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Mechanics - Work and Kinetic Energy question

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A 3kg body is at rest on a frictionless horizontal air track when a constant horizontal force F acting in the positive direction of an x axis along the track is applied to the body. The force F is applied to the body at t = 0 when the body is at the origin. A stroboscope records the position of the body at 0.5 s intervals. The readings are as follows.

    Interval (I) 1: 0.04 m
    I2: 0.2 m
    I3: 0.44 m
    I4: 0.8 m

    How much work is done on the body by the applied force F between t = 0 and t = 2s?


    2. Relevant equations
    K = 1/2mv^2
    W = [tex]\Delta[/tex]K
    s = ut + 1/2at^2
    v^2 = u^2 + 2as

    3. The attempt at a solution

    I really have no idea why I'm not getting the right answer. I can't see an error in what I've done.

    1) Find a
    u = 0

    a = 2s/t^2
    a = 2 x 0.04 / 0.5^2 = 0.32 ms^-2

    2) Find v^2

    v^2 = ((0 ms^-1)^2) + 2(0.32 ms^-1 x 0.8m)

    3) Find the change in kinetic energy and hence the work done

    [tex]\Delta[/tex]K = 1/2mv^2 = 1/2(3kg)(0.512m^2 s^-2) = 0.768 J

    ---

    The answer is meant to be 0.96J. I think I might have done this question months ago but I cannot see where I have gone wrong in the least. Please point out any glaring mistakes :P.
     
  2. jcsd
  3. Jan 18, 2009 #2

    Doc Al

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    Staff: Mentor

    You didn't make any mistakes, but the data seems to be off a bit. (The acceleration is not constant.) Try finding the acceleration using the I4 data.
     
  4. Jan 18, 2009 #3
    I did what you suggested and got the question right. Thank you Doc Al. I've a few of questions about this problem.

    If the acceleration is not constant then it follows that the force must be variable. This contradicts the first part of the question in which it is stated that the force is constant. Such a situation would require the use of calculus.

    In this instance, I did not have to use calculus and got the correct answer. I initially got the answer wrong because I use the first time interval to work out the acceleration, which I assumed was constant. Why does this method work if the acceleration is nonlinear? Is it the average acceleration that is found using this method?
     
  5. Jan 18, 2009 #4

    Doc Al

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    Staff: Mentor

    Correct. I think the explanation is simply that whoever made up this problem messed up the data. For example, if the data were:
    I1 = 0.05
    I2 = 0.2
    I3 = 0.45
    I4 = 0.8​
    then the acceleration would be a constant 0.4 m/s^2.

    What textbook is this from?
     
  6. Jan 18, 2009 #5
    The question is from "Fundamentals of Physics, 7th Edition," Halliday, Resnick and Walker.
     
  7. Jan 18, 2009 #6

    Doc Al

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    Staff: Mentor

    What chapter and problem number?
     
  8. Jan 18, 2009 #7
    Chapter 7, question 9. I sincerely hope I didn't write the distance down incorrectly, haha. I'm fairly sure I didn't.
     
  9. Jan 18, 2009 #8

    Doc Al

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    Staff: Mentor

    Looks like I don't have that particular edition, but I suspect a typo was involved.
     
  10. Jan 18, 2009 #9
    I hope so. This problem was really getting at me earlier when it shouldn't have been because of its simplicity!
     
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