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Use the work-energy theorem to derive an expression for v^2

  1. Mar 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Capture.png

    2. Relevant equations
    Work energy theorem

    3. The attempt at a solution
    .5mv2 - .5mv2 = (kx - μmg)d
    final velocity is 0

    ½mv2 =(kx - μmg)d
    solve for v2

    ((kx - μmg)d2)m
    When this was wrong I tried integrating Fx but it was still wrong
    Capture2.png
     
  2. jcsd
  3. Mar 24, 2016 #2

    BvU

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    I miss a question in this problem statement ?
    I miss what ##k## means (but I can guess)
    I also miss what ##x## means, which is very important. Note that there is a ##d## already.
     
  4. Mar 26, 2016 #3
    K is the spring constant
    X is supposed to be d, im just used to using x for position, in answers 3-4 i changed it to d
     
  5. Mar 26, 2016 #4

    haruspex

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    If x is d, your very first equation is wrong.
    Remeber that there is energy in the spring both initially and after the mass has moved distance d. What is the change in that energy?
     
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