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Two questions I need help on

  1. Feb 11, 2004 #1
    one is this:

    An exceptional standing jump would raise a person 0.8m off the ground. To do this, what force must a 66-kg person exert against the ground? Assume the person crouches a distance of 0.2m prior to jumping, and thus the upward force has this distance to act over before he leaves the ground.

    The second is this:

    A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.20 and the push imparts an initial speed of 4.0 m/s?

    On the first one, I was thinking something like F=Ma. However, I guess my problem is that there isn't any acceleration when the person is jumping (their initial speed>their final speed (0)).

    On the second one, I'm pretty much clueless. Please explain these to me.
     
  2. jcsd
  3. Feb 11, 2004 #2

    ShawnD

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    With most problems dealing with forces and velocities, you can use energy formulas like Ek = EP or W = Ep or W = Ek. Always think of trying an energy formula first; they usually work, and you won't end up doing multiple steps trying to solve for one thing then another thing. With the way you wanted to solve it, you would have to run around solving this and that; it would take a while to get an answer.
    To use F = ma, you would need to find the initial takeoff velocity using Vf^2 = Vi^2 + 2ad then solve for Vi since Vf is 0. Then using that as your take off speed you would put that into the same equation Vf^2 = Vi^2 + 2ad but this time it would be Vf. Vi would be 0 and 'a' would be what you solve for.
    The energy method is much much much easier.


    work (before) = potential energy (after)

    [tex]Fd = mgh[/tex]

    [tex]F = \frac{mgh}{d}[/tex]

    [tex]F = \frac{(66)(9.81)(0.8)}{0.2}[/tex]

    F = 2589.84N


    That question is a classic, every teacher asks it every year. There are a few ways to solve it but this is the way I do it. Remember what I said before, try energy formulas first; you don't even need to draw a FBD for this.

    kinetic energy (before) = work (after)

    [tex]\frac{1}{2}mv^2 = Fd[/tex]

    [tex]\frac{1}{2}mv^2 = 0.20mgd[/tex] see how the mass cancels out?

    [tex]\frac{1}{2}v^2 = 0.2gd[/tex]

    [tex]\frac{1}{2}(4)^2 = (0.2)(9.81)d[/tex]

    d = 4.077m
     
    Last edited: Feb 11, 2004
  4. Feb 11, 2004 #3

    HallsofIvy

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    "On the first one, I was thinking something like F=Ma. However, I guess my problem is that there isn't any acceleration when the person is jumping (their initial speed>their final speed (0))."

    Just one note: the way physics uses the term "acceleration", it means any change in speed. Going from a positive initial speed to a final speed of 0 certainly does involve a (negative) acceleration. That is of course, because the force of gravity is "negative" (downward while we were taking velocity upward to be positive).
     
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