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Elastic Potential Energy and Springs

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Elastic Potential Energy and Springs - Complete Interactive Lesson

Part 1: Hooke\'s Law

🔩 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

Displacement	Spring Force
$x > 0$ (stretched right)	$F < 0$ (pulls left)
$x < 0$ (compressed left)	(pushes right)

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

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

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:

Hooke's Law Concepts 🎯

Hooke's Law Calculations 🧮

Use $g = 10$ m/s².

A spring stretches 0.05 m when a 2 kg mass hangs from it. What is the spring constant $k$ (in N/m)?
A spring with $k = 800$ N/m is compressed by 0.03 m. What force is needed to hold it compressed (in N)?
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)?

Spring Force Analysis 🔍

Exit Quiz — Hooke's Law ✅

Part 2: Spring Constant

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

Part 3: Elastic PE Formula

🔧 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 $x_f$ :

Part 4: Work Done 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 $x$

Part 5: Springs in Series & Parallel

🔄 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

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

Part 6: Problem-Solving Workshop

🛠️ 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

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

Part 7: Synthesis & AP Review

🎓 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

Concept	Equation	Key Point
Hooke's Law	$F_s = -kx$	Restoring force; linear
Spring constant

Spring Type	$k$ (N/m)
Slinky	~1
Screen door spring	~100
Car suspension	~30,000
Atomic bond	~100

Finding $k$ Experimentally

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

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

(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

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

Round all answers to 3 significant figures.

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

Property	Detail
Units	Joules (J)
Sign	Always $\geq 0$ (because $x^2 \geq 0$ )
Maximum when	$x$ is at maximum stretch or compression

The $x^2$ Dependence

Because PE depends on $x^2$ :

Displacement	PE
$x$	$\frac{1}{2}kx^2$
$2x$	$\frac{1}{2}k(2x)^2 = 4 \times \frac{1}{2}kx^2$
$3x$	$\frac{1}{2}k(3x)^2 = 9 \times \frac{1}{2}kx^2$

Doubling the displacement quadruples the stored energy!

Same PE for Stretch and Compression

Since $PE_s = \frac{1}{2}kx^2$ and :

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

Feature	Gravitational PE	Elastic PE
Formula	$mgh$	$\frac{1}{2}kx^2$
Depends on	Height ( $h$ ) — linear	Displacement ( $x$ ) — quadratic
Reference	Chosen by you	Always $x = 0$ (equilibrium)
Can be negative?	Yes	No
Force type	Constant ( $mg$ )	Variable ( $kx$ )
Associated force	Gravity	Spring force
Conservative?	Yes	Yes

Elastic PE Concepts 🎯

Elastic PE Calculations 🧮

A spring ( $k = 500$ N/m) is stretched 0.08 m. What PE is stored (in J)?
A spring stores 18 J when compressed 0.3 m. What is the spring constant (in N/m)?
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 ✅

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

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

Process	$W_{\text{spring}}$	Sign
Releasing compressed spring ( $x_i = A, x_f = 0$ )	$+\frac{1}{2}kA^2$	Positive (spring pushes object)
Stretching from equilibrium ( $x_i = 0, x_f = A$ )	$-\frac{1}{2}kA^2$
Compressing from equilibrium ( $x_i = 0, x_f = -A$ )	$- \frac{1}{2} k^{}$
Returning from stretch ( $x_i = A, x_f = 0$ )	$+\frac{1}{2}kA^2$

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 = $\frac{1}{2} k$

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.

How much work is needed to stretch the spring from equilibrium to $x = 0.1$ m (in J)?
How much work is needed to stretch it from $x = 0.1$ m to $x = 0.2$ m (in J)?
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 force: $F = -kx$ (restoring)
Acceleration: $a = -kx/m$ (toward equilibrium)
The acceleration is not constant — it depends on position!

When released from displacement $A$ (amplitude):

Block accelerates toward equilibrium
Passes through equilibrium at maximum speed
Overshoots, compressing (or stretching) the spring
Spring slows the block, stops it at $-A$
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 $A$ :

Position	Displacement	Force	Acceleration	Speed
Maximum stretch	$+A$	$-kA$ (toward center)	$-kA/m$ (max)	0
Equilibrium	$0$	$0$	$0$	Maximum
Maximum compression	$-A$	$+kA$ (toward center)	$+kA/m$ (max)	0

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.

