Conservation of energy questions

[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!

 

[Andrew Mason]

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

 

[jamesm113]

0 did not work.

 

[Andrew Mason]

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

 

[jamesm113]

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

 

[Andrew Mason]

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]

 

[jamesm113]

I did:
9.8*.0112(.78+.49)-.5(.0112)(1.48^2) = .1271289, the same answer i had gotten before.

 

[Andrew Mason]

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

 

[jamesm113]

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