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Determining Net Force Based on Motion Diagram

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    It's webbased, so I just took a screenshot: http://img27.yfrog.com/img27/2237/pyquestion.jpg [Broken]

    2. Relevant equations

    F = ma

    3. The attempt at a solution

    Well, the way I figured it was that we could determine the force by calculating the average velocity between two time periods and subtracting to see how much it changed. Doing this I came up with that the acceleration is -75 cm/s. Converting to m/s and plugging in to F = ma (along with the 4 kg) I got an answer that is apparently wrong. I tried some other ways as well to no avail.

    Help!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 22, 2009 #2

    berkeman

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    Staff: Mentor

    They give you position as a function of time, so you need to calculate v(t) and a(t) from that. Can you post what you calculated for each of the snapshot positions shown? Is the a(t) really constant (it may be, but it's not obvious to me from looking).
     
    Last edited by a moderator: May 4, 2017
  4. Sep 22, 2009 #3
    I'm not actually sure what you mean. By v(t) do you mean velocity * time? If that's the case, I'm not sure how to find velocity other than by subtracting the position at two points and dividing by the difference in time at those two points.

    For that, I got (9-0)/(2-0) for the first two points.
     
  5. Sep 22, 2009 #4

    berkeman

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    Staff: Mentor

    Subtracting and dividing is exactly what I mean. Now do it for the other points, and list the v(t) and delta t values. Then use those to calculate the acceleration value(s).
     
  6. Sep 22, 2009 #5
    (16-9)/2

    (21-16)/2

    (24-21)/2

    (25-24)/2

    Taking two of the values to determine acceleration:

    [(1/2)-(3/2)]/(2) = -0.5 units/sec

    Since 1 unit = 75 cm, this comes to -37.5 cm/s^2. Converting to m/s we get 0.375 m/s.

    Plugging in to F = ma along with 4 kg we get a value of -150 N.

    This value is incorrect.
     
  7. Sep 22, 2009 #6
    Bump

    I know I asked late, but this is due in 30 minutes, so if anyone could help that'd be wonderful!
     
  8. Sep 23, 2009 #7

    berkeman

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    Staff: Mentor

    Bumps after one hour are not allowed. Sounds like you should have worked on your homework a bit earlier. I'll try to help tomorrow, but that sounds like it will be too late for your assignment.
     
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