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

At any position $x$ :

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

Energy at Key Positions

Position	$x$	$PE_s$	$KE$	Speed
Maximum stretch	$+A$	$\frac{1}{2}kA^2$	$0$	$0$
Equilibrium	$0$	$0$	$\frac{1}{2}kA^2$
Maximum compression	$-A$	$\frac{1}{2}kA^2$	$0$

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

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 $KE = PE$ : $\frac{1}{2}k(A^2 - x^2) = \frac{1}{2}kx^2$

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.

What is the total energy of the system (in J)?
What is the speed of the block at $x = 0.06$ m (in m/s)?
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 ✅

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

Common Problem Types

Spring launcher: Spring PE → KE ( $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$ )
Object compressing spring: KE → Spring PE ( $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$ )
Vertical spring drop: Gravitational PE → Spring PE ( $mgh = \frac{1}{2}kx^2$ )
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.

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)?
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)?
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 ✅

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.

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)?
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)?
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 ✅

- \frac{1}{2} k A^{2}

- \frac{1}{2} k A^{2}

+ \frac{1}{2} k A^{2}

\frac{1}{2} k (x_{2}^{2} - x_{1}^{2})

x = \pm \frac{A}{\sqrt{2}} \approx \pm 0.707A

Elastic Potential Energy and Springs

Typical Values

Finding $k$ Experimentally

Important Note

Defining Elastic PE

Properties

The $x^2$ Dependence

Same PE for Stretch and Compression

Graphical Connection

Elastic vs. Gravitational PE

Important Cases

Work Done ON the Spring

Key Relationship

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

Graphical Interpretation

Work = Area Under the Curve

Work Between Two Displacements

Key Behavior

Vertical Spring-Mass System

New Equilibrium Position

Oscillation About the New Equilibrium

Important Insight

Force and Acceleration Analysis

Key Relationships

Energy at Key Positions

Energy vs. Position

Energy Graphs

PE vs. Position

KE vs. Position

Total Energy vs. Position

Where Do KE and PE Equal?

Energy in Vertical Spring-Mass

Simplification

Key Point

Common Problem Types

Worked Example: Spring Launcher on a Ramp

Big Ideas

Elastic Potential Energy and Springs

Elastic Potential Energy and Springs - Complete Interactive Lesson

Part 1: Hooke\'s Law

🔩 Hooke's Law: F=−kxF = -kxF=−kx

Hooke's Law

What the Negative Sign Means

The Spring Constant kkk

Units

Force vs. Displacement Graph

Applied Force to Stretch/Compress

Part 2: Spring Constant

🌀 Elastic PE: PEs=12kx2PE_s = \frac{1}{2}kx^2PEs​=21​kx

Part 3: Elastic PE Formula

🔧 Work Done by Springs

Work Done BY the Spring

Part 4: Work Done by Springs

🏋️ Spring-Mass Systems

Horizontal Spring-Mass System

At Equilibrium (x=0x = 0x=0)

Displaced by x

Part 5: Springs in Series & Parallel

🔄 Energy in Spring-Mass Oscillations

Energy Flow During Oscillation

Total Energy

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Problem-Solving with Springs

Energy Conservation with Springs

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Key Equations Summary

Typical Values

Finding kkk Experimentally

Important Note

Defining Elastic PE

Properties

The x2x^2x2 Dependence

Same PE for Stretch and Compression

Graphical Connection

Elastic vs. Gravitational PE

Important Cases

Work Done ON the Spring

Key Relationship

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

Graphical Interpretation

Work = Area Under the Curve

Work Between Two Displacements

Key Behavior

Vertical Spring-Mass System

New Equilibrium Position

Oscillation About the New Equilibrium

Important Insight

Force and Acceleration Analysis

Key Relationships

Energy at Key Positions

Energy vs. Position

Energy Graphs

PE vs. Position

KE vs. Position

Total Energy vs. Position

Where Do KE and PE Equal?

Energy in Vertical Spring-Mass

Simplification

Key Point

Common Problem Types

Worked Example: Spring Launcher on a Ramp

Big Ideas

🔩 Hooke's Law: $F = -kx$

The Spring Constant $k$

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

At Equilibrium ( $x = 0$ )

Displaced by $x$

Finding $k$ Experimentally

The $x^2$ Dependence

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