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Spring pushing a block

  1. May 10, 2008 #1
    A spring of spring constant k = 340 Nm-1 is used to launch a 1.5-kg block along a horizontal surface by compressing the spring by a distance of 18 cm. If the coefficient of sliding friction between the block and the surface is 0.25, how far does the block slide?

    Well i have:

    W=1/2kx^2
    W=55080J
    W=U(potential energy) so i know the potential energy of the spring is 55080
    Energy is conserved..so after the spring has decompressed all the energy is in the block..which is now kinetic
    K=1/2mv^2
    55080=1/2(1.5)v^2
    so Vi=406.5m/s ?? that seems pretty darn high, this is just the initial velocity right?
    so Vf=0
    friction = 3.7N
    F=ma
    Vf^2=vi^2+2ax

    I assume i'm using these equations..but am not sure how i arrise to get accel or if i'm on the right track..
     
  2. jcsd
  3. May 11, 2008 #2
    The spring constant is given in Nm-1 so you have to convert the length of the spring to
    meters. you also multiplied by the mass (1.5kg) when you should have divided when calculating Vi.
    Calculating the speed of the block is really unnecessary, because you can do it all with potential energy. All of the potential energy of the spring will go to friction. Since the friction force is constant and in the opposite direction of the velocity. The energy dissipated by friction is Force * distance.
    (using a rounded off value of 3.7N in your calculations could make the final answer wrong)
     
  4. May 11, 2008 #3

    well..friction is 3.675N
    So, i know the potential energy of the spring is 55080J, how would i set up an equation to figure out the distance the block would travel?
     
  5. May 11, 2008 #4

    Doc Al

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    Staff: Mentor

    Hint: All the initial energy ends up dissipated by friction. What work is done by the friction force?
     
  6. May 11, 2008 #5
    The work of friction (Wf) is:
    W=Fx
    so Wf=3.675x
     
  7. May 11, 2008 #6

    Doc Al

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    Staff: Mentor

    Good. Keep going.
     
  8. May 11, 2008 #7
    the work of the spring (potential energy) (using the work-energy theorem) is 55080...
    so 55080=3.675x
    x=14987.75?
    that can't be right..i must be missing a step?
     
  9. May 12, 2008 #8

    Doc Al

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    Staff: Mentor

    Redo this calculation, paying attention to units. (Convert cm to m.)
    Once you correct your value for spring PE you should be fine. I don't see any missing steps. (Calculating KE at intermediate points is unecessary.)
     
  10. May 12, 2008 #9


    BLAGH!!
    stupid mistake on my part...
    so work now is equal to 5.5J
    so 5.5=3.675x
    x=1.49m

    answer to the problem..
     
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