## Elastic and gravitational potential energy, with friction and a box on incline.

First off, long time lurker here on his first post. Your forum seems absolutely brilliant, so I hope you can help me :)

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

A box on an elastic string is dropped down an incline and proceeds to slide the distance L before being pulled back by the string and stopping at the distance L/2. At the starting point the string is slack. Find the coefficient of friction and the spring constant.

Known data:
• Distance, L
• Angle of incline, Theta
• Mass of box, m

2. Relevant equations
$$U_{{{\it el2}}}=1/2\,k{x}^{2}$$
$$U_{{{\it grav}}}=mgy$$
$$y=\sin \left( \theta \right) L$$
$$W_{{{\it res}}}=-mg\cos \left( \theta \right) \mu\,x$$
$${\it K1}+U_{{{\it grav1}}}+W_{{{\it res12}}}={\it K2}+U_{{{\it grav2}} }+U_{{{\it el2}}}$$

3. The attempt at a solution
Right, so I've basically just changed x to L plugged in my attributes into $${\it K1}+U_{{{\it grav1}}}+W_{{{\it res12}}}={\it K2}+U_{{{\it grav2}} }+U_{{{\it el2}}}$$ and solved for my variables. I did first for when the box goes from point 1 to point 2, then 2 to 3 and finally 1 to 3. All three equations give me the same answer.

k:
$$2\,{\frac {mg \left( \sin \left( \theta \right) -\cos \left( \theta \right) \mu \right) }{L}}$$
mu:
$${\frac {-2\,mg\sin \left( \theta \right) +kL}{mg\cos \left( \theta \right) }}$$

What I don't understand is why the problem is formulated as it is (why give me the information that it springs back to L/2, if I don't need it to find the constants).
Also, I was hoping that it would be possible to get equations for the constants where the unknowns doesn't figure, i.e. an equation for the spring coefficient that doesn't have mu in it and vice versa.
I've tried with all the algebra I could think of, but nothing seems to work - is there anyone who can point me to what I'm missing?
Thanks.
Simon
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
 Recognitions: Homework Help Welcome to PF, and my compliments for the really nice diagram! The trouble is you have μ and k in both expressions, so it isn't "solved". You must have had these equations at one point: mg(sinθ + μcosθ) = .5*kL and mg(sinθ - μcosθ) = .5*kL If you subtract one from the other, you will eliminate k so you have the value of μ. Sub that into one of the original equations and solve for k.
 Thank you very much, Inkscape and Maple makes writing here a lot easier :) I found the expressions you mentioned from my own, but oddly enough I ended up with mu = 0, which I doubt is correct, giving the way the problem was formulated. It'll have to do for now though :)

Recognitions:
Homework Help

## Elastic and gravitational potential energy, with friction and a box on incline.

I agree with that! Looks like the problem was cooked up to make the answers work out nicely.

I use SmartDraw myself, but I'm downloading Inkscape to see how it works. Draw programs are wonderful!