Gravitational and Elastic Potential Energy

In 1638, under house arrest and half blind, Galileo published the observation that started all of energy physics: a ball rolling down any ramp arrives at the bottom with a speed that depends only on the height it descended — not on the slope, the shape, or the time taken. Drop it, roll it, spiral it down a helix: same height, same final speed. Nature, it turns out, keeps an account book in which position alone fixes the balance. We now call that balance potential energy, and this article is about its two workhorse entries: the energy of a raised mass, $U_g = mgy$ , and the energy of a deformed spring, $U_s = \tfrac12 kx^2$ . The two formulas could hardly look more different — one linear in position, one quadratic, one with a freely movable zero, one pinned to the spring's natural length — yet they are the same idea twice: the work you must do against a conservative force, banked in the configuration of a system and repayable in full, on demand, as kinetic energy. Between them they price almost everything in introductory mechanics: pile drivers and pendulums, dams and drawn bows, roller coasters and rock climbers' ropes. Master these two and you inherit the energy method — the trick of skipping Newton's vector equations entirely and reading speeds, turning points, and equilibria straight off a scalar function. A hydroelectric reservoir is $mgy$ at grid scale; the quartz tuning fork timing your watch is $\tfrac12 kx^2$ at the micron scale. Same account book, twenty orders of magnitude apart.

Key takeaways

Potential energy is banked work: for a conservative force, $U(\mathbf r) = -\int \mathbf F\cdot\mathrm{d}\mathbf r$ depends only on position, and the force is recovered as the downhill slope, $F = -\mathrm{d}U/\mathrm{d}x$ .
Near Earth's surface the gravitational entry is $U_g = mgy$ — linear in height, path-independent, with a zero level you may place anywhere because only differences $\Delta U$ are physical.
A Hooke's-law spring stores $U_s = \tfrac12 kx^2$ , where $x$ is measured from the natural length — this zero is not movable, and the energy is positive for stretch and compression alike.
$mgy$ is itself an approximation: expanding the exact $U = -GMm/r$ for heights $h \ll R_\oplus$ recovers $mgh$ with $g = GM/R^2$ ; the error is about $h/R$ — already 6% at ISS altitude.
Hanging a mass on a vertical spring does not change the spring physics: gravity only shifts the equilibrium by $mg/k$ , and measured from the new equilibrium the potential is again $\tfrac12 k u^2$ plus a constant.
Key idea: for a conservative force the work done is minus the change of a position-only function $U$ ; with $U_g = mgy$ and $U_s = \tfrac12 kx^2$ in hand, conservation of $E = K + U$ turns force-and-acceleration problems into one-line energy balances.

The core insight

For a conservative force the work done is minus the change in a position-only function $U$ ; setting $E = K + U$ constant turns force-and-acceleration problems into one-line energy balances, with $U_g = mgy$ and $U_s = \tfrac12 kx^2$ the two foundational examples.

Where this lives today

The world's largest batteries are still gravitational. Pumped-storage plants like Bath County in Virginia — the biggest on Earth — pump water uphill when electricity is cheap and let it fall through turbines when the grid is hungry: about $4.4\times10^{10}\ \mathrm{kg}$ of water behind the upper dam at a $380\ \mathrm{m}$ head is an $mgy$ deposit of $1.6\times10^{14}\ \mathrm{J}$ , roughly $45\ \mathrm{GWh}$ gross — the scale of tens of millions of phone batteries, stored as nothing more exotic than raised water. The catch is dilution: at $380\ \mathrm{m}$ of head, each kilogram banks only $gh \approx 3.7\ \mathrm{kJ}$ , about 250 times less than a kilogram of lithium-ion cell — which is why gravity storage needs mountains and lakes, and your phone does not. The elastic store went the other direction: miniature and precise. The quartz resonator timing your watch, the MEMS accelerometer that flips your phone screen, and the atomic-force-microscope cantilever that images single molecules are all engineered parabolic valleys — $\tfrac12 k_{\mathrm{eff}}x^2$ with spring constants tuned from thousands of newtons per metre down to thousandths. And the "any minimum is a spring" fact from the limiting-cases table is the working assumption of modern computational chemistry: near equilibrium, every chemical bond is modeled as a spring whose $k = U''(x_0)$ is read off the quantum potential-energy surface, so the vibrational spectrum of a molecule is a table of effective spring constants.

