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Center of mass question.

  1. Apr 14, 2005 #1
    Greetz :)

    I recently found this problem and have made a few attempts at answering it but still have not found an answer i am 100% confident with, I would appreciate it if anyone who is keen to attempt it could give some advice and explanations for their reasoning in the solving of this question :)

    A student of height H jumps vertically up from the squat postion. At the top point of the jump, the student's center of mass is at a height 3h/4 from the ground. Find the average force F acting on the floor prior to the moment when the student loses contact with the floor. It is known that the when the student stands on the floor, the center of mass is at a height h/2 from the floor; in the "squat" position, the center of mass is at a height h/4 from the floor. The mass of the student is m.
     
  2. jcsd
  3. Apr 14, 2005 #2
    Some initial thoughts:

    The free fall trajectory starts from the height of h/2 (when the student is upright and ready to lose contact) to 3h/4. So, in effect the CM climbs by h/4 till the height becomes zero. So you can find the initial speed and KE = 1/2Mv^2.
    The force is applied from the height of h/4 to h/2, so it is applied over a distance of h/4. Use the Work-Energy theorem:

    [tex]W = \Delta E = \Delta (KE + PE)[/tex]

    If you find work, then it's easy:

    [tex]F_{av} = \frac{W}{\Delta h}[/tex]
     
  4. Apr 14, 2005 #3
    Yeah i see what you mean and this was one of my methods but it didn't achieve a final answer.. or a definative forumla..
     
  5. Apr 14, 2005 #4
    I don't see any point where you get stuck. It's simple!
    Just to clarify something else, the work W is the sum of the positive work done by the muscles and the negative work done by (constant) gravity. So you need to specifically determine the work done by the legs of the leaping person.
     
  6. Apr 14, 2005 #5

    lol oh my...
    sorry !!! got it now thankyou ! :D
     
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