Conservative Forces & Potential Energy

A force that gives the energy back

In the last lesson you added up work along a path and got the change in kinetic energy. That works, but it can be miserable. Drop a ball from a window and the force of gravity is easy. Slide the same ball down a curved chute and the force still points straight down while the path wanders, so the integral $\int \vec F \cdot d\vec r$ looks like a mess.

It turns out not to be a mess, and the reason is worth a whole lesson. Carry a book from the floor up to a shelf a height $h$ above it. Lift it straight up: gravity pulls down $mg$ through a rise $h$, so gravity does $-mgh$. Now carry the book twice around the room, up a ladder, along the top of a bookcase and back down onto the same shelf. Every horizontal step has gravity perpendicular to the motion and contributes nothing. Every bit of rise contributes $-mg$ times that rise, every bit of fall contributes $+mg$ times the fall, and the rises minus the falls add up to $h$ no matter how you wander. Gravity does $-mgh$ again.

Now drag the book across the carpet from one corner to another and back. Sliding friction opposes the motion the whole way, so it takes energy on the outbound trip and takes energy again on the return. Bring the book back to where it started and friction has taken a lot; gravity has taken nothing.

A force is conservative when the work it does between two points depends only on the endpoints and not on the route. Equivalently, the work it does around any closed loop is zero:

$\oint \vec F \cdot d\vec r = 0$

Gravity, the spring force and the electric force of fixed charges are conservative. Sliding friction, air drag and the push of your hand are not. The distinction is not a technicality. It is exactly the condition that lets you invent potential energy at all.

Potential energy is work you can get back

Here is the trick. If the work a force does depends only on where you start and where you finish, then you can bottle it. Pick a reference point and define a function $U$ of position such that the work done by the conservative force going from point 1 to point 2 is the drop in $U$:

$W_{\text{cons}} = U_1 - U_2 = -\Delta U$

$U$ is the potential energy: the work the force is still willing to do for you, stored in the arrangement of things. This definition is only legal for a conservative force. For friction the "work still available" would depend on which road you take next, so there is no function of position to write down.

Apply it to gravity near the ground. Lifting through a height $h$ costs gravity $-mgh$ of work, so $\Delta U = +mgh$, and the gravitational potential energy is

$U_g = mgh$

with $m$ in kilograms, $g = 9.8\ \mathrm{m/s^2}$ and $h$ the height above whatever level you called zero. Go below the reference level and $h$ is negative and so is $U$. There is nothing wrong with negative potential energy.

Now a spring. The spring force is $F = -kx$, where $x$ is the stretch away from the natural length and $k$ is the stiffness in newtons per metre. The minus sign says the spring pulls back toward $x = 0$. To find $U$ you need the work done by the spring as you pull it out to $x$, and that force is not constant: it grows as you pull. Use the graph. The force you must apply is $+kx$, a straight line through the origin, and the work you do is the area under it: a triangle of base $x$ and height $kx$. So the work you put in, which is the energy the spring stores, is $\tfrac{1}{2} \cdot x \cdot kx$:

$U_s = \tfrac{1}{2}kx^2$

The square is the whole story of springs. Double the stretch and the force doubles, but the energy goes up by four.

Where you put the zero is your choice

Only differences in $U$ ever appear in a physical answer, because the definition only ever fixed $\Delta U$. You may add any constant to $U$ you like. In practice that means you choose the zero to make the algebra shortest: the floor, the lowest point of a swing, the spring's natural length.

For the book on the shelf $2.2\ \mathrm{m}$ above the floor with $m = 1.5\ \mathrm{kg}$, taking the floor as zero,

$U = mgh = 1.5 \times 9.8 \times 2.2 = 32.34\ \mathrm{J}.$

Take the shelf as zero instead and $U = 0$ there and $-32.34\ \mathrm{J}$ on the floor. Every question you could ask about the book (how fast it lands, how much work you did lifting it) comes out identical, because all of them involve a difference.

Worked example 1: a pendulum pulled aside

A $0.30\ \mathrm{kg}$ bob hangs on a $1.2\ \mathrm{m}$ string. Pull it aside until the string is $40^\circ$ from vertical. How much gravitational potential energy has it gained relative to the lowest point?

You cannot plug into $mgh$ until you know $h$, and the problem never tells you $h$; it tells you an angle. This is the setup step. With the string at angle $\theta$, the bob is a vertical distance $L\cos\theta$ below the pivot. At the bottom it is $L$ below the pivot. So the rise is

$h = L - L\cos\theta = L(1 - \cos\theta).$

Now the numbers. $\cos 40^\circ = 0.766$, so $1 - 0.766 = 0.234$, and $h = 1.2 \times 0.234 = 0.2807\ \mathrm{m}$. Then

$U = mgh = 0.30 \times 9.8 \times 0.2807 = 0.825\ \mathrm{J}.$

Write down $h = L(1-\cos\theta)$ once and you own every pendulum problem in the bank.

Force is the negative slope of the potential

So far you have gone from force to potential energy. The arrow runs the other way too, and that direction is where most of the hard problems live.