The physical picture

Lift a brick slowly to a shelf: every joule your hand supplies is deposited in the Earth–brick system, and the account pays out in full — as kinetic energy — the moment the brick falls. Draw a bow: the work of your arm is coiled into the bent limbs, and the arrow collects it on release. Neither deposit cares how it was made. Lift the brick straight up or walk it up a staircase; draw the bow fast or agonizingly slowly — the stored energy depends only on the final configuration. That indifference to history is what makes a force conservative, and it is the only license you need to define potential energy at all.

Figure

Two ways to bank energy — a raised mass stores mgy, a deformed spring stores one half k x squared

The most useful mental image is the energy landscape: plot $U$ against position and imagine the system as a ball rolling on that terrain. Gravity's landscape near the surface is a featureless ramp of constant slope $mg$ ; the spring's is a parabolic valley centered on the natural length. The force at any point is the downhill steepness, $F = -\mathrm{d}U/\mathrm{d}x$ : the ramp pushes one way everywhere, the valley always pushes back toward the bottom. Conservation of energy draws a horizontal line at height $E$ across the landscape — the system can only ever be where the terrain lies below that line, it moves fastest where the terrain is lowest, and it turns around exactly where the line meets the ground.

Figure

The two energy landscapes — a constant-slope ramp for gravity, a parabolic valley for the spring

One asymmetry between the two stores deserves attention before the algebra, because it is the most common source of wrong answers. Gravity's zero level is yours to choose — floor, tabletop, sea level — because only changes in $mgy$ ever appear in the physics. The spring's zero is not negotiable: $x$ in $\tfrac12 kx^2$ is the deformation from the natural, unstretched length, because that is the unique configuration where the spring pushes on nothing. Move that origin and the formula simply stops being true.

From work to a bank account

Potential energy is defined by the work a conservative force does. To make the idea precise we start from the general work integral and turn it around: instead of asking how much work the force does along some path, we ask what function of position alone could summarize that work once and for all. Such a function exists exactly when the work is path-independent — which both uniform gravity and the ideal spring satisfy.

U(\mathbf r) = -\int_{\mathbf r_0}^{\mathbf r} \mathbf F\cdot\mathrm{d}\mathbf r \qquad (1)

where: $U(\mathbf r)$ — potential energy at position $\mathbf r$ ; $\mathbf r_0$ — the reference configuration where $U \equiv 0$ ; $\mathbf F$ — the conservative force whose work is being banked. The minus sign implements the accounting convention: when the force does positive work (the brick falls, the spring relaxes), the account is drawn down, so $U$ decreases. Change the reference point $\mathbf r_0$ and you shift $U$ everywhere by one constant — which is why the zero is physically empty freedom. Once the account exists, the force can be read back out of it. Differentiating equation (1) undoes the integral, and in more than one dimension the derivative that appears is the gradient — the direction of steepest increase of $U$ , with the force pointing the opposite way, downhill:

\mathbf F = -\nabla U, \qquad F_x = -\frac{\mathrm{d}U}{\mathrm{d}x} \qquad (2)

where: $\nabla U$ — the gradient of the potential; $F_x$ — the one-dimensional case used throughout this article. This is the landscape picture made literal: steep terrain means strong force, flat terrain means equilibrium. It also explains instantly why adding a constant to $U$ changes nothing — the derivative of a constant is zero, so the force, and hence all the motion, is untouched. With the machinery in place, the two stars of this article are one integral each. For sizing them up quickly:

situation	use	zero point	shape
mass at height $y$ , $h \ll R_\oplus$	$U_g = mgy$	anywhere you like	straight ramp, slope $mg$
spring deformed by $x$	$U_s = \tfrac12 kx^2$	natural length only	parabola, curvature $k$
altitude not small vs $R_\oplus$	$U = -GMm/r$	infinity (convention)	$-1/r$ well
both at once (vertical spring)	$\tfrac12 ku^2 + \text{const}$	new equilibrium	same parabola, relocated

Derivation AGravitational potential energy from the work integral

We are deriving $U_g = mgy$ , the potential energy of a mass in the uniform field near Earth's surface, directly from the definition (1). The strategy is to compute the work done by the constant force $\mathbf F = -mg\,\hat{\mathbf y}$ along a completely arbitrary path from the reference point to the final position, and watch the horizontal part of the path drop out of the dot product — that disappearance is the path-independence that licenses a potential energy in the first place. The step to watch is the dot product itself: gravity is vertical, so only the vertical component of each path element survives.

