FOCUSED REVIEW
Focused Review — Work and the Work–Energy Theorem — Algebra-Based
Reinforce the highest-leverage ideas and representative problem-solving tools for Work and the Work–Energy Theorem.
TIME
Approximately 15 minutes
BEST FOR
Targeted reinforcement
FINISH WITH
A readiness check
After this focused review, you'll be able to...
reinforce the key relationships, apply them to representative problems, and identify what still needs work.
Choose how you want to review
Course Alignment
This Physics Sensei Topic Review is an independent learning resource. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.
Related Unit Review: If you need to review the complete unit material, review Work, Energy, and Power here →
RESOURCE: Physics Sensei Topic Review | TOPIC ID: MEC-U06-T01 | TOPIC: Work and the Work–Energy Theorem | PARENT UNIT: MEC-U06 — Work, Energy, and Power | COURSE LEVEL: Algebra-Based
BEST USED ✓ After learning the topic ✓ Before starting homework ✓ Before a quiz or exam
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 15 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
Key Ideas
Answer before revealing the response.
Write the constant-force work relationship and identify the angle in it.
Reveal Answers
W = Fd cos θ, where θ is the angle between force and displacement.
Why it works: The dot product reduces to magnitude times magnitude times cosine of the included angle.
ACTIVITY 2
Common Mistakes
Answer before revealing the response.
What work is done by a force perpendicular to the displacement? Explain using cosine.
Reveal Answers
0 J because cos 90° = 0.
Why it works: Perpendicular force and displacement give zero dot product.
ACTIVITY 3
Quick Application
Answer before revealing the response.
A 30 N force acts 5 m in the direction of motion. Find the work.
Reveal Answers
150 J because the angle is 0° and cos 0° = 1.
Why it works: A parallel force contributes its full magnitude along the displacement.
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
Reinforce the two highest-leverage relationships, then use them in representative situations.
KEY CONCEPT 1
Constant-Force Work
For a constant force, work is the dot product of force and displacement. In magnitude form, the cosine of the angle between them determines how much of the force contributes along the displacement.
W = Fd cos θ
Example: A 40 N force over 2 m at 60° does 40 J of work.
Sensei note: Do not use sine unless the geometry has first been converted to the angle between force and displacement.
KEY CONCEPT 2
The Work–Energy Theorem
The net work done by all forces equals the change in kinetic energy. This is often faster than solving for acceleration first when the question asks about speed or energy.
Wnet = ΔK
Example: If net work is +32 J, kinetic energy increases by 32 J.
Sensei note: Use net work in the theorem: include all forces that do work.
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
Apply the reinforced ideas to two representative situations, then use the strategy card to check your setup.
PRACTICE 1
Guided Example
Resolve the work directly with the cosine factor.
A 60 N force pulls a box 8 m at 30° above the horizontal. Find the work done by the pulling force.
Reveal Answers
About 416 J: (60 N)(8 m)cos 30°.
Why it works: Only the component of the pulling force along the displacement does work.
PRACTICE 2
Independent Check
Use net work to update kinetic energy.
A 2.0 kg object starts at 4.0 m/s. The net work on it is +20 J. Find its final kinetic energy and final speed.
Reveal Answers
Initial kinetic energy is 16 J; final kinetic energy is 36 J; final speed is 6.0 m/s.
Why it works: Net work changes kinetic energy; then kinetic energy gives speed.
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
Angled Force
Solve independently before checking the solution.
A 25 N force acts through 6 m at 90°. What work does it do?
Reveal Answers
0 J. A perpendicular force has no component along the displacement.
Why it works: The cosine factor is zero at 90°.
QUICK CHECK 2
Final Speed from Work
Solve independently before checking the solution.
A 1.0 kg object starts at 3.0 m/s and receives +8.0 J of net work. What is its final speed?
Reveal Answers
5.0 m/s. The initial kinetic energy is 4.5 J; after +8.0 J it is 12.5 J.
Why it works: Use the theorem first, then solve the kinetic-energy expression for speed.
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 essential ideas you’ll want to remember.
KEY TAKEAWAY 1
Use the Angle Between the Vectors
For constant force, use the cosine of the angle between force and displacement to obtain the energy transfer.
KEY TAKEAWAY 2
Net Work Updates Kinetic Energy
Set net work equal to the kinetic-energy change, then use kinetic energy to connect to speed when needed.
MY ONE-SENTENCE SUMMARY
In my own words, the most important idea is:
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