Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²): Worked examples

This page is a complete tour of cases. The parent note built the three formulas. Here we make sure you never meet a situation you haven't already seen solved: every sign, every reference choice, the degenerate inputs (zero height, zero stretch, infinite distance), the limiting behaviours, a real-world word problem, and an exam-style twist.

First, the toolbox we are allowed to use (all three earned in the parent):

The scenario matrix

Every problem this topic can throw is one of these cells. The Example column tells you which worked case below covers it.

Cell	What varies	Trap / edge behaviour	Example
G1	Flat gravity, going up	$Δ U > 0$	(a)
G2	Flat gravity, going down / below origin	$Δ U < 0$ , $h$ negative	(a)
G3	Full gravity, moving inward vs outward	$U < 0$ always, but $Δ U$ sign flips	(b)
G4	Reference choice ( $h = 0$ vs $r = \infty$ )	same physics, different $U$ values	(c)
G5	Limiting: $- GM m / r \to m g h$ as $h ≪ R$	approximation, not equality	(d)
S1	Spring stretched ( $x > 0$ )	$U > 0$	(e)
S2	Spring compressed ( $x < 0$ )	$U > 0$ still, symmetry	(e)
S3	Spring degenerate ( $x = 0$ )	$U = 0$ , force $= 0$	(e)
F1	Force from a general $U (x)$	sign of $- d U / d x$ , equilibria	(f)
W1	Real-world word problem (energy conservation)	multiple stores add up	(g)
X1	Exam twist (mixed spring + gravity, or graph)	which reference, don't double-count	(h)

Worked examples

Worked example (a) Flat gravity — up

and below the floor · cells G1, G2 A $2 kg$ box starts on a table at $h = 0$ . (i) It is lifted to $h = + 1.5 m$ . (ii) From the table it is lowered into a pit to $h = - 0.8 m$ . Find $Δ U$ for each ( $g = 9.8$ ).

Forecast: Guess the signs first. Up should store energy ( $+$ ); dropping below the reference should give negative PE. Do the numbers agree?

Up: $Δ U_{i} = m g h = 2 \times 9.8 \times (+ 1.5) = + 29.4 J$ . Why this step? Displacement is up, gravity is down → gravity does negative work → we bank positive energy.
Down: $Δ U_{ii} = m g h = 2 \times 9.8 \times (- 0.8) = - 15.68 J$ . Why this step? Below the reference $h$ is negative. Nothing special happens to the formula — the sign of $h$ carries the physics. PE below your chosen zero is genuinely negative.

Verify: Units: $kg \cdot (m/s^{2}) \cdot m = J$ . ✓ Case (i) positive, (ii) negative — matches the forecast. A box in a pit would speed up if released upward toward the table, consistent with lower PE at the bottom.

Worked example (b) Full gravity — inward vs outward · cell G3

Earth: $GM = 3.986 \times 1 0^{14} m^{3} / s^{2}$ , radius $R = 6.371 \times 1 0^{6} m$ . A $1000 kg$ satellite moves from $r_{1} = 7.0 \times 1 0^{6} m$ inward to $r_{2} = 6.6 \times 1 0^{6} m$ . Find $Δ U$ .

Forecast: Both $U$ values are negative (well). Moving inward means falling deeper → $Δ U$ should be negative. Guess the magnitude: order $1 0^{9} J$ ?

$U (r) = - \frac{GM m}{r}$ , so $U_{1} = - \frac{3.986 \times 1 0 ^{14} \times 1000}{7.0 \times 1 0 ^{6}} = - 5.694 \times 1 0^{10} J$ . Why this step? Reference is infinity, so bound satellites have negative $U$ .
$U_{2} = - \frac{3.986 \times 1 0 ^{14} \times 1000}{6.6 \times 1 0 ^{6}} = - 6.039 \times 1 0^{10} J$ .
$Δ U = U_{2} - U_{1} = - 6.039 \times 1 0^{10} - (- 5.694 \times 1 0^{10}) = - 3.45 \times 1 0^{9} J$ . Why this step? Deeper in the well = more negative = energy released ( $Δ U < 0$ ), which becomes kinetic energy of the fall.

Verify: $Δ U$ is negative ✓ (moved inward). Magnitude $\sim 3 \times 1 0^{9} J$ , matching the forecast. Both $U$ values negative ✓.

