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)] _(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Conservation of Energy involving Spring

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