Work–Energy Theorem
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
The net work done on an object equals its change in kinetic energy: Wnet = ΔK = ½mvf² − ½mvi². It works for any forces, constant or not, so it’s often the fastest way from a force to a speed.
1. The theorem
2. When to use it
Use it when you know forces over a distance and want a speed (or the reverse). It skips finding the acceleration and the time.
It’s just Newton’s second law integrated over position: ∫m(dv/dt) dx = ∫mv dv = ΔK.
Add up the work by every force (gravity, friction, applied forces), or find the net force first and integrate it. Don’t also add potential energy for a force you’ve already counted as work.
3. Example, then try it
A 3.0 kg object moving at 4.0 m/s has 30 J of net work done on it. Find its new speed.
Ki = ½(3.0)(4.0)² = 24 J, so Kf = 24 + 30 = 54 J.
v = √(2 × 54 ÷ 3.0) = √36 = 6.0 m/s.
A 2.0 kg object starts from rest at x = 0. The net force on it is F(x) = 10 − 2x (N). Find its speed at x = 3 m, its maximum speed, and where it comes to rest again.
Work: W = ∫0x(10 − 2x) dx = 10x − x².
At 3 m: W = 30 − 9 = 21 J = ½(2.0)v², so v = √21 ≈ 4.6 m/s.
Max speed: F = 0 at x = 5 m: W = 50 − 25 = 25 J, so v = 5.0 m/s.
At rest again: W = 10x − x² = 0 at x = 10 m.
Answer: 4.6 m/s at 3 m; 5.0 m/s max at 5 m; back to rest at 10 m.
4. Common mistakes
The net work done on an object over some interval is zero. What can you conclude about its speed?
Wnet = ΔK = 0 means K, and so the speed, is the same at the start and end. It could have changed in between.
Ready to practice?
4 questions on Work–Energy Theorem, each with a hint and an explanation for every wrong answer.