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Non-Conservative Work

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A 0.620 kg wood block is firmly attached to a spring (k = 180 N/m). It is noted that the block-spring system, when compressed 0.050 m and released, stretches out 0.023 m beyond the equilibrium position before stopping and turning back. what is the coefficient of kinetic friction between the block and the table?

    2. Relevant equations

    -(Wnc) = PEf - PEo + KEf - KEo

    3. The attempt at a solution

    1/2 kx2 = 1/2 mv2 + umgd

    This equation has two unknowns (u and v). I don't know any other equation to solve for the unknowns, so I'm stuck here... What other equation do I use?
  2. jcsd
  3. Nov 8, 2008 #2
    I would look at the block at the two times it has zero velocity (i.e. before it is released, and just as it changes direction, call these points A and B), and compare the potential energies stored in the spring. Then work done against friction in going from A to B is the difference in potential energies:

    [tex] W = \frac{k}{2}(0.05^2 - 0.023^2) [/tex]

    You know the distance travelled is the distance AB, so using the work calculated in the last line:

    [tex] W = F(0.023 + 0.05)[/tex]

    Then you need to work out the coefficient of friction. First, the reaction force R of the table on the block:

    [tex]R = mg[/tex]


    [tex]F = \mu R[/tex]

    I'm not sure whether this is correct.
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