Question

On top of a spiral spring of force constant 500Nm-1 is placed a mass of 5*10kg. If the spring is compressed downwards by the length of 0.02m and then released. Calculate the height of which the Mass is projected

Answer 1

$$PE_s=PE,$$ $$\dfrac{1}{2}kx^2=mgh,$$ $$h=\dfrac{kx^2}{2mg},$$ $$h=\dfrac{500\ \dfrac{N}{m}\times(0.02\ m)^2}{2\times5\times10^{-3}\ kg\times9.8\ \dfrac{m}{s^2}}=2\ m.$$