A particle moves along the x-axis while acted by a conservative force described

Question

A particle moves along the x-axis while acted by a conservative force described by the potential energy function as shown in the figure.
(a) What is the direction of the force on the particle at Point A? At Point C? At Point E? At Point G?
(b) If the particle is released from rest at point B, describe the motion of the particle.
(c) At what point does the particle have the most kinetic energy?
(d) What are the positions corresponding to points of stable equilibrium? Of unstable equilibrium?
(e) Describe the motion of the particle if it is instead released from rest at Point H?

Explanation / Answer

(a) Knowing that: F = -dU/dx

At point A, dU/dx is negative, so force on the particle is positive.

At point C, dU/dx is zero, so force on the particle is zero.

At point E, dU/dx is zero, so force on the particle is zero.

At point G, dU/dx is positive, so force on the particle is negative.

(b) At point B, force on the particle is positive, so the particle will move in the positive x-direction and reverse its direction somewhere along the path CDE where its PE becomes equal to its PE at B. So, the particle will oscillate between that point and point B.

(c) Since the acting force is conservative, total energy of the particle is constant. KE will be highest where PE is the lowest. So, KE is maximum at C.

(d) Points C and F are the positions of stable equilibrium since for any small displacement of the particle from these points, the acting force will lead them back to these positions.

Point E is a position for unstable equilibrium.

(e) If the particle is released form H, it will move in the negative x-direction since the force is negative and reverse its direction at the first point where its PE becomes equal to its PE at H. So, the particle will oscillate between that point and point H.

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