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Math Help - Speed down hill!

  1. #1
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    Smile I need help please!

    1.A skier coasts down a very smooth, 10-m-high slope. If the speed of the skier on the top of the slope is 5.0 m/s, what is his speed at the bottom of the slope?
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    Quote Originally Posted by babygirl
    1.A skier coasts down a very smooth, 10-m-high slope. If the speed of the skier on the top of the slope is 5.0 m/s, what is his speed at the bottom of the slope?
    This can either be done using the Work-(Kinetic) Energy Theorem or the non-conservative work equation. I'll do it both ways.

    For either method we need a zero for Gravitational PE. I'll do my usual and set that at the lowest point in the problem: the bottom of the slope in this case. As always when working with the GPE, +y is upward. Note that we don't need to know the path the object has travelled!

    Work-Energy Theorem.
    W = \Delta K
    What's doing work in this problem? The problem doesn't mention friction so 99 times out of 100 that means we ignore friction. (See also the words "very smooth" in the problem.) The object's weight is the only force here doing work. (The normal force never does work.) Since the weight is a conservative force (gravity) we can write the work done by it in terms of the (gravitational) PE: W = -mg \Delta h. (Note the "-" sign. The work done by any conservative force is always W = - \Delta PE ).

    So: W = -(mgh - mgh_0) = (1/2)mv^2 - (1/2)mv_0^2
    Here h = 0 m, h0 = 10 m, v0 = 5.0 m/s, we don't know m and we're looking for v. So solve for v:
    v^2 = \frac{(1/2)mv_0^2 + mgh_0}{(1/2)m} (Notice that the "m" cancels!)

    Finally: v = \sqrt{v_0^2+2gh_0}.

    Non-conservative Work Equation:
    W_{nc} = \Delta E

    Here Wnc is the work done by all non-conservative forces. There are no non-conservative forces in the problem (friction is non-conservative) so Wnc = 0 J. E, of course, is the total mechanical energy of the object in question, which is just E = K + PE = (1/2)mv^2 + mgh. So:
    W_{nc} = 0 = E - E_0 = \left ( (1/2)mv^2 + mgh \right ) - \left ( (1/2)mv_0^2 + mgh_0 \right )

    When we solve this equation for v we get the same answer as in the Work-Energy Theorem case.

    -Dan
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