Quick answer: The net work done on an object equals its change in kinetic energy: W_net = ΔK = (1/2)m(v² − u²). Net work includes all forces, including negative work by friction.
Worked example
How much net work accelerates a 2 kg object from 3 m/s to 5 m/s?
- Initial kinetic energy = (1/2)×2×3² = 9 J.
- Final kinetic energy = (1/2)×2×5² = 25 J.
- Net work = 25 − 9 = 16 J.
Check the result
The positive work increases the speed and kinetic energy.
Stuck on a question? Ask it here and get a step-by-step answer in seconds — pay only per answer, no subscription. Ask your doubt now →
Common mistake to avoid
Subtract squared speeds, not the square of the speed difference. Work by an individual force equals ΔK only if it is the sole contributor to net work.
Try it yourself
1. Find net work to stop a 2 kg object moving at 4 m/s. 2. If net work is zero, what happens to kinetic energy?
Answers and reasoning
1. 0 − (1/2)×2×16 = −16 J. 2. It is unchanged; the direction of motion can still change.
Need help with your own attempt? Ask a study question. AI answers require an account and paid credits. Check important steps with your textbook or teacher.
Stuck on a question? Ask it here and get a step-by-step answer in seconds — pay only per answer, no subscription. Ask your doubt now →