# Spring Problem

1. Oct 25, 2006

### cd80187

You push a 4.6 kg block against a horizontal spring, compressing the spring by 26 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 83 cm from where you released it. The spring constant is 330 N/m. What is the coefficient of kinetic friction between the block and the table?

So to start, I have figured out the potential energy of the spring when it is compressed, and it is .5 (330 N/m)(.26 m) squared = 11.154 J But after that, I am not really sure what to do. I tried something with force x distance, but I cannot remember what I did. But I am just unsure where to start on this one. Help would be great

2. Oct 25, 2006

### Hootenanny

Staff Emeritus
HINT: Think about conservation of energy.

3. Oct 25, 2006

### cd80187

I know, but I'm trying to figure out what is actually there. I know that there is the max kinetic energy when the block hits the spring, but no potential energy (I'm taking the uncompressed spring to be height 0), and when the spring compresses, it is the gravitational potential energy and spring potential energy that are not 0, right?? so in the end, it should be (1/2)(m)(v)squared = 1/2 (k)(x)squared plus (m)x(g)x (y) where y and x are the distance compressed by the spring?

4. Oct 25, 2006

### Hootenanny

Staff Emeritus
Your almost correct there, but since the spring and table are horizontal the gravitational potential of the block is constant, so now you have;

$$\frac{1}{2}mv^2=\frac{1}{2}kx^2$$

Edit: well it seems latex is down at the moment, so: 1/2mv2=1/2kx2

Do you follow?

5. Oct 25, 2006

### cd80187

Not really. So I have the maximum potential energy when the spring is compressed, which is all converted to kinetic energy when it is moved, right? So then because F x distance = Change in KE, can i simply sum of the forces work (Work of Block + Work of friction opposing it = 0?)

6. Oct 25, 2006

### Hootenanny

Staff Emeritus
Yes, so you can calculate the frictional force by dividing the change in kinetic energy by the distance.

7. Oct 25, 2006

### cd80187

So I find the force and then divide by force normal and that gives me the answer?

8. Oct 25, 2006

### cd80187

I actually got it, thank you very much for the help