Spring Energy--a block slides down a frictionless incline plane A block of mass m = 13 kg is released from rest on a frictionless incline angled of angle θ = 30°. Below the block is a spring that can be compressed 2.0 cm by a force of 270 N. The block momentarily stops when it compresses the spring by 5.8 cm. What is the speed of the block when if first touches the spring? Ug = mgh, Us = 1/2kx2, K = 1/2mv2 By K = 1/2kx^2, I know the spring constant. 270 N * 0.02m = 1/2k(0.02)2. So, k = 27,000. I solved for the h in mgh--> 1/2kx^2 = mgh, 1/2(27,000)(0.058)^2 = 13 * 9.81 * h. h = 0.356 m. The equation when it first touches the spring is 1/2mv^2 = mgh. To find this new h value, I used the 30-60-90 triangle. When h was 0.356 m, the length of the relative incline is 2 * 0.356, or 0.712. So, the length of the new incline is 0.712 - the length of compression. So, 0.712 m - 0.058 m = 0.654 m, which is the length of the incline. Using 30-60-90 triangles, the relative vertical height of the incline is 0.654/2 = 0.327 m. So, if mass cancels, 1/2(v^2) = 9.81 * 0.327. v = 2.53 m/s But my online homework thing says it's wrong??