# Block compresses spring, find coefficient of kinetic friction

1. Nov 12, 2008

### Nick-

You push a 2.5 kg block against a horizontal spring, compressing the spring by 18 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 90 cm from where you released it. The spring constant is 200 N/m. What is the block-table coefficient of kinetic friction?

Having trouble finding the initial and final mech energy (sum of kinetic and elastic potential energy - I don't know how to find the velocity so I can't find the kinetic energy, would you set up energy equations?)

Using the spring constant I found: F=-k(d) = -200N/m (.18m) = 36 N
Elastic potential energy = 1/2kx^2 = 1/2(450N/m)(.18m)^2 = 7.29 N
The change in thermal energy = Fk(d) = (Uk)(Fn)(d)
Normal Force = mg = 24.5 N
Also, initial mech energy - change in thermal energy = final mech energy
That's pretty much all I got, any help would be much appreciated, thanks

2. Nov 12, 2008

### LowlyPion

Welcome to PF.

You know how much potential energy is in the spring just before it's released.

So what force acting over the 90 cm distance will have absorbed the PE that went into the object's KE? Won't that work equal the Potential it had when it was on the spring?