FULL REVIEW
Full Review: Springs and Elastic Potential Energy — Calculus-Based
Build the complete calculus connection among position-dependent spring force, work integrals, potential energy, and mechanical energy.
TIME
45–60 minutes
BEST FOR
A complete topic review
FINISH WITH
A readiness check
After this full review, you’ll be able to...
analyze springs and elastic potential energy using the Calculus-Based treatment with confidence.
Choose how you want to review
Course Alignment
This Physics Sensei Unit Review reinforces the unit below. Use it to review core ideas, prepare for homework, or refresh before a quiz or exam.
RESOURCE: Physics Sensei Unit Review
UNIT: Springs and Elastic Potential Energy
TREATMENT: Calculus-Based
COURSE LEVEL: Introductory college physics
BEST USED
✓ After learning the unit
✓ Before starting homework
✓ Before a quiz or exam
Physics Sensei is an independent educational resource organized around physics concepts, problem-solving strategies, and guided review.
Your Review Plan
Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.
6 Stages • Approximately 45–60 minutes.
Warm-Up Check
Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.
ACTIVITY 1
Position-Dependent Force
Recognize the spring as a variable force.
For F(x) = −kx, what is dF/dx?
Reveal Answers
−k.
Why it works: The ideal spring’s force-versus-position graph is linear with constant slope −k.
ACTIVITY 2
Work Integral
Set up spring work between two positions.
What integral gives the work done by an ideal spring from xi to xf?
Reveal Answers
Ws = ∫xᵢx_f(−kx) dx.
Why it works: Work by a position-dependent force is the definite integral of F(x) over displacement.
ACTIVITY 3
Potential Gradient
Recover force from potential energy.
If U(x) = ½kx², what force follows from Fx = −dU/dx?
Reveal Answers
Fx = −kx.
Why it works: Differentiating the spring potential gives dU/dx = kx, and the force is the negative gradient.
Ready to strengthen your understanding?
You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?
Core Concepts
Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.
KEY CONCEPT 1
Spring Work from Integration
Because the spring force varies with position, calculate work with a definite integral.
Ws = ∫xᵢx_f(−kx) dx = ½kxi² − ½kxf².
Example: Moving from x = 0.10 m to x = 0 with k = 250 N/m gives Ws = +1.25 J.
Sensei Note: The signed area under the F-x graph is the work.
KEY CONCEPT 2
Potential Energy and Force
For a conservative one-dimensional force, potential energy and force are linked by differentiation.
Fx = −dU/dx; for a spring U(x) = ½kx² + C.
Example: Choosing U(0) = 0 sets C = 0 and recovers Fx = −kx.
Sensei Note: At a stable equilibrium, dU/dx = 0 and d²U/dx² > 0.
KEY CONCEPT 3
Mechanical Energy as a Function of Position
For conservative horizontal spring motion, total energy fixes the allowed speed at each position.
E = ½mv² + ½kx² = ½kA²; v(x) = √[(k/m)(A² − x²)].
Example: At x = ±A, v = 0. At x = 0, the speed is maximum.
Sensei Note: This energy result does not require solving the simple-harmonic-motion differential equation.
Ready to apply these ideas?
You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?
Guided Practice
Now it's time to apply what you've reviewed.
Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.
PRACTICE 1
Integrate the Spring Force
Evaluate a definite work integral.
A 300 N/m spring moves from x = 0.12 m to x = 0.04 m. Find Ws.
Reveal Answers
1.92 J.
Why it works: ∫(−300x)dx from 0.12 to 0.04 gives −150[(0.04)² − (0.12)²] = 1.92 J.
PRACTICE 2
Force from a Potential
Differentiate a given potential-energy function.
If U(x) = 40x² joules, find F(x) and the equivalent spring constant.
Reveal Answers
F(x) = −80x N and k = 80 N/m.
Why it works: F = −dU/dx = −80x, which matches the form −kx.
PRACTICE 3
Speed from Energy
Use total energy to find speed at an intermediate position.
A 0.40 kg block is attached to a 160 N/m spring with A = 0.15 m. Find v at x = 0.090 m.
Reveal Answers
2.40 m/s.
Why it works: v = √[(k/m)(A² − x²)] = √[400(0.0225 − 0.0081)] = 2.40 m/s.
Ready to check your understanding?
You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?
Confidence Check
You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.
QUICK CHECK 1
Slope of U(x)
Infer force direction from the potential-energy slope.
At a point where dU/dx > 0, what is the sign of Fx?
Reveal Answers
Negative.
Why it works: Fx = −dU/dx.
QUICK CHECK 2
Stable Equilibrium
Use derivatives of U to classify equilibrium.
What derivative conditions identify a stable equilibrium in one dimension?
Reveal Answers
dU/dx = 0 and d²U/dx² > 0.
Why it works: The force vanishes at an extremum, and a positive second derivative identifies a local minimum.
QUICK CHECK 3
Turning Point from Energy
Connect total energy with a potential-energy curve.
For U(x) = ½kx² and total energy E, what equation determines the turning points?
Reveal Answers
E = ½kx², so x = ±√(2E/k).
Why it works: At a turning point K = 0, so the total energy equals the potential energy.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Integrate Variable Force
W = ∫F(x)dx; for a spring this gives Ws = ½kxi² − ½kxf².
KEY TAKEAWAY 2
Differentiate Potential to Recover Force
Fx = −dU/dx and Us = ½kx².
KEY TAKEAWAY 3
Use Energy Curves to Read Motion
Turning points satisfy E = U; stable equilibrium occurs at a minimum of U.
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