Potential and Kinetic Energy: Down an incline with spring

[jti3066]

Homework Statement

A 4.1 kg block starts at rest and slides a distance d down a frictionless 26.0 deg incline, where it runs into a spring. The block slides an additional 18.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 428 N/m .

[URL]http://loncapa4.fsu.edu/enc/64/444d8fc39f68b5d50a0ae69506afa18b07b17f4630544f446b909dc06ad62e4ccad64d37e395dbf1b8bce7f718c3ca893e7291ecae8bce02527ece77d5a672e050dafc3a7d01eb0880c83b19b1ecf1e46bc159a279e62c69.gif[/URL]

What is the value of d?..(answered)

What is the distance between the point of first contact and the point where the block's speed is greatest?...(need help)

Homework Equations

U = mgh

U = .5k(x^2)

The Attempt at a Solution

I found the answer to the first question by using the above relevant equations:

U = mgh = .5k(x^2)

4.1 * 9.81 * h = .5 * 428 * (.18^2)

h = 0.172 m

then some trig..

sin(26)=.172/(.18 + d)

d = .213 m

As far as how to solve the second question I am completely lost. At the point of contact with the spring the acceleration should be zero correct. From there would I need to find the intial velocity? We know that V_f = 0, but how to use these "known" quantities in relationship to the block being slowed by the spring is confusing me. I think I'm making this problem more complicated than it really is. TIA for your help.

 

[jti3066]

I tried using:

mgh = F_net * delta(x)

0.172 = (mgsin() - .5k(x^2))(.18 - x)

x = .162 m

This approach did not work...

 

[jti3066]

Ok, I found with some help that mgsin()=kx' will get me the correct answer...however I am still confused where this came from...