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How great of a force is required?

  1. Nov 12, 2003 #1
    Haha i did it.. thanks.
    A force of 600 N stretches a certain spring a distance of 0.3m .

    What is the potential energy of the spring when it is stretched a distance of 0.3m ?

    What is its potential energy when it is compressed a distance of 4.0 m ?

    I know I am supposed to use the Spring Potential Energy equation (1/2)*k*x^2 but how do I find k? Is it 600N?? I've tried 27 and 600 and those are incrorect.
    Last edited: Nov 12, 2003
  2. jcsd
  3. Nov 12, 2003 #2


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    To lift at a constant speed, the net force must = 0 (neglecting the acceleration from rest to the final speed), so the lifting force is equal to the weight. The work done by the lifting force is that force multiplied by the given distance.

    Edit: The preceding paragraph is in reference to the question that used to be above it (which has since been edited over with a different question). I was just looking at it and thought it looked really silly...
    Last edited: Nov 15, 2003
  4. Nov 12, 2003 #3


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    The spring equation is:
    wher k is the contstant, F is the force, x is the displacement.

    You should get an answer in N/m or something similar.
  5. Nov 12, 2003 #4
    so how does the first part relates to the second where the spring is compressed .04 meters?
  6. Nov 12, 2003 #5


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    2000 N/m = k

  7. Nov 12, 2003 #6
    but the second question says its compressed .04 meters. does K change or do i just plug in .04 like usual.
  8. Nov 12, 2003 #7

    Doc Al

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    You have the spring constant k, thanks to NateTG; now use it. It doesn't change.
  9. Nov 12, 2003 #8
    yeah but this time the spring is compressed not stretched
  10. Nov 13, 2003 #9


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    Then open your textbook and read, carefully, the section where it first refers to "spring constant".
  11. Nov 13, 2003 #10
    so k is F/x =600/0.3=2000 (N/m)

    remember that potential enery when string stretched ot compressed is the same.

    potential energy(when stretched 0.3m) for string is (1/2)Fx=(1/2)*k*x^2=(0.5)*600*0.3=90 J

    to compress string to 4.0 m it needs k*x=2000*4.0=8000 N

    so potential energy(when compressed 4.0m) for string is (0.5)*8000*4= 16000 J

    alternatively you can also use (1/2)*k*x^2=(1/2)*2000*4^2 =16000 J still giving you the same answer....

    hope my answer and explanation can clear your doubt.
  12. Nov 15, 2003 #11
    the answer was 1.6J haha. you were all wrong
  13. Nov 15, 2003 #12


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    So... What's funny? Seems a peculiar way to acknowledge those who are trying to help.

    You mistyped the first question as 4.0 m instead of 0.04 m so of course you got some people to give you wrong answers. Further, the idea is not to get people to do your homework for you. Hence, hints by HallsofIvy and NateTG to get you to solve the problem yourself. Apparently, you were not up for the challenge.
  14. Nov 16, 2003 #13
    I did not make any mistakes(provided that you type your question correctly).Prove me that I am wrong....

    I am sure that either you type wrongly or you come with the wrong answer.Just these two are the options.

    I agree with your idea krab.Thanks krab.Maybe he does not know how I spent my 15 minutes away with the answer, trying to choose the words to make him understood.
  15. Nov 16, 2003 #14
    i said its compressed .04m
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