{gathered} ∆PE=∆KE \\ =\frac{1}{2}m{v}_C^2 \end{gathered}$$

Substituting the values,

$$\begin{gathered} 2=\frac{1}{2}\times 0.5\times v_C^2 \\ {v}_C=2.828m/s \end{gathered}$$

Calculate the potential energy change from point A to point D,

$$\begin{gathered} ∆PE=5-0 \\ =5J \end{gathered}$$

Calculate the particle's velocity at point D,

localid="1647864382946" $$\begin{gathered} ∆PE=∆KE \\ =\frac{1}{2}m{v}_D^2 \end{gathered}$$

Substituting the values,

$$\begin{gathered} 5=\frac{1}{2}\times 0.5\times v_D^2 \\ {v}_D=4.472m/s \end{gathered}$$