Step 1. Write the work done by gravity along an arbitrary path from the origin (our reference, $U = 0$ ) to the point $(x, y)$ , chopping the path into elements $\mathrm{d}\mathbf r = \mathrm{d}x\,\hat{\mathbf x} + \mathrm{d}y\,\hat{\mathbf y}$ :

W_{\mathrm{grav}} = \int (-mg\,\hat{\mathbf y})\cdot(\mathrm{d}x\,\hat{\mathbf x} + \mathrm{d}y\,\hat{\mathbf y}) = -mg\int_0^{y}\mathrm{d}y'

Why the horizontal motion vanishes: $\hat{\mathbf y}\cdot\hat{\mathbf x} = 0$ , so every sideways step is free — gravity neither helps nor hinders it. The integral no longer knows anything about the path's shape, only its net rise $y$ .

Step 2. Evaluate the surviving integral and apply the definition (1), $U = -W_{\mathrm{grav}}$ :

W_{\mathrm{grav}} = -mgy \quad\Longrightarrow\quad U_g(y) = mgy

Why the sign comes out right: raising the mass means gravity did negative work on it, so the account balance $U$ went up — energy was deposited.

Step 3. Check the reference-level freedom explicitly. Measuring height from a level shifted by $y_c$ replaces $y \to y - y_c$ , so $U \to mgy - mgy_c$ : the same function plus a constant. By equation (2) the force is unchanged, and in any energy balance the constant cancels between the two sides. Only $\Delta U = mg\,\Delta y$ is ever physical. For example, a 2 kg ball lifted 3 m gains $\Delta U = 2 \times 9.81 \times 3 = 58.9\ \mathrm{J}$ regardless of path — the direct route and any staircase route give the same answer.

Go deeper · Derivation B · 1 minElastic potential energy and why it is quadratic

Derivation BElastic potential energy and why it is quadratic

We are deriving $U_s = \tfrac12 kx^2$ for an ideal Hooke's-law spring, again straight from the definition (1). The route is a one-dimensional integral of the restoring force $F = -kx$ from the natural length out to deformation $x$ . Two features deserve watching: where the $\tfrac12$ comes from — the force is not constant, so the work is the area of a triangle, not a rectangle — and why the result cannot be negative, which is what pins the zero of elastic energy to the natural length with no freedom of choice.

Step 1. Integrate the definition with $F(x') = -kx'$ , taking the natural length $x' = 0$ as the reference where $U = 0$ :

U_s(x) = -\int_0^{x}(-kx')\,\mathrm{d}x' = k\int_0^{x} x'\,\mathrm{d}x' = \tfrac12 kx^2

Why the half: the opposing force grows from $0$ at the start of the pull to $kx$ at the end, so the work is the area under a straight ramp — a triangle of base $x$ and height $kx$ , area $\tfrac12(x)(kx)$ . A constant force $kx$ over the whole distance would have cost twice as much.

Step 2. Read the shape. The square makes $U_s$ an even function: compression ( $x < 0$ ) and stretch ( $x > 0$ ) of equal size store identical energy, and $U_s \ge 0$ always. The spring's account can never be overdrawn — every deformation is a deposit. Why the zero is not movable: $x$ is not a coordinate of the mass in space but the physical deformation of the spring. The configuration $x = 0$ is singled out by the hardware — it is the one shape in which the spring exerts no force. Redefining $x$ from anywhere else breaks $F = -kx$ itself, not just the potential.

Step 3. Check consistency with equation (2): $F = -\mathrm{d}U_s/\mathrm{d}x = -kx$ . The parabola's slope reproduces Hooke's law, and its curvature is the stiffness: $U_s''(x) = k$ everywhere. Doubling the stretch quadruples the stored energy — both the force you fight and the distance you fight it over have doubled. As a concrete check: a $k = 200\ \mathrm{N/m}$ spring stretched $0.10\ \mathrm{m}$ stores $\tfrac12 \times 200 \times 0.01 = 1.0\ \mathrm{J}$ ; stretched $0.20\ \mathrm{m}$ it stores $4.0\ \mathrm{J}$ — doubling the stretch quadruples the energy, as the quadratic formula demands.

