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Distance traveled by a particle using Potential energy function

  1. Nov 14, 2014 #1
    1. The problem statement, all variables and given/known data
    A 1kg particle is in a potential given in joules by U(x)=3x2+4x-5, where x is in metres. The particle is initially at position x=1m and moving with velocity v=4m/s in the +x direction.

    Find the position when the particle first comes to rest.

    2. Relevant equations
    dKe = Kf-Ki , Kf=0
    Ke=-1/2mvi^2
    F=ma
    w=F x d


    3. The attempt at a solution
    U'(x) = 6x+4
    Force at 1m = 10N
    Ke=-1/2(1kg)(4)^2 = -8J
    -8=10xd
    d=-0.8m
    d=xf-xi
    -0.8=xf-1
    xf=0.2m
    However this was wrong. i also tried using several different methods but they were all wrong ..
    Help would be appreciated !!
     
  2. jcsd
  3. Nov 14, 2014 #2

    haruspex

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    I see no need to consider forces - just work with ... work.
    What are the initial and final PEs and KEs?
     
  4. Nov 14, 2014 #3
    Alright so,
    Ui = 2J (at x=1) Uf=0J (at rest) dU=-2J
    Ki = 8J (from 1/2mv^2) Kf=0J (at rest) dK=-8J
    W= dK+dU = -2 + -8 = -10J
    -dU/dx = F = -6x-4, therefore at x=1 F=-10N
    W= F x d
    -10J = -10N x d d=1?
     
  5. Nov 14, 2014 #4

    haruspex

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    No. If it finshes at position xf, what is U there? Remember, in this question U represents only the potential energy.
    As I said, you don't need to work with forces at all.
     
  6. Nov 14, 2014 #5
    if i dont use forces, how do i find my distance ? im confused...
     
  7. Nov 14, 2014 #6

    haruspex

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    The force varies. You cannot here simply multiply force by distance to get energy. It requires an integration. So you might just as well work only with energy.
    You correctly calculated the initial potential energy. Suppose the final position is xf. Using the same method, calculate the final potential energy in terms of xf.
     
  8. Nov 14, 2014 #7
    Ahhh i see! So i solved the quadratic function and it gave me a positive and negative number.. i entered the positive number however that was wrong..? is it possible for the particle to have a negative final position?
     
  9. Nov 14, 2014 #8

    haruspex

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    That should have worked. Please post your working.
     
  10. Nov 14, 2014 #9
    Okay, my procedure was;

    U(xf) = 3(xf)^2 +4(xf) -5

    xf = (-4 +/- sqrt(4^2-(4*3*-5)))/2*3
    xf = (-4 +/- sqrt(16-(-0)))/6
    xf = (-4 + sqrt(76))/6 = 0.786m
    xf = (-4 - sqrt(76))/6 = -2.119m
     
  11. Nov 14, 2014 #10

    haruspex

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    You appear to have solved the equation U(xf) = 0. That is the wrong equation. What equation do you have for U(xf) ?
     
  12. Nov 15, 2014 #11
    The is the equation i have:

    U(xf) = 3(xf)^2 +4(xf) -5 , do i have to set it equal to the initial potential energy which was 2 ? or do i have to somehow include the kinetic energy in the equatin also?
     
  13. Nov 15, 2014 #12

    mfb

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    You have to set it equal to the initial energy. The initial energy comes both from potential and kinetic energy.
     
  14. Nov 15, 2014 #13
    Thank you so much!! it finally worked :)
     
  15. Nov 15, 2014 #14
    is the answer ( 5/3 m ) = 1.7 approximately ?
     
  16. Nov 15, 2014 #15

    haruspex

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    So, do you now understand that to solve it you needed to apply conservation of work, initial (PE+KE) = final (PE+KE)?
     
  17. Nov 21, 2014 #16
    Yes i do thank you very much :)
     
  18. Nov 21, 2014 #17
    Yes it was 1.67m
     
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