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Net Force Question

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data

    75.0kg Man decends from 3.1m above the ground. At 3.1m his feet touch the ground and he absorbs the landing through a constant acceleration of 0.6m to come to a verticle rest. What is the net force on him?

    2. Relevant equations

    Vi = √(2*a*y); a = -[(Vi)^2]/(-2*y); F = m*a; where y = -0.60m and a = -9.8m/s^2.

    3. The attempt at a solution

    Vi = √[2*a*(-0.60m)] = -7.79m/s
    a = -[(-7.79m/s)^2]/(-2*(-0.60m) = -50.63m/s^2
    F = 75.0kg*(-50.63m/s^2) = -3800N.

    I get 3800N from the ground to him, but the book has 4530N. Their answer works if we add on the absolute value of his mass multiplied by gravity, but why would they do that?
     
  2. jcsd
  3. Feb 20, 2013 #2
    We will add it because:

    Firstly ground has to apply normal reaction R equal to his weight to balance him. Secondly, in this problem the person possesses additional force ma due to the kinematic equation. To make him come to vertical rest, ground must apply net force equal to what he possessed. In this case he had net force mg+ma. By newton's third law, normal reaction should be equal and opposite to it.
     
    Last edited: Feb 21, 2013
  4. Feb 21, 2013 #3
    I get it, thanks a lot.
     
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