Hooke's law — spring force F = −kx: Question bank

This page is a concept gym for Hooke's law. No heavy arithmetic — just the traps that catch people who "know the formula" but haven't felt the physics. Read the question, cover the answer, say your reasoning out loud, then reveal.

Before we start, look at the shared picture below — every question on this page refers to it.

Figure — Hooke's law — spring force F = −kx

A horizontal spring lies on a table: one end pinned to a wall, the other end holding a block. The block's rest position is $x = 0$ — the natural length, where the spring is neither stretched nor squeezed. Stretching pulls the block right, so we call that $x > 0$ . Compressing pushes it left, so that's $x < 0$ . In the picture, notice the spring's force arrow always points back toward $x = 0$ .

Now, the second key picture — the force–displacement graph — which turns every "line vs area" trap below into something you can see.

True or false — justify

Every answer must give a reason, not just "T" or "F."

x = 0

the spring exerts zero force.

True —

F = - k (0) = 0

; at natural length there is nothing to restore, so the spring is "happy" and pushes/pulls with nothing.

A compressed spring exerts a negative force.

False — compression means

x < 0

, so

F = - k x

is positive (points right, back toward

x = 0

). The minus sign flips the force's sign, it does not make it permanently negative.

Doubling the stretch doubles both the force and the stored energy.

False — force doubles (

F \propto x

) but energy quadruples (

U \propto x^{2}

). This split (linear force, quadratic energy) is the single most-missed spring fact.

A stiffer spring stores more energy at the same stretch.

True —

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

grows with

k

, so at equal

x

the larger-

k

spring holds more energy; its

F

-vs-

x

line is steeper, giving a taller triangle of area (second figure).

If a spring's force is

10 N

when stretched and

10 N

when compressed by the same amount, the two forces are the same.

False — the magnitudes match but the directions are opposite (stretch pulls back left, compression pushes back right). "Same" must mean same vector, and these are not.

The spring constant

k

gets larger the more you stretch the spring.

False —

k

is the fixed slope of the line, a property of the coil (material, thickness, turns). Only

F

changes as you stretch, not

k

— provided you stay within the elastic limit.

The work you do stretching a spring equals the energy the spring stores.

True (for a slow, no-friction stretch) — every joule you put in against the spring force is deposited as elastic PE

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

; that is exactly why Elastic potential energy equals your work done.

Hooke's law is a special property only of metal coils.

False — Taylor-expanding any stable energy minimum gives a leading

\frac{1}{2} U^{''} (0) x^{2}

term, so every stable system (bonds, strings, pendulums for small swings) looks like

F = - k x

near rest.

Spot the error

Each item is a plausible-sounding wrong statement. Find the flaw.

"Energy stored

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

The force is not constant — it grows from

0

up to

k x

as you stretch. You must use the average force

\frac{1}{2} k x

, giving

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

; equivalently, take the shaded triangle area under the

F

-vs-

x

line, not the full rectangle (second figure).

F = - k x

tells us the spring force is always negative, so it always points left."

Wrong — the sign of

F

tracks the sign of

x

. When compressed (

x < 0

) the force is positive and points right. The minus sign encodes "opposite to displacement," never "always negative."

"Since

F = - k x

, at large

x

the spring must eventually pull infinitely hard."

The formula is linear, but real springs obey it only up to the elastic limit. Past that the coil deforms permanently and

F

is no longer proportional to

x

— the law simply stops applying.

"A spring at its natural length has maximum potential energy because it's relaxed."

Backwards — natural length (

x = 0

) is the energy minimum,

U = 0

. That's precisely why the force there is zero:

F = - \frac{d U}{d x}

and the slope of

U

is zero at its bottom.

"To find the spring force, integrate:

F = \int k x d x

Confusing force with energy. Force is read directly as

F = - k x

; you integrate force over distance to get work/energy (

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

), and you differentiate energy to recover force (

F = - d U / d x

). The two operations are inverses. See Work done by a variable force.

"For a mass hanging at rest on a vertical spring, the spring force is zero."

