1. The problem statement, all variables and given/known data A ball is dropped from a height 'h' above a tube containing oil. Describe what happens to the GPE when the ball has reached terminal velocity in the oil. 2. Relevant equations KE = 1/2mv^2 GPE = mgh 3. The attempt at a solution When the ball is dropped,if we ignore air resistance then GPE will be converted directly into KE. When it strikes the surface of the oil, the ball's velocity will decrease which means the KE will also decrease. There is also still GPE which is not going into KE. So I think that both the decrease in kinetic energy and GPE both go into doing work against the drag force. When the ball reaches terminal velocity the KE will no longer change so there will be no more contribution from the kinetic energy, however GPE is still being lost which will go into work against the drag force. I am confused as I checked the answer for this question in my book and it just says "at terminal velocity GPE of the ball is converted into thermal energy". Surely GPE is being converted into both thermal energy and energy from doing work against the drag force or are these two the same? i.e. is heat lost due to friction with the oil the same as work done against the drag force?