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Minimum value of the coefficient of static friction

  1. Oct 10, 2004 #1
    A 2.0 kg block slides along a horizontal surface with a coefficient of friction µk = 0.30. The block has a speed of v = 1.3 m/s when it strikes a massless spring head-on

    (a) If the spring has a force constant k = 120 N/m, how far is the spring compressed?

    (b) What minimum value of the coefficient of static friction, µs, will assure that the spring remain compressed at the maximum compressed position?

    (c) If µs is less than this, what is the speed of the block when it detaches from the decompressing spring? [Hint: Detachment occurs when the spring reaches its natural length (x = 0).]

    Explain why detachment occurs when the spring reaches its natural length.

    I'm able to do a) but not the others. thanks for your help.
     
  2. jcsd
  3. Oct 10, 2004 #2

    HallsofIvy

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    If you were able to do (a), show us your work on that at least.

    Actually, I think (b) is easier than e. What's the force the spring exerts when it is compressed at the maximum compressed position?
     
  4. Oct 10, 2004 #3
    a) 1/2mv^2-umgx=1/2kx^2.
    for b, how do I find Xmax?
     
  5. Oct 11, 2004 #4
    anyone.....?
     
  6. Oct 11, 2004 #5

    Pyrrhus

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    What force will pull back the string to its equilibrium position? the force -kx, is the force that brings the spring back to his position, so what minimiun value should the static friction had in order to surpass or cancel that force.
     
  7. Oct 11, 2004 #6
    Well, for part (b) you might look at what the force of the spring happens to be when it is compressed the distance x you found in part (a). For the spring to become stationary, you must assume all the forces cancel out so, "force spring" = "force friction", with of course the new force of friction being based on some new µ-static.
     
  8. Oct 12, 2004 #7
    how about c?
     
  9. Oct 12, 2004 #8

    Pyrrhus

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    Do you understand when the block hits the spring it will compress it til it slows down, then the conservative force of the spring will start pushin the block creating a speed, and when the spring reaches its equilibrium length (x=0) it won't go on, while the block will continue moving with the speed at the equilibrium point, until the friction puts it back to rest.
     
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