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One Dimensional Kinematics: Force

  1. Jan 29, 2009 #1

    TG3

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    1. The problem statement, all variables and given/known data

    A .3 kg ball is compressed a maximum of 0.6 cm when it strikes the floor at 9.29 m/s. Assuming acceleration is constant, what is the force the ball exerts on the floor?

    2. Relevant equations

    vf^2 = v0^2 + 2A(x-x0)
    Once I find A it will be easy, since
    F=MA

    3. The attempt at a solution

    0^2 = 9.29^2 + 2A (.006)
    0 = 86.3041 + .012 A
    -86.3041= .012A
    -7192= A
    F=MA
    F=.3 (-7192)
    F= -2157
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 29, 2009 #2

    Delphi51

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    Homework Helper

    That looks good to me! You have assumed constant acceleration, which probably isn't really right but which is probably a standard assumption in your course.

    I actually did it a different way. I made a sketch of a v vs t graph, a straight line going from 9.29 at time 0 to zero at time t. The area under a v vs t graph is the distance .006. Using the area formula I was able to find the time t it takes for the ball to compress and stop. Then I used the idea that the slope on the v vs t graph is the acceleration. I got the same answer you have.
     
  4. Jan 29, 2009 #3

    TG3

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    I found the "correct" answer: the computer wanted me to add the force due to gravity (.3 x 9.81) to the force exerted by the floor. This seems a bit conceptually shaky to me, but the computer said that was the correct answer. For my own future knowledge: is it, or was my first answer correct?
     
  5. Jan 29, 2009 #4

    Delphi51

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    Homework Helper

    Oh dear, the computer is right! I forgot about the weight. Sorry.
     
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