# Work on an object in equilibrium

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?

• If it is implicitly assumed that $F=-k dx$, there isn't any contradiction. – user8736288 Sep 10 '19 at 7:28

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 $$x=0$$.

• Can you elaborate a bit onto why you can define the Force with Hooke’s law? – Gary Song Sep 10 '19 at 16:57
• @GarySong It was an example. You need an external force which is in the direction of the $x=0$ position ie antsy displacement from the equilibrium position will result in a force in the opposite direction to the displacement. – Farcher Sep 10 '19 at 20:14
• what if the force wasn’t proportional to x and instead was a constant such as 8? – Gary Song Sep 10 '19 at 20:48
• @GarySong The force of $8$ is in the positiv x-direction and so when the x-position of the body is positive the force would not try and restore the body back towards the equilibrium position. – Farcher Sep 10 '19 at 20:52
• What about -8 then? – Gary Song Sep 12 '19 at 15:21

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 $$x^2$$.