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Using energy conservation to find speed

  1. Oct 17, 2006 #1
    A stone of mass m is thrown straight up into the air with speed v_o. While the flight it feels a force of air resistance of magnitude f. Determine the speed of the stone when it hits the ground. You will need to combine energy conservation equations for the upward and downward parts of the motion.

    This is what I did...

    upward

    (1/2)m(v_o)^2 - F = (1/2)(m)(v_f)^2 + mgh


    downward

    mgh - F = (1/2)m(v_o)^2

    Am I correct so far? How do I combine them?

    The correct answer is v = (v_o) SQRT(mg-f/mg+f). How do I get there?
     
  2. jcsd
  3. Oct 17, 2006 #2

    OlderDan

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    You cannot add or subtract force and energy. You have to either equate forces with forces or energies with energies (work is like energy in that it is the mechanism by which energies are changed)
     
  4. Oct 17, 2006 #3
    Thanks for pointing that out
    Now I have ...

    upward

    (1/2)m(v_o)^2 - F(h) = (1/2)(m)(v_f)^2 + mgh

    downward

    mgh - Fh = (1/2)m(v_o)^2

    Is this correct?
     
  5. Oct 17, 2006 #4

    PhanthomJay

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    That last term in the last equation...where you have v_o....that's not correct. You are trying to find its speed as it hits ground....
     
  6. Oct 17, 2006 #5

    OlderDan

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    You need your first equation to find the maximum height. In that equation v_f is zero. Then use that h in the second equation, making the correction Jay noted.
     
  7. Oct 18, 2006 #6

    OlderDan

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    That would be true without the resistance, but not when the motion is constantly being opposed.
     
  8. Oct 18, 2006 #7
    Thanks for your help.

    -Johnson
     
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