Worked example (c) Same physics, two references · cell G4

Show that the change in PE from the surface to $h = 100 km$ is identical whether you use $m g h$ (reference at surface) or $- GM m / r$ (reference at infinity). Use $m = 1 kg$ , $g = 9.8$ , $GM = 3.986 \times 1 0^{14}$ , $R = 6.371 \times 1 0^{6} m$ , $h = 1.0 \times 1 0^{5} m$ .

Forecast: The values of $U$ differ wildly (one starts at 0, the other at a huge negative). But the difference $Δ U$ must nearly match — only differences are physical.

Flat: $Δ U_{flat} = m g h = 1 \times 9.8 \times 1.0 \times 1 0^{5} = 9.80 \times 1 0^{5} J$ .
Full: $Δ U_{full} = - GM m (\frac{1}{R + h} - \frac{1}{R})$ $= - 3.986 \times 1 0^{14} (\frac{1}{6.471 \times 1 0 ^{6}} - \frac{1}{6.371 \times 1 0 ^{6}}) = 9.66 \times 1 0^{5} J$ . Why this step? Absolute $U$ depends on reference, but subtracting cancels the constant offset — the physical quantity survives.

Verify: $9.80 \times 1 0^{5}$ vs $9.66 \times 1 0^{5}$ — agree to $\sim 1.5%$ . ✓ The small gap is because $g$ weakens slightly with altitude, which the flat formula ignores. This is the reconciliation the parent note promised, made numeric.

Worked example (d) The limiting case, made honest · cell G5

The parent note wrote $- GM m / r \approx m g h$ for small $h$ . Quantify "small": at what height does the flat formula err by $1%$ ? Use $R = 6.371 \times 1 0^{6} m$ .

Forecast: The error grows with $h / R$ . A $1%$ error should need $h$ around a few percent of $R$ — tens to a hundred km.

Exact factor: full $Δ U = m g h \cdot \frac{R}{R + h}$ (algebra of $\frac{1}{R} - \frac{1}{R + h} = \frac{h}{R ( R + h )}$ , times $GM m = g R^{2} m$ ). Why this step? Writing the exact result as (flat) $\times$ a correction factor $\frac{R}{R + h}$ isolates exactly how wrong the approximation is.
Fractional error $= 1 - \frac{R}{R + h} = \frac{h}{R + h}$ . Set equal to $0.01$ : $h = \frac{0.01}{0.99} R = 6.44 \times 1 0^{4} m$ . Why this step? Solving for the threshold turns "small" into a real number.

Verify: $h \approx 64 km$ — a few percent of $R$ , matching the forecast. Below Everest ( $\sim 9$ km) the error is under $0.15%$ , so $m g h$ is excellent for everyday problems. ✓

Worked example (e) Spring — stretch, compress, and natural length · cells S1, S2, S3

A spring $k = 200 N/m$ . Find $U$ for (i) stretched $x = + 0.10 m$ , (ii) compressed $x = - 0.10 m$ , (iii) at natural length $x = 0$ . See figure.

Forecast: Since $U = \frac{1}{2} k x^{2}$ squares $x$ , (i) and (ii) should give the same positive number, and (iii) exactly zero.

Stretched: $U = \frac{1}{2} (200) (0.10)^{2} = 1.0 J$ .
Compressed: $U = \frac{1}{2} (200) (- 0.10)^{2} = 1.0 J$ . Why this step? $(- 0.10)^{2} = (+ 0.10)^{2}$ . The parabola $U (x)$ is symmetric — pushing in stores the same as pulling out (look at the blue curve, equal heights either side of $x = 0$ ).
Natural length: $U = \frac{1}{2} (200) (0)^{2} = 0 J$ ; force $F = - k x = 0$ . Why this step? This is the degenerate input — the bottom of the well. Zero stored energy and zero force (flat tangent, orange arrow vanishes).

Verify: Symmetry (i)=(ii)=1.0 J ✓, degenerate (iii)=0 ✓. Units: $(N/m) \cdot m^{2} = N\cdotm = J$ ✓.

Worked example (f) Force from a general potential — finding equilibria · cell F1

A particle has $U (x) = x^{3} - 3 x$ (joules, $x$ in metres). Find the force $F (x)$ , and locate + classify the equilibrium points. See figure.

Forecast: Equilibria are where the landscape is flat ( $F = 0$ ). A cubic $U$ has a hill and a valley, so expect two equilibria — one stable (valley), one unstable (hill).

