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$\frac{1}{2}mv^2+\frac{1}{2}kx^2=\frac{1}{2}kx_o^2$

Giancoli 6th Edition, Chapter 6, Problem 41

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Chapter 6, Problem 41 is solved.

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Transcript for this Giancoli solution

The total kinetic, and potential energy of this system with this spring being compressed a distance 'x' naught, is kinetic energy plus potential energy, is one half 'mv' squared, whatever it's speed happens to be at a certain position, plus one half 'kx' squared. So that's the elastic potential energy here, and the kinetic energy there. And this total will be, one half 'kx' knot squared, where 'x' knot is the initial starting compression position. More could have been stretched too, for that matter and it would be the same.

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