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Spring constant for a block sliding PE and KE

  1. Mar 19, 2013 #1
    1. An .8kg block is held in place against the spring by a 67 N horizontal external force. The external force is removed, and the block is projected with a v1=1.2. upon separation for the spring. The block then descends a ramp and has a velocity v2=1.9 m/s at the bottom. The track is frictionless up to that point. The block enters a rough section B, extending to E. The coefficient of friction over this section is .39. The velocity of the block is v3=1.4 m/s at C. The block moves on to D, where it stops. The spring constant of the spring is closest to:

    1100N/m 3900N/m 2600N/m 1600N/m 2000N/m




    2. It seems like a lot to take in. I have a couple thought about how to go about it but not sure if any would work. There's also a diagram of the situation included.



    3. First I started simple and thought maybe we didn't need all that and tried just using F=.5kx^2 but we don't know x. So then I proceeded trying to use Conservation of Energy. At the top there would be some PE and the KE would be .576. after the drop, the PE would be 0 and the KE goes up to .5*.8*1.9^2=1.444. Since there is no friction, the initial PE must have been 1.444-.576=868N. That gets me to the spring with the PE and KE known but I have no clue from there because there's also PE in the spring along with the gravitational PE. I'm not sure if I need to include the friction part somehow. Thanks!
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Mar 19, 2013 #2

    TSny

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    Does .5kx2 represent force or potential energy?
    If you know the KE of the block as it leaves the spring, what does that tell you about the initial PE of the spring?
     
  4. Mar 20, 2013 #3
    Think I'm getting there but the x has me stuck

    I see the correlation now. PE of held spring = KE of v1. So PE=.576 or .5kx^2=.576. We still have two variables though. Can I just solve for x and substitute? Or do I need to use the 67-N force that's holding the spring in place?
     
  5. Mar 20, 2013 #4

    TSny

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    Two unknown variables: k and x. Need another equation. What is the relation between force and k and x?
     
  6. Mar 20, 2013 #5
    Not sure? You mean like k-2F/x2 or x= sqrt(2F/k)?
     
  7. Mar 20, 2013 #6

    TSny

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    Did you cover "Hooke's law"? It relates the force of a spring to the spring constant k and displacement x.
     
  8. Mar 20, 2013 #7
    Yes. F=1/2kx2 you mean?
     
  9. Mar 20, 2013 #8
    Oh wait no.....that's F=kx.....AHHHHHHH :surprised I see

    So...maybe

    PE=.576
    .5kx2=.576

    .5F2=.576

    .5*492.... Oh wait nevermind:mad:
     
  10. Mar 20, 2013 #9

    TSny

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    Yes. So.....? :smile:
     
  11. Mar 20, 2013 #10
    F=1/2mv2

    F=.576
     
  12. Mar 20, 2013 #11

    TSny

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    :bugeye: How can force equal kinetic energy? From PE = (1/2)kx2 and F = kx you need to figure out how to get k. It's all come down to algebra.
     
  13. Mar 20, 2013 #12
    Bum bum bum. ALGEBRA. So wait maybe I was right and just stopped too soon.

    so F = kx
    so k=F/x

    So 1/2(F/x)x2=PE

    So 1/2 * Fx2/x

    1/2* Fx =PE ?
     
  14. Mar 20, 2013 #13
    No.


    x=F/k

    1/2k(F/k)2

    PE=1/2F2/k ?

    Force if 67 so 1/2*67^2/k

    I think I've just confused myself by working too many different problems....
     
    Last edited: Mar 20, 2013
  15. Mar 20, 2013 #14

    TSny

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    This is great. Can you use this to find x? Then you can find k.
     
  16. Mar 20, 2013 #15

    TSny

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    This will work too! :smile:
     
  17. Mar 20, 2013 #16
    But what does this equal? PE. So 1/2*672/k=PE

    k= 672/ 2*PE ??? That doesn't make any sense to me though
     
  18. Mar 20, 2013 #17
    Quote by Zsmitty3 View Post

    Bum bum bum. ALGEBRA. So wait maybe I was right and just stopped too soon.

    so F = kx
    so k=F/x

    So 1/2(F/x)x2=PE

    So 1/2 * Fx2/x

    1/2* Fx =PE ?

    This is great. Can you use this to find x? Then you can find k.



    So to solve for x. Fx=2PE. x=2PE/F
    ????
     
  19. Mar 20, 2013 #18

    TSny

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    Good, that's it. Then get k.
     
  20. Mar 20, 2013 #19
    Not sure what to do from there though. Plug that value of x into 1/2kx2 again? And set that equation equal to what? PE? or .576?

    So

    1/2K(2PE/F)2=PE ?
    K*(2PE/F)2=2PE
    K=(2PE)/(2PE/F)2?
     
  21. Mar 20, 2013 #20

    TSny

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    What value did you get for x?
     
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