When mgy runs out — the universal potential

The ramp landscape cannot go on forever: $U = mgy$ predicts unlimited energy gain with altitude, yet a rocket that climbs far enough plainly feels gravity weaken. The honest potential energy of a mass $m$ at distance $r$ from Earth's center comes from Newton's law of universal gravitation, taking the conventional zero at infinite separation. It is worth displaying before we show that $mgy$ hides inside it:

U(r) = -\frac{GMm}{r} \qquad (3)

where: $G$ — the gravitational constant; $M$ — Earth's mass; $r$ — distance from Earth's center (not altitude). The minus sign says every finite separation is an energy debt relative to infinity: you must pay $GMm/r$ to remove the mass completely. Near the surface we never see the debt itself — only its tiny variations with height, and those variations are exactly $mgh$ .

Go deeper · Derivation C · 1 minRecovering mgh from the universal potential

Derivation CRecovering mgh from the universal potential

We are deriving the near-surface formula $U \approx \text{const} + mgh$ as the leading approximation to the exact potential (3), and identifying $g = GM/R^2$ along the way. The strategy is a Taylor expansion of $-GMm/r$ about the surface $r = R$ in the small parameter $h/R$ , where $h$ is altitude. This derivation earns its keep twice: it tells you why $mgh$ works, and — through the size of the first neglected term — exactly when it stops working.

Step 1. Substitute $r = R + h$ into equation (3) and factor out the surface value:

U(R+h) = -\frac{GMm}{R+h} = -\frac{GMm}{R}\,\frac{1}{1 + h/R}

Why this form: the physics of "near the surface" is precisely the statement $h/R \ll 1$ , so we want the small ratio $h/R$ sitting alone where an expansion can reach it.

Step 2. Expand the geometric factor to first order, $(1 + h/R)^{-1} \approx 1 - h/R$ for $h \ll R$ :

U(R+h) \approx -\frac{GMm}{R} + \frac{GMm}{R^2}\,h

Why we may stop at first order: the next term is smaller by another factor of $h/R$ — about $10^{-3}$ for a 6 km mountain flight, since $R = 6371\ \mathrm{km}$ .

Step 3. Read off the pieces. The first term is a constant — by the reference-level freedom established above, it drops out of every energy balance. The coefficient of $h$ in the second term is a force per unit mass evaluated at the surface:

g = \frac{GM}{R^2} \approx 9.81\ \mathrm{m/s^2} \quad\Longrightarrow\quad U \approx \text{const} + mgh

Step 4. Price the approximation. The fractional error of $mgh$ is of order $h/R$ : about 1% by $h \approx 64\ \mathrm{km}$ , and at the International Space Station's altitude ( $h = 400\ \mathrm{km}$ ) the formula overestimates the true energy difference by about 6% — use equation (3) there. The numbers bear this out: $GM = 3.986\times10^{14}\ \mathrm{m^3/s^2}$ and $R = 6.371\times10^6\ \mathrm{m}$ give $GM/R^2 = 9.82\ \mathrm{m/s^2}$ , and the exact per-kilogram energy to ISS altitude is $GM/R - GM/(R+h) = 3.70\ \mathrm{MJ/kg}$ versus the flat-Earth estimate $gh = 3.92\ \mathrm{MJ/kg}$ — a 6% overshoot.

Combining the two — the hanging spring

Real systems rarely offer one store at a time: a mass hanging from a vertical spring banks gravitational and elastic energy simultaneously, and the bookkeeping of the two together produces one of the tidiest results in mechanics. Measure $x$ downward from the spring's natural length, so the elastic term keeps its legal zero. The total potential is then the sum of a parabola and a ramp:

U_{\mathrm{tot}}(x) = \tfrac12 kx^2 - mgx \qquad (4)

where: the elastic term uses the true deformation $x$ ; the gravitational term is $-mgx$ because moving down (increasing $x$ ) lowers the mass and withdraws gravitational energy. Adding a ramp to a parabola cannot change its shape — it can only slide the valley. Where the new bottom lands, and what the landscape looks like around it, is the next derivation.

