Spring potential energy — derivation: Worked examples

The scenario matrix

Every spring-energy problem is one (or a blend) of these cells. We will hit all of them.

#	Cell	What makes it different	Example
A	Pure stretch, $x > 0$	Baseline plug-in	Ex 1
B	Pure compression, $x < 0$	Sign must vanish (squared)	Ex 2
C	Zero displacement, $x = 0$	Degenerate — energy is exactly $0$	Ex 2 (aside)
D	Change of state $Δ U$	Energy difference $U_{2} - U_{1}$ , not $U$	Ex 3
E	Spring → kinetic (launch)	Energy conservation, Conservation of mechanical energy	Ex 4
F	Spring → height (vertical)	Elastic PE ↔ Gravitational potential energy	Ex 5
G	Solve backwards for $x$ or $k$	Undo the square → , two roots	Ex 6
H	Limiting / scaling behaviour	"double $x$ ", "double $k$ " — the $x^{2}$ law	Ex 7
I	Exam twist: partial release	Energy only from $x_{1}$ to $x_{2}$ , friction	Ex 8

Everything below is anchored to the force–displacement triangle — energy is always the area under the line $F = k x$ .

Figure — Spring potential energy — derivation

The worked examples

Worked example Ex 2 — Cells B & C · compression and zero

The same spring ( $k = 300 N/m$ ) is now compressed by $0.05 m$ , i.e. $x = - 0.05 m$ . Then, separately, ask: what is $U$ at $x = 0$ ?

Forecast: Will compression give a negative energy? Answer before reading.

$U = \frac{1}{2} k x^{2} = \frac{1}{2} (300) (- 0.05)^{2}$ . Why this step? The formula is identical — nothing special for compression.
The square kills the sign: $(- 0.05)^{2} = + 0.0025$ . Why this step? Squaring any real number gives $\geq 0$ ; this is why a compressed spring stores positive energy.
$U = \frac{1}{2} (300) (0.0025) = 0.375 J$ — identical to Ex 1. Why this step? Look at the figure: the triangle to the left of $x = 0$ has the same area as the one to the right. Same $∣ x ∣$ , same energy.
Zero case (Cell C): at $x = 0$ , $U = \frac{1}{2} k (0)^{2} = 0 J$ . Why this step? The relaxed spring stores nothing — the triangle has zero base.

Verify: Stretch and compression by equal amount give equal $U$ ( $0.375 = 0.375$ ) ✓. The lowest possible $U$ is $0$ , reached only at equilibrium ✓.

Worked example Ex 3 — Cell D · change of stored energy

A spring ( $k = 400 N/m$ ) is stretched from $x_{1} = 0.10 m$ to $x_{2} = 0.20 m$ . How much extra energy did you add?

Forecast: The stretch grew from $0.10$ to $0.20$ . Did the stored energy double? Guess.

Energy at each state: $U_{1} = \frac{1}{2} k x_{1}^{2}$ , $U_{2} = \frac{1}{2} k x_{2}^{2}$ . Why this step? The work you supply equals the difference $Δ U = U_{2} - U_{1}$ , not $U_{2}$ alone — you started already stretched.
$U_{1} = \frac{1}{2} (400) (0.10)^{2} = \frac{1}{2} (400) (0.01) = 2 J$ .
$U_{2} = \frac{1}{2} (400) (0.20)^{2} = \frac{1}{2} (400) (0.04) = 8 J$ . Why this step? $x$ doubled, and since $U \propto x^{2}$ , $U$ went up 4× ( $2 \to 8$ ).
$Δ U = 8 - 2 = 6 J$ . Why this step? This is the extra area — the trapezoid strip of the triangle between $x_{1}$ and $x_{2}$ .

Verify: $Δ U = 6 J$ . Sanity: it is more than $U_{1}$ itself, because the force is bigger over that outer slice ✓.

Worked example Ex 5 — Cell F · vertical launch (spring → height)

A spring ( $k = 1000 N/m$ ) is compressed $0.06 m$ under a $0.1 kg$ ball, then fired straight up. How high does the ball rise above the release point? Use $g = 10 m/s^{2}$ .

Forecast: Above or below $2 m$ ? Guess.

Elastic energy: $U = \frac{1}{2} (1000) (0.06)^{2} = \frac{1}{2} (1000) (0.0036) = 1.8 J$ . Why this step? This is the whole energy budget.
At the top the ball is momentarily still, so all $U$ became gravitational PE: $U = m g h$ . Why this step? Elastic PE $\to$ Gravitational potential energy; kinetic energy is zero at the peak.
$1.8 = (0.1) (10) h = 1 \cdot h$ .
$h = 1.8 m$ . Why this step? Solve the single linear equation for height.

