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Work done by a constant force question

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A 9.9 kg body is at rest on a frictionless horizontal air track when a constant horizontal force [tex]\vec{F}[/tex] acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in Fig. 7-26. The force [tex]\vec{F}[/tex] is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force [tex]\vec{F}[/tex] between t = 0.71 and t = 1.1 s?

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/fig07_26.gif

    2. Relevant equations



    3. The attempt at a solution

    Firstly, unless my numbers are wrong, the acceleration for object this is not constant.

    From t = 0 to t = 0.5, [tex]\delta s = 0.04[/tex]. So then [tex]v = 0.08[/tex]

    From t = 0.5 to t = 1.0, [tex]\delta s = 0.16[/tex]. So then [tex]v = 0.24[/tex]

    From t = 1.0 to t = 1.5, [tex]\delta s = 0.24[/tex]. So then [tex]v = 0.48[/tex]

    But notice that for the first two intervals, [tex]\delta v = 16[/tex] while for the second, it is [tex]\delta = 0.24[/tex]. So the acceleration is not constant on the interval t = 0 to t = 1.5.

    However, I did manage to boil down the answer to the solution to this equation:

    [tex]d = (x_f - x_i) = \frac{1}{2}a*t_f^2 - \frac{1}{2}a*t_i^2 = \frac{1}{2}a*(t_f^2 - t_i^2)[/tex]

    Solving for work:

    [tex]W = Fd \cos(\phi) = m*a*d = \frac{1}{2}m*a^2*(t_f^2 - t_i^2)[/tex]

    Subbing the values:

    [tex]W = (0.5)*(9.9)*(1.1^2 - 0.71^2)*a = 3.4947*a[/tex]

    Where a is whatever value that works, given the data. Note that I've tried a = 0.48, the acceleration that I found from t = 0 to t = 1.5, a = 0.32 which is the discrete average acceleration from t = 0 to t = 2.0, a = 0.426, which is the arithmetic average of the three values for a I have. Suffice to say none of them work.

    Kinda feel dumb for being hung up on such a simple problem, but whatever..
     
  2. jcsd
  3. Oct 9, 2009 #2
    If the force applied and external forces are all constant, what gives you the impression that acceleration isn't?
     
  4. Oct 9, 2009 #3
    The data! I should post up the chart I made to show you what I mean, but I don't know if the forums accept tables.
     
  5. Oct 9, 2009 #4
    I plotted the points on my calculator and got the linear expression a = 0.4v - 0.1333333...

    This seems like a strange problem to give as homework. Also, how did you come up with that expression for acceleration?
     
  6. Oct 9, 2009 #5
    There are 5 data points, which gives 4 position deltas:

    [tex]\delta s_1 = 0.04 - 0 = 0.04[/tex]
    [tex]\delta s_2 = 0.20 - 0.04 = 0.16[/tex]
    [tex]\delta s_3 = 0.44 - 0.20 = 0.24[/tex]
    [tex]\delta s_4 = 0.80 - 0.44 = 0.36[/tex]

    Which in tern gives you 3 velocity deltas: (using *2 instead of / 0.5)

    [tex]\delta v_1 = 2*(0.16 - 0.04) = 0.24[/tex]
    [tex]\delta v_2 = 2*(0.24 - 0.16) = 0.16[/tex]
    [tex]\delta v_3 = 2*(0.36 - 0.24) = 0.24[/tex]

    And from there you can just * 2 again for the average velocity on that interval.

    [tex]a_1 = 2*0.24 = 0.48[/tex]
    [tex]a_2 = 2*0.16 = 0.32[/tex]
    [tex]a_3 = 2*0.24 = 0.48[/tex]

    From there, I decided to try out 0.48 since it included t = 1.1 and t = 0.71. Then I tried to find the acceleration between the first and last velocities

    [tex]a_{avg} = \frac{v - v_0}{t - t_0} = \frac{0.72 - 0.8}{2} = 0.32[/tex]

    When that didn't work, I tried averaging the 3 values together to get:

    [tex]a = \frac{0.48 + 0.32 + 0.48}{3} = 0.426[/tex]

    These are the three that were provided by the text book, although it's possible I overlooked one, but I doubt it.
     
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