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Conservation with Springs

  1. Nov 13, 2006 #1
    A 500g block is attached to a 60cm long massless spring with spring constant of 40N/m and is at rest with the back of the block at point A on a frictionless table. A horizontal force of 20N is applied to pull the block towards the right. What is the block's speed when the back of the block reaches point B which is 25cm to the right of point A?

    Spring Potential=.5(40N/m(.0625m^2)=1.25J
    Kinetic Energy=Fx=20N(.25m)=5J
    ME=KE+PE = 6.25J=(.5kg(v^2))/2 + 1.25J
    v=4.47m/s

    Does this look right?
     
  2. jcsd
  3. Nov 13, 2006 #2

    Hootenanny

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    Where did this value come from?
     
  4. Nov 13, 2006 #3
    .25m^2, the distance from A to B
     
  5. Nov 13, 2006 #4

    Hootenanny

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    Indeed, but this would not give you the initial potential energy of the system before the spring was released.
     
  6. Nov 13, 2006 #5
    There are two points of interest in this problem. The point before the release of the spring (after the force is applied) and the point .25 m to the right of A. The energy at both of these points is equal. One will have only potential energy, the other will have a combination of potential and kinetic energy.

    I think you meant to write 0.625 for your spring potential calculation, not 0.0625. As I read the problem 0.60 m is the equilibrium point of the spring, not the stretched length.

    Dorothy
     
  7. Nov 13, 2006 #6
    I wrote my reply before I saw Hootenanny's comment. So disregard my guess about the origins of the 0.0625 m. You do need to find the potential energy of the system before the spring is released.
     
  8. Nov 13, 2006 #7
    The block has yet to be released. I'm assuming that it is asking for the speed of the block as it is being pulled along with the 20N force. I used the sum of work to pull the block along + the springs potential to give me the total ME for the system. Am I looking at this the wrong way?
     
  9. Nov 13, 2006 #8

    Hootenanny

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    Ahh, my apologies, I misinterpreted the question.
     
  10. Nov 13, 2006 #9
    For the second part of the question it asks.
    When the block reaches B, you let it go back. How close does it get to the wall where the spring is attached.

    Ki + Ui = Kf + Uf
    6.25J=0+1/2(kx^2)
    .3125m^2=x^2
    x=.56m
    it gets .29m close to the wall (Equilibrium+Stretch=.85-.56)
     
  11. Nov 13, 2006 #10
    Yes. Me too. Sorry about that.
     
  12. Nov 13, 2006 #11
    I appreciate and need all the help I can get. Hoot, Dorothy, thanks for taking the time to reply. I interpreted it exactly the same way you guys did the first time I looked at it. This was a test question I'm trying to correct. Does everything else look ok?
     
  13. Nov 13, 2006 #12

    OlderDan

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    If I am interpreting the problem correctly, your first part is not correct. I take it to mean that the block is connected to the spring and initially at rest with the spring initially unstretched, so the energy of the block/spring system is initially zero. The 20N force is then applied to the block and you are looking for the speed after it hs moved 25 centemeters. The 20N is an additional force to the force applied by the spring. So looking at your original post, I see the following difficulty

    If my interpretation in the first part is correct, you will need to adjust the total energy on the LHS and find a new x. But .85m is not the equilibrium length of the spring. That is the length of the stretched spring when you stop applying the 20N force.
     
    Last edited: Nov 13, 2006
  14. Nov 13, 2006 #13
    Thanks for your graciousness, Ronnin. Well, I'm not batting 1000 at this problem, but it seems to me that conservation of energy would imply that kx^2/2 would be the same at both positions of the spring, assuming that the extra 20N force is not present when you let the spring go back, of course.

    Dorothy
     
  15. Nov 13, 2006 #14
    Ummm. So I guess I was wrong again. Whoops. The x on the left wouldn't equal the x on the right, because of inertia, I guess? Once the force is removed, the spring continues to move a bit to the right, so you need to account for this extra kinetic energy in calculating the amount the spring is compressed, in other words?

    Dorothy
     
  16. Nov 13, 2006 #15

    OlderDan

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    The maximum distance left and right is equal, but as you say the mass is not finished moving right when the 20N force stops at 25cm. It contiunes to move right until all the energy is spring energy, then reverses and goes the same distance past equilibrium on the other side.
     
  17. Nov 14, 2006 #16
    Wow, i didn't even think about the inertia. I'm going to confirm the exact parameters of the question because this seems a bit out of scope. Dan, if I am pouring 5J into this system then that is my total energy. Part goes into KE and part into PE? So only 3.75J is going into KE because the spring is absorbing the rest? Am I following this correctly?
     
  18. Nov 14, 2006 #17

    OlderDan

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    Yes. And that is of course going to keep things moving to the right for some distance after the force is gone.
     
  19. Nov 14, 2006 #18
    V=3.87m/s. How does that look
     
  20. Nov 14, 2006 #19

    OlderDan

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    Looks OK. Did you find the distance from the wall?
     
  21. Nov 14, 2006 #20
    5J=1/2(.5kg)(3.87m/s)^2+1/2(40N/m)x^2
    x=.2506m extra travel past the B point

    But this doesn't seem right because U at this new distance of .5006m of stretch give me more than 5J of energy. Did I set this up wrong?
     
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