Work-Energy Theorem
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
The net work done on an object equals its change in kinetic energy: Wnet = ΔK. For a system with potential energy, the work done by outside forces changes the system’s total energy: Wext = ΔK + ΔUg + ΔUs.
1. Work in, energy out
2. Choose the system first
Object alone (say, a ball): gravity is an outside force, so you count the work done by gravity, and there’s no Ug.
Object + Earth: gravity is inside the system, so you use ΔUg instead of gravity’s work.
Either way gives the same answer. Just don’t count gravity twice: as work and as ΔUg.
If outside forces do negative work (like air resistance or friction from outside the system), the system’s mechanical energy goes down.
3. Example, then try it
A 5.0 kg block moving at 2.0 m/s is pushed 2.0 m along the floor with a 20 N force while 10 N of friction acts on it. Find its final speed.
Ki = ½(5.0)(2.0)² = 10 J.
Wpush = (20)(2.0) = +40 J; Wfriction = −(10)(2.0) = −20 J.
Kf = 10 + 40 − 20 = 30 J → ½(5.0)v² = 30 → v ≈ 3.5 m/s.
A 0.50 kg ball is thrown straight down at 4.0 m/s from 10 m above the ground. Because of air resistance, it lands at 12 m/s. How much work did air resistance do? (System: ball + Earth.)
ΔK: ½(0.50)(12)² − ½(0.50)(4.0)² = 36 − 4 = +32 J.
ΔUg: mgΔy = (0.50)(10)(−10) = −50 J.
Wext = ΔK + ΔUg = 32 + (−50) = −18 J.
Check: without air resistance, all 50 J would become kinetic energy. Only 32 J did; air resistance took the other 18 J.
Answer: air resistance did −18 J of work on the ball.
4. Common mistakes
The speed of a 2.0 kg cart increases from 2.0 m/s to 4.0 m/s. How much net work was done on it?
Wnet = ΔK = ½(2.0)(4.0²) − ½(2.0)(2.0²) = 16 − 4 = 12 J. You can’t just use the change in speed: K depends on v².
Ready to practice?
5 questions on Work-Energy Theorem, each with a hint and an explanation for every wrong answer.