No — at rest the spring force must balance gravity, so

k x = m g \neq = 0

. The net force is zero, but the spring itself is stretched and pulling up with force equal to the weight.

Why questions

Explain the reason, not just the fact.

Why is there a minus sign in

F = - k x

Because the force always points back toward equilibrium: whenever

x

is positive the force is negative and vice versa, and only a minus sign makes

F

and

x

carry opposite signs automatically.

Why is spring energy

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

and not

k x^{2}

Because force starts at

0

and rises to

k x

, so the average force during the stretch is only

\frac{1}{2} k x

; work

=

average force

\times

distance

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

. Geometrically it's the shaded triangle, not the rectangle (second figure).

Why does every stable system behave like a spring for small displacements?

Near a stable minimum the energy's first derivative vanishes and the first surviving Taylor term is

\frac{1}{2} U^{''} (0) x^{2}

; differentiating gives

F = - U^{''} (0) x = - k x

. This is why Interatomic forces act like tiny springs.

Why does

F = - k x

produce oscillation rather than a single push?

A restoring force pointing home accelerates the mass through

x = 0

, but the mass overshoots by inertia, gets pulled back from the other side, and repeats — that endless "always toward home" is exactly Simple Harmonic Motion.

Why can we call the spring force conservative?

Because it comes from a potential,

F = - \frac{d U}{d x}

with

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

; the work it does depends only on start and end positions, so energy swaps cleanly between KE and PE — see Conservative forces and potential energy.

Why does the

F

-vs-

x

graph being a straight line matter so much?

A straight line through the origin means force is strictly proportional to displacement; its slope is the constant

k

, and the tidy triangle under it is what makes the energy exactly

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

(second figure).

Edge cases

Boundaries, zeros, and limits — where sloppy intuition breaks.

What is the force and energy exactly at

x = 0

Both are zero:

F = - k (0) = 0

and

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

. Natural length is simultaneously the force-free point and the energy floor.

Is the stored energy ever negative?

No —

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

contains

x^{2} \geq 0

, so energy is zero at rest and positive for any displacement, stretched or compressed alike.

Does compressing store the same energy as stretching by the same distance?

Yes — because

U

depends on

x^{2}

, the sign of

x

is erased;

x = + d

and

x = - d

store the identical

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

What happens to the force in the limit

k \to 0

(at fixed displacement

x

The spring becomes infinitely floppy:

F = - k x \to 0

and

U = \frac{1}{2} k x^{2} \to 0

, so at any fixed stretch it offers no resistance and stores no energy — effectively not a spring at all.

What happens as

k \to \infty

if the displacement $x$ is held fixed?

Both force and stored energy blow up (

F = - k x \to \infty

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

); it takes ever more work to reach the same

x

, which is why a truly rigid rod resists moving at all.

What happens as

k \to \infty

if instead the force $F$ is held fixed?

The displacement collapses:

x = - F / k \to 0

, and the stored energy

U = \frac{1}{2} \frac{F ^{2}}{k} \to 0

. So under a fixed load a stiffer spring moves less and banks less energy — the opposite conclusion to the fixed-

x

case, which is exactly why the "what's held fixed?" question must be stated.

At the instant a mass passes through

x = 0

during oscillation, is its energy zero?

No — the spring PE is zero there, but all the energy has become kinetic; total mechanical energy is conserved throughout, per Conservation of mechanical energy.

What if you stretch past the elastic limit?

Hooke's law fails — the material deforms permanently,

F

is no longer proportional to

x

, and

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

no longer describes the (now partly lost) energy.

Recall One-line summary to carry away

Force lives on the line ( $F = - k x$ , changes sign with $x$ ), energy lives in the area ( $\frac{1}{2} k x^{2}$ , always positive, grows as the square), and $k$ is the slope that never moves until the spring breaks.

Connections

Parent: Hooke's law $F=-kx$
Work done by a variable force · Elastic potential energy · Conservative forces and potential energy · Simple Harmonic Motion · Conservation of mechanical energy · Interatomic forces