Expert: Daniel Mazur - 11/21/2008

Question
a 1.5 block sliding on a rough horizontal surface is attached to One end of a horizontal spring (k=2000 N/m) which has its other end fixed. If this system is displaced 20 cm horizontally from the equilibrium postion and released from rest, the block first reaches the equilibrium postion with a speed if 2.0 m/s. What is the coefficient of kintetic friction between the block and the horizontal surface on which it slides.

Answer
Hi,
this looks like a test or homework question, but for once I can answer it. It will be in symbolic terms only, though, because there are units missing in "a 1.5 block".
The potential energy at full extension is
Ep = 0.5*k*x^2  .
This potential energy transforms into kinetic energy and heat, the latter is through friction. The remaining kinetic energy at the equilibrium position is
Ekrem = 0.5*m*v^2  .
The difference between these two energy values was lost in dissipation (friction), which has this dependence
Ep - Ekrem = Ef = Ff*x = f*m*g*x  ,
where Ff is friction force (constant over the whole length x) and f is the coefficient of friction. The other symbols are obvious, I hope.
Expressing the coefficient of friction gives

f =(Ep - Ekrem)/(m*g*x)=(k*x^2 + m*v^2)/(2*m*g*x)

This is the result, use it well. I hope next time I will be allowed to help you with some deeper physics.
Cheers,
Daniel