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Spring compression distance

  1. Dec 7, 2008 #1
    Given: Load of sacks have total weight of 2250 N and each sack weighs 225 N, how far will each of 2 springs in one system be compressed (spring coeff = 2200 N/m) when a sack is put on it?


    Energy init +work=final Energy

    2250=1/2(4400)x^2
    1.02=x^2
    x=1.01

    Not sure where to go from here.
     
  2. jcsd
  3. Dec 7, 2008 #2

    turin

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    I wouldn't use energy. There is another (even simpler) spring relation that I would use. BTW, I don't know what the problem means by "one system".
     
  4. Dec 7, 2008 #3
    Then using Us=1/2kx^2
    I have 1/2(2200)x^2=2250
    x=1.43

    One system meaning that there are 2 springs working identically to each other.
     
  5. Dec 8, 2008 #4

    turin

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    Is there a reason why you can't use Hooke's law?



    I still don't know what that means.
     
  6. Dec 8, 2008 #5
    Using Hooke's law Fs=-kx
    2250N=(-2200N/m)x
    x=1.02m
    1.02m/2 springs =0.5 m
     
  7. Dec 8, 2008 #6

    turin

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    What is the meaning of this x value? Can you interpret it physically?
     
  8. Dec 9, 2008 #7
    The x value is the total amount that the springs would be compressed by a force of 2250 N.
     
  9. Dec 9, 2008 #8

    turin

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    And is that the condition that the problem is asking for?
     
  10. Dec 9, 2008 #9

    LowlyPion

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  11. Dec 9, 2008 #10

    turin

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  12. Dec 9, 2008 #11
    I didn't post the picture of the springs. It shows them as being parallel to each other with a platform on top. Sorry. My bad. I should be using them together. The link helped as well.

    Now, using Fspring=-kx
    2250=(-)(2200+2200)x
    x=0.5 m
     
  13. Dec 9, 2008 #12

    turin

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    I ask again, is that the x that the problem is asking for? Hint: it has the same interpretation as the last time I asked.
     
  14. Dec 9, 2008 #13
    Yes, they are asking for x.
     
  15. Dec 10, 2008 #14

    turin

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    Nope.
     
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