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Maximum force exerted by floor?

  1. Oct 24, 2008 #1
    1. A 220 g ball is dropped from a height of 2.2 m, bounces on a hard floor, and rebounds to a height of 1.7 m. The figure shows the impulse received from the floor.

    What maximum force does the floor exert on the ball?


    2. Relevant equations
    To find the velocity, I used vi=[tex]\sqrt{}2gh[/tex] and vf=[tex]\sqrt{}2gh[/tex]. To find the impulse, I used F= [tex]\stackrel{}{}m(vf-vi)t[/tex]

    3. The attempt at a solution
    Vi = 6.57 m/s (I used 2.2m for h)
    Vf = 5.77 m/s (I used 1.7 m for h)
    t= .005 s
    F = [tex]\stackrel{(220)(5.77-6.57)}{.005}[/tex] = -35,200
    Last edited: Oct 24, 2008
  2. jcsd
  3. Oct 24, 2008 #2


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    Homework Helper

    Mind your units
    [tex]F_{avg} = \frac {(.220)(5.77-6.57)}{.005} = -35.2N[/tex] This is the Average Force.

    Your Force Graph indicates that it is Triangular, hence the Maximum Force is twice the Average.

    Fmax = 70.4 N
    Last edited: Oct 24, 2008
  4. Nov 7, 2008 #3
    I found this post while googling my problem just to see how others had attempted it, and I noticed another mistake here. You have to consider the direction of the momentum. One of the velocities should be negative, because they are going in opposite directions. Therefore we have 5.77 + 6.57 for Jx rather than 5.77 - 6.57. Fmax should actually be 1085.92 N here.
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