A 21.4 kg mass starts from rest and slides a distance d down a frictionless 35.6 degree incline where it contacts an uncompressed spring of negligible mass, as shown. The mass slides an additional distance 0.2413 m as it is brought momentarily to rest by compressing the spring (force constant k = 4245 N/m). Find the initial seperation d between the mass and the spring. i used 1/2 kx^2 to find the PE of the block after being compressed to be 123.584 J. then i used mgh to find h. so 123.584/(35.6*21.4) = 1.011 m... but its not right. what did i do wrong? thanks.