Hooke's law — spring force F = −kx: Worked examples

Reminders of the two tools we lean on (both fully built in the parent note):

The scenario matrix

Before solving anything, let's list every kind of situation a Hooke's-law problem can be. Each row is a "cell"; the worked examples below are tagged with the cell they fill.

Cell	What makes it different	Example
A. Stretch (x>0)	positive displacement, force pulls back	Ex 1
B. Compression (x<0)	negative displacement, force pushes out — sign trap	Ex 2
C. Zero displacement (x=0)	degenerate case: force and energy both zero	Ex 3
D. Energy vs force scaling	tripling $x$ : force ×3, energy ×9	Ex 4
E. Vertical hanging mass	gravity balances spring, find $k$	Ex 5
F. Energy → speed (motion)	spring PE converts to kinetic energy	Ex 6
G. Elastic limit exceeded	limiting/degenerate: law breaks down	Ex 7
H. Combined springs (twist)	series & parallel effective $k$	Ex 8
I. Work over an interval	work between two non-zero positions	Ex 9

We cover every sign of $x$ , the zero case, the "force is not constant" trap, real-world gravity, energy↔motion conversion, the failure boundary, an exam-style combination, and a variable-force interval. That is the whole territory.

Ex 1 — Cell A: a plain stretch

Worked example Stretched spring

A spring with $k = 150 N/m$ is stretched by $x = 0.08 m$ . Find (a) the spring force and (b) the stored energy.

Forecast: Guess before computing — is the force positive or negative? Bigger or smaller than 10 N?

Step 1 — Force. Apply $F = - k x$ . $F = - (150) (0.08) = - 12 N$ Why this step? Direct substitution; the minus sign is built into the law.

Step 2 — Read the sign. $x > 0$ (stretched right) gives $F < 0$ (pulls left). The spring pulls back toward home with magnitude $12 N$ . Why this step? The number is direction-encoded; we translate sign into a physical arrow.

Step 3 — Energy. Apply $U = \frac{1}{2} k x^{2}$ . $U = \frac{1}{2} (150) (0.08)^{2} = \frac{1}{2} (150) (0.0064) = 0.48 J$ Why this step? Energy is the area under the F–x line, not $F \cdot x$ .

Verify: Units: $(N/m) (m) = N$ ✓ and $(N/m) (m^{2}) = N\cdotm = J$ ✓. Sanity: $F \cdot x = 0.96$ J, and our energy is exactly half of that — correct, because force grew from 0. ✓

Ex 2 — Cell B: compression (the sign trap)

Worked example Compressed spring

The same spring ( $k = 150 N/m$ ) is now compressed by $4 cm$ , i.e. $x = - 0.04 m$ . Find the force and energy.

Forecast: Will the force be negative (like Ex 1) or positive? Will the energy be negative?

Step 1 — Force. Substitute the negative $x$ : $F = - k x = - (150) (- 0.04) = + 6 N$ Why this step? Two minus signs multiply to a plus. This is the whole point of Cell B.

Step 2 — Interpret. $x < 0$ (pushed left) gives $F > 0$ (pushes right). The spring shoves outward, back toward home — opposite direction to Ex 1, exactly as "restoring" demands. Why this step? The minus sign does NOT mean "force is always negative"; it means "force opposes displacement."

Step 3 — Energy. Square kills the sign: $U = \frac{1}{2} (150) (- 0.04)^{2} = \frac{1}{2} (150) (0.0016) = 0.12 J$ Why this step? $x^{2} \geq 0$ always, so compressing stores energy just like stretching. Energy is never negative here.

The picture below plots the applied force $k x$ against displacement — a straight line through the origin. Follow the red vertical lines dropped at $x = + 0.04$ m (stretch, Ex 1's spring) and $x = - 0.04$ m (compression, this example). The two shaded triangles are the stored energies. Because energy is $\frac{1}{2} k x^{2}$ and the square erases the sign, the two triangles have identical area (0.12 J) — stretching and compressing by the same distance cost exactly the same energy, even though the forces point in opposite directions.

Verify: Force sign flipped vs Ex 1 (as it must for opposite displacement) ✓. Energy positive ✓. In the figure, the left and right red triangles are mirror images of equal area ✓.

Ex 3 — Cell C: the zero (degenerate) case

Worked example Spring at natural length

A spring sits at its natural length: $x = 0$ . Find $F$ and $U$ .

Forecast: Both zero? One zero? Trick question?

Step 1 — Force. $F = - k (0) = 0$ . Why this step? At natural length there is nothing to restore — equilibrium means zero net spring force.

Step 2 — Energy. $U = \frac{1}{2} k (0)^{2} = 0$ . Why this step? No displacement means no stored work.

Step 3 — Why this cell matters. Look at the figure from Ex 2: the straight force line passes exactly through the origin. At $x = 0$ the line is at zero height, so the force is zero, and the triangle whose area is the energy has shrunk to nothing — so the energy is zero too. Nudge the spring even a tiny bit either way and the line immediately rises (stretch) or falls (compression), producing a restoring force that pushes it back. This "always pushed back toward $x = 0$ " is exactly the seed of oscillation in Simple Harmonic Motion.

Verify: Both zero regardless of how stiff $k$ is ✓. On the $F$ – $x$ line, the origin is the single point where force is zero ✓.

