Potential energy

Potential energy ($U$) is energy stored by where something sits. Near the ground a mass at height $h$ stores $U = mgh$; an ideal spring stretched a distance $x$ stores $U_s = \tfrac{1}{2}kx^2$. Mass is in kilograms, height in meters, and the energy comes out in joules (J).

The idea that runs through this whole topic is that $U$ is always measured from a chosen zero. You decide where $U = 0$, then heights and separations are measured from there. Move the zero and every value of $U$ shifts with it.

So only the change $\Delta U$ from one place to another is physical; the bare number depends on the choice. Three habits cost points: treating $U$ as an absolute number and dropping its sign (§2), reading the force as the value of $U$ instead of its slope (§3), and giving a potential energy to forces like friction that do not have one (§4).

Throughout this lesson, $g = 10$ m/s$^2$ where it is needed.

Near the ground, $U = mgh$ rises as you go up. Above the chosen zero $h > 0$ and $U > 0$; below it $h < 0$ and $U < 0$. The zero is wherever you decide to put it, often the floor or a tabletop.

For gravity over large distances the natural zero is at infinity, and there $U = -\tfrac{GMm}{r}$. With that choice $U$ is negative for every finite separation, for every pair, bound or not; bringing the masses closer (smaller $r$) makes $U$ more negative. As $r$ grows, $U$ rises back toward $0$.

So a negative $U$ does not mean the system is bound. A probe on a hyperbolic flyby has $U < 0$ at every instant of the encounter and still escapes. What settles it is the total mechanical energy $E = K + U$: $E < 0$ is bound (circular or elliptical), $E = 0$ is the marginal escape case (parabolic), and $E > 0$ is unbound (hyperbolic). Bound simply means the kinetic energy is not enough to reach infinity, $K < |U|$, since $U \to 0$ as $r \to \infty$.

The minus sign is part of the formula, not optional. Dropping it and writing $U = +\tfrac{GMm}{r}$ flips the sign of $U$ everywhere and wrecks every energy total you build from it, and that is the most common slip on this topic.

Because the zero is a choice, the bare value of $U$ is not by itself meaningful; only the change is fixed. Near the ground $\Delta U = mg(y_2 - y_1)$, and the same holds for $U = -\tfrac{GMm}{r}$: report how much $U$ changes between two points, not the value at one.

On a graph of $U$ versus position, the force is the negative slope: $F = -\tfrac{dU}{dx}$. How steep the graph is sets the size of the force; the direction is downhill, toward lower $U$.

The value of $U$ at a point is not the force. A point where $U$ is large can have zero force if the graph is flat there, and a point where $U = 0$ can have a large force if the graph is steep. Read the slope, not the height.

For example, take $U(x) = 2x^2$. The slope is $\tfrac{dU}{dx} = 4x$, so the force is $F = -4x$. At $x = 1.5$ m the force is $F = -6$ N, pointing back toward $x = 0$, which is where $U$ is lowest. The force always pushes toward lower potential energy.

The trap is to grab the value of $U$ and call it the force, or to keep the plus sign. The force is minus the slope, $F = -\tfrac{dU}{dx}$.

Equilibrium is where the slope is flat, $\tfrac{dU}{dx} = 0$, not where $U = 0$. At the bottom of a valley in the graph the slope is zero, so the force is zero, even though $U$ there can be any value. For $U(x) = 2x^2$ the slope $4x$ is zero at $x = 0$, so that is the equilibrium.

Not every force has a potential energy. Only conservative forces, gravity and an ideal spring, store energy you can get back, and only those have a $U$ whose slope gives the force.

Friction and air drag do not. They oppose the motion and turn energy into heat, so there is no single $U$ that describes them. You can still find the work they do, but you cannot assign them a potential energy.

A potential energy also belongs to an interaction, not to one object alone. Gravitational $U$ is shared by the two masses (the ball and the Earth together), and spring $U$ belongs to the spring and block as a pair. Saying "the ball's potential energy" is just shorthand for the pair's.

A potential-energy question usually tests where you set the zero, how you read a force off the energy graph, or which forces earn a potential energy at all.

The sign and the zero. Pick where $U = 0$, then use $U = mgh$ near the ground or $U = -\tfrac{GMm}{r}$ with the zero at infinity, where $U < 0$ at every finite separation; boundness is read off the total, $E = K + U < 0$, not off the sign of $U$. Report the change $\Delta U$, not the bare value at one point.

The force from the graph. The force is minus the slope, $F = -\tfrac{dU}{dx}$: downhill toward lower $U$, and zero where the slope is flat. The value of $U$ is not the force.

Which forces qualify. Only conservative forces, gravity and ideal springs, have a potential energy; friction and drag do not, and the $U$ belongs to the interaction, not to one object.

Setup. (a) A $2$ kg book sits $1.5$ m above a tabletop; taking the tabletop as the zero ($g = 10$ m/s$^2$), find its gravitational potential energy. (b) A particle has $U(x) = 2x^2$ (joules, $x$ in meters); find the force at $x = 1.5$ m and the equilibrium point. (c) While a block slides, friction does $-8$ J on it; find the potential energy stored by friction.

(a) Sign and reference. The book is above the chosen zero, so $h = +1.5$ m and $U = mgh = (2)(10)(1.5) = 30$ J, positive. A different zero would change the number, but a move between two heights gives the same $\Delta U$ either way.

(b) Force from the slope. The slope is $\tfrac{dU}{dx} = 4x$, so the force is $F = -\tfrac{dU}{dx} = -4x$. At $x = 1.5$ m, $F = -6$ N, pointing toward $x = 0$. Equilibrium is where the slope is flat: $4x = 0$ gives $x = 0$, not the point where $U$ is smallest in size.

(c) Which forces qualify. Friction is not conservative, so it stores no potential energy. The $-8$ J is energy removed as heat, not a change in any $U$. There is simply no spring or gravity term to write for it.

The traps to avoid. Do not treat $U$ as an absolute number or drop its sign (report $\Delta U$, and judge boundness from $E = K + U$, not from the sign of $U$), do not read the force as the value of $U$ (it is minus the slope, $-6$ N here), and do not hand friction a potential energy.