Compressed spring on the ceiling.

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SUMMARY

The discussion centers on a physics problem involving a mass (M = 1.0 kg) attached to a spring (k = 112.0 N/m) that is initially compressed to half its relaxed length (L = 28.0 cm). The objective is to determine the distance below the ceiling where the box comes to rest after being released. The relevant equation used is mgh = mgy2 + 0.5k(y2)2, which relates gravitational potential energy to spring potential energy. The user expresses difficulty in progressing with the calculations, indicating a need for clarification on the application of the energy conservation principle in this context.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with gravitational potential energy equations
  • Knowledge of energy conservation principles in mechanics
  • Basic algebra for solving equations
NEXT STEPS
  • Review the application of energy conservation in spring-mass systems
  • Study the derivation of the equation mgh = mgy2 + 0.5k(y2)2
  • Practice similar problems involving springs and gravitational forces
  • Explore the concept of equilibrium positions in spring dynamics
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of spring dynamics in real-world applications.

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Homework Statement


A box of mass M = 1.0 kg is attached to the ceiling by a spring with spring constant k = 112.0 N/m. The spring has a relaxed length L = 28.0 cm. Initially the spring is compressed to a length L/2. If the box is released, at what distance below the ceiling will the box first be brought to rest by the spring?


Homework Equations


I don't know if this is correct, but I attempted to use an equation given by my teacher for an object falling onto a spring, which is like opposite of this problem.
mgh = mgy2 + .5k(y2)2


The Attempt at a Solution


I used the equation but pretty much went to a dead end >.<
 
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