Ball dropped into oil: change in internal energy

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jamesm113
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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!
 

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jamesm113 said:
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.
Energy is conserved. There is no change in energy. Just a change in the form of energy. The answer is 0.

AM
 
0 did not work.
 
jamesm113 said:
0 did not work.
Then the question is assuming potential energy is not included in "internal" energy. What is your definition of internal energy?

AM
 
temperature rise, or kinetic energy in the wake of the ball, sound etc.
 
jamesm113 said:
temperature rise, or kinetic energy in the wake of the ball, sound etc.
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]<br /> <br /> The total height is 78 + 49 cm<br /> <br /> AM[/tex]
 
I did:
9.8*.0112(.78+.49)-.5(.0112)(1.48^2) = .1271289, the same answer i had gotten before.
 
jamesm113 said:
I did:
9.8*.0112(.78+.49)-.5(.0112)(1.48^2) = .1271289, the same answer i had gotten before.
In significant figures, this is .13 Also, what are the units?

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