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Potential energy curve, turning points

  1. Jul 6, 2015 #1
    1. The problem statement, all variables and given/known data
    The figure shows a plot of potential energy U versus position x of a 0.220 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9 J, UC = 20 J and UD = 24 J. The particle is released at the point where U forms a “potential hill” of “height” UB = 12 J, with kinetic energy 7.00 J. What is the speed of the particle at (a)x = 3.5 m and (b)x = 6.5 m? What is the position of the turning point on (c) the right side and (d) the left side?

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/q40.jpg

    2. Relevant equations
    K=E-U

    3. The attempt at a solution
    I know the turning point is when K=0 but not sure how to find that from this graph.
     
  2. jcsd
  3. Jul 7, 2015 #2

    Nathanael

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    Homework Helper

    Kinetic energy is zero when the potential energy is equal to the total energy.
     
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