Mechanical energy can become thermal energy
You will be able to: Balance mechanical loss and thermal gain without claiming energy disappears.
Where does the energy go when a sliding object stops?
A 1 kg block moving at 4 m/s has 8 J of kinetic energy. It slides to rest on a rough floor. The motion’s energy becomes internal energy in the block and floor, with possible sound; total energy has not vanished.
A useful starting point: Trade height for speed on a smooth track →
Words and symbols before equations
- Thermal-energy increase ΔE_th
- Increase in microscopic internal energy associated with frictional dissipation, measured in J.
- Kinetic friction f_k
- For this ideal level-floor model, μ_k mg, opposing sliding.
- Stopping distance
- Distance traveled before K reaches zero under the specified resisting force.
- Energy accounting boundary
- Including the floor lets the dissipated energy be counted within the enlarged system.
What this picture assumes
A block slides on a fixed level floor with no other horizontal force; g = 10 m/s². The enlarged block–floor–Earth system retains the dissipated energy. Actual sliding distance is capped at the stop; distribution of heat between objects is not modeled.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Friction = 5 N; stop at 5 m. Actual sliding distance 3 m; K = 10 J; thermal gain 15 J; speed 3.162 m/s. The block is still sliding at this distance.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For a particle sliding on a fixed level floor, friction work on the particle is W_f = −μ_k mgd while sliding persists. The work–energy theorem gives K_f = K_i − μ_k mgd. This is the mechanical result; it does not specify exactly how the thermal gain is split between floor and block.
For block + floor + Earth treated as an isolated system, K_i = K_f + ΔE_th when potential energy does not change and other transfers are negligible. The thermal increase equals μ_k mgd. Friction is internal to this enlarged system, but it still reduces mechanical energy.
The stop occurs at d_stop = v_i²/(2μ_k g). After the block stops on a level floor with no applied force, kinetic friction no longer continues draining energy. The model caps actual sliding distance at the stop rather than displaying negative K.
A worked example, step by step
A 2 kg block starts at 5 m/s on a level floor with μ_k = 0.25 and g = 10 m/s². Find speed after 3 m and stopping distance.
- K_i = ½(2)(5²) = 25 J. Normal force is 20 N, so friction magnitude is 5 N.
- After 3 m, friction has dissipated 15 J. K_f = 25 − 15 = 10 J.
- Speed is √(2×10/2) = √10 = 3.16 m/s.
- At rest, all 25 J has left translation: d_stop = 25 J / 5 N = 5 m. Beyond that, the assumed continuing slide is invalid.
Mechanical energy is not conserved during frictional sliding. Total energy can still be conserved when all relevant forms and transfers are included.
Does twice the mass mean twice the stopping distance in this model?
Compare with an explanation
No. Both K and friction force double, so the distance needed to remove K is unchanged.
Predict. Change one thing. Explain.
Keep μ and initial speed fixed. Increase the requested travel distance through the stop. Compare actual distance, remaining K and thermal gain. Then double mass and test whether stopping distance changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Friction = 5 N; stop at 5 m. Actual sliding distance 3 m; K = 10 J; thermal gain 15 J; speed 3.162 m/s. The block is still sliding at this distance.
A block slides on a fixed level floor with no other horizontal force; g = 10 m/s². The enlarged block–floor–Earth system retains the dissipated energy. Actual sliding distance is capped at the stop; distribution of heat between objects is not modeled.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant work, system boundary, energy or calculus 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.
Original written challenge
4 points · self-check · not an official AP questionA 1 kg block at 6 m/s slides on a fixed horizontal floor with μ_k = 0.3 and g = 10 m/s². Find friction, stopping distance and thermal gain, and describe why −f d cannot be extended past the stop.
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Compare with the answer and four-point rubric
- 1 point: Friction magnitude is μmg = 3 N while sliding.
- 1 point: Initial K = 18 J, giving stopping distance 18/3 = 6 m.
- 1 point: Thermal gain of the block–floor system is 18 J at rest under the stated assumptions.
- 1 point: After stopping, sliding ceases. Continuing the same formula would falsely produce negative K.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Can internal forces change a system’s mechanical energy?
Yes. Internal friction can convert it to thermal energy.
RECALL 2Which distance belongs in frictional dissipation?
The actual relative sliding distance for the specified model.
RECALL 3What does the simple model not tell us?
How the dissipated thermal energy is divided between the contacting objects.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Mechanical energy can become thermal energy
- On a fixed level floor: f_k = μ_k mg.
- K_i = K_f + ΔE_th, with ΔE_th = f_k d during sliding.
- d_stop = v_i²/(2μ_k g), for μ_k > 0 and no other horizontal forces.
Remember: Mechanical energy is not conserved during frictional sliding. Total energy can still be conserved when all relevant forms and transfers are included.
Conditions: A block slides on a fixed level floor with no other horizontal force; g = 10 m/s². The enlarged block–floor–Earth system retains the dissipated energy. Actual sliding distance is capped at the stop; distribution of heat between objects is not modeled.
Refresh Kid · AP Physics C: Mechanics Unit 3 (official Unit 3) · Objectives 3.4.B; 3.4.C · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.4, objectives 3.4.B; 3.4.C. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 3: Work, Energy, and Power. The unit covers Topics 3.1–3.5. Calculus connects work to force integrals, force to potential-energy derivatives, and power to the rate of energy transfer. Models distinguish object-only and multi-object systems; translational models exclude rotational energy unless explicitly noted. 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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