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Spring acceleration

  1. Oct 7, 2004 #1
    A spring hanging vertically is stretched by 0.14 Meters when a mass of 77.4 kg is hung from it. The mass is then placed on a smooth horizontal surface and pulled using the same spring. If the spring is now stretched by 0.02 Meters, what is the acceleration of the mass (in M/s2).

    I used the formula F=ma and got 9.81. I got it wrong. What am suppose to do with the spring length change?
     
  2. jcsd
  3. Oct 7, 2004 #2

    Pyrrhus

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    In the first case what forces are acting on the block?
     
  4. Oct 7, 2004 #3

    Diane_

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    Remember Hooke's Law.
     
  5. Oct 7, 2004 #4
    F or T on the box: T=mg = 77.4 * 9.81 = 759.294
    Hookes law: F = -kx so, 759.294/(0.14-0.02) = -6327.45
    how do I find the acceleration of the mass from here?
     
  6. Oct 7, 2004 #5

    Pyrrhus

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    buffgilville, do you know what the x in Hooke's Law mean?
     
  7. Oct 7, 2004 #6
    x = direction moved
     
  8. Oct 7, 2004 #7

    Pyrrhus

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    Do you see a problem with what you did?
     
  9. Oct 7, 2004 #8
    I know I did the problem wrong, but I don't know what? I don't know how to find the acceleration for this type of problem. Can you please explain it to me?
     
  10. Oct 7, 2004 #9

    Pyrrhus

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    Write what you need, and then read all i said, and see if you can understand. If not try to read Hooke's Law again, and see what [itex] \vec{F} = k \vec{x} [/itex] means.
     
  11. Oct 7, 2004 #10

    Diane_

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    Maybe this will help - in your solution, you subtracted the .02 m from the .14 m. The .14 m came from when the mass was hung from the spring. The .02 m came from when you dragged the mass along a surface. Those were two completely different situations - why would the two distances have anything to do with each other?

    On the other hand, there were some things about the two situations that remained exactly the same between them - those are the things that you'll use from the first problem to solve the second. I realize that may be a little cryptic, but ask yourself - what did not change from the first part to the second?
     
  12. Oct 7, 2004 #11

    Doc Al

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    First use the information provided when the mass is hanging vertically to find the spring constant. Then use Hooke's law to find the force on the mass when it's on the horizontal surface. Finally, use Newton's 2nd law to find the acceleration of the mass.
     
  13. Oct 7, 2004 #12
    Thanks a million for all of your help!
     
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