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!