Figure

Adding gravity to a spring slides the parabolic valley down by mg over k without changing its shape

Go deeper · Derivation D · 1 minA hanging mass on a spring — gravity only moves the zero

Derivation DA hanging mass on a spring — gravity only moves the zero

We are deriving the equilibrium position of a mass on a vertical spring and showing that, measured from that new equilibrium, the combined gravitational-plus-elastic potential is again a pure spring parabola, $\tfrac12 ku^2$ , up to an ignorable constant. The method is completing the square on equation (4) — no new physics, just algebra — and the payoff is the standard exam license to ignore gravity entirely in vertical spring problems, provided displacements are measured from the hanging equilibrium. Watch how the linear terms cancel identically: that cancellation is gravity being absorbed into the choice of origin.

Step 1. Locate the equilibrium from equation (2): set $U_{\mathrm{tot}}'(x) = kx - mg = 0$ , giving the static stretch

x_0 = \frac{mg}{k}

Why this is familiar: it is just force balance — the spring force $kx_0$ holding up the weight $mg$ . The landscape view and the force view agree, as equation (2) guarantees they must.

Step 2. Complete the square in equation (4) about $x_0$ , substituting $u = x - x_0$ for the displacement from equilibrium:

U_{\mathrm{tot}} = \tfrac12 k(x - x_0)^2 - \tfrac12 kx_0^2 = \tfrac12 ku^2 - \frac{m^2g^2}{2k}

Why the cross terms vanish: the linear part of the parabola's expansion, $kx_0 u$ , exactly cancels the gravitational ramp $-mgu$ — by construction, since $x_0$ was chosen to zero the net force. Gravity has been completely absorbed into the shift.

Step 3. Discard the constant $-m^2g^2/2k$ by the reference-level freedom. What remains is a pure Hooke's-law valley of the same curvature $k$ , centered on the hanging equilibrium. The oscillation frequency about the new bottom is therefore unchanged: $\omega = \sqrt{k/m}$ , exactly as for a horizontal spring. As a numerical illustration: for $m = 2.0\ \mathrm{kg}$ and $k = 1960\ \mathrm{N/m}$ , the static sag is $x_0 = 2.0 \times 9.81/1960 = 1.00\ \mathrm{cm}$ , and the oscillation frequency $\sqrt{1960/2.0} = 31.3\ \mathrm{rad/s}$ is identical with or without gravity — exactly as the completed-square result predicts.

Limiting cases

case	result	meaning
$h \ll R_\oplus$	$-GMm/r \to \text{const} + mgh$ , $g = GM/R^2$	the uniform-field ramp is the first Taylor term of the true $-1/r$ well
$h \sim R_\oplus$ and beyond	$mgh$ overestimates; use $-GMm/r$	error $\sim h/R$ : 1% at 64 km, 6% at ISS altitude
small deformation	$F = -kx$ , $U = \tfrac12 kx^2$	Hooke's linear regime; real springs go anharmonic, then yield
any smooth $U$ near a stable minimum	$U \approx U(x_0) + \tfrac12 U''(x_0)(x - x_0)^2$	every gentle valley is a spring: $k_{\mathrm{eff}} = U''(x_0)$
vertical spring, $u$ from hanging equilibrium	$U = \tfrac12 ku^2 + \text{const}$	gravity shifts the valley by $mg/k$ but does not reshape it
$g \to 0$ or $k \to 0$	$U = \text{const}$	no force, no store: motion is uniform (Newton's first law)

Common trapThe spring's zero is wherever it is convenient — no, it is the natural length, always

Gravity's reference level really is yours: floor, bench, sea level — pick what simplifies the arithmetic, because only $\Delta(mgy)$ is physical. Students then extend the same courtesy to the spring, measuring $x$ from a stretched starting position or from the hanging equilibrium while still writing $\tfrac12 kx^2$ — and the energy audit breaks. The elastic formula is only true with $x$ the deformation from the natural length, because that is the configuration the hardware itself singles out as force-free. The one legal shortcut is the completed-square result above: from the hanging equilibrium you may write $\tfrac12 ku^2$ only if you also drop gravity from the books entirely — the two errors cancel by construction. Mixing conventions ( $\tfrac12 ku^2$ plus $mgy$ terms) double-counts gravity and is the single most common wrong answer in vertical-spring problems.

