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Kinetic Energy

  1. Feb 20, 2008 #1
    A 7.0 kg bowling ball falls from a 2.0 m high shelf. Just before hitting the floor, what will be its kinetic energy? (g = 9.8 m/s² and assume air resistance is negligible)

    a. 14 J
    b. 19.6 J
    c. 29.4 J
    d. 137 J

    Now I tried using the normal Kinetic Energy KE equation KE = 1/2mv² but when I plug 7.0 in for mass, and (9.8 m/s²)(2.0m) for velocity and square it, my answer is much larger than any of the choices.

    Like 4,705 so I figure I am doing something wrong.
  2. jcsd
  3. Feb 20, 2008 #2
    i am guessing there's no initial velocity.
    use the equation that has all the knowns and final velocity to find out what the velocity is before hitting the floor.
    with that velocity, you can use KE equation.
    my answer is one of those and you should get the same
    good luck!
  4. Feb 20, 2008 #3
    There is an easier way to do that.
    What happens to all the potential energy that your system has initially when falling?
  5. Feb 20, 2008 #4

    So am I using the wrong equation? What equation do you mean that has all of the knowns?

    I am up for an easier way if there is one. Doesn't all of the potential energy change to kinetic engergy since it is falling and actually moving?
  6. Feb 20, 2008 #5
    That's correct. So do you know the equation for gravitational potential energy on earth?
    You had the right idea at first I think, but velocity does not equal [tex]ad[/tex].
  7. Feb 20, 2008 #6
    yeah you are on right track and the equation that i was talking about is...

    V^2 = Vo^2 - 2g(change in height)

    that will get you V and you can use KE.
  8. Feb 20, 2008 #7
    I think I found it in my book. Is it PE = mgy ?

    y = 0 right since it is the ground? But won't that just make potential energy equal to 0? I guess I am still a little confused. Thanks for all the help though.
  9. Feb 20, 2008 #8
    PE = mgh

    you are asked to find KE before it hits the ground so you are looking for PE at initial so y (or h) is NOT 0 ;)

    increase in PE, decrease in KE
    increase in KE, decrease in PE
  10. Feb 20, 2008 #9
    Okay so PE = 137.2 right. Now that I know the PE how do I use the equation you mentioned above to obtain the KE?

    The one I am referring to is: V^2 = Vo^2 - 2g(change in height)?
  11. Feb 20, 2008 #10
    yeah you get the samething at the end.
  12. Feb 20, 2008 #11
    So are you saying that the PE = KE? Which would mean KE = 137.2?

    Just out of curiosity if that is so, what does Vo^2 = ?
  13. Feb 21, 2008 #12
    Yes, in the end your initial potential energy will equal your final kinetic energy.

    And if you want to find [tex]v_f[/tex] now, you need only to solve for it in your kinetic energy equation, but you can still just use the equation krnhseya posted.

    [tex]v_f^2 = v_i^2 + 2a\Delta{y}[/tex]

    Alternatively you could also get velocity by solving using,
    [tex]s = v_it + \frac{1}{2}at^2[/tex]
    [tex]v_f = v_i + at[/tex]
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