# Compression of spring

1. Feb 14, 2014

### s.dyseman

1. The problem statement, all variables and given/known data

A 15.0kg stone slides down a snow-covered hill, leaving point A with a speed of 12.0m/s . There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.30N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

How far will the stone compress the spring?

Other relevant information:

Height of A = 20m
Distance between A and B (horizontal): 15m

2. Relevant equations

f=ma
Ke=.5mv^2
.5kx^2
Fr=umg

3. The attempt at a solution

I found the velocity at point b = 23.2 m/s or a KE of 4036.8 J. I then subtracted work done by friction and set it all equal to the spring compression equation:

.5*15*23.2^2-umg(100+x)=.5(2.3)x^2
4036.8-2943+29.43x-1.15x^2

However, this is off by quite a distance. Any help on this? Thanks in advance!

2. Feb 14, 2014

### collinsmark

That's generally the right approach. So you're doing well. But you do have a problem with rounding errors or something. The 4036.8 J figure is a little off. Make sure to keep plenty of significant figures in the intermediate steps (and/or keep everything in variable form until the final step).

After fixing the rounding errors, check your setup prior to the quadratic formula. See the text in red above. Are you sure that's the right sign?

3. Feb 14, 2014

### s.dyseman

Thank you very much, I found the answer at 20.4 m! You were a big help. The incorrect sign was a glaring error, how did I miss that?