Box with mass attached to spring and board

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SUMMARY

The discussion centers on a physics problem involving a 0.580 kg wood block attached to a spring with a spring constant of 160 N/m. The block is compressed 4.0 cm and subsequently stretches 0.9 cm beyond its equilibrium position. Participants clarify the application of the conservation of mechanical energy and the determination of the coefficient of friction, emphasizing that the choice of positive and negative distances is arbitrary as long as a consistent convention is followed.

PREREQUISITES
  • Understanding of mechanical energy conservation principles
  • Familiarity with Hooke's Law and spring constants
  • Knowledge of friction and its coefficients
  • Ability to convert units (e.g., cm to m)
NEXT STEPS
  • Study the application of conservation of mechanical energy in oscillatory systems
  • Learn about the calculation of coefficients of friction in various scenarios
  • Explore Hooke's Law in-depth, particularly in relation to energy stored in springs
  • Practice problems involving spring dynamics and energy transformations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to springs and friction.

12boone
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Homework Statement



A 0.580Kg wood block is firmly attached to a very light horizontal spring. It is noted that the block-spring system (k=160N/m), when compressed 4.0cm and released, stretches out 0.9 cm beyond the equilibrium position before stopping and turning back.

Homework Equations



Conservation of Mechanical Energy

KEf+PEf=KEi+PEi

and

1/2kx^2

The Attempt at a Solution



Ok, What I tried to do is solve for the work of friction using

(coefficient of friction)(Normal Force)(d)=1/2k(xf-xi)^2. However, I do not understand what the problem is asking as far as distances. I converted the distance to m but I still do not know what they mean by compressed for 4 cm. Does that mean they pushed it 4 cm forward. Let it go and it went past its resting point .9 cm. If so is .9 negative? or is 4 cm negative?
 
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12boone said:
However, I do not understand what the problem is asking as far as distances. I converted the distance to m but I still do not know what they mean by compressed for 4 cm. Does that mean they pushed it 4 cm forward. Let it go and it went past its resting point .9 cm. If so is .9 negative? or is 4 cm negative?
That would be correct.

However, you have failed to state what the question is asking you to find.
 
what I am looking for is the coefficient of friction.
 
12boone said:
what I am looking for is the coefficient of friction.
Ahh right. Looking back at the question, is doesn't matter which (extension/compression) you choose as positive and which you choose as negative provided that one is positive and one is negative.

Your approach is correct by the way.
 

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