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Calculus and Dynamics question

  1. Aug 4, 2014 #1
    1. The problem statement, all variables and given/known data

    A block of mass m is at rest at the origin at t = 0. It is pushed
    with constant force Fo from x = 0 to x = L across a horizontal
    surface whose coefficient of kinetic friction is μk = μ0(1 - x/L).
    That is, the coefficient of friction decreases from μ0 at x = 0 to
    zero at x = L.

    Prove that:

    ax = vx dvx/dx

    Then, find an expression for the block’s speed as it reaches position L.

    2. Relevant equations

    F = ma
    Fk = μkn

    3. The attempt at a solution

    I actually have no idea how to even approach this question, it was in the challenge problem set of my class year 1 physics textbook.

    I know that vx = dx/dt, and if I was to sub that into the equation I have to prove, it makes sense, but I do not know how to actually prove it. For the second part, I actually have no idea whatsoever. Thanks for helping the poor new engineering student.

    -Question taken from Physics for Scientists and Engineers 3rd Ed, Knight-
     
  2. jcsd
  3. Aug 5, 2014 #2

    haruspex

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    ... and ax = dvx/dt.
    What do you get if you combine the two equations so as to eliminate dt?

    Note that if you integrate the target equation and multiply both sides by the mass you will get a conservation of energy equation. You can use this to answer the question.
     
  4. Aug 5, 2014 #3
    Thanks, I think I got it. Sorry for the bad quality photos, my webcam is the only camera I have when doing late night physics :P

    1.
    https://www.physicsforums.com/attachment.php?attachmentid=71883&stc=1&d=1407216234
    2.
    https://www.physicsforums.com/attachment.php?attachmentid=71884&stc=1&d=1407216234
    3.
    https://www.physicsforums.com/attachment.php?attachmentid=71885&stc=1&d=1407216234

    Can you tell me if my final answer is right? There is no answer for this question in the textbook. The true final answer (after that factoring error I have) is:

    vx = root(L(Fo/m - uog))
     

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