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Work, Elastic and Kinetic Energy

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data

    The only force acting on a 2.4 kg body as it moves along the positive x axis has an x component Fx = - 6x N, where x is in meters. The velocity of the body at x = 3.0 m is 8.0 m/s.
    - What is the velocity of the body at x = 4.0 m?

    2. Relevant equations
    W = [tex]\Delta[/tex]0.5*m*v^2
    that is for kinetic

    W = [tex]\Delta[/tex]0.5kx^2

    3. The attempt at a solution

    I set the two equations equal to each other since no outside force is acting:
    [tex]\Delta[/tex]0.5*m*v^2 = -[tex]\Delta[/tex]0.5kx^2 . . .
    so . . . .
    .5(2.4)(v^2) - .5(2.4)(8^2) = - .5(6)(3^2) - .5(6)(4^2)

    is that right for the spring constant or is k = 6*x . . . .?
     
  2. jcsd
  3. Apr 27, 2008 #2

    alphysicist

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    Hi KMjuniormint5,

    I don't see a spring in this problem so you don't need the spring expression. You have the formula that shows the effect of the work on the kinetic energy. Now what is the formula for the work done by a force?
     
  4. Apr 27, 2008 #3
    w = F*d where d is the distance but it is only in the x direction. . let me try that
     
  5. Apr 27, 2008 #4
    so (-6)(4^2)-(-6)(3^2) = .5(2.4)(v^2)-.5(2.4)(8^2) and solve for v?
     
  6. Apr 27, 2008 #5

    alphysicist

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    That would apply for a constant force where the force and motion are in the same direction.

    However, this force varies with x, so you need to use the integral form.
     
  7. Apr 27, 2008 #6
    take the intergal of F*d? . . .
     
  8. Apr 27, 2008 #7

    alphysicist

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    For a particle moving in the x direction, the work done is:

    [tex]
    W \equiv \int\limits_{x_i}^{x_f} F_x \ dx
    [/tex]
     
  9. Apr 27, 2008 #8
    so the force is going to .5*(-6)(x)^2 with xi = 3 and xf = 4
     
  10. Apr 27, 2008 #9

    alphysicist

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    (You said force, but I'm sure you meant work.) Yes, that will be the work done from xi to xf once you evaluate it with the limits. Once you have that you can relate it to the change in kinetic energy.

    On looking back at this thread, I have to apologize. That force is the force of a spring, of course, so what we are doing will lead back to something like what you were doing in your first thread.

    The work for a spring is:

    [tex]
    W = -\frac{1}{2} k x_f^2 + \frac{1}{2}k x_i^2
    [/tex]

    which is the kind of idea you had for the work done originally. You did have a sign error in that equation, but if you use the above you should get the correct sign.
     
  11. Apr 27, 2008 #10
    ok ya. . .i was just confused about the spring part because they said to use the spring equation. . .all i had wrong was the sign. . . thanks a lot!
     
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