[SOLVED] conservation of E w/ friction a block is released from rest at height d = 58 cm and slides down a frictionless ramp and onto a first plateau, which has length d and where the coefficient of kinetic friction is 0.54. If the block is still moving, it then slides down a second frictionless ramp through height d/2 and onto a lower plateau, which has length d/2 and where the coefficient of kinetic friction is again 0.54. If the block is still moving, it then slides up a frictionless ramp until it (momentarily) stops. Where does the block stop? If its final stop is on a plateau, state which one and give the distance L from the left edge of that plateau. If the block reaches the ramp, give the height H above the lower plateau where it momentarily stops. (a)The block stops on the: 1 - First plateau; 2 - Second plateau; 3 - Ramp. Give the number of the correct answer. (b) Give L or H, which ever is appropriate. Part a) I know it will not stop on the ramp and if it was frictionless on the horizontal surfaces it would end up on the lower plateau. I don't know how to figure the kinetic friction into the equation. The mech energy is constant, so ke and pe will change based on the height and angle and the kf will factor in there somewhere. Initial PE is mgd I have no mass, or angle??? deltaEth = Fkd = PEmgd Emech2=Emech1-Fkd KE = deltaPE + PEmgd = mg(d+PEd)? I am not understanding how to set this up to find where it stops except it is where all energy has been lost to friction.