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Force and potential energy of a particle

  1. Dec 19, 2006 #1
    1. The problem statement, all variables and given/known data

    A conservative force F(x) = bx + a acts on a 2.61 kg particle, where x is in meters, b = 6.3 N/m, and a = 4 N. Calculate the change in potential energy of the particle as it moves along the x-axis from x1 = .749 m to x2 = 4.81 m.

    2. Relevant equations

    work-energy theorem, PE = mgh

    3. The attempt at a solution

    My first thought was simply 0, since it was going along the x-axis and therefore no change in PEg and no springs or such, but that was wrong. I solved that the work done by the force = 87.3556 J. I also solved initial kinetic energy if the final speed was 16.8 m/s, which was 280.9676. KEi + W = KEf. Where does the potential fall in?
     
  2. jcsd
  3. Dec 19, 2006 #2

    radou

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    For a conservative force, what is the relation between the force and its potential?
     
  4. Dec 19, 2006 #3
    I remember doing potential energy diagrams...and force is the negative derivative of one. So potential is the integral of force? That's what I did to determine the total work done, though.
     
  5. Dec 19, 2006 #4

    radou

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    Right, that's all you have to do. Find the potential and calculate the change.
     
  6. Dec 19, 2006 #5
    So it is the same as the work done?
     
  7. Dec 19, 2006 #6

    radou

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    Yes, it is.
     
  8. Dec 19, 2006 #7
    Ack, but negative. Got it now.
     
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