Initial kinetic energy = (1/2)kX^2
= 120*0.12^2 = 1.728 J
Set that equal to the maximum kinetic energy, and solve for Vmax.
(1/2)MVmax^2 = 1.728 J
Maximum speed is attained when the spring is unstressed (equilibrium position).
Suppose that a 200g mass (0.20kg) is oscillating at the end of a spring upon a horizontal surface that is essentially friction-free. The spring can be both stretched and compressed and have a spring constant of 240N/m. It was originally stretched a distance of 12cm (0.12m) from its equilibrium (un-stretched) position prior to release. What is the maximum velocity that the mass will reach in its oscillation? Where in the motion is this maximum reached?
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