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Chapter 12: Problem 25

A particle's potential energy as a function of position is given by \(U=2 x^{3}-2 x^{2}-7 x+10,\) with \(x\) in meters and \(U\) in joules. Find the positions of any stable and unstable equilibria.

Short Answer

Expert verified
The positions of the equilibria are \(x=-1\) (unstable) and \(x=7/3\) (stable).

Step by step solution

01

Differentiate the potential energy function

Firstly, we take the derivative of the given function \(U'=3x^{2}-4x-7\).
02

Find the positions of equilibria

Secondly, set the derivative equal to zero and solve for \(x\). This will give the positions of equilibrium. \(3x^{2}-4x-7 = 0\). Solving this quadratic equation gives \(x = -1, 7/3\).
03

Determine the nature of equilibria

Lastly, take the second derivative of the potential energy. The second derivative is \(U''=6x-4\). Plug in the values of \(x\) obtained in the previous step. If the second derivative is positive, then the point is a stable equilibrium. If the second derivative is negative, then the point is an unstable equilibrium. For \(x=-1, U''=-10\) which is negative, therefore it's an unstable equilibrium. For \(x=7/3, U'' = 10/3\) which is positive, therefore it's a stable equilibrium.

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