🔩 Hooke's Law: $F = -kx$

Part 1 of 7 — Elastic Potential Energy

Springs are everywhere — in car suspensions, trampolines, mattresses, and even at the atomic level. Understanding how springs store and release energy starts with Hooke's Law, which describes the force a spring exerts.

Hooke's Law

A spring that is stretched or compressed from its natural length (equilibrium position) exerts a restoring force:

$F_s = -kx$

where:

• $F_s$ = spring force (N)
• $k$ = spring constant (N/m) — stiffness of the spring
• $x$ = displacement from equilibrium (m)
• The negative sign means the force is opposite to the displacement

What the Negative Sign Means

The spring always tries to return to its natural length — this is why we call it a restoring force.

The Spring Constant $k$

The spring constant measures how stiff a spring is:

• Large $k$ → stiff spring → large force for a small stretch
• Small $k$ → soft spring → small force for a large stretch

Units

$[k] = \frac{\text{N}}{\text{m}} = \frac{\text{kg}}{\text{s}^2}$

Typical Values

Finding $k$ Experimentally

Hang a mass $m$ from a vertical spring. At equilibrium, the spring stretches by $x$:

$mg = kx \Rightarrow k = \frac{mg}{x}$

Force vs. Displacement Graph

The $F$ vs. $x$ graph for a spring is a straight line through the origin with slope $k$ (or $-k$ for the spring force):

Applied Force to Stretch/Compress

To hold a spring stretched by $x$, you must apply:

$F_{\text{applied}} = kx$

(positive — equal and opposite to spring force)

This graph is a line with slope $k$:

• At $x = 0$: $F = 0$
• At $x = 0.1$ m with $k = 200$ N/m: $F = 20$ N
• At $x = 0.2$ m: $F = 40$ N

Important Note

Hooke's Law is valid only for small displacements. Beyond the elastic limit, the spring deforms permanently and the linear relationship breaks down.

Hooke's Law Concepts 🎯

Hooke's Law Calculations 🧮

Use $g = 10$ m/s².

1. A spring stretches 0.05 m when a 2 kg mass hangs from it. What is the spring constant $k$ (in N/m)?

2. A spring with $k = 800$ N/m is compressed by 0.03 m. What force is needed to hold it compressed (in N)?

3. A spring with $k = 250$ N/m has a 5 kg mass resting on it vertically. By how much is the spring compressed (in m)?

Round all answers to 3 significant figures.

Spring Force Analysis 🔍

Exit Quiz — Hooke's Law ✅

🌀 Elastic PE: $PE_s = \frac{1}{2}kx^2$

Part 2 of 7 — Elastic Potential Energy

A compressed or stretched spring stores energy — energy that can be released to launch a projectile, close a door, or bounce a ball. This stored energy is elastic potential energy, and it depends on the spring constant and the displacement.

Defining Elastic PE

The elastic potential energy stored in a spring displaced by $x$ from equilibrium:

$PE_s = \frac{1}{2}kx^2$

where:

• $k$ = spring constant (N/m)
• $x$ = displacement from equilibrium (m)
• $PE_s$ = elastic PE (Joules)

Properties

The $x^2$ Dependence

Because PE depends on $x^2$:

Doubling the displacement quadruples the stored energy!

Same PE for Stretch and Compression

Since $PE_s = \frac{1}{2}kx^2$ and $x^2 = (-x)^2$:

A spring stretched by 5 cm has the same PE as a spring compressed by 5 cm.

Graphical Connection

On a $F$ vs. $x$ graph (for the applied force $F = kx$):

The area under the curve from $0$ to $x$ = work done = energy stored:

$\text{Area} = \frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(x)(kx) = \frac{1}{2}kx^2$

This is a triangle — the PE formula comes from the area of a triangle!

