# Elastic Potential Energy of a glider

1. Sep 4, 2007

### Edwardo_Elric

1. The problem statement, all variables and given/known data
A glider with mass 0.200kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 5.00N/m. You pull on the glider, stretching the spring 0.100m and then release it with no initial velocity. The air track is turned off, so there is a kinetic friction force acting on the glider. What must be the coefficient of kinetic friction uk between the air track and the glider so that the glider reaches x = 0 position with zero speed.

2. Relevant equations
U = 1/2kx^2

3. The attempt at a solution
U1 + U2 = K1 + K2
0 + 1/2(kx^2) + Wother = 0 + 0
-1/2(5.00N/m)(.100m)^2 = Ff (0.100m)
-1/2(5.00N/m)(.100m) = Ff
Ff = -0.25N

uk = 0.25N / (0.200kg*9.8m/s^2)
uk = 0.128

Last edited: Sep 4, 2007
2. Sep 4, 2007

### learningphysics

Yes, that looks right to me... for the beginning I'd write something like:

Wother = Mech.energy_final - Mech.energy_initial

Wother = (U2 + K2) - (U1 + K1)

(where K represents kinetic energy, and U represents elastic potential energy)

3. Sep 4, 2007

thank you

4. Sep 4, 2007

no prob.