# Help with spring compression

1. Nov 16, 2004

### Markrosoft

having trouble with this problem... any ideas?

A 2.798-kg block is on a horizontal surface with muk = 0.170, and is in contact with a lightweight spring with a spring constant of 775 N/m which is compressed. Upon release, the spring does 0.969 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed. What is the velocity of the mass as it breaks contact with the spring?

i need this equation: 1/2 kx^2 ... but i'm not sure where to go from here...

Last edited: Nov 16, 2004
2. Nov 16, 2004

### Sirus

What are some universal equations for work you could use here? Think about how energy is transfered from potential to kinetic when the spring is released and allowed to return to equilibrium.

3. Nov 18, 2004

### Markrosoft

i tried this:
1/2 k x^2=E .....0.969 J = 1/2 (775) (x)^2 ....x = 0.05m

but it's wrong... does anyone know what i'm doing wrong?

4. Nov 19, 2004

### Sirus

That would be correct if the surface were frictionless. Try to factor in the work done by friction. Remember that $W=Fd\cos\theta$ and $F_{kinetic~friction}=\mu_{kinetic}F_{normal}$.

Last edited: Nov 19, 2004