06.4 A particle moves along the x-axis while acted on by a conservative force de
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Question
06.4 A particle moves along the x-axis while acted on by a conservative force described by the potential energy function shown in the figure below.
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(a) At which labeled points does the force act to the left? To the right? At which labeled points is the force zero?
(b) What are the positions corresponding to points of stable equilibrium? Of unstable equilibrium?
(c) If the particle is released from rest at point B, describe the motion of the particle. At what point does the particle have the most kinetic energy?
(d) Describe the motion of the particle if it is instead released from rest at Point L? At what point does the particle have the most kinetic energy?
Explanation / Answer
(a) f= - d(U)/dx
so force= negative slope of potential energy graph
so the force will be towards left at the points where the slope of graph is positive and the force will be towards right at the points where the slope of the graph is negative. the force is zero at the points where the slope is zero
so force is is towards left at points D,J,K,L .The force is towards right st points A,B,F,G. the force is zero at points C,E,H.
(b)stable equilbrium are points of minimum potential energy . so they are C and H
unstable equilbrium are points of maximum potential energy and that is E.
(c) if the particle is released from rest at point B, it will start to execute simple harmonic motion between B and D. the particle will have the maximum kineticenergy at point C because the potential energy is minimum there.
(d) if the particle is released from rest at point L, it will start to execute simple harmonic motion because of the conservative forces. the particle will have the maximum kinetic energy at point H because the potential energy is minimum there.
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