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) at point A,there is a negative slope, so the force would be going to the right or positive x direction.
At point C,slope value is zero,so,there is an unstable equilibrium and so there are no forces acting on it.
Same is the case with point E.
At point G,the slope is positive and so the particle is slowing down.Hense,there is a force going to the left or negative x direction.

B) At point B,the slope is negative. It showes that kinetic energy is increasing which means, that, particle is accelerating in positive X-direction.

C) At point c, since there is least potential energy.
D) There is no point where particle is in stable equilibrium,since there is no point at which net potential energy is zero.
Particle is in unstable equilibrium at points C,E,F.
E) Then its speed would reduce and the force would work in negative X-direction.

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