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Find length of spring in its equilibrium position

  1. Sep 23, 2015 #1
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    As shown in Figure Q1a, a frictionless, massless, piston, supported by two springs, A and B, is held by a pin in a vacuum inside a rigid-walled container. The properties of the springs are: spring constants kA = 3,859 and kB = 3,090 (in N/m), natural lengths LA0 = 0.03 and LB0 = 0.1 (in m). When the piston is pinned to the wall, the lengths of the springs are LA1 = 0.1 and LB1 = 0.1 (in m), respectively. Both springs possess dissipative properties so that after the pin is pulled, the piston eventually comes to rest at an equilibrium position. Find the length of spring A in this equilibrium position (LA2) in units of m.

    Equations so far:
    Models:
    Elastic Energy Constituitive Relation: (E2 - E1) = (k/2)(x22 - x12)

    First Law of thermodynamics: (E2 - E1) = (Q1-2) - (W1-2)
    Second Law of thermodynamics...

    Don't really know where to start? Help appreciated



    Im assuming the rigid wall is the boundary, so would work be equal to O? as there are no external forces acting or pushing on the boundary? I don't understand how to incorporate the second law of thermodynamics re entropy here? No information about heat transfer?
     
  2. jcsd
  3. Sep 23, 2015 #2

    gneill

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    Staff: Mentor

    Would it not be simpler to consider the forces acting on the piston at equilibrium?
     
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