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A falling man

  1. Mar 31, 2008 #1
    [SOLVED] A falling man

    1. The problem statement, all variables and given/known data
    a 90kg man falls (freefalls) 10m, when he lands, his legs buckle and he goes down another 23cm. What was the average force he exerted on the ground?


    2. Relevant equations

    M = 90k
    a = 9.8 m/s^2


    3. The attempt at a solution

    from freefall to ground

    F(g) = 90kg (9.8m/s^2)
    F(g) = 882N

    buckling

    F(g) = 90kg (9.8m/s^2)
    F(g) = 882

    (882 + 882)/2 = 882

    average force = 882?

    Im confused, does the distance have anything to do with the question? Because F = ma, so as long as its accelerating but the distance does not matter.

    but this seems a bit too easy, so im pretty sure i did something wrong =/
     
  2. jcsd
  3. Mar 31, 2008 #2
    Have you tried to solve this by using the law of conservation of energy? Gravitational potential energy can be transferred to work. You know that [tex]W=Fs[/tex]. F is the force that is stopping the man and he exerts the same force on the ground (third Newton's law).
     
  4. Mar 31, 2008 #3
    we havent learned that yet...

    is it possible to only solve this problem with F=ma and the kinematic equations
     
  5. Apr 1, 2008 #4
    What you calculated as the average force is actually just his weight in newtons.
    How fast is he falling when his feet hit the ground?
    Then he accelerates back to zero m/s over a distance of 23cm.
     
  6. Apr 1, 2008 #5
    ahh, ic. Thank you montoyas and Dr. Jekyll. I understand it now.
     
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