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Spring compression problem.

  1. Sep 2, 2013 #1
    A child slides a block of mass 2 kg along a slick kitchen floor. If the initial speed is 4m/s and the block hits a spring with spring constant 6 N/m, what is the maximum compression of the spring? What is the result if the block slides across 2m of a rough floor that ha μk = 0.2?


    (second part of the problem)

    getting to the spring..

    vf2=v02+2as

    F(friction)=μ(mg) ------------> a=μg

    vf2=v02+2(μg)s

    vf2= 8.152m/s


    at the spring..

    vf2=v02+2as

    vf2=0

    v02=8.152

    [itex]\Sigma[/itex]F=F(friction)+F(spring)=-m|μg|-|kx|

    a=-|μg|-(|kx|/m)

    v02=2(-|μg|-(|kx|/m))s

    but when i plug all the numbers in.. i get a quadratic equation and it ends up having an imaginary component because the "b^2" term, within the quadratic formula ends up being less than the "4ac" term. i COULD go through and just solve it.. but i don't know what to do because i don't know what to do with the imaginaries.
     
    Last edited: Sep 2, 2013
  2. jcsd
  3. Sep 2, 2013 #2

    Doc Al

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    Is this 2 m stretch of rough surface before it gets to the spring?

    Do you think the rough surface acts to speed up the block? (Careful with signs!)
     
  4. Sep 2, 2013 #3
    oops, i corrected my mistake, but i end up with the same problem. can you point any more errors please
     
  5. Sep 2, 2013 #4

    Doc Al

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    Sure.

    That's a kinematic formula for constant acceleration. Not applicable at the spring, since the force and the acceleration vary. (Think in terms of energy.)
     
  6. Sep 2, 2013 #5
    so would it be this?

    (1/2)mv2=(1/2)kA2+ ∫(μmg)ds

    (1/2)mv2=(1/2)kA2+ μmgs

    since A=s

    (1/2)mv2=(A/2)(kA+μmg)

    (1/2)mv2-(1/2)kA2+(1/2)μmgA

    ?
     
  7. Sep 2, 2013 #6
    i still get a negative number for the "b2-4ac" term....... this is frustrating...
     
  8. Sep 2, 2013 #7

    Doc Al

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    I don't see how. Since "a" is negative and "c" is positive.

    In any case, are you to assume that the friction still applies once the block reaches the spring? If so, why did they specify 2 m?
     
  9. Sep 2, 2013 #8
    didn't pay attention to the signs again...

    well there was a previous part to this question, i just deleted the first part of the question since i already did it. But i assumed that they meant 2m of rough floor and then some more rough floor during the spring plays in. it wouldn't make sense to me if they made it frictionless right there. plus better practice..
     
  10. Sep 2, 2013 #9
    the answer in the back of the book 1.1m.. what the heck...
     
  11. Sep 2, 2013 #10

    Doc Al

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    Sure. Makes sense to me.

    (But you shouldn't have to assume. If you want, you can post the entire problem, word for word, if it's not clear enough.)
     
  12. Sep 2, 2013 #11
    i posted the original problem back at the top
     
  13. Sep 2, 2013 #12

    Doc Al

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    That's ambiguous to me. I would have assumed that it's just a rough patch 2 m wide, then frictionless again. Solve it both ways!
     
  14. Sep 2, 2013 #13
    i've decided i don't like this author ($400 text book..)
     
  15. Sep 2, 2013 #14
    nope, for the frictionless scenario i get 2. something and with friction present i get 1.7, can't see where my mistake is.
     
  16. Sep 2, 2013 #15

    Doc Al

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    What textbook are you using?
     
  17. Sep 2, 2013 #16
    "Classical Dynamics of Particles and Systems" by Thornton and Marion fifth edition
     
  18. Sep 2, 2013 #17

    Doc Al

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    Check your arithmetic. I just did the frictionless scenario and got about 1.65 m. (I could easily have made an error myself though.)
     
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