Figure 8-19 gives the potential energy function of a particle. (a) Rank regions AB, BC, CD, and DE according to the magnitude of the force on the particle, greatest first. What value must the mechanical energy${\mathit{E}}_{\mathbf{m}\mathbf{e}\mathbf{c}}$of the particle not exceed if the particle is to be (b) trapped in the potential well at the left, (c) trapped in the potential well at the right, and (d) able to move between the two potential wells but not to the right of point H? For the situation of (d), in which of regions BC, DE, and FG will the particle have (e) the greatest kinetic energy and (f) the least speed?

A figure which shows the potential energy function of a particle

Here, the relation between force and potential energy can be used to rank the regionsaccording to the magnitude of the force on the particle. Also, it involves the principle of conservation of energy which states that the total energy of an isolated system remains constant. Using the principle of conservation of energy, the value of mechanical energyof the particle can be used which must not exceed if the particle is to be trapped in the potential well at the left, at right and to be able to move between the two potential wells but not to the right of point

We can use the same principle to find the greatest kinetic energy and the least speed in the regionsfor the situation given in d.

Formula:

The force is given by,

$F=-\frac{dU}{dx}$

Mechanical energy is given by,

$E_{mec}=d\left(K.E\right)+dU$

Where U is the potential energy

The force is given by,

$F=-\frac{dU}{dx}$

Here, the force is the negative of the slope of U-x graph. Therefore, the greater the slope, the more will be the force acting on the particle

Therefore,

AB region = highest = largest force

CD region = second largest slope = second largest force

regions BC = DE no slope = no force

Therefore, the rank of regions AB, BC, CD, and DE according to the magnitude of the force on the particle, greatest first is$\mathrm{AB}>\mathrm{CD}>\mathrm{BC}=\mathrm{DE}$

The mechanical energy is given by,

$E_{mec}=d\left(K.E\right)+dU$

Particle will get trapped within the potential well only if the mechanical energy at point D is due to potential energy

Therefore,

$$\begin{gathered} E_{mec}=0\mathrm{J}+5\mathrm{J} \\ =5\mathrm{J} \end{gathered}$$

Therefore, the value of the mechanical energy $E_{mec}$ of the particle which must not exceed if the particle is to be trapped in the potential well at the left is 5 J.

$E_{mec}=d\left(K.E\right)+dU$

Particle will get trapped within the potential well only if the mechanical energy at point E is due to potential energy

Therefore,

$$\begin{gathered} E_{mec}=0\mathrm{J}+5\mathrm{J} \\ =5\mathrm{J} \end{gathered}$$

Therefore, the value of the mechanical energy $E_{mec}$of the particle which must not exceed if the particle is to be trapped in the potential well at the right is 5 J.

$E_{mec}=d\left(K.E\right)+dU$

For the particle to move between two potential wells, mechanical energy at point H is due potential energy at point H.

So,

$$\begin{gathered} E_{mec}=0\mathrm{J}+6\mathrm{J} \\ =6\mathrm{J} \end{gathered}$$

Therefore, the value of the mechanical energy $E_{mec}$of the particle which must not exceed if the particle is to be able to move between the two potential wells but not to the right of point H is 6 J

$E_{mec}=d\left(K.E\right)=dU$

According to this principle, kinetic energy is maximum when potential energy is minimum.

As potential energy is minimum in the FG region, so kinetic energy is maximum in the region.

Therefore, the greatest kinetic energy is in FG region

$E_{mec}=d\left(K.E\right)=dU$

According to this principle, kinetic energy is minimum when potential energy is maximum.

As potential energy is maximum in DE region, kinetic energy, and hence, speed is minimum in the DE region

Therefore, the least speed is in the DE region .