# Block on Spring with Friction

1. Oct 4, 2008

### Awwnutz

http://img48.imageshack.us/img48/7973/showmepley5.gif [Broken]

A relaxed spring with spring constant k = 50 N/m is stretched a distance di = 63 cm and held there. A block of mass M = 6 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/10.

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

I know i need to use the Work-Energy theorem.

I'm not really sure what i need to do?

The kinetic energy is 0, but i don't know how to find the work done by friction.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 3, 2017
2. Oct 4, 2008

### Hootenanny

Staff Emeritus
Welcome to Physics Forums,

You're spot on with using the work-energy theorem. What does that theorem state and how can we apply it to this problem?

Last edited by a moderator: May 3, 2017
3. Oct 4, 2008

### Staff: Mentor

What's the definition of work?

4. Oct 4, 2008

### Awwnutz

Thanks,
The work-energy theorem is: change in Kinetic Energy=the work done by friction.
I believe the change in kinetic energy is 0, but its the work done by friction i can't seem to find.

Work is a force over a distance, correct?
So would the work done by friction mean its
(Coefficient of friction-M)(Fnormal)(distance)=Work done by friction?

5. Oct 4, 2008

### Staff: Mentor

Make that the change in total mechanical energy (kinetic energy plus spring potential energy).
Exactly.

6. Oct 4, 2008

### Awwnutz

So the kinetic energy = 0
Work done by friction = M(58.86N)(.567m)
The potential energy = -Force(x2-x1)???

I'm not too familiar with potential energy yet. Is this the right formula i need to use?
If so what is the force i need? Is it F = -50N/m(x)?
If so what distance will i use, the distance it was initially stretched or the distance it comes to rest at?

7. Oct 4, 2008

### Staff: Mentor

Good.
No. (Look up the formula for the energy stored in a stretched spring. Or derive it yourself.)

Since the force is not constant, you can't just use FΔx.

8. Oct 4, 2008

### Awwnutz

I used Elastic Potential Energy Formula: (1/2)kx^2

So i set it up like this:
Work by Friction_______Potential Energy Final____Potential Energy Initial__Kinetic
M(58.86N)(0.567m) = (1/2)(50N/m)(.063)^2 - (1/2)(50N/m)(.63)^2 + 0

I solved for M = -.29, but its saying this is the wrong answer and i thought i had it right.

9. Oct 4, 2008

### Staff: Mentor

Good.

The work done by friction is negative, since it opposes the motion.

10. Oct 4, 2008

### Awwnutz

I see, i had the right answer it just needed to be positive and have a more exact answer. It was 0.294! Thanks alot for your help. I'm beginning to understand this now. I wish i had you as my teacher, i don't seem to learn too much from the one i have right now. Thanks again.