Elastic vs. Gravitational PE

Elastic PE Concepts 🎯

Elastic PE Calculations 🧮

1. A spring ($k = 500$ N/m) is stretched 0.08 m. What PE is stored (in J)?

2. A spring stores 18 J when compressed 0.3 m. What is the spring constant (in N/m)?

3. A spring ($k = 800$ N/m) stores 4 J of PE. What is the displacement from equilibrium (in m)?

Round all answers to 3 significant figures.

Elastic PE Analysis 🔍

Exit Quiz — Elastic PE ✅

🔧 Work Done by Springs

Part 3 of 7 — Elastic Potential Energy

The work done by a spring force is special because the force varies with displacement. You can't simply multiply force times distance — you need calculus (or the PE formula). The relationship between spring work and elastic PE is central to energy problems.

Work Done BY the Spring

The work done by the spring force as the displacement changes from $x_i$ to $x_f$:

$W_{\text{spring}} = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2 = -\Delta PE_s$

This is analogous to gravity: $W_g = -\Delta PE_g$.

Important Cases

Work Done ON the Spring

To stretch or compress a spring, you must apply a force against the spring force. The work YOU do:

$W_{\text{you}} = \frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2 = \Delta PE_s = -W_{\text{spring}}$

Key Relationship

$W_{\text{you}} = -W_{\text{spring}}$

• When you stretch a spring: you do positive work, spring does negative work
• When you let it snap back: spring does positive work, you don't need to do work

Why Can't We Use $W = Fd$?

The spring force is not constant — it increases as you stretch more. Using $W = Fd$ would require knowing the average force:

$F_{\text{avg}} = \frac{0 + kx}{2} = \frac{kx}{2}$

$W = F_{\text{avg}} \cdot x = \frac{kx}{2} \cdot x = \frac{1}{2}kx^2 \checkmark$

This works because the force increases linearly (the average is the midpoint).

Graphical Interpretation

On an $F$ vs. $x$ graph:

Work = Area Under the Curve

The work done by the applied force equals the area under the $F = kx$ line:

• From $x = 0$ to $x = A$: Area = triangle = $\frac{1}{2}(A)(kA) = \frac{1}{2}kA^2$
• From $x = x_1$ to $x = x_2$: Area = trapezoid = $\frac{1}{2}k(x_2^2 - x_1^2)$

Work Between Two Displacements

If a spring is already stretched from $x_1$ to $x_2$:

$W_{\text{you}} = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2$

This is the area of the trapezoid between $x_1$ and $x_2$ on the $F$-$x$ graph.

Work by Springs Concepts 🎯

Spring Work Calculations 🧮

Spring constant $k = 400$ N/m for all problems.

1. How much work is needed to stretch the spring from equilibrium to $x = 0.1$ m (in J)?

2. How much work is needed to stretch it from $x = 0.1$ m to $x = 0.2$ m (in J)?

3. The spring is compressed by 0.15 m and released. How much work does the spring do on the attached object as it returns to equilibrium (in J)?

Round all answers to 3 significant figures.

Spring Work Analysis 🔍

Exit Quiz — Work by Springs ✅

🏋️ Spring-Mass Systems

Part 4 of 7 — Elastic Potential Energy

A mass attached to a spring is one of the most fundamental systems in physics. When displaced and released, the mass oscillates back and forth. Understanding the forces and energy in this system is essential for AP Physics 1.

Horizontal Spring-Mass System

A block of mass $m$ is attached to a spring ($k$) on a frictionless horizontal surface.

At Equilibrium ($x = 0$)

• Spring force = 0
• Acceleration = 0
• Speed is maximum (if oscillating)

Displaced by $x$

• Spring force: $F = -kx$ (restoring)
• Acceleration: $a = -kx/m$ (toward equilibrium)
• The acceleration is not constant — it depends on position!

