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Find the distance the spring compresses

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data

    a block of mass m slides along ahrizontal table with speed v. At s = 0, it hits a spring with spting constat k and begins to experience a frictin oforce. the coefficient of frictio is u, find the distance the spring compresses when it momentairly comes to rest.

    2. Relevant equations



    3. The attempt at a solution

    i belive the frictin part is umgs, is this correct

    .5mv2 = .5ks2 + ugms

    if this is correct, how do i get the s alone
     

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  3. Nov 8, 2009 #2

    Doc Al

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    Staff: Mentor

    Re: springs

    It's a quadratic equation. Put it in standard form and solve it.

    (That attachment doesn't seem to match the problem you described.)
     
  4. Nov 8, 2009 #3
    Re: springs

    s = -mg + or - (((mg)2 - 4(.5k)(mgh))/k).5
     
  5. Nov 8, 2009 #4

    Doc Al

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    Re: springs

    How did you get this from the equation you gave in your first post?

    I suspect you're solving a different problem.
     
  6. Nov 8, 2009 #5
    Re: springs

    haha ya i had a very similar one sorry about that


    s = -umg + or minus (((umg)2 - 4(.5k)(-.5mv2))/k).5
     
  7. Nov 9, 2009 #6

    Doc Al

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    Re: springs

    That's almost right. (Redo it more carefully. Each term must have the same units.)

    Also, only one of the + or - choices will make sense for this problem.
     
  8. Nov 9, 2009 #7
    Re: springs

    i redid and came up with the same

    .5ks2 + umgs - .5mv2 = 0

    a = .5k
    b = umg
    c = -.5mv2

    -b - [tex]\sqrt{b^2-4ac}[/tex]/2a

    -umg - [tex]\sqrt{()^2 - 4(.5k)(-.5mv^2)}[/tex]/2(.5k)

    s = -umg - [tex]\sqrt{()^2 +kmv^2}[/tex]/k
     
  9. Nov 9, 2009 #8

    Doc Al

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    Re: springs

    Careful with how you do the division by 2a. Everything is divided by 2a, not just the square root term. (Use parentheses to show this.)

    And don't forget that it's + or -, not just -.
     
  10. Nov 9, 2009 #9
    Re: springs

    ok

    i think it should be minus because the spring is being compressed, do you concure
     
  11. Nov 9, 2009 #10

    Doc Al

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    Re: springs

    No...
     
  12. Nov 9, 2009 #11
    Re: springs

    s = -umg + (((umg)2 - 4(.5k)(-.5mv2)).5/k

    s = -umg + (((umg)2 + kmv2).5/k
     
  13. Nov 9, 2009 #12

    Doc Al

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    Re: springs

    You still haven't corrected the "division by 2a" problem from post #8.
     
  14. Nov 9, 2009 #13
    Re: springs

    = (-umg + ((umg)2 + kmv2).5)/k

    the answer how ever is Wnc = -m(v0)^2/2 (1 + k/bmg)^-1 after simplifying
     
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