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Physics - conservative forces

  1. Nov 1, 2005 #1
    2. at t=0 the particle is at rest x=12. after is release it can move under influence of conservative force whose potential energy is shown in the diagram. what is the position of the left turning point for this particle?

    ans: At the moment, Im not sure how to convert his graph to visualize the position of the particle, in order to find the left turning point. All i see is at x=12, the U is 4J. what can i do with this?

    attached is the diagram.
     

    Attached Files:

  2. jcsd
  3. Nov 2, 2005 #2

    ZapperZ

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    Use conservation of energy. At x=12, it has all of its energy in the form of PE, and no KE. Upon release, as it is moving through the different positions, it will have a combination of PE and KE (think of a rollercoaster). As long as PE (U in that diagram) is less than the original value, there will be KE and the particle will continue to move. There is only one location in that graph where the particle will regain all of the original PE. At this point, it will no longer have any KE and it will momentarily stop, before reversing its motion. That's the turning point.

    I don't think I need to tell you where this point is, do I?

    Zz.
     
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