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Homework Help: Confused about springs

  1. Oct 24, 2005 #1
    My physics class is intro more or less (it's calc based; 111 NOT 101). I'm a bit confused when it comes to springs..specifically, finding the maximum compression a spring will have.

    Here are the given variables in a problem in my textbook:

    *spring has negligible mass
    *force constant(k) = 1600n/m
    *spring is placed virtically with one end on the floor
    *1.2kg book is dropped from a height of .80m above the top of the spring

    maximum distance the spring will be compressed?

    what confuses me specifically are the signs (+/-) and the distances involved. Here is what I am thinking:

    I need to use K1 + U1 = K2 + U2 (k1 = zero)

    now when it comes to potential energy, i don't know whether to use MGY or 1/2kx^2..can someone point me in the right direction? thanks!
     
  2. jcsd
  3. Oct 24, 2005 #2

    Doc Al

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    Staff: Mentor

    Good. (What will K2 be when the the spring is maximally compressed?)
    Since the height of the book changes, you need gravitational PE; since the spring is compressed, you need spring PE. Use both!
     
  4. Oct 24, 2005 #3
    So are you saying that K2=zero as well? i guess that would make sense

    So..I need U1 (spring PE) + U2 (grav PE) = U1 (spring PE) + U2 (grav PE)

    -or-

    1/2MV1^2 + mgy1 = 1/2MV2^2 + mgy2 ?

    EDIT: Another question; I'm having trouble differentiating between the Y's..where should I solve for Y and where should I plug in .8?
     
    Last edited: Oct 24, 2005
  5. Oct 24, 2005 #4

    Doc Al

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    If K2 wasn't zero, the spring could not be fully compressed--the book would still be moving.
    That will work.
    This won't work--you left out spring PE.
    Call the amount that the spring compresses "x". Then pick a reference point for gravitational PE (may as well call the unstretched position of the spring to be y = 0) and set up the equation. Give it a shot.
     
  6. Oct 24, 2005 #5
    sorry - should have 1/2KX1^2 + MGY1 = 1/2KX2^2 + MGY2

    x is how much the spring is compressed. 0 is where the spring is at rest (so -x would be the compression)

    1/2(1600)x^2 + (1.2)(9.8)(.8) = 1/2(1600)x^2 + (1.2)(9.8)(.8)

    ....it's the same on both sides :confused: would cancel out..
     
  7. Oct 24, 2005 #6

    Doc Al

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    Initially the spring is uncompressed, so x1 = 0. Initially the book is at y1 = 0.8; but where is it when the spring is compressed by amount "x"?
     
  8. Oct 24, 2005 #7

    ...-x?

    so here's what i have now:

    mgy1 = 1/2kx2^2 + mg(-x)

    or

    9.408 = 800x2^2 - 11.76x <--looks like quadratic
     
  9. Oct 24, 2005 #8

    Doc Al

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    Good! Now you're cooking. Solve that equation.
     
  10. Oct 24, 2005 #9
    somehow i usually tend to screw the quadratic equation up, but here is what i got:

    (11.76 +/- sqrt(30243.9))/1600

    which ends up being: .116m or 11.6cm

    the answer in the book was 12cm, i don't think the sig figs required that type of rounding
     
  11. Oct 24, 2005 #10

    Doc Al

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    One of those solutions should be negative. That one can be discarded as physically irrelevant.

    (Your initial data was only good to two sig figs; so your answer should be rounded off accordingly.)
     
  12. Oct 24, 2005 #11
    so how do i get the answer if this equation is discarded?
     
  13. Oct 24, 2005 #12

    Doc Al

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    Don't discard the equation; discard one of the solutions.

    As often happens with quadratic equations for physical systems, only one of the solutions makes any physical sense. In this case, a negative answer doesn't make sense so you know that the real answer is the other one.
     
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