Conservation of energy of three spring system.

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SUMMARY

The conservation of energy in a system consisting of three springs A, B, and C with stiffness values k, 2k, and 0.5k, and natural lengths l, 0.5l, and 2l respectively, is expressed by the equation E = K + U_{gravity} + U_{springA} + U_{springB} + U_{springC}. The kinetic energy K is defined as K = (1/2)mv^2, while the gravitational potential energy is U_{gravity} = -mgx. The spring potential energies are calculated as U_{springA} = (1/2)k(x-l_0)^2, U_{springB} = k(x-l_0)^2, and U_{springC} = (1/4)k(2l_0-x)^2, confirming the individual energies have been derived correctly.

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1. Find the equation for conservation of energy in system. System consists of three springs A,B and C with stiffness k, 2k and 0.5k respectively and natural lengths l, 0.5l and 2l respectively.



Homework Equations


Equation for conservation of energy [tex]E=K+U_{spring}+U_{gravity}[/tex]
[tex]K=\frac{1}{2}mv^2[/tex]
Datum taken at point A, [tex]U_{gravity}=-mgx[/tex]
[tex]U_{spring}=\frac{1}{2}k(x-l_0)^2[/tex]
Where [tex]K[/tex] is kinetic energy and [tex]k[/tex] is the spring stiffness and [tex]l_0[/tex] is the natural length.

The Attempt at a Solution


By analysing the diagam
[tex]U_{springA}=\frac{1}{2}k(x-l_0)^2[/tex]
[tex]U_{springB}=\frac{1}{2}*2k(x-\frac{1}{2}l_0-\frac{1}{2}l_0)^2=k(x-l_0)^2[/tex]
[tex]U_{springC}=\frac{1}{2}*\frac{1}{2}k(4l_0-x-2l_0)^2=\frac{1}{4}k(2l_0-x)^2[/tex]
[tex]K=\frac{1}{2}mv^2[/tex]
[tex]U_{gravity}=-mgx[/tex]

My question is, is the answer [tex]E=K+U_{gravity}+U_{springA}+U_{springB}+U_{springC}[/tex] and have I derived the individual energies correctly?
 
Last edited:
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Any ideas anyone?
 

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