# Compression of spring

**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

Quad eq: x=46.2m

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

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