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Find the mass dropped onto the spring

  1. Jul 25, 2011 #1
    1. The problem statement, all variables and given/known data

    A block is dropped onto a spring with k. The block has a speed of v just before it strikes the spring. If the spring compresses an amount d before bringing the block to rest, what is the mass of the block?


    3. The attempt at a solution

    [tex]\int_{x_0 = 0}^{x_f = d} -kx\;dx + -mg(0 - d) = \Delta K[/tex]

    [tex]\frac{-kd^2}{2} + mgd = \frac{1}{2}mv^2[/tex]

    [tex]\frac{-kd^2}{2} = m\left(\frac{1}{2}v^2 - gd \right)[/tex]

    [tex]\frac{-kd^2}{v^2 - 2gd} = m[/tex]
     
  2. jcsd
  3. Jul 25, 2011 #2
    Wait v = 0 after because it comes to a rest.

    So it should be

    [tex]\frac{-kd^2}{-2gd} = \frac{kd}{2g} = m[/tex]
     
  4. Jul 25, 2011 #3

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    Hmm, you seem to have made it difficult for yourself and there are a few signs wrong here.
    Btw, your last remark is incorrect, because you have completely eliminated the initial speed v.
    That can not be right, because the involved kinetic energy has to go somewhere.

    Why don't you try it this way:

    What is the total energy of the system just before the spring compresses?
    And what is the total energy of the system when the spring is fully compressed?

    These two need to be equal.

    At any given time the total energy would be given by:
    Energy = Gravitational energy + Kinetic energy + Spring energy
     
  5. Jul 26, 2011 #4
    I thought there is no initial velocity because it was dropped?
     
  6. Jul 26, 2011 #5

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    The height from which it was dropped is not given.
    However, the speed just before compression is given...
     
  7. Jul 26, 2011 #6
    Why can't you take intial velocity = 0 from when it was dropped? I thought these are path independent?
     
  8. Jul 26, 2011 #7

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    Sure you can, but... what is the initial height?
     
  9. Jul 26, 2011 #8
    I thought only the change in height mattered.
     
  10. Jul 26, 2011 #9

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    Huh? :uhh:
    Mattered for what?
     
  11. Jul 26, 2011 #10
    the energy calculations, why can't we say the change in KE is 0? I mean it dropped with 0 velocity and it came to a rest?
     
    Last edited: Jul 26, 2011
  12. Jul 26, 2011 #11

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    Please elaborate?

    You can say that the change in gravitational energy is determined by the change in height.
    That is correct.
    What about it?
     
  13. Jul 26, 2011 #12
    You asked me about the initial height. Sorry I was trying to say it didn't matter right?
     
  14. Jul 26, 2011 #13

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    We're not getting anywhere...

    You don't know the initial height.
    And it does matter.

    What's the total energy at the various moments?
     
  15. Jul 26, 2011 #14
    Well at the top it is mgh.

    When it is dropped before touching the spring it is KE - PE = E?
     
  16. Jul 26, 2011 #15

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    Yes! :smile:

    Note that v=0, so KE=0, and the spring is not compressed yet, so the spring-energy=0 as well.


    The proper formula would be KE + PE = E.

    Where do you put your origin for the height?
    What is KE and what is PE in this formula?

    And what is the energy when the spring is fully compressed?
    What is the formula for the energy of a compressed spring for that matter?
     
  17. Jul 26, 2011 #16
    Oh I see, I assumed that i dropped it at the equilibrium point of the spring, but really it could have been dropped ABOVE the equilibrium point of the spring.

    So the ball from whatever height had a "constant" speed of v when it is coming down and touching x = 0 the equilibrium point of the spring. In that case can i do this?

    [tex]\int_{x_0 = 0}^{x_f = d} -kx\;dx + -mg(0 - d) = \Delta K[/tex]

    [tex]\int_{x_0 = 0}^{x_f = d} -kx\;dx + -mg(0 - d) = -\frac{1}{2}mv_0 ^2[/tex]

    EDIT: damn it doesn't work....

    is this unsolvable?
     
  18. Jul 26, 2011 #17
    Let me do this again

    [PLAIN]http://img88.imageshack.us/img88/6199/unleduqh.jpg [Broken]

    [tex]\int_{x = 0}^{x = -x_0} -kx dx - mg(l -(|x_0| + h) - l) = \Delta K[/tex]

    [tex]\frac{-k}{2}(x_0^2) + mg[|x_0| + h|] = \frac{-1}{2}mv_0 ^2[/tex]

    Okay when it was still in the air and uncontacted with the spring

    The following (so trivial!) took place

    [tex]\frac{1}{2}mv_0^2 = mgh[/tex]

    [tex]h = \frac{v_0^2}{2g}[/tex]

    So subsitute everythign back in

    [tex]\frac{-kx_0^2}{2} + mg[|x_0| + \frac{v_0^2}{2g}] = \frac{-1}{2}mv_0 ^2[/tex]

    Messed with a lot of algebra and eventually i got

    [tex]\frac{kx_0^2}{2v_0^2 + 2g|x_0|} = m[/tex]

    If this is wrong, I am thinkiing of dropping out of physics and switching to pure math lo

    EVERYTHING IS WRONG IN THIS POST
     
    Last edited by a moderator: May 5, 2017
  19. Jul 26, 2011 #18
    [PLAIN]http://img217.imageshack.us/img217/6425/unledjex.jpg [Broken]

    Reuploaded

    The change in height is killing me

    Initial height = l

    Final height = d

    l - (h + x0) = d

    [tex]\Delta height = (l - (h + x0) - l) = - (h + x0)

    Does that look right??
     
    Last edited by a moderator: May 5, 2017
  20. Jul 26, 2011 #19
    Can somebody smart please login...
     
  21. Jul 26, 2011 #20
    Sigh

    [tex]l = d + |x_0| + \Delta h[/tex]

    Before contact of the spring and about to touch the spring's equilibrium point

    [tex]mg(|x_0| + d - (\Delta h + |x_0| + d)) = \Delta K[/tex]

    [tex]-mg(\Delta h ) = \frac{1}{2}mv^2[/tex]

    I assume [tex]\Delta h < 0[/tex] because we are going down and we are losing potential? (got that idea from voltage lol)



    [tex]E_i = mgl[/tex]

    [tex]E_f = \frac{1}{2}kx_0^2 + mgd[/tex]

    [tex]mgl = \frac{1}{2}kx_0^2 + mgd[/tex]

    [tex]m = \frac{kx_0^2}{v^2 + 2g|x_0|}[/tex]
     
    Last edited: Jul 26, 2011
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