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.5kg block attached to light horiz. spring with spring constant=200N/m.If the system is stretched out .05m.?
What is the velocity as it passes though the equilibrium point the spring? Ignore friction.
1 Answer
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- oldprofLv 73 years ago
When the spring is stretched its PE = 1/2 kdX^2 and that all converts into KE = 1/2 mv^2 as it passes through dX = 0 stretch.
So we write PE = 1/2 kdX^2 = 1/2 mv^2 = KE and solve for v = sqrt(k/m)dX = sqrt(200/.5)*.05 = 1 m/s. ANS.
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