Study guides & original practice
Work and energy practice problems
A force does work when it has a component along an object’s displacement. The net work on an object equals its change in kinetic energy.
By ExamCat · Original study lesson · Published September 24, 2026
Understand the method
For a constant force, W = Fd cos θ. A force parallel to motion does positive work; an opposite force does negative work. Kinetic energy is K = ½mv², so changing speed has a squared effect.
Worked example, step by step
A 2 kg cart speeds up from 3 m/s to 5 m/s. Find the net work done on it.
- Initial kinetic energy: ½ × 2 × 3² = 9 J.
- Final kinetic energy: ½ × 2 × 5² = 25 J.
- Net work is the difference: 25 − 9 = 16 J.
- The positive result matches an increase in speed. It does not reveal the work of each individual force unless more information is supplied.
A mistake to watch for
Do not square the change in speed to calculate the change in kinetic energy. The correct difference uses vfinal² − vinitial², not (vfinal − vinitial)².
Try it yourself: three questions
Write an answer before opening the explanation. For calculations, keep your units and intermediate steps; for concepts, state the reason for your choice.
1. A 10 N force pushes parallel to displacement for 3 m. Find work.
Show answer and explanation
30 J. Both force and displacement point the same way.
2. A 4 N friction force opposes 5 m of travel. Find its work.
Show answer and explanation
−20 J. The sign indicates energy transfer that reduces mechanical energy in this simple model.
3. A force remains perpendicular to displacement. How much work does it do?
Show answer and explanation
Zero, because cos 90° = 0. A force can affect direction without doing work on the object.
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Further reading
OpenStax Physics provides textbook background. The worked scenario and practice questions on this page were written independently for ExamCat.
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