Minimum value of the coefficient of static friction

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A 2.0 kg block with a coefficient of kinetic friction of 0.30 slides at 1.3 m/s into a massless spring with a force constant of 120 N/m. To determine the maximum compression of the spring, the energy conservation equation is used, equating kinetic energy and spring potential energy. For the spring to remain compressed, the minimum coefficient of static friction must equal the ratio of the spring force to the normal force at maximum compression. If the static friction is insufficient, the block will detach from the spring when it returns to its natural length, as the spring's restoring force will no longer be able to counteract the block's motion. Understanding these dynamics is crucial for solving the problem effectively.
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A 2.0 kg block slides along a horizontal surface with a coefficient of friction µk = 0.30. The block has a speed of v = 1.3 m/s when it strikes a massless spring head-on

(a) If the spring has a 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 remain compressed at the maximum compressed position?

(c) 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).]

Explain why detachment occurs when the spring reaches its natural length.

I'm able to do a) but not the others. thanks for your help.
 
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If you were able to do (a), show us your work on that at least.

Actually, I think (b) is easier than e. What's the force the spring exerts when it is compressed at the maximum compressed position?
 
a) 1/2mv^2-umgx=1/2kx^2.
for b, how do I find Xmax?
 
anyone...?
 
What force will pull back the string to its equilibrium position? the force -kx, is the force that brings the spring back to his position, so what minimiun value should the static friction had in order to surpass or cancel that force.
 
Well, for part (b) you might look at what the force of the spring happens to be when it is compressed the distance x you found in part (a). For the spring to become stationary, you must assume all the forces cancel out so, "force spring" = "force friction", with of course the new force of friction being based on some new µ-static.
 
how about c?
 
Do you understand when the block hits the spring it will compress it til it slows down, then the conservative force of the spring will start pushin the block creating a speed, and when the spring reaches its equilibrium length (x=0) it won't go on, while the block will continue moving with the speed at the equilibrium point, until the friction puts it back to rest.
 

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