Conservation of energy questions

1. Jun 5, 2006

jamesm113

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. Jun 5, 2006

Andrew Mason

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

AM

3. Jun 5, 2006

jamesm113

0 did not work.

4. Jun 6, 2006

Andrew Mason

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

AM

5. Jun 6, 2006

jamesm113

temperature rise, or kinetic energy in the wake of the ball, sound etc.

6. Jun 6, 2006

Andrew Mason

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

7. Jun 6, 2006

I did:

8. Jun 6, 2006

Andrew Mason

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

AM

9. Jun 6, 2006

jamesm113

Units are in joules. We've never had to do sig figs before.