A 0.321-kg mass is attatched to a spring with force constant of 13.3 N/m. If the mass is displaced 0.256 m from the Equalibrium and released, what is its speed when it is 0.128 m from equalibrium?

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\[ \frac{1}{2}mv^2=\frac{1}{2}kx_{total}^2-\frac{1}{2}kx^2\] where k is the spring constant and \[ x_{total}\] is the total displacement from equilibrium. \[x \] is the distance 0.128 m.

Law of conservation of energy yo. Kinetic energy = (potential energy of spring at total displacement) - (potential energy of spring at whatever point you're trying to find the velocity at)

okay

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