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Force using velocity graph

  1. Sep 23, 2007 #1
    1. The problem statement, all variables and given/known data

    http://i241.photobucket.com/albums/ff4/alg5045/Ex5-07.gif

    That figure shows the velocity graph of a 5.5 kg object as it moves along the x-axis. What is the net force acting on this object at the given times? a.) 3s b.) 4s and c.) 7s

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    For 3 seconds, I used F= m (change in velocity/change in time), and I got 22N, which was the right answer.

    For 4 seconds, I got zero.

    I am stuck on how to find the 7 second one.
     
  2. jcsd
  3. Sep 23, 2007 #2

    learningphysics

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    What is the acceleration at 7s?
     
  4. Sep 23, 2007 #3
    By looking at the graph, I assumed the velocity was 6m/s. I got the acceleration to be .857 m/s^2. Then I did the same thing that I did for part a, and I got 4.714, but that was wrong.
     
  5. Sep 23, 2007 #4

    learningphysics

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    What is the slope of the v-t graph at t = 7s? The slope is the same all the way from 6 to 8s.
     
  6. Sep 23, 2007 #5
    Yeah, about that slope business. I haven't had a math class using slope for well over three years. I don't remember. I just remember that it's rise over run
     
  7. Sep 23, 2007 #6

    learningphysics

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    Yeah, you can do the first part like that... slope = rise/run = change in velocity/change in time.

    Just do it the same way you did the first part... the acceleration is the same all the way from 6s to 8s... what is the change in velocity from 6s to 8s... what is the change in time?
     
  8. Sep 23, 2007 #7
    The change in velocity is 12 m/s and the change in time is 2 s. So the acceleration is 6m/s^2. Then the force would be 33. I tried that, but it was wrong.
     
  9. Sep 23, 2007 #8

    learningphysics

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    The change is -12m/s.
     
  10. Sep 23, 2007 #9
    Ohhhhh.....thank you for pointing that out.
     
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