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Another Spring Problem

  1. Oct 26, 2004 #1
    Hello,

    I have a problem that I thought I knew how to figure out but I'm getting a bit stuck. Here's the problem:

    A 0.82kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from unstrained length increases by a factor of 8. What is the mass of the second block?

    My problem is that there isn't a spring constant given here.

    I would guess that once the second block has been added, the two blocks aren't going up and down or oscillating. So that would make this problem an equilibrium one.

    In that case I know that F = W ------ F = kx and W = mg

    I's assuming that x is 8 but that doesn't make sense to me since that isn't 8 really isn't a distance. At least I don't think it is....

    Anyway, I'm just lost..Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 26, 2004 #2

    Pyrrhus

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    Homework Helper

    Make the formulas for the data in case 1, and then in case 2.
     
  4. Oct 26, 2004 #3
    Ok, lets see...

    The first case the block is just hanging there. Mass is 0.82. That's all I know about the first case. If I drew a force diagram I would have the weight force pointing down and the spring force pointing up. I just mentioned earleir that [itex]F = W[/itex] Which is also written as [itex]kx = mg[/itex]. If I drew my origin at the point where the spring and the block meet that would make x = 0...right?? So the only thing in play for the first case is the blocks mass times gravity.....

    In the second case a new block of mass m is added to the already hanging block of mass = 0.82kg. The spring now stretches by a factor of 8. <----(I gotta admit I have no idea what that means.) So if I drew a new force diagram for the second case I would have two weight forces pointing down and still one spring force pointing up. It's here that I don't know where to go next.
     
  5. Oct 26, 2004 #4
    Am I on the right track here? I'm not sure where I need to be going after the first case.
     
  6. Oct 26, 2004 #5
    the first equation is correct. i will rewrite it again here.
    [tex]m_{1}g=kx[/tex]
    case 2 :
    [tex](m_1+m_2)g=k(8x)[/tex] assuming that this phrase 'the spring stretches from unstrained length increases by a factor of 8' indicates the spring stretches by a factor of 8 from the previous one.
     
  7. Oct 26, 2004 #6
    OK, for case two, though, how in the world do I solve for that?? I don't know what k or x is and I'm trying to find m2 so how do I solve for an equation which has 3 unknown variables.
     
  8. Oct 26, 2004 #7
    divide the two equations to get rid of x and k.
     
  9. Oct 26, 2004 #8
    Oh....ok. Thanks for telling me...But I think it would be important for me to know why I would be dividing the equations? I mean, I probably would never have come to that point on my own....well obviously or else I wouldn't be here. lol

    Also wouldn't the m1 also cancel out?

    Thanks for helping me out.
     
  10. Oct 26, 2004 #9
    as what i mentioned just now, you have to divide them to get rid of x and k to find m2. m1 won't be cancelled out. you will get :
    [tex]\frac{m_1}{m_1+m_2}=\frac{1}{8}[/tex]

    or you can see this way : substitute equation 1 into equation 2 and k and x will be gone.
     
    Last edited: Oct 26, 2004
  11. Oct 26, 2004 #10
    Wow, ok now I am lost....I get now to divide the two equations to get rid of x and k, but where did that 1 come from?
     
  12. Oct 26, 2004 #11
    or you can see this way : substitute equation 1 into equation 2 and k and x will be gone.
     
  13. Oct 26, 2004 #12
    Oh, ok making a substitution makes a little more sense to me. It makes it a bit easier to understand.

    Thanks again for your help.
     
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