An inclined plane of angle θ = 20.0° has a spring of force constant k = 525 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 = 2.31 kg is placed on the plane at a distance d = 0.312 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? I used equation 1/2mv^2+mgh=1/2kx^2 and got x=.084, which was incorrect.