Worked examples

Example 1 — banking the same energy two ways (confidence builder). A crane lifts a $120\ \mathrm{kg}$ steel beam $18\ \mathrm{m}$ from the ground to a rooftop. How much energy is banked, and what compression of a stiff industrial buffer spring ( $k = 4.0\times10^5\ \mathrm{N/m}$ ) would store the same amount?

Step 1. The gravitational deposit is $\Delta U_g = mg\,\Delta y = 120 \times 9.81 \times 18 = 21{,}190\ \mathrm{J} \approx 21.2\ \mathrm{kJ}$ . Path irrelevant: swinging on the way up changes nothing.

Step 2. Demand the same balance from the spring: $\tfrac12 kx^2 = 21{,}190\ \mathrm{J}$ , so $x = \sqrt{2 \times 21{,}190 / (4.0\times10^5)} = \sqrt{0.106}\ \mathrm{m}$ .

Step 3. Evaluate: $x = 0.325\ \mathrm{m}$ . A third of a metre of compression on a spring stiff enough to stop a truck, against eighteen metres of lift — a first taste of how dense elastic storage is over short distances, and how dilute gravity's ramp is.

Answer: $\Delta U_g \approx 2.1\times10^4\ \mathrm{J}$ ; the buffer spring must compress $\approx 0.33\ \mathrm{m}$ .

Example 2 — a block dropped onto a spring (exam level). A $2.0\ \mathrm{kg}$ block is released from rest $0.50\ \mathrm{m}$ above the top of a relaxed vertical spring with $k = 1960\ \mathrm{N/m}$ . Find the maximum compression of the spring, and compare it with the static equilibrium compression.

Step 1. Choose the moment of maximum compression as the target: there the block is momentarily at rest, so all the energy is potential. Put the gravitational zero at the top of the relaxed spring and measure compression $x$ downward — the block's total descent is then $h + x$ with $h = 0.50\ \mathrm{m}$ .

Step 2. Balance the books between release (at rest, height $h$ above the zero) and maximum compression (at rest, depth $x$ below it): $mg(h + x) = \tfrac12 kx^2$ . Both stores appear: gravity has been drawn down by the full descent, the spring has banked all of it.

Step 3. Insert numbers: $2.0 \times 9.81 \times (0.50 + x) = 980\,x^2$ , i.e. $980\,x^2 - 19.62\,x - 9.81 = 0$ .

Step 4. Solve the quadratic: $x = \dfrac{19.62 + \sqrt{19.62^2 + 4 \times 980 \times 9.81}}{2 \times 980} = \dfrac{19.62 + 197.1}{1960} = 0.111\ \mathrm{m}$ . (The negative root is a phantom above the spring — discard it.)

Step 5. Sanity-check the energy audit: descent $= 0.611\ \mathrm{m}$ gives $mg(h+x) = 12.0\ \mathrm{J}$ ; the spring holds $\tfrac12 \times 1960 \times 0.111^2 = 12.0\ \mathrm{J}$ . Balanced. The static equilibrium compression is far smaller: $x_0 = mg/k = 19.62/1960 = 1.0\ \mathrm{cm}$ .

Answer: maximum compression $\approx 0.111\ \mathrm{m}$ — eleven times the static compression $x_0 = 1.0\ \mathrm{cm}$ . At the lowest point the spring pushes with $kx = 217\ \mathrm{N}$ , about 11 times the block's weight — which is why the block does not stop there but rebounds hard: maximum compression is a turning point, not an equilibrium.

The people behind the idea

Origins · 1 minGalileo Galilei and how this was discovered

Galileo Galilei

1564–1642 · Pisa and Padua — the height-alone insight that became potential energy

Galileo never wrote the word "energy," but he discovered its accounting rule. Rolling balls down inclines of every slope in his Padua workshop — with water clocks and his own pulse for timing — he established that the speed acquired depends only on the vertical drop, and that a pendulum released from a given height climbs back to the same height even when a nail interrupts the string. Height in, height out: the account always balances. Published in the Discorsi (1638) while he was under house arrest in Arcetri, this "height-alone" principle is precisely the path-independence that makes gravity conservative — the license behind $U = mgy$ . What Galileo established with ramps and pendulums, later generations formalized: the account book he opened is the one every energy method still balances today.

