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Conservation of energy questions

  1. Jun 5, 2006 #1
    A ball of mass 11.2 g is dropped from rest at a height of 78 cm above the surface of oil that fills a barrel to a depth 49 cm. The ball reaches the bottom of the barrel with a speed of 1.48 m/s. What is the change in the internal energy of the system ball + oil?
    So, I used the change in Eint = -change in U - change K.

    For change in U, I did mg(Hf)-mg(Hi)=.0112(9.8)(0)-.0112(9.8)(.49) = -.0537824
    For change in K, I did m(Vf)^2/2-m(Vi)^2 = .0112(1.48)^2-.0112(3.909987)^2 = -.07334655

    I then subtracted did -change in U - change in K = --.0537824--.07334655 = .12712895. However, this is not the correct answer. :-/

    Thanks!
     
  2. jcsd
  3. Jun 5, 2006 #2

    Andrew Mason

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    Energy is conserved. There is no change in energy. Just a change in the form of energy. The answer is 0.

    AM
     
  4. Jun 5, 2006 #3
    0 did not work.
     
  5. Jun 6, 2006 #4

    Andrew Mason

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    Then the question is assuming potential energy is not included in "internal" energy. What is your definition of internal energy?

    AM
     
  6. Jun 6, 2006 #5
    temperature rise, or kinetic energy in the wake of the ball, sound etc.
     
  7. Jun 6, 2006 #6

    Andrew Mason

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    Then the increase in energy of the oil + ball is the change in potential energy of the ball less its kinetic energy:

    [tex]\Delta E = mg(h_{air} + h_{oil}) - \frac{1}{2}mv^2[/itex]

    The total height is 78 + 49 cm

    AM
     
  8. Jun 6, 2006 #7
    I did:
    9.8*.0112(.78+.49)-.5(.0112)(1.48^2) = .1271289, the same answer i had gotten before.
     
  9. Jun 6, 2006 #8

    Andrew Mason

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    In significant figures, this is .13 Also, what are the units?

    AM
     
  10. Jun 6, 2006 #9
    Units are in joules. We've never had to do sig figs before.
     
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