Question

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)?

Answer 1

$kx_1^2/2=mv^2/2+kx_2^2/2\to$

$v=\sqrt{(kx_1^2/2-kx_2^2/2)\frac{2}{m}}=\sqrt{(7\cdot 0.125^2/2-7\cdot 0.05^2/2)\frac{2}{0.1}}=0.96\ (m/s)$ .Answer