1. The problem statement, all variables and given/known data A freight car loaded with ore is sitting on a 5o incline when its brakes fail. After traveling 150 m on a frictionless track, it reaches level ground where it strikes a massive safety spring with a spring constant k = 18,500 N/m. When its velocity has decreased to zero, an automatic latch will catch the car. The equation that can be solved for the car's kinetic energy just before striking the spring is a. (1/2)mv2f + mghf - (1/2)kx2f = (1/2)mv2i + mghi - (1/2)kx2i. b. (1/2)mv2f + mghf = (1/2)mv2i + (1/2)kx2max. c. (1/2)mv2f + mghf - (1/2)kx2max = +mghi. d.(1/2)mv2f + mghf = (1/2)mv2f + mghf - (1/2)kx2f = (1/2)mv2i. e. (1/2)mv2f + mghf = (1/2)mv2i + mghi. 2. Relevant equations U = mgh Potential energy of a spring = (1/2)kx2 KE = (1/2)mv2 3. The attempt at a solution I'm hoping it is right to assume the answer of interest to the question specifically is (1/2)mv2f as that would be the car's kinetic energy just before striking the spring? (assuming that vf here is referring to the velocity of the car right before it hits the spring and nothing else) I'm not quite understanding where to relate the potential energy of the spring into this. I would assume the max potential energy of the spring from compression would be less than the potential energy from the top of the incline. Wouldn't (1/2)mv2i be 0 assuming the velocity here is the initial velocity when the car's brakes fail? Or is this problem simple with the answer being E since the spring is at the bottom of the incline and it would make sense for (1/2)mv2f to be equal to mghi.