1. The problem statement, all variables and given/known data A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at angle θ = 40 °. A 1.6 kg block is projected up the plane, from an initial position that is distance d = 0.80 m from the end of the relaxed spring. With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by 0.50 m? 2. Relevant equations Ui + Ki = Uf + Kf U(elastic)=.5kx2 U(gravitational)= mgy 3. The attempt at a solution So I assumed that Ui=0 because I'm assuming the block starts at the bottom of the plane. I also assumed that because the block is momentarily stopping that Kf=0 so that leaves me with Ki=Uf and solving for Uf will give me Ki Uf= U(elastic) + U(gravitational) Uf = .5(200N/m)(.5m)2 + (1.6kg)(9.8m/s2)([.5m+.8m] sin 40) Uf= 38.1026J therefore Ki should equal 38.1026 However, that answer is refused so I would like to know what I am doing wrong. Thankyou.