$F = - \frac{d U}{d x} = - (3 x^{2} - 3) = 3 - 3 x^{2}$ . Why this step? Force is minus the slope of $U$ — it points downhill.
Equilibria: $F = 0 \Rightarrow 3 - 3 x^{2} = 0 \Rightarrow x = \pm 1$ . Why this step? No force means no slope — a resting configuration.
Classify with the shape of $U$ : at $x = - 1$ , $U = (- 1) - 3 (- 1) = 2$ (a peak, unstable); at $x = + 1$ , $U = 1 - 3 = - 2$ (a valley, stable). Why this step? A ball nudged from the peak rolls away (unstable); nudged from the valley it rolls back (stable). Look at the green valley vs red peak in the figure.

Verify: $F (\pm 1) = 3 - 3 (1) = 0$ ✓. $U (- 1) = 2 > U (+ 1) = - 2$ , so $x = + 1$ is the lower (stable) point ✓, matching the forecast of one hill + one valley.

Worked example (g) Real-world word problem — three energy stores · cell W1

A $0.5 kg$ ball sits on a spring ( $k = 200 N/m$ ) compressed $x = 0.10 m$ . Released, it flies straight up. Ignoring air, how high above the release point does it rise? ( $g = 9.8$ )

Forecast: All the spring's stored energy becomes gravitational PE at the top (speed = 0 there). Expect a height of a few tens of centimetres.

Spring PE released: $U_{s} = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.10)^{2} = 1.0 J$ . Why this step? This is the total energy budget for the flight.
Energy conservation: at the highest point all of it is gravitational, $m g h = 1.0$ . Why this step? No friction → the recoverable spring PE fully converts. At the peak KE $= 0$ .
$h = \frac{1.0}{m g} = \frac{1.0}{0.5 \times 9.8} = 0.204 m$ .

Verify: $h \approx 0.20 m$ ✓ (tens of cm, as forecast). Units: $J / (kg \cdot m/s^{2}) = m$ ✓. Subtlety: strictly the ball also rises by the compression $0.10$ m while still on the spring; here $h$ is measured from the release point, so we count only spring→gravity conversion.

Worked example (h) Exam twist — spring + gravity, no double-counting · cell X1

A $0.5 kg$ bead on a vertical spring ( $k = 200 N/m$ ) is pushed down, compressing it by $x = 0.10 m$ below the natural length, then released. Find its speed at the instant it passes back through the natural-length position.

Forecast: Two PE stores change: spring (releases $\frac{1}{2} k x^{2}$ ) and gravity (the bead rises by $x$ , which costs $m g x$ ). Net energy → KE. Speed should be a bit less than the pure-spring $2 m/s$ from example (c) in the parent, because gravity steals some.

Spring energy released: $\frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.10)^{2} = 1.0 J$ . Why this step? Spring goes from compressed to natural length → all its PE is freed.
Gravity PE gained (bead rose height $x$ ): $m g x = 0.5 \times 9.8 \times 0.10 = 0.49 J$ . Why this step? Rising costs energy; we must subtract it, not add — the classic exam trap of double-counting or wrong sign.
KE at natural length: $\frac{1}{2} m v^{2} = 1.0 - 0.49 = 0.51 J$ , so $v = \frac{2 ( 0.51 )}{0.5} = 1.428 m/s$ .

Verify: $v \approx 1.43 m/s$ , less than the horizontal-case $2 m/s$ ✓ (gravity took $0.49$ J). Units check: $J / kg = m/s$ ✓.

Recall Quick self-test on the matrix

Which cell has $U < 0$ yet $Δ U$ can be either sign? ::: G3 — full gravity: $U$ is always negative, but moving inward gives $Δ U < 0$ , moving outward $Δ U > 0$ . Why does compressing a spring store the same energy as stretching it equally? ::: $U = \frac{1}{2} k x^{2}$ depends on $x^{2}$ , so sign of $x$ doesn't matter (cell S2 = S1). In example (h), why subtract $m g x$ ? ::: The bead rose, so gravitational PE increased — that energy is unavailable to KE (cell X1, avoid double-counting).

Connections

1.3.05 Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²) (Hinglish)
Force from potential — F = -dU/dx
Conservation of mechanical energy
Hooke's Law
Gravitation — Newton's law
Escape velocity
Work done by a force
Conservative vs non-conservative forces