Ex 4 — Cell D: scaling (force vs energy)

Ex 5 — Cell E: hanging mass (find k)

Ex 6 — Cell F: energy becomes speed

Worked example Spring launches a block

A spring with $k = 800 N/m$ is compressed by $x = 0.05 m$ and released, pushing a frictionless block of mass $m = 0.2 kg$ . Find the block's launch speed.

Forecast: A few m/s? Tens of m/s?

Step 1 — Stored energy. Compressed spring holds $U = \frac{1}{2} k x^{2} = \frac{1}{2} (800) (0.05)^{2} = \frac{1}{2} (800) (0.0025) = 1.0 J .$ Why this step? This is the energy reservoir (Elastic potential energy) available to do work.

Step 2 — Convert to kinetic energy. Frictionless, so all PE becomes KE (Conservation of mechanical energy): $\frac{1}{2} m v^{2} = U .$ Why this step? No energy leaks out, so PE lost = KE gained.

Step 3 — Solve for $v$ . $v = \frac{2 U}{m} = \frac{2 ( 1.0 )}{0.2} = 10 \approx 3.16 m/s .$ Why this step? Rearranging the KE formula isolates speed.

Verify: Units $J / kg = m^{2} / s^{2} = m/s$ ✓. Recompute KE: $\frac{1}{2} (0.2) (3.16)^{2} \approx 1.0$ J $= U$ ✓.

Ex 7 — Cell G: beyond the elastic limit (law breaks)

Worked example When Hooke's law fails

A spring with $k = 500 N/m$ obeys $F = - k x$ only while $∣ x ∣$ is below its elastic limit $x_{m a x} = 0.20 m$ . It is stretched to $x = 0.35 m$ . Can we still use $F = - k x$ and $U = \frac{1}{2} k x^{2}$ ?

Forecast: Yes, just plug in? Or no?

Step 1 — Check the limit. $x = 0.35 m > x_{m a x} = 0.20 m$ . Why this step? Every formula on this page assumes the spring is still elastic — a straight $F$ – $x$ line.

Step 2 — Conclusion. Beyond $x_{m a x}$ the spring deforms permanently; the $F$ – $x$ relation is no longer a straight line, so $F = - k x$ and $U = \frac{1}{2} k x^{2}$ do not apply. You cannot compute a reliable number at $x = 0.35$ m. Why this step? This is the boundary of the model. Recognising it is itself the "answer."

Step 3 — What's still true. Up to $x_{m a x} = 0.20 m$ the formulas hold. The maximum energy this spring can store while still elastic is $U_{lim} = \frac{1}{2} k x_{m a x}^{2} = \frac{1}{2} (500) (0.20)^{2} = \frac{1}{2} (500) (0.04) = 10 J .$ Why this step? We push the valid model to its edge, not past it, and get a concrete number.

Verify: $0.35 > 0.20$ is true, so the law is out of range ✓. $U_{lim} = 10$ J at the limit ✓.

Ex 8 — Cell H: exam twist, combined springs

Ex 9 — Cell I: work over an interval (variable force)

Worked example Stretching from one point to another

A spring with $k = 400 N/m$ is stretched from $x_{1} = 0.02 m$ to $x_{2} = 0.05 m$ . How much work do you do against the spring?

Forecast: Is it $\frac{1}{2} k (x_{2} - x_{1})^{2}$ ? Or something else?

Step 1 — Work is an energy difference. The work you do equals the change in stored energy (Work done by a variable force): $W = U (x_{2}) - U (x_{1}) = \frac{1}{2} k x_{2}^{2} - \frac{1}{2} k x_{1}^{2} .$ Why this step? Force is not constant, so we cannot use force × distance; instead we take the difference of the triangular areas.

Step 2 — Compute. $W = \frac{1}{2} (400) [(0.05)^{2} - (0.02)^{2}] = 200 (0.0025 - 0.0004) = 200 (0.0021) = 0.42 J .$ Why this step? Plug the two endpoints into the energy formula and subtract.

Step 3 — Trap check. The naive $\frac{1}{2} k (x_{2} - x_{1})^{2} = 200 (0.03)^{2} = 0.18$ J is wrong — the spring is already partly stretched, so the force starts at $k x_{1}$ , not zero. Why this step? Subtracting areas, not treating the interval as a fresh stretch from zero.

Verify: $U (0.05) = 200 (0.0025) = 0.5$ J, $U (0.02) = 200 (0.0004) = 0.08$ J, difference $= 0.42$ J ✓.

Recall Scenario checklist — can you place each in a cell?

A spring pushes a puck: which cell? ::: Cell F (energy → speed) A spring compressed 3 cm: which cell, and is $F$ positive or negative? ::: Cell B; $F > 0$ (pushes out) Stretch tripled, energy factor? ::: Cell D; ×9 Mass hangs and stretches spring, find $k$ : which cell? ::: Cell E ( $k x = m g$ ) Work from $x_{1}$ to $x_{2}$ (both nonzero): which formula? ::: Cell I; $\frac{1}{2} k (x_{2}^{2} - x_{1}^{2})$ , NOT $\frac{1}{2} k (x_{2} - x_{1})^{2}$

Connections

Work done by a variable force — why interval work (Ex 9) is a difference of areas.
Elastic potential energy — the reservoir released in Ex 6.
Conservation of mechanical energy — PE ↔ KE in Ex 6.
Simple Harmonic Motion — Ex 3's energy valley is where oscillation begins.
Conservative forces and potential energy — $F = - d U / d x$ underlies every energy step.