Move a small distance $dx$ along the $x$ axis. The work done by the force is $F_x\,dx$, and by definition that equals $-dU$. Divide by $dx$:

$F_x = -\frac{dU}{dx}$

In words: the force points downhill on the potential energy curve, and its magnitude is the steepness. If $U(x) = 5x$ joules then $F = -5\ \mathrm{N}$, pointing toward $-x$ everywhere, because the hill slopes up to the right. If $U(x) = 40x^2$ then $F = -80x$, which at $x = 0.30\ \mathrm{m}$ has magnitude $80 \times 0.30 = 24\ \mathrm{N}$, pulling back toward the origin.

Running it backwards, if you are handed a force you integrate to get the potential:

$U(x) = -\int F(x)\,dx$

with the arbitrary constant set by wherever you declared $U = 0$. Take $F(x) = -kx + \beta x^2$ with $k = 20\ \mathrm{N/m}$ and $\beta = 30\ \mathrm{N/m^2}$, and $U(0) = 0$. Then

$U(x) = -\int_0^x (-kx' + \beta x'^2)\,dx' = \tfrac{1}{2}kx^2 - \tfrac{1}{3}\beta x^3.$

At $x = 0.50\ \mathrm{m}$: the first term is $\tfrac{1}{2}(20)(0.25) = 2.5\ \mathrm{J}$, the second is $\tfrac{1}{3}(30)(0.125) = 1.25\ \mathrm{J}$, and $U = 2.5 - 1.25 = 1.25\ \mathrm{J}$.

Three things follow from $F = -dU/dx$ immediately, and they are the reason a $U(x)$ graph is worth more than a page of algebra.

Equilibrium is a flat spot. $F = 0$ exactly where $dU/dx = 0$. So finding equilibria means differentiating and setting to zero, nothing more. For $U(x) = 2x^3 - 12x$, $dU/dx = 6x^2 - 12 = 0$ gives $x^2 = 2$, so the positive equilibrium sits at $x = \sqrt{2} = 1.414\ \mathrm{m}$.

A minimum is stable, a maximum is unstable. At a minimum, push the particle right and the curve rises, so the force points back left. It gets shoved home. At a maximum, any nudge puts the particle on a downslope and the force drives it further away. A ball in a bowl against a ball on a dome.

Near any smooth minimum, everything is a spring. Zoom in on the bottom of a well and the curve looks like a parabola $U \approx U_0 + \tfrac{1}{2}U''(x_0) (x-x_0)^2$. Compare with $\tfrac{1}{2}kx^2$ and the effective stiffness is $k_{\text{eff}} = U''(x_0)$, the second derivative at the minimum. Small oscillations therefore have

$\omega = \sqrt{\frac{U''(x_0)}{m}}.$

For a $0.50\ \mathrm{kg}$ particle in $U(x) = 6x^2 + 2x^4$: $U' = 12x + 8x^3$ and $U'' = 12 + 24x^2$, which at the minimum $x = 0$ is $12\ \mathrm{N/m}$. Then $\omega = \sqrt{12/0.5} = \sqrt{24} = 4.899\ \mathrm{rad/s}$. The $x^4$ term does nothing at all for small swings.

Reading a curve: turning points and barriers

Draw a horizontal line at the total energy $E$. Since $K = E - U$ and kinetic energy can never be negative, the particle lives only where $U(x) \le E$. Where the line meets the curve, $K = 0$: the particle stops and turns around. That point is a turning point, and the gap between the curve and the line at any $x$ is the kinetic energy there.

Worked example 2: turning point and barrier height

A $2.0\ \mathrm{kg}$ particle moves in $U(x) = 4x^2 - x^4$ joules, with $x$ in metres, and has total energy $E = 3.0\ \mathrm{J}$, starting near $x = 0$. Find the positive turning point, and the height of the barrier it is trapped behind.

Turning point first. Set $U = E$:

$4x^2 - x^4 = 3.$

Let $u = x^2$. Then $u^2 - 4u + 3 = 0$, which factors as $(u-1)(u-3) = 0$, so $u = 1$ or $u = 3$. Both are mathematical solutions, so you have to think about which one the particle actually reaches. Starting at $x = 0$ and moving out, it hits $x = \sqrt{1} = 1.0\ \mathrm{m}$ first and stops there. It never gets to $\sqrt{3}$. The answer is $x = 1.0\ \mathrm{m}$.

Barrier height next. The particle escapes only if $E$ clears the top of the hump, so find the maximum of $U$ for $x > 0$:

$\frac{dU}{dx} = 8x - 4x^3 = 4x(2 - x^2) = 0.$

The roots are $x = 0$ (the minimum) and $x = \sqrt{2}$. At $x = \sqrt{2}$, $x^2 = 2$ and $x^4 = 4$, so

$U = 4(2) - 4 = 4.0\ \mathrm{J}.$

The barrier is $4.0\ \mathrm{J}$ tall. With $E = 3.0\ \mathrm{J}$ the particle rattles between $\pm 1.0\ \mathrm{m}$ forever. Give it more than $4.0\ \mathrm{J}$ and it clears the hump, and then $U$ drops without limit and it flies off. Drag the slider in Figure 3 past 4.0 and you will see exactly that.

