Asked by Kayla
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?
Answers
Answered by
drwls
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).
= 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).
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