Solving problems usng conservation of energy

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SUMMARY

The discussion focuses on calculating the spring constant (k) for a block of mass 1.5 kg released from a compressed spring. The block travels 0.9 m with a velocity of 14.5 m/s while experiencing a kinetic friction coefficient of 0.4. The correct calculation yields a spring constant of approximately 321 N/m, derived from the conservation of energy principles. Participants confirm the method used is valid but suggest re-evaluating the numerical inputs for accuracy.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with kinetic and potential energy equations
  • Knowledge of friction and its impact on motion
  • Basic algebra for solving equations
NEXT STEPS
  • Review the derivation of the spring constant using Hooke's Law
  • Learn about energy loss due to friction in mechanical systems
  • Explore the relationship between mass, velocity, and kinetic energy
  • Investigate different methods for calculating work done by friction
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to clarify concepts related to springs and friction in motion.

tutojean
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Homework Statement


A block of mass 1.5 kg is placed against a horizontal spring of force constant k that is compressed .20 m. the spring is then released and the block travels to the right along a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.4. After the block has traveled 0.9 m to the right, measured from its initial position against the compressed spring, its velocity is 14.5 m/s. Calculate the force constant k of the spring.


Homework Equations


Kinetic/gravitational/elastic/friciton (Usage varies for different situations.


The Attempt at a Solution



[0+0+Es] Initial = [K+0+0+Ef]

1/2kv^(2) = 1/2mv^(2) = uNd

N=fg=mg so this replaces N essentially!

plugging in...

1/2k(0)^(2) = 1/2(1.5)(14.5)^(2) + (.4)(1.5)(9.8)(.9)

k=320.667
k=321 Nt

Is this correct? Just needing full proof assistance thank you so much.
 
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The method is correct but you need to redo the numbers because they don't agree with mine. What number did you put here (in red)?
tutojean said:
1/2k(0)^(2) = 1/2(1.5)(14.5)^(2) + (.4)(1.5)(9.8)(.9)

k=320.667
k=321 Nt
 

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