1. The problem statement, all variables and given/known data Starting from rest an 11kg block slides 11.9m down a frictionless ramp (inclined at 30° from the floor) to the bottom. The block then slides an additional 26.3m along the floor before coming to a stop. The acceleration of gravity is 9.8 m/s^2. Find the speed of the block at the bottom of the ramp. 2. Relevant equations Vf^2=vi^2+2a(d-do) Conservation of energy: Ebefore=Eafter KE=.5mv^2 PE=mgh 3. The attempt at a solution I tried solving this two differant ways, and both gave me the same answer, which was wrong. First I tried kinematics. Vf^2=vi^2+2a(d-do) Vf=sqrt(2*9.8/sin30*11.9) (I used 9.8/sin30 because 9.8 is the vertical component, so 9.8/sin30 is the acceleration down the ramp) Vf=21.598 The other way I tried used conservation of energy. h=11.9/sin30=23.8 PE=KE mgh=.5mv^2 gh=.5v^2 v=sqrt(9.8*23.8/.5) v=21.598 I don't see where I went wrong in my logic, so I would greatly appreciate some guidance. Thank you!