# Energy and a spring

1. Oct 11, 2005

### ViewtifulBeau

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.

2. Oct 11, 2005

### zwtipp05

You found the vertical distance h that is traversed, what they are asking for the the distance d along the incline it travels. You have the y-component of the distance, now you have to calculate the hypoteneuse which is d.

You did everything right, you just didn't take the final step.

3. Oct 11, 2005

### ViewtifulBeau

so i divided 1.011 by sin 35.6 but it is still not right. (1.7367)