•An equilibrium point ~xe is stable if U (~xe) is a minimum. A mass hanging at rest from a string is at a stable equilibrium at the bottom.

•An equilibrium point ~xe is unstable if U (~xe) is a maximum. A pencil balanced on its point (if you can ever manage such a feat) is in unstable equilibrium – the slightest disturbance and it will fall.

•An equilibrium point ~xe is neutral if U (~xe) is flat to either side, neither ascending or descending. A disk placed on a perfectly level frictionless table is in neutral equilibrium – if it is place at rest, it will remain at rest no matter where you place it, but of course if it has the slightest nonzero velocity it will coast until it either reaches the edge of the table or some barrier that traps it. In the latter sense a perfect neutral equilibrium is often really unstable, as it is essentially impossible to place an object at rest, but friction or drag often conspire to “stabilize” a neutral equilibrium so that yes, if you put a penny on a table it will be there the next day, unmoved, as far as physics is concerned...

3.1: Work and Kinetic Energy

If you’ve been doing all of the work assigned so far, you may have noticed something. In many of the problems, you were asked to find the speed of an object (or, if the direction was obvious, its velocity) after it moved from some initial position to a final position. The solution strategy you employed over and over again was to solve the equations of motion, solve for the time, substitute the time, find the speed or velocity. We used this in the very first example in the book and the first actual homework problem to show that a mass dropped from rest that falls a height H hits the ground at speed v = √2gH, but later we discovered that a mass that slides down a frictionless inclined plane starting from rest a height H above the ground arrives at the ground as a speed √2gH independent of the slope of the incline!

If you were mathematically inclined – or used a di erent textbook, one with a separate section on the kinematics of constant acceleration motion (a subject this textbook has assiduously avoided, instead requiring you to actually solve the equations of motion using calculus repeatedly and then use algebra as needed to answer the questions) you might have noted that you can actually do the algebra associated with this elimination of time once and for all for a constant acceleration problem in one dimension. It is simple.

If you look back at week 1, you can see if that if you integrate a constant acceleration of an object twice, you obtain:

as a completely general kinematic solution in one dimension, where v0 is the initial speed and x0 is the initial x position at time t = 0.

Now, suppose you want to find the speed v1 the object will have when it reaches position x1. One can algebraically, once and for all note that this must occur at some time t1 such that:

We can algebraically solve the first equation once and for all for t1:

and substitute the result into the second equation, elminating time altogether from the solutions:

Many textbooks encourage students to memorize this equation as well as the two kinematic solutions for constant acceleration very early – often before one has even learned Newton’s Laws – so that students never have to actually learn why these solutions are important or where they come from, but at this point you’ve hopefully learned both of those things well and it is time to make solving problems of this kinds a little bit easier.

However, we will not do so using this constant acceleration kinematic equation even now! There is no need! As we will see below, it is quite simple to eliminate time from Newton’s Second Law itself once and for all, and obtain a powerful way of solving many, many physics problems – in particular, ones where the questions asked do not depend on specific times – without the tedium of integrating out the equations of motion. This “time independent” formulation of force laws and motion turns out, in the end, to be even more general and useful than Newton’s Laws themselves, surviving the transition to quantum theory where the concepts of force and acceleration do not.

One very good thing about waiting as we have done and not memorizing anything, let alone kinematic constant acceleration solutions, is that this new formulation in terms of work and energy works just fine for non-constant forces and accelerations, where the kinematic solutions above are (as by now you should fully appreciate, having worked through e.g. the drag force and investigated the force exerted by springs, neither of which are constant in space or in time) completely useless and wrong.

Let us therefore begin now with this relatively meaningless kinematical result that arises from eliminating time for a constant acceleration in one dimension only – planning to use it only long enough to ensure that we never have to use it because we’ve found something even better that is far more meaningful :

where x is the displacement of the object x1 − x0.

If we multiply by m (the mass of the object) and move the annoying 2 over to the other side, we can make the constant acceleration a into a constant force Fx = ma:

We now define the work done by the constant force Fx on the mass m as it moves through

The work can be positive or negative.

Of course, not all forces are constant. We have to wonder, then, if this result or concept is as fragile as the integral of a constant acceleration (which does not “work”, so to speak, for springs!) or if it can handle springs, pendulums, real gravity (not near the Earth’s surface) and so on. As you might guess, the answer is yes – we wouldn’t have bothered introducing and naming the concept if