 # Why is the gravitational potential energy of ideal uniform massive spring mgx/2, not mgx? Israel Hale 2022-07-23 Answered
Why is the gravitational potential energy of ideal uniform massive spring $mgx/2$, not $mgx$?
$L=T-V=\frac{1}{2}\frac{m}{3}{\stackrel{˙}{x}}^{2}+\frac{1}{2}M{\stackrel{˙}{x}}^{2}-\frac{1}{2}k{x}^{2}-\frac{mgx}{2}-Mgx$
where $mgx/2$ refers to gravitational potential energy of ideal spring. But I do not get why it's $mgx/2$ instead of $mgx$. What is the reason?
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$x$ is the extension of the spring. If you extend the spring by a distance $x$, keeping the other end fixed, its centre-of-mass moves by $\frac{x}{2}$
###### Not exactly what you’re looking for? Livia Cardenas
If the spring is stretching by an amount of $x$, its centre of mass is displaced by an amount of $x/2$