Finding x for a spring using energy equations

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SUMMARY

The discussion focuses on determining the displacement of a spring when a mass is hung on it, using energy equations instead of force equilibrium. The key equations involved are the gravitational potential energy (Ug = mgh) and the elastic potential energy (Us = kx²/2). The user initially equated these energies to find the displacement but realized that this method calculates the maximum stretch rather than the equilibrium position. The correct approach requires understanding that the mass will have velocity as it passes the equilibrium point, indicating that energy conservation must account for kinetic energy as well.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Knowledge of gravitational force (w = mg)
  • Familiarity with potential energy concepts (Ug and Us)
  • Basic principles of energy conservation in mechanics
NEXT STEPS
  • Study the relationship between kinetic and potential energy in oscillatory motion.
  • Learn about the concept of equilibrium in spring-mass systems.
  • Explore the dynamics of simple harmonic motion (SHM) and its equations.
  • Investigate energy conservation principles in mechanical systems.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to clarify concepts related to spring dynamics and energy equations.

cbasst
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Homework Statement



This isn't a homework question, just something I've been thinking about. If given a situation in which mass m is hung on a spring with constant k, I know you can set the spring force equal to the weight of the mass to find how far it stretches. Is there a way to find how far it stretches using energy equations? It seems like it should be doable, but I can't figure out how.

Homework Equations



w = mg
F = -kx
ΔUg = mgh
Us = kx2/2

The Attempt at a Solution



As I said, setting mg = kx can solve for x. How do I do it with energy though?
Ug = Us
mgx = kx2/2

But now I get a different answer for x. What did I do wrong with the energy equations?
 
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The energy approach will give you the farthest the mass on the spring will fall. This is not the equilibrium position. Think about it, if I let the mass drop, then when it passes the equilibrium position, it will not have zero velocity.
 
Ah. This makes sense now! Thanks!
 

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