History

1638

— Galileo's Discorsi establishes that a falling body's speed depends only on the height descended, and a pendulum reclaims its release height — path-independence before the concept existed

1676

— Hooke publishes the anagram ceiiinosssttuv, unscrambled two years later as ut tensio, sic vis: the linear spring law behind the elastic store

1687

— Newton's Principia gives the inverse-square law of universal gravitation, fixing the true potential that mgy approximates

1807

— Thomas Young introduces the word energy for vis viva; Coriolis and Poncelet make work equal force times distance a quantitative tool in the 1820s

1853

— Rankine coins the term potential energy, completing the modern vocabulary of stored-versus-moving energy

Going further

Texts. Morin, Introduction to Classical Mechanics, ch. 5 (Conservation of energy and momentum) · Taylor, Classical Mechanics, ch. 4 (Energy) · Kleppner & Kolenkow, An Introduction to Mechanics, ch. 5 · Landau & Lifshitz, Mechanics, §§6–7, 21. The key results can be traced directly to the boxed formulas: $U_g = mgy$ and $U_s = \tfrac12 kx^2$ follow from integrating $F = -mg$ and $F = -kx$ via $U = -\int \mathbf F\cdot\mathrm{d}\mathbf r$ ; the expansion $-GMm/r \to \text{const} + mgh$ with $g = GM/R^2 \approx 9.81\ \mathrm{m/s^2}$ carries a $\sim h/R$ error (6% at $h = 400\ \mathrm{km}$ ); and the vertical-spring completed square gives $x_0 = mg/k$ , $U = \tfrac12 ku^2 + \text{const}$ , with unchanged $\omega = \sqrt{k/m}$ . The worked-example numbers ( $\Delta U_g = 21.2\ \mathrm{kJ}$ and $x = 0.325\ \mathrm{m}$ for the beam-and-buffer; $x_{\max} = 0.111\ \mathrm{m}$ vs $x_0 = 1.0\ \mathrm{cm}$ , peak force $217\ \mathrm{N} \approx 11mg$ , energy balance $12.0\ \mathrm{J} = 12.0\ \mathrm{J}$ for the dropped block; Bath County $1.6\times10^{14}\ \mathrm{J} \approx 45\ \mathrm{GWh}$ gross; water at 380 m head $\approx 3.7\ \mathrm{kJ/kg}$ ) are illustrative arithmetic drawn from those same formulas.

Builds on: Newton's Second Law · Gravitational Force Near Earth's Surface · Position, Displacement, and Reference Frames · Vectors in Kinematics and Coordinate Systems. Leads to: Circular Motion and Centripetal Acceleration · Projectile Motion.

If I can put the zero of gravitational PE anywhere, how can potential energy be a real physical thing?

Because the thing is never $U$ itself — it is $\Delta U$ , and differences are immune to where you put the zero. Every measurable consequence (speeds from energy balances, forces from slopes, turning points from $E = U$ ) involves either a difference or a derivative of $U$ , and both operations kill an added constant. It is exactly like altitude: "300 m" is meaningless until someone says above what — sea level, the runway — yet "the plane climbed 300 m" is a hard physical fact under every convention. The spring is the instructive contrast: there the hardware itself defines a special configuration (the natural length, where the force vanishes), so nature, not you, fixes that zero.

Why does the spring energy have a one-half but mgh has none?

Because gravity is the same $mg$ during the whole lift, while the spring fights back harder the further you pull. Work is force times distance only for a constant force; lift a brick by $h$ and you pay $mg \times h$ , no discount, no surcharge. The spring starts at zero force and ends at $kx$ , so on a force-versus-position graph the work is a triangle — base $x$ , height $kx$ — whose area is $\tfrac12 kx^2$ . A rectangle versus a triangle: that is the entire story of the $\tfrac12$ . The same logic explains the average-force shortcut: the spring's mean opposition over the pull is $\tfrac12 kx$ , and $(\tfrac12 kx) \times x$ gives the half automatically.

Where is the energy actually stored — in the brick, in the Earth, in the spring metal?

