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LESSON 05 / 14 · TOPIC 3.2

Net work and the change in speed

You will be able to: Use the sum of all forces’ work to determine a physically possible final speed.

Free study resourceReview editionTeacher review pending

How much work is needed to change an object’s speed?

A cart starts with 4 J of kinetic energy. A push adds 12 J while resistance removes 4 J. It finishes with 12 J: the energy change is the net transfer, not just the push’s contribution.

A useful starting point: Kinetic energy →

Words and symbols before equations

Net work W_net
Sum of the work done by all forces on the particle.
ΔK
Change in kinetic energy: K_final minus K_initial.
Initial / final
Before / after the chosen interval; often marked i and f.
Energy account: final K = initial K + net work0 J4 JInitial K12 JNet work16 JProposed KBar lengths share one energy scale; labels give exact values.
Read this model snapshot. Final K=16 J; final speed=4 m/s. Direction is not determined by energy alone.
What this picture assumes

Particle mass 2 kg; initial speed 2 m/s; initial kinetic energy 4 J. The work value represents all forces combined. Negative final kinetic energy flags an impossible proposed endpoint.

Connect the picture to the physics

For a particle or suitable nonrotating rigid object, W_net=ΔK. Find each force’s work with its sign, add them, then use K_f=K_i+W_net.

To find final speed, rearrange K_f=½mv_f²: v_f=√(2K_f/m). The result gives speed magnitude. Direction needs other information about the motion.

K_f cannot be negative. If a resistance force would remove more than the available kinetic energy over a proposed distance, the object stops before traveling that far under the assumed motion. For extended deforming systems, energy can also become internal; use a full system energy balance.

A worked example, step by step

A 2 kg cart starts at 2 m/s. Over 3 m, a 6 N push acts forward and a 2 N resistance acts backward. Find final speed.

  1. Initial K=½(2)(2²)=4 J.
  2. Push work =18 J; resistance work =−6 J; net work =12 J.
  3. Final K=4+12=16 J.
  4. Final speed =√(2×16/2)=4 m/s. The cart speeds up because net work is positive.
Common mix-up

Net work equals a change in kinetic energy, not automatically the final kinetic energy.

CHECK THE IDEA

Can negative work leave an object still moving?

Compare with an explanation

Yes. It may remove only part of the initial kinetic energy.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Keep mass 2 kg and initial speed 2 m/s. Change net work through positive, zero and negative values. Identify the stopping limit and explain why more-negative inputs cannot describe the proposed endpoint.

Energy account: final K = initial K + net work0 J4 JInitial K12 JNet work16 JProposed KBar lengths share one energy scale; labels give exact values.

Final K=16 J; final speed=4 m/s. Direction is not determined by energy alone.

Particle mass 2 kg; initial speed 2 m/s; initial kinetic energy 4 J. The work value represents all forces combined. Negative final kinetic energy flags an impossible proposed endpoint.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use a work, energy or power relationship to justify your prediction.

Apply the idea to a fresh problem Practice →

Show what you understand.

Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.

1. Initial K=10 J and net work=−6 J. Final K is…

Show answer and reasoning

4 J. K_f=K_i+W_net=4 J.

2. A calculation predicts K_f=−2 J. This means…

Show answer and reasoning

The assumed endpoint is not physically reachable under that model. K cannot be negative; revisit the assumed motion and stopping point.

Original written challenge

4 points · self-check · not an official AP question

A 1 kg cart starts at 6 m/s and experiences a constant 3 N force opposite its motion. (a) Find initial K. (b) Find work over 4 m before stopping. (c) Find its speed there. (d) Find the stopping distance.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: 18 J.
  2. 1 point: −12 J.
  3. 1 point: K_f=6 J, so v=√12≈3.46 m/s.
  4. 1 point: 3d=18 gives d=6 m.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1What changes kinetic energy?

Net work by all forces in the particle model.

RECALL 2Does zero net work mean zero final speed?

No; final and initial speeds have equal magnitudes.

RECALL 3Why reject a negative final K?

It violates the nonnegative kinetic-energy model.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

Net work and the change in speed

  • W_net=K_f−K_i; K_f=½mv_i²+W_net.
  • Use all working forces and require K_f≥0.

Remember: Net work equals a change in kinetic energy, not automatically the final kinetic energy.

Conditions: Particle mass 2 kg; initial speed 2 m/s; initial kinetic energy 4 J. The work value represents all forces combined. Negative final kinetic energy flags an impossible proposed endpoint.

Refresh Kid · Unit 3 · Objectives 3.2.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 3.2, objectives 3.2.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.

Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.

Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.

Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.

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