The same recipe handles any named potential. The Lennard-Jones potential between two atoms is $U(r) = A/r^{12} - B/r^6$. Its minimum is where $dU/dr = -12A/r^{13} + 6B/r^7 = 0$, so $12A/r^{13} = 6B/r^7$, giving $r^6 = 2A/B$. With $A = 2.0$ and $B = 4.0$, $r^6 = 1$ and $r = 1.0\ \mathrm{m}$.

Worked example 3: the bungee jumper

A $70\ \mathrm{kg}$ jumper steps off a bridge on a cord of unstretched length $20\ \mathrm{m}$ and stiffness $60\ \mathrm{N/m}$. How far below the bridge is the lowest point?

Call the total drop $d$. At the bottom the jumper is momentarily at rest, and they started at rest, so the kinetic energy is zero at both ends of the trip. That means all the gravitational energy given up has gone into the cord.

Gravity gives up $mgd$. The cord only stretches for the part of the fall past its natural length, so its stretch is $d - 20$, and it stores $\tfrac{1}{2}k(d-20)^2$. Setting them equal:

$mgd = \tfrac{1}{2}k(d-20)^2.$

Numbers: $mg = 70 \times 9.8 = 686\ \mathrm{N}$ and $\tfrac{1}{2}k = 30$. So

$686\,d = 30(d-20)^2 = 30d^2 - 1200d + 12000.$

Bring everything to one side: $30d^2 - 1886d + 12000 = 0$. The quadratic formula gives

$d = \frac{1886 \pm \sqrt{1886^2 - 4(30)(12000)}}{60} = \frac{1886 \pm \sqrt{3\,556\,996 - 1\,440\,000}}{60}.$

The square root is $\sqrt{2\,116\,996} = 1455.0$. The minus sign gives $d = 431/60 = 7.2\ \mathrm{m}$, which is less than the $20\ \mathrm{m}$ of slack and so is not a real solution: the cord was not stretched at all there, and the equation you solved does not apply. Keep the plus sign:

$d = \frac{1886 + 1455.0}{60} = \frac{3341.0}{60} = 55.68\ \mathrm{m}.$

Two springs at once

Problems love combining springs, and both cases fall straight out of what you already have.

Side by side (parallel). Both springs stretch the same $x$, so the total stored energy is $\tfrac{1}{2}k_1x^2 + \tfrac{1}{2}k_2x^2 = \tfrac{1}{2}(k_1 + k_2)x^2$. The stiffnesses add. With $k_1 = 250$, $k_2 = 150\ \mathrm{N/m}$ and $x = 0.10\ \mathrm{m}$: $\tfrac{1}{2}(400)(0.01) = 2.0\ \mathrm{J}$.

End to end (series). Now the same force $F$ runs through both, and each stretches by its own amount $x_i = F/k_i$. It is easier to write each spring's energy in terms of force: $U = \tfrac{1}{2}kx^2 = F^2/(2k)$. So the total is $F^2/(2k_1) + F^2/(2k_2)$. With $k_1 = 300$, $k_2 = 600\ \mathrm{N/m}$ and $F = 90\ \mathrm{N}$: $F^2 = 8100$, so the energy is $8100/600 + 8100/1200 = 13.5 + 6.75 = 20.25\ \mathrm{J}$. Equivalently the pair behaves like one spring with $1/k = 1/300 + 1/600$, so $k = 200\ \mathrm{N/m}$, and $8100/400 = 20.25\ \mathrm{J}$.

Note that $U = F^2/(2k)$ is the fastest route whenever a problem hands you a force rather than a stretch: a spring with $k = 800\ \mathrm{N/m}$ pushing back with $60\ \mathrm{N}$ stores $3600/1600 = 2.25\ \mathrm{J}$.

How to tell this is the technique

Reach for potential energy when you see any of these.

The problem gives you a height or a stretch and asks for energy, or the reverse. That is $mgh$ or $\tfrac{1}{2}kx^2$ directly.

A curved or complicated path, with only endpoints mentioned. Ramps, chutes, loops, pendulums. The path is there to scare you; only the height difference matters.

A $U(x)$ or $U(r)$ is written down explicitly. Then everything is calculus on one function. Force: $-dU/dx$. Equilibrium: $dU/dx = 0$. Stable or unstable: sign of $U''$. Small oscillations: $\omega = \sqrt{U''/m}$. Turning points: solve $U(x) = E$. Barrier height: the value of $U$ at the maximum. Six questions, one derivative.

A force law given as a function of position, and the word "conservative". You are being asked to integrate to get $U$.

Something momentarily at rest at both ends. Bungee cords, blocks launched by springs, pendulums released and caught. Both kinetic energies vanish, so the whole problem is one potential energy equals another.

If instead the problem features friction, drag, or a person pushing, potential energy alone will not close the books: you need the full energy accounting, which is exactly where Conservation of Energy picks up next.