For gravity, strictly in the configuration of the Earth–brick pair (in field language: in the gravitational field). Saying "the brick has PE" is a harmless shorthand because Earth barely moves, but only the pair has a well-defined $mgy$ — a brick alone in empty space stores nothing. The spring is more concrete: the energy lives in the elastically strained bonds of the metal itself. Stretch steel and interatomic separations shift off their equilibrium; each distorted bond is a microscopic parabolic valley holding its share of $\tfrac12 kx^2$ . That is also why the elastic store has an absolute zero (bonds relaxed, natural length) while the gravitational one does not.

Can potential energy be negative, and does that mean anything is wrong?

Nothing is wrong — negative $U$ just means "below your chosen zero." Put $y = 0$ at the tabletop and the floor sits at $U = -mg \times 0.8\ \mathrm{m}$ ; all predictions come out identical to any other choice. One negative sign is meaningful though: the universal potential $-GMm/r$ is negative everywhere by the convention $U(\infty) = 0$ , and there the sign encodes being bound — you owe $GMm/r$ of work to escape to infinity. A satellite with total $E < 0$ is trapped in the well; $E \ge 0$ escapes. Elastic energy, by contrast, is genuinely never negative: $\tfrac12 kx^2 \ge 0$ because any deformation, either direction, is a deposit.

For a vertical spring, why am I allowed to completely ignore gravity if I measure from the hanging equilibrium?

Because completing the square has already spent gravity on relocating the valley. The combined potential $\tfrac12 kx^2 - mgx$ is algebraically identical to $\tfrac12 ku^2 + \text{const}$ with $u$ measured from the sag point $x_0 = mg/k$ — the linear gravity term cancels exactly against the linear part of the shifted parabola, by the very equation ( $kx_0 = mg$ ) that defines equilibrium. So "spring-only bookkeeping from the new equilibrium" is not an approximation; it is the same function renamed. The frequency $\omega = \sqrt{k/m}$ is untouched because a ramp has no curvature to add. The only sin is mixing conventions — writing $\tfrac12 ku^2$ and an $mgy$ term double-counts gravity.

Does the falling ball somehow know the path does not matter, even before it lands?

The ball knows only the local force; path-independence is a property of that force field, not a computation the ball performs. Uniform gravity is vertical, so each step of any route is billed only for its vertical component — horizontal excursions are free of charge, and every route between the same two points contains the same net rise. The honest caution: this is special, not automatic. Add air drag, which always opposes the motion, and longer paths genuinely cost more — no potential energy can be defined for drag, and the energy method loses its license. "Conservative force" is precisely the certificate that the path-blindness holds.

Is the potential energy of a pendulum gravitational or elastic — and what about a bouncing ball?

A pendulum on a rigid rod is purely gravitational: $U = mgy$ with $y$ the bob's height; the rod's tension does no work (always perpendicular to the motion). A bouncing ball is the two stores taking turns: gravitational on the way down, then — during the millisecond of contact — the ball deforms and the energy goes elastic, stored in the squashed rubber as an effective $\tfrac12 k_{\mathrm{eff}}x^2$ , before being repaid as kinetic energy on the rebound. An imperfect bounce is an honest audit of the elastic stage: whatever fraction the deformation fails to return has leaked to heat and sound, and the ball reclaims correspondingly less height on each cycle.

My textbook says any potential minimum acts like a spring. Why is that, and why should I care?

Taylor-expand any smooth $U(x)$ about a stable equilibrium $x_0$ : the constant is ignorable, the linear term vanishes because $x_0$ is an equilibrium ( $U'(x_0) = 0$ ), and the first surviving term is $\tfrac12 U''(x_0)(x - x_0)^2$ — formally identical to a spring with $k_{\mathrm{eff}} = U''(x_0)$ . Every gentle valley, whatever its global shape, is parabolic at the bottom. That is why small oscillations are universally simple harmonic — pendulums, molecules, buildings, quartz crystals — with $\omega = \sqrt{U''(x_0)/m}$ , and why the humble $\tfrac12 kx^2$ you learn here reappears in every physics course you will ever take, up to and including quantum field theory.

Gravitational and Elastic Potential Energy

Where this lives today

The physical picture

From work to a bank account

When mgy runs out — the universal potential

Combining the two — the hanging spring

Limiting cases

Worked examples

The people behind the idea

Going further

Before this

Where this leads