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Conservative force and potential energy

  1. Oct 8, 2014 #1
    Problem:

    A conservative force F is acting on a particle that moves vertically. F can be expressed as (3y-6) j head N, y is in m and j head is a unit vector along vertical direction.

    a) Calculate the potential energy associated with F, with the potential energy set to zero at y=0.

    b) At what values of y will the potential energy be maximum?

    Relevant equations:

    U(y) = - integral of F(y)

    Attempt at the solution:

    a) When F = 0, y = 2

    U = - [integrate 0 to 2] (3y-6) = 6J

    b) PE is maximum when F = 0. From (a), when F = 0, y = 2.


    I am not familiar with this kind of questions. Could someone please tell me if it's correct?
     
  2. jcsd
  3. Oct 8, 2014 #2

    rude man

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    In (a) you have not expressed U as a function of y.
    In (b) you solved for max. U in a clever way, realizing that U will start to go down when F goes positive.
     
  4. Oct 8, 2014 #3
    U = - integrate [0 to 2] F(y) = - integrate [0 to 2] (3y-6) = (3/2)y^2 - 6y.

    Substituting the limits I get 6J. Is that correct?
     
  5. Oct 8, 2014 #4

    rude man

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    Yes. More convetionally, y=2 for max. U is not known, so you would go

    U = - ∫(3y - 6)dy = -3y2/2 + 6y + constant
    But U = 0 when y = 0 so
    0 = 0 + constant
    constant = 0
    Then set dU/dy = 0:
    6 - (3/2)2y = 0
    y = 2
    You can check that U is a max, not a min, by computing d2U/dy2

    with y = 2 to show that the second derivative is negative, meaning U is a max.
     
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