I heard somewhere that when an object is in equilibrium on the flat bottom of a hill that it will require work proportional to $dx^2$ when moved the small distance of $dx$. I thought that it would be proportional to $dx$ because $dw=F\,dx$. What makes the bottom of the hill so special?
Assume a one dimensional case with $\hat x$ being the unit vector in the positive x-direction.
Suppose a body is the system under consideration and it is at position $x=0$.
To be in a position of stable equilibrium a displacement of the body needs to result in an external force acting in a direction towards the position $x=0$.
An example of such an external force is $\vec F = - k \,\vec x\Rightarrow F\,\hat x = - k\, x\, \hat x \Rightarrow F= -k\,x$.
If such an external force is acting then the work done by the external force in displacing the body a distance $x$ is $\displaystyle \int _0^x(-k\,x)\,dx = - \frac 12 k x^2$
Now negative work done by an external force on a body can be thought of as positive work done by the body.
This is where the idea, that the body has to do work proportional to the change in position squared to move to a new position, comes from; ie work done by body $= +\frac 12 k (\Delta x)^2 \propto (\Delta x)^2$
So a body at position $x=0$ might be given a "kick" which gives it some kinetic energy.
The body then starts moving away from the position $x=$ but in dong so has to do work at the expense of its kinetic energy.
Eventually all the kinetic energy is used up and the body stops.
However it has a force acting on it which makes the body start moving back towards the $x=0$ position with the external force doing work on the body and the body gaining kinetic energy.
With no "lossy" forces eg friction acting the body will overshoot the position $x=0$ and undergo oscillatory motion about that position.
If friction does act then the amplitude of oscillation of the body will decrease with time until the body eventually stops at position $x=0$.
The bottom of a hill is a position of stable equilibrium.If you will displace a point mass by a distance $x$ then the restoring forces behave like spring force.It is just an analogy.The work done in stretching the spring is proportional to $x^2$,hence the work done to displace a point mass is also proportional to $x^2$.