Conservation of Energy involving Spring

AI Thread Summary
The discussion focuses on a physics problem involving a 30kg mass attached to a spring, where the maximum extension is calculated using the Conservation of Energy principle. The initial gravitational potential energy converts to elastic potential energy, leading to a maximum extension of 0.6m, while the equilibrium extension is found to be 0.3m. The discrepancy of 45J in potential energy is attributed to energy lost as heat due to the spring's oscillations and air resistance, which performs negative work on the system. The conversation emphasizes the difference between theoretical and practical scenarios, highlighting that in reality, energy is dissipated rather than conserved entirely. Overall, the discussion illustrates the complexities of energy conservation in dynamic systems.
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Given a problem : A 30kg mass is attached to the lower end of an almost massless spring with force constant k=1000N/m. The mass is then released to rest. Ignoring air resistance, what is the maximum extension of the spring? Note : The spring is said to obey Hooke's Law [F=-kx; Elastic Potentaial Energy=(1/2)(k)(x)(x)] _

I use Conservation of Energy to solve this problem whereby the original Gravitational Potential Energy mgh is converted 100% to the Elastic Potential Energy (1/2)(k)(h)(h) when the spring is monentarily at rest, where h is the max extension of the spring. I get h=0.6m.

When the mass reaches its final equilibrium position, the tension of the spring is equal to mg, i.e. khe=mg where he is the extension length at equilibrium. Therefore he=0.3m. The Elastic Potential Energy stored in the spring is then (1/2)(1000)(0.3)(0.3)=45J. This is half the change in Gravitational PE from before the mass is released which is mgh=30(10)(0.3)=90J [take g=10m/s^2]. How does this reconcile with the Principle of Conservation of Energy?
 
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I believe I can answer this one.

When you used the conservation of energy method to find when the energies equal each other, that is the MAXIMUM displacement. When you drop something onto a spring, you'll notice that the spring will initially compress a lot, and when it reaches equilibrium (stops bouncing), the spring will be compressed a lot less than it was when you first dropped the object.
 
But where has the 45J of PE gone to & how?

Hi Shawn,

Thks for your reply. I understand about the oscillations the spring undergoes before coming to its equilibrium rest, that's why I use the Conservation of Energy method to find the maximum extension. But where has the 45J of PE gone to & how?`
 
In reality, it goes into heat in the spring.
In theory, it doesn't go anywhere at all. The spring will oscillate forever.

Take a piece of metal, and bend it back and forth a lot; the metal will get very hot.
You can get a nice piece of metal from inside your computer where the drive bays are (unless you've already taken those out).
 
Spring slows down because of air resistance?

Thks for your prompt reply, Shawn. I got the same answer from other websites after posting the question too. Just to confirm, the spring slows down because of air resistance in reality? Which is doing negative work done thus reducing the total mechanical energy.
 
Yep that's correct.
 
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