Calculating Spring Compression for Stone Sliding Down Hill

[s.dyseman]

Homework Statement

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

Homework Equations

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

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!

 

[collinsmark]

Homework Statement

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

Homework Equations

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

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

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).

4036.8-2943+29.43x-1.15x^2
Quad eq: x=46.2m

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?

 

[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?