A spring of spring constant k = 340 Nm-1 is used to launch a 1.5-kg block along a horizontal surface by compressing the spring by a distance of 18 cm. If the coefficient of sliding friction between the block and the surface is 0.25, how far does the block slide? Well i have: W=1/2kx^2 W=55080J W=U(potential energy) so i know the potential energy of the spring is 55080 Energy is conserved..so after the spring has decompressed all the energy is in the block..which is now kinetic K=1/2mv^2 55080=1/2(1.5)v^2 so Vi=406.5m/s ?? that seems pretty darn high, this is just the initial velocity right? so Vf=0 friction = 3.7N F=ma Vf^2=vi^2+2ax I assume i'm using these equations..but am not sure how i arrise to get accel or if i'm on the right track..