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Single conservative force acting on a particle

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-10-25_16-55-4.png

    2. Relevant equations
    U(final)(x)= (-) Integral F dx + U(initial)
    Integration from (x-initial to x-final)


    3. The attempt at a solution

    U(final)(x)= (-) integral (-Ax+Bx^2)dx

    Not sure on limits of integration
     
  2. jcsd
  3. Oct 25, 2015 #2

    andrewkirk

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    You have to start the integration at a place where U is known. What place is that?

    If you want to give the value of U at location coordinate x, what is the upper limit of integration?
     
  4. Oct 25, 2015 #3
    Not sure. Lower = 0. upper x=2.00 & x=3.00?
     
  5. Oct 25, 2015 #4

    andrewkirk

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    I was referring to part (a). There is no 2.00 or 3.00 in part (a). Once you have solved part (a), part (b) is just a question of substitution and subtraction.
    I think you may be confusing yourself by mislabelling your integration variables.
    You wrote:

    U(final)(x)= (-) Integral F dx + U(initial)

    That can't be right because the LHS depends on X while the RHS does not. By using x as the integration variable you make it disappear once the integral is taken. Instead use y as your integration variable and x as one of your limits of integration. Which one?
     
  6. Oct 25, 2015 #5
    So, my potential energy function is U(x) = (-) integral F dx = (-) integral (-Ax+Bx^2) dx = 1/2Ax^2 + 1/3 BX^3

    U=0 @ x=0

    Though I am seeing this as a lower limit
     
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