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Potential Energy of a single particle

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data

    A single conservative force acting on a particle varies as Fvec = (-Ax + Bx5)i-hat N, where A and B are constants and x is in meters. Accurately round coefficients to three significant figures.

    (a) Calculate the potential energy function U(x) associated with this force, taking U = 0 at x = 0. (Use A, B, and x as appropriate.)

    U = ____________

    (b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 1.20 m to x = 3.80 m. (Use A, B, and x as appropriate.)

    ΔU = ____________

    ΔK = ____________


    2. Relevant equations

    Not sure if there are relevant equations but PE = mgh

    3. The attempt at a solution

    I really have never been this confused about a physics problem... I think the answer needs to be in an equation form? Im really not sure on this one guys, I tried to understand it, its just not coming to me. I did all the rest of the hw completely fine other than this one.
     
  2. jcsd
  3. Mar 1, 2009 #2
    To get you started: Potential energy is just the integral of the force.
     
  4. Mar 1, 2009 #3
    So, Like -Ax^2+Bx^6?
     
  5. Mar 1, 2009 #4
    Just so. Don't forget to divide by the power though. Since the question gives you some initial conditions you can also find the constant of integration.
     
  6. Mar 1, 2009 #5
    ok so for the first question the formula would be -Ax^2+B/5x^6 i think. But then they give you U and x are = 0 ... So i guess you cant just plug x in because that would be 0. How do you solve for potential energy?
     
  7. Mar 1, 2009 #6
    Try to use the LaTeX for equations, it's much clearer. For the potential you use,

    [tex]\int F dx= U[/tex]

    For any indefinite integral you will have a constant of integration, for example,

    [tex]\int x^{n} dx= \frac{x^{n+1}}{n+1} +C[/tex]

    When it gives you the value of U when x is zero it is basically saying plug these vaules into your P.E. equation to find your integration constant.
     
  8. Mar 1, 2009 #7
    ok well i found part a.. but when i go to plug in 3.8 and 1.2 to get the answer, I come out with 7.94A-502.3B , and it comes up incorrect..
     
  9. Mar 1, 2009 #8
    Maybe you have a minus out of place? Is this what you are evaluating?

    [tex][\frac{-Ax^{2}}{2}+\frac{Bx^{6}}{6}]_{1.2}^{3.2}[/tex]
     
  10. Mar 1, 2009 #9
    Im evaluating A/2x^2-B/6x^6 , Im sorry, i tried using latex but ive never used it before and i just couldnt figure it out.. But yea thats what im evaluating, you had to multiply the whole thing by -1 because of the formula. So with A/2x^2-B/6x^6 i am taking (3.8) - (1.2), and i get what i told you before
     
  11. Mar 1, 2009 #10
    Check your calculation, I think you added the A's rather than subtracted the second from the first.
     
  12. Mar 1, 2009 #11
    aahhhh, thank you sir, very wise!
    haha alright well thanks again man and have a great night.
     
  13. Mar 1, 2009 #12
    Thanks and your welcome.
     
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