Spring oscillation with friction

AI Thread Summary
The discussion revolves around a spring-mass system experiencing oscillation with friction. The initial compression of the spring is determined using the equilibrium position, where gravitational and spring forces balance. The confusion arises around sign changes in the energy conservation equation, particularly when accounting for friction. The correct approach emphasizes that the equilibrium position is where vertical forces equal zero, clarifying the calculations. Overall, understanding the equilibrium position is crucial for solving the problems related to spring oscillation with friction.
mmht579
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I think I'm getting really confused by sign changes... can someone help me with this?

Spring with spring constat K stands vertically, and mass m is placed on top. Mass is gradually lowered to its euqilirium position. With the spring held at this compression length, the system is rotated to horizontal position. Left end of the spring is attached to a wall, and the mass is placed on a table with coefficient of kinetic friction=1/8. The mass is released.

(a) What is the initial compression of the spring?
(b) How much does the maximal stretch or compression of the spring decrease after each half oscillation?
(c) How many times does the mass osciallate back and forth before coming to rest?

(a)mgh= (1/2)kh^2
h= 2mg/k

(b) this is where I am not sure with signs...

conservation of energy:
spring potential energy initial= work friction +spring potential energy
.5k(2mg/k)^2= mg( (2mg/k) + x)/8 + .5kx^2 (?)
i get x= (sq rt.97 -1)mg/(8k)

that seems wrong...
 
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Welcome to PF!
(a) You are asked for the EQUILIBRIUM position, not the position in which all gravitational energy has been converted into spring potential energy!
That is, the equilibrium position of the system is where the sum of all (vertical) forces acting on the mass is 0.
See how this helps you along..
 
oops.. i forgot that i posted...thanks for the help though!
 
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