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Spring Force and Acceleration with Friction

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data

    A light spring with force constant 3.85 N/m is compressed by 8.00 cm as it is held between a 0.250-kg block on the left and a 0.500-kg block on the right, both resting on a horizontal surface. The spring exerts a force on each block, tending to push them apart. The blocks are simultaneously released from rest. Find the acceleration with which each block starts to move, given that the coefficient of kinetic friction between each block and the surface is (a) 0, (b) 0.100, and (c) 0.462

    2. Relevant equations

    Fx = -kx
    F = ma
    W = the integral of Fxdx from initial to final position
    W = FΔR
    Ffriction = umg

    3. The attempt at a solution

    Here's my attempt so far, but I don't think it's right.

    Since Fx = -kx and F = ma then ma=-kx and -kx/m = a
    So for the first block (-3.85N/m)(0.04m)/(0.250kg) = 1.232 m/s2
    And for the second block (-3/85N/m)(0.04m)/(0.500kg) = 0.616 m/s2

    I chose 0.04 m, because that is the amount the spring is being compressed on each side, maybe I am supposed to use 0.08 m in the equation anyways though.

    For the cases with friction I believe I would have to subtract the force of friction in the F = ma and somehow put it in the equations I used in part b and c.

    Anyway, I don't even know if I started the question correctly, but any help would be awesome.

    Last edited: Oct 28, 2007
  2. jcsd
  3. Oct 28, 2007 #2


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    Good start, but you used the wrong delta x, and you've got to watch your direction of the acceleration for each block. The spring is compressed 8 cm, and that is the distance you must use in calculating the spring force, which is the same on each block. For part b, yes, use newton 2 after identifying the value and direction of the spring and friction forces.
  4. Oct 28, 2007 #3
    So this would be the correct answer for the first part then?

    Since Fx = -kx and F = ma then ma=-kx and -kx/m = a
    So for the first block (-3.85N/m)(0.08m)/(0.250kg) = -1.232 m/s2
    And for the second block (-3.85N/m)(-0.08m)/(0.500kg) = 0.616 m/s2

    I had actually calculated both of the accelerations with the 0.08m last time, but I included the proper directions this time.

    And for part b would it look something like this?

    Fx = -kx and F = ma +/- umg then ma = -kx +/- umg and a = [(-kx)/m] +/- ug
    The reason I have +/- umg is because friction is acting in a different direction in both cases
    So for the first block [-1.232m/s2] + [(0.100)(9.81m/s2)] = -0.251m/s2
    And for the second block [0.616m/s2] - [(0.100)(9.81 m/s2) = -0.365m/s2

    I don't think I have the part b right now that I look at it, if somebody could tell me where I went wrong that would help a lot.

    Thanks again
  5. Oct 28, 2007 #4
    I think that for part b and c if you just see for the magnitudes then the following would be right.

    [tex] kx - \mu}mg = ma [/tex]
  6. Oct 28, 2007 #5


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    Part a looks good. In part b, the acceleration is not the same as in part a, so don't use it. For the left block, you've fot the spring force acting left and the friction force acting right. Use newton's law. Careful using the different friction coefficients that are given. In some cases, Newton 1 may apply.
  7. Oct 28, 2007 #6

    When I do that I get the same numbers as above and that doesn't make sense, especially with friction because the larger block has a higher acceleration, could you possibly try out your way with the numbers so I can see if I actually understand what you're saying?
  8. Oct 28, 2007 #7
    So it will be net force divided by the mass to get acceleration

    First block: -(kx - umg)/m = a = [(-3.85N/m)(0.08m)+(0.100)(0.250kg)(9.81m/s2)]/0.250kg = -.0251m/s2

    Second block [(-3.85N/m)(0.08m)-(0.100(0.250kg)(9.81m/s2)]/0.500kg = a negative number indicating that the friction force is too high and the block won't move?
  9. Oct 28, 2007 #8
    Just apply

    [tex] kx - \mu}mg = ma [/tex]

    And that would be the answers.:smile:
  10. Oct 28, 2007 #9
    When I do that I obtain the same answers I got in the post above, so I assume they were right. And for part c I get answers that suggest both blocks aren't moved when the coefficient of friction is that strong.
  11. Oct 28, 2007 #10

    So for part c are you getting zero for a.
  12. Oct 28, 2007 #11
    Well, I'm getting answers where the force of friction is greater than kx meaning the blocks won't be able to go anywhere.

    Thanks a bunch for all the help, I think I've got it all now.
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