Verify: $m g h = 0.1 \times 10 \times 1.8 = 1.8 J$ ✓. Below the $2 m$ guess. (Strictly the ball also rises the $0.06 m$ of spring travel while gaining energy, but "above release point" means we measure from there.)

Worked example Ex 7 — Cell H · scaling laws (limiting behaviour)

A spring stores $U_{0} = 2 J$ at some stretch $x_{0}$ . (a) If you triple the stretch, what is the new energy? (b) If instead you keep $x_{0}$ but use a spring with double the $k$ , what is the new energy?

Forecast: For (a) will it be $6 J$ or $18 J$ ? Guess before reading.

(a) $U \propto x^{2}$ , so tripling $x$ multiplies $U$ by $3^{2} = 9$ . Why this step? The $x$ inside the square carries the whole scaling.
New energy $= 9 \times 2 = 18 J$ . Why this step? Not $6 J$ — the trap is thinking energy is linear in $x$ .
(b) $U \propto k$ (linearly), so doubling $k$ doubles $U$ : $2 \times 2 = 4 J$ . Why this step? $k$ appears to the first power — a stiffer spring stores proportionally more at the same stretch.

Verify: With $k = 100, x_{0} = 0.2$ : $U_{0} = \frac{1}{2} (100) (0.2)^{2} = 2 J$ ✓; triple $x$ : $\frac{1}{2} (100) (0.6)^{2} = 18 J$ ✓; double $k$ : $\frac{1}{2} (200) (0.2)^{2} = 4 J$ ✓.

Worked example Ex 8 — Cell I · exam twist · partial release with friction

A spring ( $k = 600 N/m$ ) is compressed to $x_{1} = 0.20 m$ . It pushes a $1 kg$ block and lets go when the compression is only $x_{2} = 0.10 m$ (the spring's natural length carries the block no further). Friction does $1.5 J$ of work against the block. Find the block's speed at release.

Forecast: Above or below $3 m/s$ ? Guess.

Energy given up by the spring is the difference of stored energies, $Δ U = U_{1} - U_{2}$ . Why this step? At release the spring still holds $U_{2}$ ; only the drop between $x_{1}$ and $x_{2}$ was delivered (Work done by a variable force).
$U_{1} = \frac{1}{2} (600) (0.20)^{2} = \frac{1}{2} (600) (0.04) = 12 J$ .
$U_{2} = \frac{1}{2} (600) (0.10)^{2} = \frac{1}{2} (600) (0.01) = 3 J$ .
Energy released $= 12 - 3 = 9 J$ . Why this step? The trapezoid strip between $x_{2}$ and $x_{1}$ — the same " $Δ U$ " idea as Ex 3.
Subtract friction, rest becomes kinetic: $\frac{1}{2} m v^{2} = 9 - 1.5 = 7.5 J$ . Why this step? Energy accounting — friction removed $1.5 J$ as heat, breaking pure mechanical conservation.
$\frac{1}{2} (1) v^{2} = 7.5 \Rightarrow v^{2} = 15 \Rightarrow v = 15 \approx 3.87 m/s$ .

Verify: $\frac{1}{2} (1) (3.873)^{2} \approx 7.5 J$ , plus friction $1.5 J$ , gives $9 J = Δ U$ ✓. Above $3 m/s$ .

Recall Cell-by-cell self-check

Which cell tests the square killing a sign? ::: Cell B (compression, Ex 2). Which cell needs $Δ U = U_{2} - U_{1}$ ? ::: Cells D and I (Ex 3, Ex 8). Which cell forces a $\pm$ square root? ::: Cell G (Ex 6). Tripling $x$ multiplies $U$ by what? ::: $9$ (Cell H, Ex 7). In Ex 8, why isn't mechanical energy conserved? ::: Friction removes $1.5 J$ as heat.

Connections

1.3.12 Spring potential energy — derivation — the parent derivation these examples exercise.
Hooke's Law — the $F = - k x$ line whose triangle we keep measuring.
Work done by a variable force — the $Δ U$ (trapezoid strip) reasoning in Ex 3 & 8.
Conservative forces and potential energy — why work against the spring equals stored $U$ .
Conservation of mechanical energy — Ex 4, 5, 8.
Gravitational potential energy — Ex 5, the elastic-to-height swap.
Simple Harmonic Motion — the same $\frac{1}{2} k x^{2}$ that these launches release.