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)