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F = MA 2012 Exam # 19, 20 (Using Graphs)

  1. Jan 26, 2013 #1
    Woops, it's the 2010 exam -- my bad.

    1. The problem statement, all variables and given/known data
    They have diagrams that I'm not sure how to display; so find them at the following link:
    These are #'s 19 and 20 and they refer to the PE graph in 18.

    2. Relevant equations
    Not sure, perhaps:
    E_mec = K + U

    3. The attempt at a solution
    I'm not at all sure how to use the PE graph and a position vs time graph to find total energy or describe the motion. Also, I'm not sure how a potential energy vs position could give rise to position vs time. If anyone could direct me to the concept or link that I'm missing -- that'd be great.
  2. jcsd
  3. Jan 26, 2013 #2

    Simon Bridge

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    Use the graphs separately.
    What is the motion of the particle in the potential?
    (You did Q18 OK didn't you? So you know the forces?)

    i.e. Will the particle have a constant velocity or just hold it's position, or accelerate or what?

    Then look at the position time-graphs. What kind of motion do each describe?
    Which one matches up?

    You can do it the other way around ... look at the first graph: what is the particle doing? What happens to it's position?
    Now look at the potential vs position graph - does that make sense? (This is probably easier.)

    For Q20 - it is pretty much the same ... what is the motion of the particle? Describe it in words - is it sitting still, moving at a constant velocity, oscillating between limits, what? At what kinetic energy would the particle have t have in order to have that motion?
  4. Jan 26, 2013 #3
    So for numeral I, it makes sense since net force is 0 at 15m so acceleration is zero and this can be represented by a constant position. TRUE
    For numeral II, there should be a positive acceleration but the position vs time graph is constant; clearly FALSE.
    Numeral III, the velocity is constant - thus the acceleration is zero. This agrees with the graph. TRUE
    So 19 is both I and III. Thanks

    Here's a thought for 20:
    E_mec = K + U
    E_mec = 1/2mv^2 + U
    Let's make it easy on ourselves by finding where v = 0 (where dx/dt = slope = 0)
    Two points of this are: x = 5 m and x = -5m
    Here K = 0 so
    E_mec = U
    Tracking these points on the graph both yield:

    Wow, those questions were actually kind of easy; thanks for your insight. I guess I should really pay more attention to what I am given.
    Last edited: Jan 26, 2013
  5. Jan 26, 2013 #4

    Simon Bridge

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    Reading graphs is a skill that many people resist learning because it looks harder than it is. It is actually a good shortcut to solutions.

    i.e. #20 the graph shows the particle oscillating sinusoidally between -5 and +5 ... to find the kinetic energy that does that, draw a horizontal line through the potential energy graph and see where it intersects.

    All the difficult part is in knowing which lines to draw.
    Anyway - well done.
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