1. The problem statement, all variables and given/known data Hello, folks. I know this same problem has been answered before here but the previous responses are just not doing it for me, sorry! An inclined plane of angle θ = 20.0° has a spring of force constant k = 495 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.37 kg is placed on the plane at a distance d = 0.318 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest? Here is an image of such a setup. http://imgur.com/5JF0F 2. The attempt at a solution So basically I want to find the displacement in the spring given these values. What I do first is get the velocity at which the box hits the spring so I can get it's kinetic energy at that point. It starts at .750m/s but it's acceleration due to gravity is 9.8*sin(20 degrees) since it's on an incline and it has to travel .318m to hit the spring so I've got: x_final=x_initial+v_initial*t+1/2*a*t^2 .318=0+0.75*t+1/2*(9.8*sin(20 degrees))*t^2 t=0.26951 seconds now that we have time let's get final velocity. v_final=v_initial+a*t v_final=0.75+9.8*sin(20 degrees)*0.26951 v_final=1.653 m/s Now let's get the kinetic energy of the block when it hits the spring. Kinetic=m*v^2/2 2.37*1.653^2/2=3.238J Now let's get the potential energy on the spring U_spring=1/2*k*(x_final^2-x_initial^2) U_spring=1/2*495*(x^2-0) 495/2*x^2 J now let's solve 3.238=495/2*x^2 x=0.114m. This is incorrect. Where have I gone wrong? Thanks!