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farleyknight
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Homework Statement
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
Homework Equations
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..