A spring with a force constant of 7.0 N/m is hung vertically and then compressed by 0.125 m. If a 0.100 kg mass is attached to the spring while compressed by 0.125 m, what will the speed of the mass be when it is passing a point 0.175 m below the compression point (i.e. 0.050 m or 5.0 cm below the equilibrium point)?
"kx_1^2\/2=mv^2\/2+kx_2^2\/2\\to"
"v=\\sqrt{(kx_1^2\/2-kx_2^2\/2)\\frac{2}{m}}=\\sqrt{(7\\cdot 0.125^2\/2-7\\cdot 0.05^2\/2)\\frac{2}{0.1}}=0.96\\ (m\/s)" .Answer
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