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Unformatted text preview: and determine if they are stable/unstable equilibrium points. ( 29 6 104 3 == x x dx x dU The roots of the cubic equation are x = 0, ± 0.24 m The equilibrium points are found by setting the first derivative of U(x) = 0: 7 The 2 nd derivative of the potential energy is ( 29 6 312 2 2 2= x dx x U d At x = 0 the second derivative is less than 0 and at ± 0.24 m it is > 0. x = 0 is a point of unstable equilibrium x = ± 0.24 m are points of stable equilibrium Example continued 80.10.05 0.05 0.1 0.15 0.20.410.210.01 0.19 0.39 x (m) U (J) A plot of U(x) 9 Summary ( 29 dx x dU F x= The work done by a conservative force may be expressed as a change in potential energy. r F d dU ⋅=...
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 Spring '11
 Staff
 Energy, Force, Potential Energy, dx, equilibrium points, kx dx dx

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