**Unformatted text preview: **:now much energy can be stored in a spring with j,,= 320 Njm if the maximum allowed stretch is 126 CHAPTER8 where the spring has its natural length the unstretchedposition (A). (a) At equilibrium (B), kx = mg, so the potential energy of just the spring is Us(B) = !kx2 = !k(mgjk)2 = 0.5(3 kgx9.8 m/s2)2 + (240 N/m) = 1.80 J. (b) The change in just the gravitational potential energy between the equilibrium and unstretehed positions is dU:A = Ug(A) Ug(B) = mgx = mg(mg/k) = 3.60J. (c) The corresponding change in the spring's potential energy is L:i.U{3A= Us(A) - Us (E.) = 0 - 1.80 J. Although the change in the total potential energy,L:i.UBA = AU:A + AU{3A = 3.60 J -'-1.80 J = 1.80 J, is not zero, there is no discrepancy. In order to move the fish slowly upward, you would have to exert an upward applied force (Fa mg - kx) that would do work W!A == mgx kx2 = dU BA, as required by the work-energy theorem. Solution
For a one-dimensional force, one can use Equation to find U(x) - U(O) = _fJ5cm(-kx - cx3)dx! = 8-2a l!kx 2 + ~cx41~5cm=
3 i(3.6 kN/m )(0.15 could be recaptured from the spring...

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