Spring potential energy — derivation: Question bank

Spot the error

Each line contains a flawed argument. Name the flaw in one sentence.

"Force at full stretch is

k x

and distance is

x

, so work

= k x \cdot x = k x^{2}

The flaw is using the final (maximum) force over the whole path; force started at zero, so you must use the average

\frac{k x}{2}

, giving

\frac{1}{2} k x^{2}

"When compressed,

x < 0

, so

U = \frac{1}{2} k x^{2}

is negative."

The flaw is forgetting

x

is squared —

x^{2} \geq 0

always, so

U

is never negative regardless of stretch or compression sign.

"The spring pulls back, so its force does the storing — the force is

+ k x

The flaw is the wrong sign: the spring's restoring force is

- k x

and does negative work while you stretch, so it is your applied force that stores the energy.

"I plugged in the spring's total length of

0.30 m

for

x

The flaw is using absolute length;

x

is the displacement from the natural relaxed length, not the whole length of the spring.

U = \frac{1}{2} k x^{2}

is like

\frac{1}{2} m v^{2}

, so the ½ comes from the same reason."

The flaw is assuming a shared origin; the spring's ½ is the triangle area / average-force factor, while kinetic energy's ½ comes from integrating

m v d v

— coincidence of form, not cause.

"Since force is constant during a tiny slice, work over the whole stretch is

F x

The flaw is that force is constant only within one infinitesimal slice; across the full stretch

F

changes, so you must sum (integrate) the slices, not multiply once.

"The spring force is conservative, so it does zero net work on any path."

The flaw confuses conservative with zero-work; conservative means work depends only on endpoints (zero around a closed loop), not zero on every path.

Explain the reason, not just the fact. (Note on notation: when a derivation sums over the stretch, it uses a dummy variable written $x^{'}$ — read "x-prime" — for the moving position during the pull, so that the fixed final displacement can stay named $x$ ; $d x^{'}$ is just a tiny step of that moving position.)

Why can't we use

W = F \times d

with a single

F

for a spring?

Because the force

- k x

changes with position (grows linearly), so no single value of

F

describes the whole stretch — we must add up tiny constant-force slices via an integral.

Why does the derivation produce a factor of

\frac{1}{2}

The force rises linearly from

0

k x

, so its average is $\frac{k x}{2}$ ; work = average force × distance =

\frac{k x}{2} \cdot x = \frac{1}{2} k x^{2}

, equivalently the triangle's half.

Why does making each slice infinitesimally small (

d x^{'}

) let us use

W = F d

on it?

Over an infinitely thin slice the moving position

x^{'}

barely changes, so the force

k x^{'}

is effectively constant, and the simple constant-force work rule applies exactly.

Why is work done against the spring equal to the potential energy gained?

Because the spring force is conservative — no energy leaks to heat, so every joule you push in against it is stored and fully recoverable. See Conservative forces and potential energy.

Why does the same formula

\frac{1}{2} k x^{2}

appear again in Simple Harmonic Motion?

Because SHM is mass-on-a-spring; the stored

U = \frac{1}{2} k x^{2}

and its trade with

\frac{1}{2} m v^{2}

drive the oscillation, continually swapping between the two. See Simple Harmonic Motion.

Why is spring PE quadratic in

x

while gravitational PE

m g h

is only linear in height?

Gravity gives a constant force (

m g

), so its work is force × distance (linear); a spring's force grows with $x$ , so integrating a rising line gives a quadratic. See Gravitational potential energy.

Edge cases

Boundary and degenerate situations where sloppy thinking breaks.

What is

U

when

x = 0

(spring at natural length)?

Exactly zero —

\frac{1}{2} k (0)^{2} = 0

; the relaxed spring holds no elastic energy, which is the reference point for measuring

U

For a mass oscillating on a spring, where is

U

maximum and where is kinetic energy maximum?

U

is maximum at the turning points (largest

∣ x ∣

, momentarily at rest) and kinetic energy is maximum at equilibrium (

x = 0

, fastest); they trade off with constant total. See Conservation of mechanical energy.

What happens to

\frac{1}{2} k x^{2}

k \to 0

(an infinitely floppy spring)?

U \to 0

for any finite

x

— a spring with no stiffness exerts no force, so no work is needed to move it and nothing is stored.

If a spring is stretched past its elastic limit, does

U = \frac{1}{2} k x^{2}

still hold?

No — Hooke's law breaks down (force is no longer

- k x

), some energy goes into permanent deformation/heat, so the neat triangle formula no longer applies.

Compare energy stored stretching from

0

x

versus from

x

2 x

They are not equal:

0 \to x

stores

\frac{1}{2} k x^{2}

, but

x \to 2 x

stores

\frac{1}{2} k (2 x)^{2} - \frac{1}{2} k x^{2} = \frac{3}{2} k x^{2}

— three times as much, because

U

grows quadratically.

What is the net work by the spring over one full oscillation cycle (back to start)?

Zero — the spring force is conservative, so over a closed path returning to the same

x

the net work is exactly zero.

Connections

Hooke's Law — the linear force law these traps repeatedly test.
Work done by a variable force — why integration, not multiplication.
Conservative forces and potential energy — the "work against = stored" logic.
Conservation of mechanical energy — the PE↔KE trade-off edge cases.
Simple Harmonic Motion — where $\frac{1}{2} k x^{2}$ reappears.
Gravitational potential energy — the linear-vs-quadratic contrast.
Spring potential energy — derivation — parent note with the full build.

Question bank — Spring potential energy — derivation

True or false — justify

Spot the error

Why questions

Edge cases

Connections