Apply net work to kinetic energy on a level path
Add driving and resisting work with their signs, recover speed and verify a constant-force energy result with Newton’s law.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.5. Use g = 10 m s⁻² throughout these problems. Masses are positive and constant. Work by a constant force is Fd cos θ, with θ between force and displacement; scalar-product notation is not required. Kinetic energy is ½mv² and gravitational potential-energy change is mgΔh. State the system and count gravitational work or potential-energy change once, not both. Mechanical energy is conserved only under the stated loss-free model; friction and supplied resistance transfer mechanical energy to other forms. Smooth curved-slide examples use overall energy changes and assume contact throughout, rather than requiring a normal-reaction calculation. Average power is work divided by elapsed time; P = Fv applies to a force along the motion. Constant driving power generally gives variable force and acceleration; instantaneous calculations do not justify constant-acceleration formulae over a whole journey. Check string, contact and motion-state changes before continuing a model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate signed work by parallel constant forces.
- Use kinetic energy and rearrange for speed.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A drive does +100 J and resistance does −40 J on a level particle. What is the net work?
What you will learn
- Apply total work = change in kinetic energy for a stated particle model.
- Find a stopping distance or final speed and check the interval remains physical.
Use all the signed work contributions
On a horizontal path, weight and normal reaction do no work. If a 4 kg particle starts at speed 3 m s⁻¹, a 20 N drive does +100 J over 5 m while an 8 N resistance does −40 J. Net work is +60 J, so final K = initial K + 60 = 18 + 60 = 78 J. Therefore ½ × 4 × v² = 78 and final speed is √39 m s⁻¹, about 6.24.
The work-energy equation uses the sum of work by all forces on the selected particle. It is not a rule that the driving work alone equals ΔK when resistance acts. Energy gives speed rather than direction, so use the stated forward motion and any stopping event to interpret the root. A negative proposed final kinetic energy means the assumed travel cannot occur without a changed force or motion stage.
Stopping distance can be found without travel time
A 5 kg particle at speed 6 m s⁻¹ has K = 90 J. With only a constant 15 N opposing force on the level path, −15d = 0 − 90 gives d = 6 m to first rest. Continue neither the same resistance direction nor the negative-energy prediction beyond that rest event without a new model. The amount transferred from kinetic energy is 90 J, while the signed resisting work is −90 J.
For constant along-path forces, Newton’s law offers an independent check: a = (20 − 8)/4 = 3 in the first example. Then v² = 3² + 2 × 3 × 5 = 39, agreeing with energy. This cross-check uses a constant acceleration justified by constant resultant and positive mass. It is not valid for an entire constant-power journey where the driving force changes with speed.
Pause and explain
If a forward-travel calculation predicts negative final kinetic energy, what should you do?
Worked example
A 4 kg particle moves 5 m forward on a level path, starting at speed 3 m s⁻¹ under drive 20 N and opposing resistance 8 N. Find its final speed.
Show the worked solution
- Initial K = ½ × 4 × 3² = 18 J.
- Net work is (20 − 8) × 5 = 60 J, so final K is 78 J.
- Thus 2v² = 78, giving speed √39 m s⁻¹; Newton’s law gives the same squared speed.
Answer Final K = 78 J; speed √39 m s⁻¹, approximately 6.24.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 2 kg particle starts from rest and travels 3 m forward on a level path under constant drive 12 N and resistance 4 N. Find final kinetic energy and speed.
Give me a hint
Find net work before substituting into ½mv².
Compare my reasoning
- Net work is (12 − 4) × 3 = 24 J.
- Initial kinetic energy is zero, so final K = 24 J.
- Since ½ × 2 × v² = 24, final speed is 2√6 m s⁻¹.
Final K = 24 J; speed 2√6 m s⁻¹.
Look for these in your work
- Subtract the work lost to resistance.
- Choose the non-negative speed root.
A 5 kg particle moves forward at 6 m s⁻¹ on a level path. A constant 15 N force opposes it until first rest. Find its stopping distance and signed work by that force.
Give me a hint
At first rest, final kinetic energy is zero.
Compare my reasoning
- Initial K = ½ × 5 × 36 = 90 J.
- The balance is −15d = −90, so d = 6 m.
- The opposing force does −90 J; the positive amount transferred out of kinetic energy is 90 J.
Stopping distance 6 m; resisting work −90 J.
Look for these in your work
- Use the signed work of the opposing force.
- End the calculation at first rest.
A 4 kg particle starts at speed 3 m s⁻¹ and travels 5 m forward under constant drive 20 N and resistance 8 N. Find final speed using energy, then check using acceleration.
Give me a hint
The same constant resultant allows an independent Newton-plus-kinematics check.
Compare my reasoning
- Initial K = 18 J and net work = 60 J, so final K = 78 J and v² = 39.
- Newton’s law gives a = (20 − 8)/4 = 3 m s⁻².
- Then v² = 9 + 2 × 3 × 5 = 39, so both methods give speed √39 m s⁻¹.
Speed √39 m s⁻¹; independent check a = 3 m s⁻² and v² = 39.
Look for these in your work
- Include both force works in the energy method.
- Justify constant acceleration before applying the motion formula.
Common mistakes
- Add signed resistance work to the drive work.
- Energy determines speed magnitude, not a velocity direction by itself.
- A stopping event can end the stated sliding-force model.
What can you explain now?
Can the drive’s work alone be equated to ΔK when an opposing resistance also acts?
Compare with the explanation
No; the resistance’s negative work must also be included.
The kinetic-energy change equals total signed work on the selected particle.
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Make it stick
Total signed work changes kinetic energy; check every contribution and stop before an impossible negative-energy stage.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.