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Spring Problem (Urgent)

  1. Oct 6, 2008 #1
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    Please Help some1??

    To be honest I don't what to do........
     
  2. jcsd
  3. Oct 6, 2008 #2

    Hootenanny

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    Saying that you have no idea doesn't count as an attempted solution. Have you read your class notes and/or text?
     
  4. Oct 6, 2008 #3
    my professor doesn't go in details and book is to consusing but still I will give a try:

    attempt: First I thought to use

    W = delta KE + delta PE

    but I was confused b/c i did not see a way to put those spring forces in there

    so i was kindly asking for a help
     
  5. Oct 6, 2008 #4

    Hootenanny

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    Thank you for trying, your threads will generally get answered a lot quick if you show an attempted solution.

    You're right to approach the problem using conservation of energy. You know that the work done by friction and the 400 N/m spring must be equal to the energy stored in the N/m spring.

    Can you go from here?
     
  6. Oct 6, 2008 #5
    is it like this:

    F x S = Ef - Ei
     
  7. Oct 6, 2008 #6

    Hootenanny

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    We'll take it a step at a time. What is the energy stored in the 200N/m spring before the block is released?
     
  8. Oct 6, 2008 #7
    = .5k(x.x) + .5mv.v + mgh
     
  9. Oct 6, 2008 #8

    Hootenanny

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    Could you explain how each of the terms arise, what does each term represent?
     
  10. Oct 6, 2008 #9
    1st is spring force , 2nd is kinetic energy and 3rd is potential enery, which is 0.
     
  11. Oct 6, 2008 #10

    Hootenanny

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    Good. What about the kinetic energy, is that non-zero before the block is released?
     
  12. Oct 6, 2008 #11
    i think it is zero since Vi is 0
     
  13. Oct 6, 2008 #12

    Hootenanny

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    Correct. So, the total energy of the system is simply the energy stored by the spring, can you calculate it?

    Now, when the block reaches the 400N/m spring is compresses it. So how much energy is stored in this spring? Can you also calculate this value?
     
  14. Oct 6, 2008 #13
    at intial = .5(200N/m)(.1 x .1) = 1N

    at Final = .5(4000N/m)(.05 x .05) = .5 N
     
  15. Oct 6, 2008 #14
    do i just add them together?
     
  16. Oct 6, 2008 #15

    Hootenanny

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    No.

    You started off with this much energy:
    And ended up with this much energy:
    Where did the rest of the energy go?

    You also need to be careful with your units, the Newton is not a measure of energy.
     
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