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Homework Help: Work and Energy

  1. Jan 11, 2010 #1
    A light horizontal spring has a spring constant
    of 108 N/m. A 3.05 kg block is pressed against
    one end of the spring, compressing the spring
    0.111 m. After the block is released, the block
    moves 0.204 m to the right before coming to
    The acceleration of gravity is 9.81 m/s2 .
    What is the coefficient of kinetic friction
    between the horizontal surface and the block?

    This is how I solved it, although I don't think it's correct.

    When the spring is compressed, it has some potential energy and kinetic energy at this point is 0:
    spring compressed:

    I don't know the acceleration in the horizontal acceleration so I can't figure out the velocity.
    The problem also says the block moves .204m before coming to stop.
    When the block comes to stop, all of the potential energy has been converted into kinetic energy and some has been lost to friction.
    W' (work done by friction)
    =0+Uinitial, when compressed =Fk*d
    =0+ .67J=mu k*N*d (N=mg for obj. on flat surface)
    .67J=mu k (29.89N)(.204m)
    muK= .11

    any suggestions??
  2. jcsd
  3. Jan 11, 2010 #2
    As you compress this block against your spring you aren't going to have any initial kinetic energy. Why? Because the block hasn't started moving yet.

    See if you can come up with a rough formula to describe the conservation of energy throughout this system.

    HINT: Use [tex] KEnergy, PEnergy[/tex] and [tex] \Delta Eth [/tex]
  4. Jan 11, 2010 #3

    ok, if the energy is conserved then
    ΔE=ΔU+ΔKe=constant or 0
  5. Jan 11, 2010 #4
    I'm not sure why you're using Delta's for all the other quantities of energy. Initially all your energy is stored in the spring. In the final state all you energy has been destroyed by friction. So,

    [tex]Initial Potential Energy[/tex] = [tex]\Delta Thermal Energy[/tex]

    Can you define both types of energy?
  6. Jan 11, 2010 #5
    I used deltas to define the initial and final energies of the entire syste.
    Ei = Ui + KEi
    = 1/2kx^2 + 0
    So, the initial energy = 1/2kx^2 only b/c the block has not been released yet.
    Ef = (Uf + KEf) - work done by friction
    This is how it makes sense to me.
    my mcat book, however, says to use the formula:
    W' = U + Ke = Fk*d
  7. Jan 11, 2010 #6
    In your final state you have no kinetic or spring energy so its safe to say that,

    [tex]Uf=0[/tex] and [tex]Kf=0[/tex].


    [tex]\frac{1}{2}kx^{2} = \Delta Eth[/tex]

    Simply define [tex]\Delta Eth[/tex]
  8. Jan 11, 2010 #7
    Ok, got it! Thank you so much!
    W'(work done by friction) = delta E = Fk*d
  9. Jan 11, 2010 #8
    No problem, I'm assuming you take things from here.
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