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Homework Help: Spring with friction (nonconservative force)?

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A 5 kg mass slides to the right on a surface having a coefficient of friction 0.48. The mass has a speed of 7 m/s when contact is made with a spring that has a spring constant 138 N/m. The mass comes to rest after the spring has been compressed a distance d. The mass is then forced toward the left by the spring and continues to move in that direction beyond the unstretched position. Finally the mass comes to rest a distance D to the left of the unstretched spring. The acceleration of gravity is 9.8 m/s2. Find the compressed distance d.

    2. Relevant equations

    Wnet = WC + WNC

    WNC = Ffrdcos[tex]\theta[/tex]

    [tex]\Sigma[/tex]Ebefore = [tex]\Sigma[/tex]Eafter + WNC

    3. The attempt at a solution

    I'm just not sure if I should separate this into two parts where I find the the final energy just before it touches the spring and then the second part, find the distance with the final energy found in part one?

    And I'm really confused as to incorporating the nonconservative force in this equation. I've looked at my book and there is no example of this or anything =( They only have it in relation to potential energy in the y-direction.
  2. jcsd
  3. Oct 19, 2009 #2


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    Homework Helper

    hint: What is the kinetic energy of the mass when the spring is compressed a distance d?

    hint: What is θ?

    caution: Be careful about minus signs. They are arbitrary, but you should assign them consistently. Work done on the mass should have an opposite intrinsic sign to work done by the mass. This is in addition to the sign that you calcuate from cosθ.
  4. Oct 19, 2009 #3
    KE (when compressed) = 1/2mvf2 + 1/2kd2 - Ffrd

    [tex]\theta[/tex] = 180

    Right track??
  5. Oct 19, 2009 #4
    I figured it out! Thank you so much!

    1/2mvi2= 1/2kd2 + Ffrd

    0 = 1/2kd2 + Ffrd - 1/2mvi2

    solve with quadratic ...

    (I was getting stumped on that sign for the frictional force... I think that's what you were hinting at and I just wasn't putting 2 and 2 together!)
  6. Oct 20, 2009 #5


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    I'm sorry, I wasn't online this evening. Yes, that looks correct. (Of course, you still need to determine Ffr, but I suppose you know how to do that.) Good job.
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