# Equilibrum position of an extensible spring

A stone of mass $m$ is attached at one end of a vertical spring whose another (upper) end is fixed at horizontal surface. During the motion, a spring takes vertical direction. Force in spring is proportional to extension.

How can I determine equilibrium position?

Will it be in equilibrium when forces that act on it are equal by intensity?

• The equilibrium will be achieved when forces on the stone(Spring force and gravity) exactly cancel out, i.e. net force is zero. – udiboy1209 Aug 7 '13 at 16:47
• So, if I take $y$-axis directed vertically downwards, forces that are acting on the stone are $F_g=mg\vec j$ and $F=-k(y-a)\vec j$, so the it will achieve equilibrium position when $mg=k(y-a)$, i.e. $y=\frac{mg}{k}+a$, where $a$ is natural length of spring and $k$ is constant of extension? – gov Aug 7 '13 at 16:54
• that is perfectly right! – udiboy1209 Aug 7 '13 at 16:55