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Conservative Force Problem

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data
    A mass is placed at the top of a frictionless incline track. The bottom of the track goes into a loop. At what minimum height does the block with mass m have to be released above the ground in order to reach point b (the top of the loop).


    2. Relevant equations


    mg(hmin)=mvb2/2 +mg2R.

    3. The attempt at a solution

    The above formula shows that the minimum height has to be 2R because vb=0. I understand that Ugravity= mgh. But I don't understand why mv^2/2 was added on the right side. I am sitting here study my textbook and can't make sense of it.
     
  2. jcsd
  3. Nov 15, 2011 #2

    Doc Al

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    Staff: Mentor

    How fast must the block be moving at the top of the loop in order to maintain contact? (vb = 0 won't work.)
     
  4. Nov 15, 2011 #3
    That is from the answer key of a past exam. Maybe I stated it incorrectly. It seems right. If the block is just barely reaching that point that means that it is also stopping there which would make the velocity zero.
     
  5. Nov 15, 2011 #4

    Doc Al

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    There's a minimum speed required at the top (greater than zero) otherwise the block will leave the track before ever reaching the top.

    There's nothing wrong with that formula: it's just energy conservation. But if the answer key says that vb = 0, that's incorrect.
     
  6. Nov 15, 2011 #5
    Okay, well more importantly is that equation an example of the Mechanical Energy= Kinetic Energy+ potential energy?
     
  7. Nov 15, 2011 #6

    Doc Al

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    Staff: Mentor

    Sure. But if you actually wanted to solve for the minimum height, you'd need to input a minimum value for the kinetic energy term. (You'd solve for that term by applying Newton's 2nd law at the top of the loop.)
     
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