# Stretched spring -find coeff of friction

1. Aug 8, 2007

stretched spring --find coeff of friction

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 = 66 cm and held there. A block of mass M = 10 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/7

2. Relevant equations
7What is the coefficient of kinetic friction µk between the block and the floor?

3. The attempt at a solution
This problem is a problem for me.
i know about spring constants and friction but cant visualize where to start?

I know:
M= 10kg
df = .094m
Fs = .09 * 70 = 6.3N
N = 9.8*10= 98N

dont know
µk (µk = Fk/N)

Also i know (obviously) that block stops because µk force must be greater than the spring constant force.

thanks for any help

2. Aug 8, 2007

### Staff: Mentor

Fun problem. I don't know the answer right off the bat, but the way I'd approach a problem like this is to initially break it into understandable segments. Like, you know the initial force on the block from the spring and its mass, so what would be the possible accelerations for the block initially based on the mu values? And if you were going to set up a differential equation for the instantaneous velocity of the block based on the things that you are given, what would that equation look like? And given that instantaneous differential equation, how would you integrate or solve for the motion versus time, based on the mu value? And given that and the answer that you are given for the final V=0 position, how can you then solve for the mu?

3. Aug 8, 2007

### learningphysics

I'd use the work-energy theorem... The final energy - initial energy = work done by non-conservative forces (ie work done by friction).

4. Aug 9, 2007

this is what i did but its wrong sumhow.

Fsp = -K delta x
Fk = mu N

so i equated equations

-K delta x = mu N

mu = -K delta s / N

mu = -70(.094) / (10*9.8)

mu = .067

=wrong????

5. Aug 9, 2007

### learningphysics

That gives the coefficient of static friction, not kinetic.

6. Aug 9, 2007

coeff of static and kinetic use same formula dont they?
Fk = mu*N
Fs = mu*N

7. Aug 9, 2007

oh so whole picture is wrong
i am looking at it like : equate forces, then when they become equal (and opposit) that is mu-max for static.\
should be..

8. Aug 9, 2007

### learningphysics

There are 2 different mu's... mustatic and mukinetic

9. Aug 9, 2007

so is kinetic coeff, --get that from KE=.5mv^2 somehow?

10. Aug 9, 2007

### learningphysics

Yes. To get mu-kinetic, you need to use work-energy...

11. Aug 9, 2007

### learningphysics

You don't need that... what's the initial mechical energy of the system (includes all kinetic and potential energies)... what's the final mechanical energy of the system... It starts at rest and ends at rest so the kinetic energies are 0. There is kinetic energy in between, but we don't need to worry about that... we just need to worry about the initial and final states...

work done by friction = final mechanical energy - initial mechanical energy

From the left hand side you can get mu-kinetic.

Last edited: Aug 9, 2007
12. Aug 9, 2007

OK so i am equating PE of spring with force of spring??

I know its these 3 equations :

U=1/2kx^2
Fsp = -K delta x
Fk = muN

But Fsp and Fk are in in newtons and U is in Joules.
I want everything in joules to use conserv of energy??

How can i get both equations in J?

13. Aug 9, 2007

### learningphysics

Just use the first equation for now... what's the initial energy, and what's the final energy?

14. Aug 9, 2007

### learningphysics

Work done = Force * distance... which is N*m which is the same as J.