(adsbygoogle = window.adsbygoogle || []).push({}); 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 = 1/2kx^{2}, K = 1/2mv^{2}

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??

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Spring Energy-a block slides down a frictionless incline plane

**Physics Forums | Science Articles, Homework Help, Discussion**