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Work with varying forces. Work energy theory

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A net force along the x-axis that has x-component F(x)= -12.0N + (0.300 N/m^2)x^2 is applied to a 3.70 kg object that is initially at the origin and moving in the -x-direction with a speed of 8.80 m/s .


    2. Relevant equations

    w= f * s

    w= ∫F(x)dx (varying forces)

    w = Δk

    k = 1/2 mv^2



    3. The attempt at a solution

    I am stuck on how to approach this question. I have seen many of these types of questions answered with potential and kinetic energy, but we have not covered that in class so far. We have only covered kinetic energy and the work energy theory. I realize that the force is varying so that It will eventually turn the object around in the positive x direction so that I can calculate the speed at 7m, but I am not sure what to do with the information that I have. Any help or pointers to help me see what I am missing would be appreciated.


    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 15, 2012 #2

    SammyS

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    What's the question you're trying to answer ?
     
  4. Mar 15, 2012 #3

    rock.freak667

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    If you need to get the speed at 7 m then you need to use the fact that a = dv/dt to get the relationship for speed. (you will need to use the chain rule to change dv/dt to something else involving dv/dx).
     
  5. Mar 15, 2012 #4
    Yes that part is important.

    What is the speed of the object with it reaches the point x = 7 m?
     
  6. Mar 15, 2012 #5
    Rock.freak i am not sure how to get the acceleration. It starts out going -8.8 m/s, so its kinetic energy is 1/2*m*v^2 or 143.264 J. So I need to apply 143.264 J just to stop it, then the force function will move it in the positive direction. So if I integrate the force function, can I set the 143 J equal to that function, solve for x, and then find the distance between it and +7m?

    Does that make any sense?
     
  7. Mar 15, 2012 #6

    SammyS

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    Use the work-energy theorem.

    If all you need is the velocity at x = 7m, then there's no need for you to find the acceleration.
     
  8. Mar 16, 2012 #7
    write f as ma and a as dv/dt. multiply both sides by dx. then it becomes ..vdv=...dx
    integrate both sides
     
  9. Mar 16, 2012 #8

    SammyS

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    The result of doing this is the Work-Energy theorem for this particular force.
     
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