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Real and Point Particle Systems of a jumper

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A jumper of mass m=71 kg starts in a crouched position with the jumper's center of mass at a position y1=0 m. Just before the jumper leaves the floor, his or her center of mass is at y2=0.5 m. From another measurement, you have found that the velocity of the jumper's center of mass has only a y component of v2=1.8 m/s as the jumper's feet leave the floor.

    Use your results from this lab to calculate how strong the floor must be to support this jump.

    Ffloor = ?


    Now assume Ffloor changes while the jumper goes from the crouched to the lift-off position. While the jumper's feet are in contact with the floor assume the force decreases linearly or

    Ffloor(y) = -C[y-(y2-y1)],

    where y is the changing position of the center of mass, and C is a positive constant. Note that when y=y2-y1 this force goes to zero and, of course, once the jumper's feet leave the floor this force of the floor on the jumper must be zero.

    For the initial conditions above, find the maximum value of Ffloor and so the required strength of the floor.

    Ffloormax = ?

    Hint: You will have to integrate the force over the displacement to find the work and from that and the energy principle you can find the constant C. Once you know C it is straightforward to find the maximum value of Ffloor.


    2. Relevant equations
    W=-mg(y2-y1)
    Ktrans=1/2mv^2
    F=(mg(y2-y1)/(2y-y1)) + mg.....don't know if this is right?

    3. The attempt at a solution
    Thought it was just m*g, but it's not.
     
  2. jcsd
  3. Mar 29, 2009 #2

    LowlyPion

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    For one thing the energy that needs to be imparted from the work of jumping is not only the change in potential energy, but also the change in kinetic energy isn't it?

    W = m*g*h + 1/2*mv2
     
  4. Mar 30, 2009 #3
    I added in the kinetic energy into the solution and it said it was wrong for the force of the floor.
    W =71*9.8*.5+.5*71*1.8^2=462.92N...isn't this right?
     
  5. Mar 30, 2009 #4

    LowlyPion

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    That's the work, but what about the Force? Force is Newtons. Work is Joules.
     
  6. Mar 30, 2009 #5
    Wouldn't force just be m*g*h?
     
  7. Mar 30, 2009 #6

    LowlyPion

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    Wrong units.

    Joules are the units of work

    Work = F*d
     
  8. Mar 31, 2009 #7
    Ffloormax=C * y2
    C= ((.5*m*v2+m*g*y2)/(y22))*2
     
    Last edited: Mar 31, 2009
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