Key Behavior

When released from displacement $A$ (amplitude):

1. Block accelerates toward equilibrium
2. Passes through equilibrium at maximum speed
3. Overshoots, compressing (or stretching) the spring
4. Spring slows the block, stops it at $-A$
5. Process repeats — simple harmonic motion

Vertical Spring-Mass System

When a mass hangs from a spring:

New Equilibrium Position

The spring stretches by $x_0$ where gravity is balanced:

$kx_0 = mg \Rightarrow x_0 = \frac{mg}{k}$

Oscillation About the New Equilibrium

If displaced from this new equilibrium by $x$:

• The net force is still $F = -kx$ (measured from the new equilibrium)
• The oscillation is identical to the horizontal case!
• Gravity shifts the equilibrium point but doesn't change the oscillation behavior

Important Insight

For oscillation problems, measure displacement from the equilibrium position (where the net force is zero), not from the spring's natural length.

Force and Acceleration Analysis

For a spring-mass system oscillating with amplitude $A$:

Key Relationships

• Maximum force/acceleration occur at maximum displacement (endpoints)
• Zero force/acceleration occur at equilibrium (center)
• Maximum speed occurs at equilibrium
• Zero speed occurs at endpoints

Force/acceleration and speed are out of phase — when one is max, the other is zero.

Spring-Mass System Concepts 🎯

Spring-Mass Calculations 🧮

Use $g = 10$ m/s².

1. A 2 kg mass hangs from a spring ($k = 100$ N/m). How far does the spring stretch at equilibrium (in m)?

2. A 0.5 kg block oscillates on a spring ($k = 200$ N/m) with amplitude 0.1 m. What is the maximum acceleration (in m/s²)?

3. Same system as problem 2. What is the maximum speed of the block (in m/s)?

Round all answers to 3 significant figures.

Spring-Mass Analysis 🔍

Exit Quiz — Spring-Mass Systems ✅

🔄 Energy in Spring-Mass Oscillations

Part 5 of 7 — Elastic Potential Energy

As a spring-mass system oscillates, energy continuously transforms between kinetic and elastic potential energy. Understanding this energy flow is crucial for predicting the motion and solving AP problems.

Energy Flow During Oscillation

For a horizontal spring-mass system oscillating with amplitude $A$ (no friction):

Total Energy

$E_{\text{total}} = \frac{1}{2}kA^2 = \text{constant}$

At any position $x$:

$E_{\text{total}} = KE + PE_s = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

Energy at Key Positions

Energy vs. Position

At any displacement $x$:

$KE = \frac{1}{2}k(A^2 - x^2)$ $PE_s = \frac{1}{2}kx^2$ $v = \sqrt{\frac{k}{m}(A^2 - x^2)}$

Energy Graphs

PE vs. Position

The elastic PE graph is a parabola (upward-opening): $PE_s = \frac{1}{2}kx^2$

• Minimum ($PE = 0$) at $x = 0$
• Maximum ($PE = \frac{1}{2}kA^2$) at $x = \pm A$

KE vs. Position

The KE graph is an inverted parabola: $KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)$

• Maximum at $x = 0$
• Zero at $x = \pm A$

Total Energy vs. Position

A flat horizontal line at $E = \frac{1}{2}kA^2$.

Where Do KE and PE Equal?

Set $KE = PE$: $\frac{1}{2}k(A^2 - x^2) = \frac{1}{2}kx^2$ $A^2 - x^2 = x^2$ $x = \pm \frac{A}{\sqrt{2}} \approx \pm 0.707A$

Energy in Vertical Spring-Mass

For a vertical spring-mass system, both gravitational and elastic PE are involved:

$E_{\text{total}} = KE + PE_g + PE_s$

Simplification

If we measure displacement from the equilibrium position (where $mg = kx_0$), the problem reduces to the horizontal case. The oscillation energy:

$E = \frac{1}{2}kA^2$

where $A$ is measured from the equilibrium position.

Key Point

Don't mix up:

• Natural length of spring (no mass attached)
• Equilibrium position (mass attached, at rest)
• Amplitude (measured from equilibrium, not natural length)

Oscillation Energy Concepts 🎯

Energy in Oscillations Calculations 🧮

A 0.5 kg block oscillates on a horizontal spring ($k = 200$ N/m) with amplitude $A = 0.1$ m.

