Hey guys. We were given a problem, which was to consider bv^2 as the force acting upwards on a body falling in water. We were asked to find the depth at which a body entering the water at 5 * ve (terminal speed) would end up with a speed of 1.1 * ve. Starting from -bv^2 + mg = ma, the first step is to linearize the non-linear expression bv^2 at the equilibrium point (ma = 0). The equation at the chosen equilibrium point becomes b(ve)^2 = mg The linearization gives: b(ve)^2 + 2b(ve)(Δv). Substitution into the original equation gives -b(ve)^2 - 2b(ve)(Δv) + mg = ma. We know that at the chosen equilibrium point, b(ve)^2 = mg. Thus, mg - b(ve)^2 = 0. I can remove these 2 expressions in the above equation to get the linearized equation at the equilibrium point ve: Linearized upward force at terminal speed = 2b(ve)(Δv) = ma Also our teacher showed us that the 2b(ve) expression must be used to find the value of the linearized ve point. Thus, using the equilibrium equation above but substituting b(ve)^2 by the new constant coefficient 2b(ve) which multiplies ve, we get 2b(ve)*ve = mg -> ve = sqrt(mg/2b) Given the above equation in bold, I was able to solve for height h by using the following substitution: a = v * dv / dh Using this substitution I could integrate on both sides of the above equation and solve for h pretty easily. I confirmed with my teacher that our answer was correct. Here's my question: I would like to solve this problem by using the conservation of energy equation instead of solving the F = ma equation. Before my teacher confirmed that we had the right answer, I wanted to solve the problem using this other method because I could've verified my answer by myself. However, I don't know how to solve this. I put the initial energy + the work on the left hand side, and the final energy on the right hand side. Assuming the object falls into the water at a height h at an initial speed of 5ve, V1 = mgh T1 = m/2 * (5ve)^2 Work = 2b(ve)(Δv) * h V2 = 0 T2 = m/2 * (1.1ve)^2 V1 + T1 - Work = V2 + T2 I don't know what to do with the expression in bold, for the work of the linearized force. Obviously an integral is needed, but if I integrate as I specified above, I have to divide both sides by Δv since dv is on the right side of the equation. I end up with 2b(ve)h = stuff. The left side of this equation is NOT equal to the Force times acceleration because the force has a Δv on the left side. So, how can I find the "correct" work to put into the equation?