homework and exercises - If string is stretched just by weight, where does the gravitational potential energy goes if only half is converted to elastic potential energy?

If a spring is stretched by a weight of mass m, (so the extension is $\Delta x$) then $ k = \frac W{\Delta x} = \frac {mg}{\Delta x}$. So $ k\Delta x= mg $.

When the Spring is stretched by distance $\Delta x$ (by the weight) then it looses gravitational potential enegry. ($ \Delta GPE= mg\Delta x$)

But, when we calculate the change in elastic potential energy, We get
$ U = \frac 12 k(\Delta x)^2 $. Since $ k\Delta x = mg$, $U=\frac 12 mg\Delta x$

At, equilibrium we have no kinetic energy $E_k = 0$

Doesn't that violate Energy Conservation?

Where does the rest of lost GPE goes?

Here's a illustration

Answer

Look at the force $F$ vs extension $x$ graph for a spring.
It is a straight line graph through the origin of gradient $k$ the spring constant and $F=kx$.
The work done by the external foce $F$ to extend the spring from being unextended, $x=0$ until it has an extension $x_o$ is $\displaystyle \int_0^{x_o} F dx = \int_0^{x_o} kx \;dx = \frac 1 2 k x^2$
Put another way.
The average force during the extension is $\dfrac {kx_o} {2}$ and so the work done by the external force is $\dfrac {kx_o} {2} \; x_o = \dfrac {kx_o^2} {2}$

Now when you add a mass $m$ to the end of the spring that mass has a constant weight $mg$ and so potentially can exert a constant force on the spring.
You can replicate the analysis done above with a force on the spring which changes with the extension of the spring by applying an upward force $F_{up}$ on the mass such that the net force exerted on the spring $F = mg - F_{up}$ and you will then get that the energy stored in the spring is $\dfrac 1 2 k x^2$ as the work done on the spring is $\displaystyle \int_0^{x_o} (mg - F_{up}) \; dx = mg\;x_o + \left[ - \int_0^{x_o} F_{up} \; dx \right ]$

The first term being the work done by the gravitational force and the second term being the work done on the force $F_{up}$

If the force $F_{up}$ is not present then the $mg$ again does work $mgx_o$ but now the mass $m$ is accelerating since $mg > kx$ and it carries on accelerating until $mg=kx_o$ when the net force on the mass is zero.

However although this is the static equilibrium condition in terms of forces the mass is moving having gained kinetic energy $\dfrac 1 2 k x_o^2$ during its descent will continue onwards until it finally stops when the extension $x = 2 x_o$.
In terms of energy, the spring has stored in it a potential energy of $\dfrac 1 2 k (2x_o)^2 = 2 kx_o^2$ in it and the work done by the gravitational force is $mg\;2x_o = 2 kx_o^2$.

So no energy has been lost.

If the spring mass-system were left alone and there were no dissipative forces acting then the mass would oscillate about the static equilibrium position $x=x_o$ for ever.

In practice with frictional forces present the mass would undergo damped harmonic motion and eventually finish up stationary at the static equilibrium position with energy $\dfrac 12 k x_o^2$ dissipated as heat.