1. The problem statement, all variables and given/known data (a)A spring is attached to a ceiling, and has a relaxed length of 25 cm. When a mass m = 0.80 kg is attached to the spring it stretches to an equilibrium length of L0 = 34 cm. (b) I lift the mass until the spring returns to its relaxed length, and then release it. When the mass returns to the equilibrium length, what is its speed? (c) After I release the mass and it falls, what is the length of the spring when the mass reaches its lowest point? 2. Relevant equations F=k*deltay .5mv^2=kinetic energy mgy=potential energy 3. The attempt at a solution .80*9.81=K*(.34m-.25m) so K=87.2 for part A Sorry for the missleading title but i realized im not sure how to do part b or c, i believe for part b i have to set mgy+.5mv^2=mgy+.5mv^2 but im not entirely sure Want to make sure im doing this right before my quiz tommorow, thanks.