1. The problem statement, all variables and given/known data An inclined plane of angle θ = 20.0° has a spring of force constant k = 460 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 = 2.51 kg is placed on the plane at a distance d = 0.324 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? 2. Relevant equations K = 1/2 mv^2 Elastic Potential energy of spring = 1/2 kx^2 3. The attempt at a solution So first I found the acceleration of the object: gsin(θ) = a Then I found the velocity of the mass at the time it hits the spring d = vt +1/2 at^2 t is about .269593 so velocity when the object hits the spring = v+at which is about 1.654 Then I had: 1/2 mv^2 = 1/2 kx^2 solving for x, I got about 0.122 meters However, the website to send the answer says I'm wrong. I have looked this up online, and I have substituted my values into other peoples equations and I still get 0.122 Thanks in advance!