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Potential vs. Kinetic Energy

  1. Nov 2, 2005 #1
    Basically, an object moves along a straight line (starting at rest) and reaches a maximum kinetic energy at .7 m. At 0 meters, the potential energy is 4 J and at .7 meters, the potential energy is -2 J. What is the maximum kinetic energy?

    I know that the answer is 4 J. However, it would seem to me that the change in kinetic energy is equal to the negative potential energy.

    [tex]\Delta K=K_f-K_i=K_f[/tex]

    [tex]-U=\Delta K=2[/tex]

    This seems that the kinetic energy is 2. Where am I going wrong?

    Thanks for the help.
     
  2. jcsd
  3. Nov 2, 2005 #2

    Doc Al

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    Staff: Mentor

    Assuming mechanical energy is conserved, the change in KE equals the negative change in PE.

    (What makes you think that the answer is 4 J?)
     
  4. Nov 2, 2005 #3
    It is a web-assign type of thing and it took 4 J as the correct answer. The graph is attached if you want to take a look.

    Thank you.
     

    Attached Files:

  5. Nov 2, 2005 #4
    Also, on the same graph, it asks for stable and unstable eq. points. The answer is that stable eqs. occur at x=.7 and x=1.9 while unstable eqs. occur at x=1.5. I know what a stable / unstable equilibrium is for a differential equation, but how does that apply here?

    Thanks again.
     
  6. Nov 2, 2005 #5

    Doc Al

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    Staff: Mentor

    Looking at the graph, I don't see how "At 0 meters, the potential energy is 4 J ...". Looks to me like the graph doesn't even show the potential at 0 meters.
     
  7. Nov 2, 2005 #6

    Doc Al

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    In mechanics, a particle is in equilibrium if the force is zero. (That will be when the slope of the potential curve is zero.) If you nudge the particle from its equilibrium position, and the force tends to return it to equilibrium, then that point is one of stable equilibrium. On the other hand, if the force tends to send the particle away from the equilibrium position, it's unstable. (Think of a marble sitting in a valley versus being balanced at the top of a hill. Which position do you think is stable? Unstable?)
     
  8. Nov 2, 2005 #7
    I see now. For some reason I was just assuming the particle began at 0. I guess the problem requires the assumption that the particle begins at point A (at least, that's what seems to make sense).

    Also, thanks for the clarification on equilibrium. That makes more sense now.

    Thanks again!
     
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