How Is the Coefficient of Friction Calculated in a Spring Block System?

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In a spring block system, the coefficient of friction (μ) can be calculated using energy conservation principles. A 1 kg block compresses a spring by 0.2 m, storing potential energy of 2 J. Upon release, the block travels 1.0 m before stopping, indicating that the work done against friction equals the initial potential energy. The kinetic energy at the point of release is also 2 J, allowing for the calculation of the frictional force. The discussion emphasizes the relationship between potential energy, kinetic energy, and friction in determining μ.
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Homework Statement


A block of mass 1 kg is forced against a horizontal spring of negligible mass, compressing the spring an amount x1= 0.2m. When released, the block moves on a horizontal tabletop a distance x2 = 1.0m before coming to rest. The spring constant k is 100 N. What is the coefficient of friction, μ, between the block and the table?


Homework Equations


K1 + U1 = K2 + U2
fs= -kx
Kinetic energy = 1/2kv2
Potential energy = mgh


The Attempt at a Solution


At x1 kinetic energy is 0 and potential energy is 1/2kΔx2
So the total energy would be = 40.
However, I do not know how to tie this in with the coefficients of friction.
 
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potential energy of a linear spring = \frac{1}{2} kx^2

therefore potential energy of this spring at this compression = \frac{1}{2} (100) (.04) = 2J

conservation of momentum E_k = E_p

so kinetic energy also = 2J at the moment that the spring reaches full length. \frac{1}{2}mv^2 = 2

Now solve for velocity at the instant it leaves the spring. Calculate the force due to gravity on the block.
Let me know how you go from there, it should be clear how to finish the question.
 
Last edited:
Okay I understand now, thank you!
 
southernbelle said:
Okay I understand now, thank you!

No worries, I just realized that there was an error in the equation I wrote out on the second line of my post, however the result is still the same. I have corrected it now. apologies.
 
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