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Transformation of Kinetic and Gravitational Potential Energy:Conservation of Energy.

  1. Nov 20, 2007 #1
    1. The problem statement, all variables and given/known data

    How far would a 1.00 Kg ball have to fall freely to reach a speed of 100 Km/h ?

    2. Relevant equations

    [tex] \Delta E_g = \Delta E_k [/tex]

    3. The attempt at a solution

    [tex] \Delta E_g = \Delta E_k [/tex]

    i then expanded to:

    [tex] mgh' - mgh = \frac{1}{2}m(v')^2 - \frac{1}{2}mv^2 [/tex]

    i got rid of the 1/2's and the mass

    and 0'd out the equatiosthat would equal zero…

    [tex] 2gh' = -(v')^2 [/tex]
    [tex] h' = \frac{-(v')^2}{2g} [/tex]

    subbed in the variables and ot 5.1 x 10^2 J.. which is wrong..

    appreciate any help... thanks.
  2. jcsd
  3. Nov 20, 2007 #2
    you got height in terms of Joules???
  4. Nov 21, 2007 #3
    the answer i got was 5.1 x 10^2 Joules, if thats what your asking....
  5. Nov 21, 2007 #4
    Joules is the unit of work, not height (the answer you have to calculate in #1 is height). Even if you talking about the energy content of the ball, its not 5.1 x 10^2 J.
  6. Nov 21, 2007 #5
    sorry i meant meters, im actually not sure if its is meters... m/s / N/kg?

    anyways, so lets go with meters. stil cant solve the answer..
  7. Nov 21, 2007 #6
    :confused: still I don't understand why you are not getting the answer. You have the equation with you, just plugin the values in your last equation and get the answer (I got 39.4 m)
    Last edited: Nov 21, 2007
  8. Nov 21, 2007 #7
    howd you get that? i still get 510....
  9. Nov 21, 2007 #8

    [tex] h' = \frac{-(v')^2}{2g}[/tex]
  10. Nov 21, 2007 #9
    Check your units... you plugged in v as 100 kph and g as 9.8 m/s[tex]^{2}[/tex]
  11. Nov 21, 2007 #10
    oh shii, unit conversion... sorry....
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