1. The problem statement, all variables and given/known data A 2.1 kg is released from rest from the bottom of a 45 degree inclined ramp. The package is attached to an ideal spring K = 35 N/m that is attached to the top of the incline causing the package to be launched up the incline. The coefficients of friction between the package and the surface are Us = 0.35 and Uk = 0.25 When the package is released, the spring is elongated 1.3m from its equilibrium position just before going up the ramp. A. What is the speed of the package when the spring reaches its equilibrium position on the way up? B. What will be the maximum compress of the spring? 2. Relevant equations Kf + Ui = Ki + Ui + Other forces Fs = 1/2kx^2 Ug = mg(sin45) Us = 1/2kx^2 3. The attempt at a solution I been trying to figure out how to set up the problem because i am unsure where Uspring and Ugravity are replacing in the conservation of energy. But here i go: Kf + Ui = Ki + Ui + Other forces 1/2mv^2 + mg(sin45) = 0 + 1/2kx^2 - fkD (.5)(2.1kg)(v^2) + (2.1kg)(9.8 m/s^2)(sin 45) = (.5)(35 N/m)(1.3m)^2 - (0.25)(2.1kg)(9.8 m/2^2)(cos45)(1.3) also once you get the vf from part A you can find Part B's maximum compress which is K by using conservation of energy and substituting vf in and solving for K correct?