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Finding Kinetic Energy From A Graph

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2.0 kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x axis along the surface. Beginning at time t = 0, a steady wind pushes on the lunchbox in the negative direction of the x axis. Figure 7-50 shows the position x of the lunchbox as a function of time t as the wind pushes on the lunchbox. From the graph, estimate the kinetic energy of the lunchbox at (a)t = 1.0 s and (b)t = 5.0 s. (c) How much work does the force from the wind do on the lunchbox from t = 1.0 s to t = 5.0 s?
    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/fig07_50.gif


    Fig. 7-50


    2. Relevant equations
    Change in Y/ Change X

    K = .5mv^2
    3. The attempt at a solution

    I tried finding the slope for each part bc that would be the velocity and then I plug in the velocity into the K equation but it did not work. I am not sure what else to try after that.
     
  2. jcsd
  3. Oct 7, 2009 #2

    kuruman

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    Your method is correct. Show us exactly what you did and how you put numbers in the equations. We can't pinpoint what didn't work without knowing what you did.
     
  4. Oct 7, 2009 #3
    For part a) I used the the coordinate (0,0) and (1,1) to get the slope and it was just 1
    Then I plug it into K = (1/2)(2)(1^2) = 1

    For part b ) I used the coordinates (0,0) and (5,2.5) and found the slope for that to be (1/2)
    then K = (1/2)(2)(1/2^2) = .25

    I think for part 3 you have to subtract the first two answers to get the total but I can't get those yet
     
  5. Oct 7, 2009 #4

    kuruman

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    That's about right.

    You can't do that. The slope is a line tangent to the graph at the appropriate point. At 5.0 s, the line tangent to the graph is parallel to the time axis. What is the slope at t = 5.0 s?
     
  6. Oct 7, 2009 #5
    The slope would just be 0 then.

    I also tried the answer for a and it was wrong. Is there something I missed there?
     
  7. Oct 7, 2009 #6

    kuruman

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    Not really, but you may wish to refine the estimate for the velocity. Read the position at 0.5 s and at 1.5 s, take the difference and divide by (1.5 - 0.5) s. You should get a better value that way.
     
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