# Spring and slide block problem

1. Oct 20, 2008

### Bones

1. The problem statement, all variables and given/known data
A 5.0 kg block slides along a horizontal surface with a coefficient of friction µk = 0.30. The block has a speed v = 2.1 m/s when it strikes a massless spring head-on (as in the figure).
http://www.webassign.net/gianpse4/8-18.gif
(a) If the spring has force constant k = 120 N/m, how far is the spring compressed?
(b) What minimum value of the coefficient of static friction, µs, will assure that the spring remains compressed at the maximum compressed position?
If µs is less than this, what is the speed of the block when it detaches from the decompressing spring? [Hint: Detachment occurs when the spring reaches its natural length (x = 0).]
2. Relevant equations

3. The attempt at a solution
a) -1/2(5.0kg)(2.1m/s)^2 + 1/2(120N/m)(x^2) = (0.30)(5.0kg)(9.8m/s^2)(cos 180)(x)
x=0.323m
b) I took u(5.0kg)(9.8m/s^2)(0.323m)(cos 180) = -1/2(120N/m)(0.323m)^2 which is not correct. Can someone help me get the correct equation??

2. Oct 20, 2008

### Bones

3. Oct 22, 2008

### Bones

Any help at all would be appreciated ;)

4. Oct 22, 2008

### LowlyPion

For a) I would write your equation as the KE of the mass = the work to compress the spring and the work against friction:

mV2/2 = kx2/2 + u*m*g*x

That yields for me 600x2 + 14.7x - 11.025 = 0

Using the quadratic formula that yields a different answer than you suggest.

5. Oct 22, 2008

### LowlyPion

For b) draw a force diagram.

The F = kx needs to balance the us*m*g

6. Oct 22, 2008

### Bones

How did you get 600x^2 from 120x^2/2??

7. Oct 22, 2008

### LowlyPion

Sorry. Of course it is 60x2.

And that yields your .323 m

8. Oct 22, 2008

### Bones

I am still not getting part B.

9. Oct 22, 2008

### LowlyPion

If it is at equilibrium then the frictional force (using the static coefficient) must be equal to or greater than the kx from the spring detent:

F = kx needs to balance the us*m*g

10. Oct 22, 2008

### LowlyPion

You were attempting to use the work relationship when you were asked what condition needed to be met for the forces to balance.

You would use the work energy relationship figuring the transfer of potential in the spring back to the kinetic and friction for the outward rebound.

Last edited: Oct 22, 2008
11. Oct 22, 2008

### Bones

So umgx=1/2kx^2?

12. Oct 22, 2008

### LowlyPion

No. Not quite.

The excess of that is the kinetic energy remaining in the block.

mV2/2 = kx2/2 - u*m*g*x

13. Oct 22, 2008

### LowlyPion

Don't forget to answer the static coefficient part of the problem.

14. Oct 22, 2008

### Bones

How do I figure that part out?

15. Oct 22, 2008

### LowlyPion

I've already told you. Look back at the earlier posts.

16. Oct 22, 2008

### Bones

The F = kx needs to balance the us*m*g

So umg=1/2kx^2

17. Oct 22, 2008

### LowlyPion

No. 1/2 k*x2 is WORK. Units are N-m

u*m*g is Force. Units are N.

The force of a spring is given as F = kx. Units are N.

So it's u*m*g = k*x

18. Oct 22, 2008

### Bones

Thank you!!

19. Oct 22, 2008

### LowlyPion

No problem then.

Cheers.