- #1
Exuro89
- 34
- 1
Homework Statement
A particle moves along the x-axis while acted on by a single conservative force parallel to the x. axis. The force corresponds to the potential-energy function graphed in Fig. 7.45. The particle is released from rest at point A.
[PLAIN]http://k.min.us/ijze8K.png
(a) What is the direction of the force on the particle when it is at point A?
(b) At point B?
(c) At what value of x is the kinetic energy of the particle a maximum?
(d) What is the force on the particle when it is at point C?
(e) What is the largest value of x reached by the particle during its motion?
(f) What value or values of x correspond to points of stable equilibrium?
(g) Of unstable equilibrium?
Homework Equations
None.
The Attempt at a Solution
Hello everyone, this is the last section in the chapter I'm studying. I wanted to make sure I was understanding this correctly. So I'll list my answer/reason to each and have you note if I have something incorrect.
(a)So at point A there is a negative slope, so the force would be going to the right or positive x direction.
(b)At point B the slope is positive and so the particle is slowing down which means there is a force going to the left or negative x direction.
(c)Kinetic energy is greatest when there is no other energy in the object so in the graph the lowest point would yield the greatest kinetic energy so .7 on the x axis.
(d)At point C there is an unstable equilibrium and so there are no forces acting on it.
(e)Because the particle begins at A, conservation of energy would show that it would reach that same height, so it will be about 2.2 or so on the x axis.
(f)Stable equilibrium would be at .7 and 1.8 or so. If you were to put the particle there and move it, it would stay in that same equilibrium.
(g)Unstable equilibrium is at point C or 1.45 on the x axis. It's unstable because if you were to push it it would move away from the equilibrium, and not return, thus unstable.
Is this correct?
Last edited by a moderator: