A 0.321-kg mass is attached to a spring with a force constant of 12.3 N/m. If the mass is displaced 0.256 m from equilibrium and released, what is it's speed when it is 0.128 m from equilibrium?

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Use the potential energy equation of  E = (1/2)kx^2. If you find the potential energy at both points, you get E = 0.40305J at 0.256m and E = 0.1008J at 0.128m.  Subtracting these two you get what potential energy has been lost (E = 0.3023J).  Since there is no friction, all of this potential has been turned to kinetic energy E = (1/2)mv^2.  Solving this for v I get 1.37237m/s

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