1. What is the total energy of the system (in J)?

2. What is the speed of the block at $x = 0.06$ m (in m/s)?

3. At what displacement is the speed half of $v_{\max}$ (in m, to 3 significant figures)?

Energy Graph Interpretation 🔍

A spring-mass system oscillates with amplitude $A$.

Exit Quiz — Energy in Oscillations ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Elastic Potential Energy

This workshop brings together Hooke's Law, elastic PE, work by springs, and spring-mass energy concepts. These multi-step problems mirror AP exam difficulty.

Problem-Solving with Springs

Energy Conservation with Springs

$KE_i + PE_{g,i} + PE_{s,i} = KE_f + PE_{g,f} + PE_{s,f} + E_{\text{thermal}}$

$\frac{1}{2}mv_i^2 + mgh_i + \frac{1}{2}kx_i^2 = \frac{1}{2}mv_f^2 + mgh_f + \frac{1}{2}kx_f^2 + |W_f|$

Common Problem Types

1. Spring launcher: Spring PE → KE ($\frac{1}{2}kx^2 = \frac{1}{2}mv^2$)
2. Object compressing spring: KE → Spring PE ($\frac{1}{2}mv^2 = \frac{1}{2}kx^2$)
3. Vertical spring drop: Gravitational PE → Spring PE ($mgh = \frac{1}{2}kx^2$)
4. Spring on incline: Mix of all three energy forms

Worked Example: Spring Launcher on a Ramp

A spring ($k = 500$ N/m) compressed by 0.2 m launches a 0.5 kg ball up a frictionless $30°$ ramp. How far up the ramp does the ball travel?

Energy equation (reference: spring position):

$\frac{1}{2}kx^2 = mgh = mgd\sin(30°)$

$\frac{1}{2}(500)(0.04) = 0.5(10)(d)(0.5)$

$10 = 2.5d$

$d = 4 \text{ m up the ramp}$

Height gained: $h = 4\sin(30°) = 2$ m

Check: $PE_{s,i} = 10$ J. $PE_{g,f} = 0.5(10)(2) = 10$ J. ✓

Workshop Problems 🎯

Workshop Calculations 🧮

Use $g = 10$ m/s².

1. A spring ($k = 480$ N/m) compressed by 0.05 m launches a 0.3 kg ball on a frictionless horizontal surface. What is the ball's speed (in m/s)?

2. A 4 kg block moving at 5 m/s on a frictionless surface hits a spring ($k = 2500$ N/m). What is the maximum compression (in m)?

3. A spring ($k = 500$ N/m, compressed 0.2 m) launches a 1 kg block across a rough surface ($\mu_k = 0.25$). How far does the block slide before stopping (in m)?

Round all answers to 3 significant figures.

Problem Type Identification 🔍

Exit Quiz — Spring Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Elastic Potential Energy

This final lesson integrates Hooke's Law, elastic PE, spring work, spring-mass systems, and energy in oscillations. These AP-level questions will test your mastery of the entire unit.

Key Equations Summary

Big Ideas

• Spring force is conservative → PE can be defined
• PE is quadratic in $x$ → doubling $x$ quadruples PE
• Energy oscillates between KE and PE (no friction)
• The spring force provides variable acceleration ($a \propto x$)

AP-Style Conceptual Questions 🎯

AP-Style Calculations 🧮

Use $g = 10$ m/s².

1. A 0.25 kg ball is launched by a spring ($k = 400$ N/m) compressed 0.1 m on a frictionless surface. What is the ball's speed (in m/s)?

2. A 2 kg block oscillates with $k = 800$ N/m and amplitude 0.05 m. What is the speed when $x = 0.03$ m (in m/s)?

3. A block is attached to a spring ($k = 300$ N/m). The block oscillates with maximum speed 6 m/s and maximum acceleration 120 m/s². What is the mass (in kg)?

Round all answers to 3 significant figures.

AP Review — True/False 🔍

Final AP Exit Quiz — Elastic PE ✅