1. The problem statement, all variables and given/known data The radius of curvature of a loop-the-loop for a roller coaster is 14.0 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.35mg. Find the speed of the roller coaster at the top of the loop. (Enter your answers from smallest to largest.) (This question has two answers, one of which, I found = the speed at the top of the loop. I'm guessing the 2nd answer is the speed at the bottom of the loop?) 2. Relevant equations ∑Fr = mar Fsp + mg = m(v2/R) a = v2/R 3. The attempt at a solution Fsp = m[(v2/R) - g] .35mg = m[(v2/R) - g] .35g = (v2/R) - g 1.25g = v2/R v_top = √(gR1.35) ≈ 13.62 m/s Now to find the 2nd answer, I think I need to calculate a (since I believe this is constant): a = v2/R ≈ 13.25 m/s2 But this can't be correct I'll just get the same answer for velocity. Does anyone know how I am supposed to find the 2nd answer?