Amassless spring has unstretched length l₀ and force constant k. one end is now attached to the ceiling and a mass m is hung from the other. the equilibrium length of the spring is now l₁.
(a) write down the condition that determines l₁.
suppose now the spring is stretched a further distance x beyond its new equilibrium length. show that the net force (spring plus gravity) on the mass is f = -kx. that is, the net force obeys hooke's law, when x is the distance from the equilibrium position — a very useful result, which lets us treat a mass on a vertical spring just as if it were horizontal.
(b) prove the same result by showing that the net potential energy (spring plus gravity) has the form u (x) = const + 1/2 kx².