# Block on Spring with Friction

1. Sep 24, 2008

### jspek9

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

A relaxed spring with spring constant k = 70 N/m is stretched a distance di = 63 cm and held there. A block of mass M = 7 kg is attached to the spring. The spring is then released from rest and contracts, dragging the block across a rough horizontal floor until it stops without passing through the relaxed position, at which point the spring is stretched by an amount df = di/9.

What is the coefficient of kinetic friction µk between the block and the floor?

The formula I'm trying to use is Ff = μFn

What I cant understand is Ff. Its defined as "the force exerted by friction". But what does that mean and how can I solve for it? The problem wants me to use the Work Energy Theorem but I cant piece it all together in my head and I just cant seem to understand constants as a whole anyways so I could really use some insight here.

Thanks as always

2. Sep 24, 2008

### ||spoon||

firstly, the "force exerted by friction" always acts opposite to the direction of motion.

To solve the problem think about how much potential energy the spring had initially and how much it has finally. Where has the extra energy gone? (remember energy is conserved)

-Spoon

3. Sep 24, 2008

### jspek9

OK I've been looking at my potential energy formulas and I don't see the connection.
I cant figure out what in the world I'm missing.

On a side note just curious but is Spoon for Spoon Sports?

4. Sep 25, 2008

### ||spoon||

Ok, the change in energy of the system should be negative, meaning you should have less potential energy than when you started. Where did the energy go, because it must be conserved?

Well the system had to overcome the frictional force to move, I.e the system did work.

Remember that work = force * distance (applied).

And no, I've never heard of spoon sports, just a nickname I acquired a few years ago :)

Hope that helps without giving it all away

-spoon