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Homework Help: Potential Energy Function

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations
    Conservation of energy

    3. The attempt at a solution
    Here are my solutions. Please feel free to correct me. Any help will be appreciated.
    (a) I think that the particle is moving in the negative direction because the potential energy decreases (?)
    (b) Use conservation of energy (total energy=4J, from the graph at x=0.5m): min potential energy=1J => max kinetic energy=3J => max speed=12.2m/s
    (d)kinetic energy=4-3=1J => speed=7.1m/s
    (e)I think the particle will change direction at the points where the derivative is 0. (I'm thinking about gravitational potential energy).
    (f)Again, I think the answer is at points where the derivative is 0 since dU/dx=-Fx

    By the way, I think this graph confuses me a bit. I tried without success to imagine the situation, and the teacher told me that it was going to be complicated since the graph was related to molecular bonds. Also, the problem did not state whether it involves only conservative force or not. Can anyone please make a comment on this graph? Thank you very much!

    Love Phys

    Attached Files:

  2. jcsd
  3. Feb 24, 2013 #2
    (a) If potential energy decreases in the positive x direction, what can be said about the force acting on the particle? What can be said about the motion of the particle under this force if it is at rest initially?

    (e) To change direction, the particle must momentarily have zero velocity. Where would that happen?

    (f) So where are the points where the derivative is zero?
  4. Feb 24, 2013 #3
    Thank you, voko!

    (a) I think the force is acting in the positive direction? Therefore, the particle will move in the positive direction.
    (b) Is it at x=4.5m? I think that's where potantial energy=4J, so kinetic energy must be 0J.
    (c) I reckon it's at x=2.5, and x=5.25?
  5. Feb 24, 2013 #4
    All correct.

    Is there anything still unclear?
  6. Feb 24, 2013 #5
    For (a), the way I came up with the answer is I took the derivative dU/dx<0, therefore Fx=-dU/dx must be positive. But if I think back of how gravitational potential energy is defined in high school, I am confused again. Assume that I have 2 points A and B on the y axis (y(A)<y(B)) (the positive direction is up), then if I move a particle from B to A, then the gravitational potential energy decreases. And because of the way I choose the positive direction, the gravitational force is acting in the negative direction. So, the particle will move in the negative direction (!?) Sorry this was the kind of picture I had in mind when I answered this question, and obviously it was wrong. Can you please help me understand this? Thank you very much!
  7. Feb 24, 2013 #6
    If you sketch this in the same way, the value of the gravitational potential energy on the vertical axis, and the y distance on the horizontal axis, what will the graph of the gravitational potential energy look like?
  8. Feb 24, 2013 #7
    I think it's going to be a straight line which passes through the origin with gradient = mg. Then if I want to move from B to A, then I'm doing negative work, but the gravitational force is doing positive work (this is true since vector Fg and vector y are in the same direction).
  9. Feb 24, 2013 #8
    The essential feature is that the gradient is positive for gravitation as you go up; in the original problem, however, the gradient in the vicinity of 0.5 m is negative, so the effect of the force is opposite in direction.
  10. Feb 24, 2013 #9
    Thank you very much, voko! I think I understand it now. Well, sometimes it's easier to follow the Math logic, instead of trying to imagine what's going on.
  11. Feb 24, 2013 #10
    I think it could be helpful to keep in mind that a system always tends to go where the potential energy is lower. If you have a plot of potential energy, imagine it is made of wire with a little bead riding on it. Then the bead's motion can be used to model the system.
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