Spring and slide block problem

[Bones]

Homework Statement

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).]

Homework Equations

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??

 

[Bones]

Please help, I am really stuck!

 

[Bones]

Any help at all would be appreciated ;)

 

[LowlyPion]

Homework Statement

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).]

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??

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.

 

[LowlyPion]

For b) draw a force diagram.

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

 

[Bones]

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

 

[LowlyPion]

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

Sorry. Of course it is 60x².

And that yields your .323 m

 

[Bones]

I am still not getting part B.

 

[LowlyPion]

I am still not getting part B.

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 u_s*m*g

 

[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.

 

[Bones]

So umgx=1/2kx^2?

 

[LowlyPion]

So umgx=1/2kx^2?

No. Not quite.

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

mV²/2 = kx²/2 - u*m*g*x

 

[LowlyPion]

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

 

[Bones]

How do I figure that part out?

 

[LowlyPion]

How do I figure that part out?

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

 

[Bones]

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

So umg=1/2kx^2

 

[LowlyPion]

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

So umg=1/2kx^2

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

 

[Bones]

Thank you!

 

[LowlyPion]

Thank you!

No problem then.

Cheers.