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When accelerating upward off a trampoline...

  1. Aug 9, 2016 #1
    1. The problem statement, all variables and given/known data
    When accelerating upward off a trampoline, how would you depict this on an acceleration-time graph? I know kinematics is the study of motion without a strict regard to the forces that are capable of changing motion, but according to Newton's 2nd law, an object accelerates in the direction of the net force. Thus, when you accelerate upward off the trampoline at the very beginning in a very small fraction of time, this force would have to overcome the person's force of weight right? So my question is this: would you display this initial acceleration as being greater than 9.8 m/s^2 or would you show the net acceleration on the acceleration-time graph (that is, would you put a number like 10 m/s^2 or 0.2 m/2^ respectively). I guess the same applies when you land back on the trampoline and accelerate in the positive direction to slow down... I would be grateful if someone could explain this to me. Thanks!

    2. Relevant equations
    F=ma and kinematic equations for free fall.

    3. The attempt at a solution
    I can't decide on the acceleration offhand...
     
  2. jcsd
  3. Aug 9, 2016 #2

    jbriggs444

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    What is "the very beginning" of the event? Is it when you are moving downward and your feet are just barely touching the trampoline? Is it when the dent your feet are making in the trampoline at its maximum depth and you have just come to a stop before starting back upward? Or is it when your feet are about to leave the trampoline?

    Pick one and then figure out what forces apply at that event and what their relative magnitudes are.
     
  4. Aug 9, 2016 #3
    When your feet are about to leave the trampoline and you accelerate upward.
     
  5. Aug 9, 2016 #4

    haruspex

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    What force is causing you to accelerate upwards when you are about to lose contact with the trampoline?
     
  6. Aug 9, 2016 #5
    This is very similar to what happens with a mass and spring in simple harmonic motion. Do you know how to analyze that situation? If so, when the mass is at its lowest point and has stopped moving downward, in terms of the spring constant k and the displacement x from the equilibrium position, what is the net force on the mass (and in what direction)? If m is the mass, what is its acceleration?
     
  7. Aug 9, 2016 #6
    Im not sure exactly what force, but Im guessing the elastic force of the trampoline when you push down initially. This is not a question that requires an exact acceleration, as it is meant to be a simple sketch of an approximation of what the graph should look like. I was just curious whether the acceleration you would put on the graph should be higher or less then the acceleration of gravity (i.e., the total acceleration without regard to gravity or the net acceleration). Sorry if this is a bad question, as it isn't really asked in the original question. I was just curious. I'm still in high school so my physics skills are not robust.
     
  8. Aug 9, 2016 #7

    rcgldr

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    When the trampoline and the person is at the lowest point, the acceleration can be 3 g upwards or more (so the force corresponds to 4+ g). When the person is just about to leave the surface of the trampoline, the upwards force from the trampoline is less than the force from gravity, so the acceleration transitions to downwards a bit before a person leaves the surface of a trampoline.
     
  9. Aug 9, 2016 #8

    haruspex

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    If you are just about to lose contact with the trampoline, it is amost flat, so there will be hardly any force. Meanwhile, another force is acting in you. Which way would the net force be at this point?
     
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