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Homework Help: Particle Motion with Forces

  1. Feb 16, 2007 #1
    Just need a bit of help with this question and to make sure im solving it properly. Thanks

    A single conservative force acts on a 5.00 kg particle.
    The equation F_x = (2x + 4) N describes this force, where x is in meters.
    As the particle moves along the x axis from x = 1.40 m to x = 6.20 m, calculate the following:

    (a) work done by the force
    (b) change in the potential energy of the system
    (c) kinetic energy of the particle at x=6.20m
    if its speed is is 3.00 m/s at x =1.40m

    Ek = 0.5mv^2
    Ep = mgh


    Ok I started by integrating the force as ∫ 2x +4
    = x^2 + 4x
    I then evaluated the difference in area between the two bounds, namely
    ((6.2)^2 + 4*6.2 - (1.4)^2 + 4*1.4) which gave 55.68
    Since this was the force I multiplied by the change in distance 6.2 - 1.4 = 4.8
    So 55.68 * 4.8 = 267.264


    I figured that since the particle is always on the x-axis, the potential energy difference should be zero.


    At 1.4m, speed is 3m/s.
    Ek = 0.5(5)(3)^2
    = 22.5

    They want Ek at 6.2m.
    And now I can't decide what to do. Because I think I need the speed at 6.2m in order to find the energy at 6.2m, but I don't kno why the Ek at 1.4m was given. Or I might need to use work and total energy. Im really not too sure.

    Any help to verify this would be appreciated.
    Last edited: Feb 16, 2007
  2. jcsd
  3. Feb 16, 2007 #2
    For part A, the integral finds the area under the curve. What are the units of the area? (look at the units for the width of the interval times the units for the height or F_x value). Since the F_x value is force, and you're multiplying that by a distance, the units don't agree with what you think you found (a force.) You're definitely on the right track to take the integral, but I think you might see what I'm talking about if you look back at your work.

    I agree with your part B.

    Therefore, for part C, you started with a certain amount of kinetic energy when x was 1.40 meters. What happened to all the work that you did, since none of it was "changed" into potential energy (and the force was conservative)?
    Last edited: Feb 16, 2007
  4. Feb 16, 2007 #3


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    Good :approve:
    Not so good. What is the defintion of the work done by a force?
    Last edited: Feb 16, 2007
  5. Feb 16, 2007 #4
    What happened to the 4? :confused:
  6. Feb 16, 2007 #5


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    Good catch neutrino, I missed that ... :rolleyes:
  7. Feb 16, 2007 #6
    oops lol, let me fix that
  8. Feb 16, 2007 #7
    ok so for part A should I have multiplied again by the distance since it was integrated?
  9. Feb 16, 2007 #8
    Hootenany, why does the work have to go to a change in potential energy for part B? (Vs. a change in mechanical energy, of which there's another possibility in this problem)
  10. Feb 16, 2007 #9


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    Lets just pause a moment and think. Write down for my the defintion of work done by a variable force F(x).
  11. Feb 16, 2007 #10
    1st, what do you multiply together to find work?
    Then, draw a quick graph with Force on the vertical axis and displacement on the x-axis. Make a little rectangle somewhere on the graph. Now, if you were trying to find the area of that rectangle (using a formulas from geometry instead of an integral), what would you be multiplying together?
  12. Feb 16, 2007 #11


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    Oops edit, my bad, misread the question, for some reason I thought we were in the vertical plane :redface:
  13. Feb 16, 2007 #12
    Isnt work just the force by the distance. And since the integrated area gave the total force in Nm shouldnt it just be that?
  14. Feb 16, 2007 #13


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    No, not quite. The defintion of work done (W) by a variable force (F) is gived by;

    [tex]W = \int^{x_1}_{x_0}F(x)dx[/tex]
  15. Feb 16, 2007 #14
    By integrating that wouldnt it just be an extra x term, meaning that its divided by the distance to get rid of it?
  16. Feb 16, 2007 #15


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    :bugeye: I don't quite follow your logic there. Lets take your function above;

    [tex]W = \int^{x_1}_{x_0} (2x+4)dx = \left[ x^2 + 4x \right]^{x_1}_{x_0}[/tex]

    [tex]W = \left(x_1^2+4x_1\right)-\left(x_0^2+4x_0\right)[/tex]
  17. Feb 16, 2007 #16
    ok, i evaluated that already but im just confused about what to do with that result
  18. Feb 16, 2007 #17


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    If you evaluated the integral correctly, it will have returned the value of the work done by the force as the particle moved between x0 and x1
  19. Feb 16, 2007 #18
    so by integrating the equation and evaluating w.r.t the endpoints i am finding the work done by the force?
  20. Feb 16, 2007 #19
    Yep. Newton-meters are not a unit of force.

    Maybe it'll help to go back to something simpler.

    For a similar problem, lets do this:
    On the vertical axis, put velocity. On the horizontal axis, let's put time. Now, here's an equation. v_t= 5.
    Find the displacement from t=1 to t=7.
    Note: if you graph this, it's just a constant function (a horizontal line).

    If we integrate, we'd have [tex]displacement = \int^{7}_{1}5dt[/tex]
    This integral is 5t evaluated from 1 to 7, or 5*7 - 5*1 = 30 (units??)
    Now instead, notice that it's just a rectangle. You can find the area in a rectangle by a simple length * width. The length is 6 seconds (from 1 second to 7 seconds) and the width is 5 m/s. Thus, you're multiplying meters/second by seconds, and the units cancel to give you an answer in meters.

    The area under a velocity vs time curve is the displacement.
    Likewise, the area under a Force vs. distance curve IS the work.
    Thus, as Hootenanny said, Work = the integral...
    (the integral is just a way to find the area under a curve.)
  21. Feb 16, 2007 #20
    That *is* the work. There's no need to multiply by anything.

    (Don't you wish we would have just told you that you went too far on your first step the first time you did it? :tongue2: ) Sometimes it's tough to help someone understand the process and what they're actually doing vs. just rote memorization of steps that you might end up confusing later on.
  22. Feb 16, 2007 #21
    great, thanks for clarifying that
    i was beggining to get very confused

    so we kno the work is the integrated equation evaluated and that the change in potential is zero (since it is on the x-axis),

    but for C)
    since we kno the work is 55.68
    and that Ek at 1.4m is 22.5
    for Ek at 6.2 is it simply 55.68 - 22.5 = 33.18
    Last edited: Feb 16, 2007
  23. Feb 16, 2007 #22
    You did it again! No, the force isn't the integrated equation evaluated. The *WORK* is the integrated equation evaulated.
  24. Feb 16, 2007 #23
    o ya ya sry i actually meant to put work, ill fix it
  25. Feb 16, 2007 #24
    You're on the right track. Now, you did work on the car (or whatever it was), by exerting a force in the positive direction. The car was originally moving in the positive direction. Would you want to add the energy (from the work) to the kinetic energy? Or do you want to subtract?
  26. Feb 16, 2007 #25
    im not too sure because wouldnt the initial kinetic be a part of the total work while the total kinetic would equal the work

    so W = Ek1 + Ek2
    and then Ek2 would be W - Ek1

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