Classical-Mechanics · Unit 7 · Video 1 · Interactive Practice

Hooke's Law: The Force Law of the Spring

IKey Formulas

Formula	Name	What it tells you
$F_x = -kx$	Hooke's Law	Linear restoring force
$\|F\| = k\,\|x\|$	Force magnitude	Proportional to displacement
$k = -(\text{slope})$	Spring constant	Read off the $F_x$ -vs- $x$ graph
$\|a\| = \dfrac{k\,\|\Delta x\|}{m}$	Predicted acceleration	Force law + Newton's 2nd law

Key Insight: The minus sign in $F_x = -kx$ is the whole story — the force always drives the block back toward $x = 0$ , so it is a restoring force.

IIVisualization 1 — The Restoring Force

A spring force $F_x = -kx$ always points back toward equilibrium, and grows with the displacement.

IIIVisualization 2 — The Force–Displacement Graph

The spring's force–displacement graph is a straight line through the origin whose slope is exactly $-k$ .

💡 Challenge: make this spring twice as stiff as the dashed reference ( $k = 2$ ) — its line should be twice as steep.

IVVisualization 3 — Predict, Then Measure

The force law forecasts the block's acceleration $|a| = k\,|\Delta x|/m$ with no measurement of it.

Your prediction for

|a|

(m/s²) 4.0

💡 The reckoning: if a real measurement of $|a|$ disagreed with this forecast, the force law itself would need revision — that is how models grow.

VQuiz Questions

Problem 1 · Apply the Force Law

Given: a spring with $k = 50\ \text{N/m}$ is stretched so the block sits at $x = +0.2\ \text{m}$ . Find the spring force $F_x$ .

✅ Correct! The force is

-10\ \text{N}

— negative, pointing back toward

x = 0

❌ Check the sign. For a stretch (

x > 0

) the restoring force points in the

-x

direction, so

F_x

is negative.

❌ Not quite. Multiply, don't divide:

F_x = -kx = -(50)(0.2)

Show solution

Apply Hooke's Law directly with $k = 50\ \text{N/m}$ and $x = +0.2\ \text{m}$ :

F_x = -kx = -(50)(0.2) = -10\ \text{N}

The negative sign means the force points in the $-x$ direction — back toward equilibrium. That is what makes it a restoring force.

Problem 2 · Compression and Signs

Given: a spring with $k = 40\ \text{N/m}$ is compressed so the block sits at $x = -0.3\ \text{m}$ . Find $F_x$ . ⚠️ Watch the double negative!

✅ Correct! The compressed spring pushes the block in the

+x

direction, back toward equilibrium.

❌ Close — mind the signs. Two negatives multiply to a positive:

-k\cdot(-0.3) = +

❌ Not quite. Keep the decimal:

(40)(0.3) = 12

, not

120

Show solution

With $k = 40\ \text{N/m}$ and $x = -0.3\ \text{m}$ :

F_x = -kx = -(40)(-0.3) = +12\ \text{N}

The two minus signs combine to a plus. A compressed spring pushes outward in the $+x$ direction — again pointing back toward $x = 0$ .

Problem 3 · Predict the Acceleration

Given: a spring with $k = 60\ \text{N/m}$ holds a block of mass $m = 3\ \text{kg}$ , stretched by $\Delta x = 0.5\ \text{m}$ . Using the force law alone, predict the magnitude of the force and the acceleration at the instant of release.

What is the force magnitude $|F|$ ?

What is the acceleration $|a|$ ?

✅ Correct! You forecast

|a| = 10\ \text{m/s}^2

from the law alone — no acceleration measured.

❌ Check the force. Use both factors:

|F| = k\,|\Delta x| = (60)(0.5)

❌ Check the acceleration. Newton's 2nd law divides by mass:

|a| = |F|/m

Show solution

Step 1 — Force from the law (no acceleration measured):

|F| = k\,|\Delta x| = (60)(0.5) = 30\ \text{N}

Step 2 — Acceleration from Newton's 2nd law:

|a| = \frac{|F|}{m} = \frac{30}{3} = 10\ \text{m/s}^2

This is the payoff of Newtonian induction: the force law lets us forecast the acceleration before touching the apparatus.

Problem 4 · A Pre-Tensioned Spring

Given: a tightly-wound spring needs a threshold force $F_0 = 4\ \text{N}$ before it begins to stretch, so for $x > 0$ the law becomes $F_x = -F_0 - kx$ . With $k = 20\ \text{N/m}$ , find $F_x$ at $x = +0.5\ \text{m}$ .

✅ Excellent! The pre-tension adds a constant

4\ \text{N}

on top of the usual

-kx = -10\ \text{N}

❌ Don't drop the pre-tension. For a stretch the law is

F_x = -F_0 - kx

, not just

-kx

❌ Not quite. Keep both terms and both minus signs:

-(4) - (20)(0.5)

Show solution

For a stretch ( $x > 0$ ) the pre-tensioned law adds the threshold $F_0$ :

F_x = -F_0 - kx = -(4) - (20)(0.5) = -4 - 10 = -14\ \text{N}

The pre-tension $F_0$ shifts the ordinary $-kx$ line downward by a constant — the graph no longer passes through the origin. Both $F_0$ and $k$ are pinned down by experiment.

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