A massless spring has unstretched length $/0$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 $/1.($a$)$ Write down the condition that determines $11.$ 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 ikx$2.$