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Finding spring constant and friction

  1. Sep 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Josh pushes a box of mass m which then travels on a horizontal surface. There is a coefficient of
    kinetic friction μ between the box and the surface. The box has speed v when it reaches x = 0 and
    encounters a spring. The box compresses the spring, stops, and then recoils and travels in the opposite direction. When the box reaches x = 0 on its return trip, it stops.
    Determine k, the spring constant, in terms of μ, m, g, and v.


    2. Relevant equations



    3. The attempt at a solution
    attachment.php?attachmentid=72955&stc=1&d=1410348224.png

    is the diagram I have drawn correct?

    If it is, I can use F=ma in x & y direction.
    However, how can I get V into this equation?
     

    Attached Files:

  2. jcsd
  3. Sep 10, 2014 #2

    gneill

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    Staff: Mentor

    Your diagram is okay as far as the forces involved. Have you considered a conservation of energy approach?
     
  4. Sep 10, 2014 #3
    Would it go like this:

    Intial Kinetic energy + Potential energy (0.5*k*x^2) - work (µ*nc*x) = final kinetic energy (0 as it is in rest in final position)

    is this correct? I don't understand the value for x though?
     
    Last edited: Sep 10, 2014
  5. Sep 10, 2014 #4

    gneill

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    Staff: Mentor

    You have two unknowns here. One is the spring constant and the other is the compression distance, which is your "x". Having two unknowns implies you'll need two equations, right?

    You know what the total energy is to begin with (it's all kinetic and you have the mass and velocity as givens). You should also know what the final energy is and where it's "located". Write energy balance equations for the times when the KE is known to be zero...
     
  6. Sep 10, 2014 #5
    Is the equation correct though?

    (0.5mv^2) + (0.5kx^2) - ((coefficient of frict) * (normal contact) * x) =0

    What will the second equation be?
     
  7. Sep 10, 2014 #6

    gneill

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    No, that equation doesn't look correct when the spring compression is x (fully compressed), although it's close. The system starts out with all the energy as KE. That's your 0.5mv^2 term. When the spring is fully compressed the new KE is zero but the spring has stored PE and some energy has been lost due to friction. Write the energy balance accordingly: OLD = NEW. So something like

    KEi = PEs + Wf

    Where KEi is the initial KE, PEs is the spring PE, Wf is the work done by friction. Use the appropriate expressions for the terms.

    Write the second equation for when the block has returned to the starting point from the fully compressed position. For that portion of the trip the initial energy was all stored in the spring as PE, so that's your "OLD" part of the balance. Where's all the "NEW" energy at the end of the trip?
     
  8. Sep 12, 2014 #7
    So the first equation would be:
    0.5mv^2 = 0.5kx^2 - (µ*nc*x)
    (work has to be negative as energy goes out of the system right?)

    the second equation would be:
    0.5kx^2 = (µ*nc*x)
    There is no kinetic energy in the final position as object comes to rest.

    Is this correct?
     
  9. Sep 12, 2014 #8

    gneill

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    Yup, looks good.

    [edit: see my correction below]
     
    Last edited: Sep 12, 2014
  10. Sep 12, 2014 #9
    Thank you so much. You're a good man
     
  11. Sep 12, 2014 #10

    gneill

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    Staff: Mentor

    Oops. Lets me correct what I said! Interchange the positions of the KE and spring PE in your equation (or equivalently, change the sign of the friction energy). When the block comes to rest at the instant the spring is fully compressed, all the remaining energy will be in the spring. So the sum of the spring energy and the friction energy should total to the original energy.

    (Clearly I need another coffee this morning!)
     
  12. Sep 13, 2014 #11
    0.5mv^2 - (µ*nc*x) = 0.5kx^2

    Is that correct?
     
  13. Sep 13, 2014 #12

    gneill

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    Staff: Mentor

    Yup. That's good.
     
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