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Homework Help: Sum of Energies

  1. Dec 5, 2009 #1
    Ernie is holding a 2 kg ball at rest at the opening (top) of a 10 m deep well. He releases the ball so that it falls into the well. Assume the bottom of the well represents zero potential energy position. Immediately before the ball hits the bottom, the sum of it's kinetic energy and potential energy is:

    * 0
    * 196 J
    * -196 J
    * 392 J
    * -392 J

    The only 2 equations I see usable for this are:

    Umg = mgy
    K = .5mv2

    What I was thinking of doing was getting the kinetic energy, but I realized there is no velocity given, so I am not sure if I should find that, or if I don't need to bother with it. We know the initial velocity is zero.

    I also thought that using the Potential energy equation, we would get:
    Umg = (2 kg)(9.8 m/s2)(10 m) = 196 J

    but I am unsure because I don't know if the 10 m is negative because the well is under ground, or if it stays positive.

    I could use some help starting this problem. I need it by tomorrow night, but I would like to get it done a.s.a.p. Thanks for any help.
  2. jcsd
  3. Dec 5, 2009 #2
    Simply use energy conservation. The problem defines the zero potential E point as the bottom of the well. So all of the energy must be of what type when the ball hits bottom? OK, when the ball is at rest, all of the energy must be of what type? Calculate it. Finally, can kinetic energy be negative?
  4. Dec 5, 2009 #3
    When it hits bottom, I guess it would be kinetic since potential is zero?

    When the ball is at rest, it should be potential because there is no motion involved yet.

    And I don't think kinetic energy can be negative.
  5. Dec 5, 2009 #4
    Good answer. So use one of the formulas above to calculate potential energy.

    PS: This is a great reminder of why words are so important and at times misleading/confusing.
  6. Dec 5, 2009 #5
    Thanks, but I am still confused. I don't understand how I can calculate anything yet.
  7. Dec 5, 2009 #6
    The kinetic energy is zero when the ball is at rest. The total energy is then given by mgy.
  8. Dec 5, 2009 #7
    So my answer is just mgh, which is:

    mgh = (2 kg)(9.8 m/s2)(10 m) = 196 J

    That's it?
  9. Dec 5, 2009 #8
    Thats it. :cool:
  10. Dec 5, 2009 #9
    Okay. Thanks a lot. It seems so easy after you helped me, but I was really stumped on that one.
  11. Dec 5, 2009 #10
    You're welcome. Physics and math is often like that--the first time you do a problem of a certain type, it is hard. Hundredth time, much, much easier.
  12. Dec 5, 2009 #11
    True. And since I am a fan of neither, it makes it that much harder.
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