Work
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
Work is force times displacement, counting only the part of the force along the motion. For a changing force, add it up: W = ∫F · dr, which in one dimension is the area under the F(x) graph.
1. The definition
2. Signs and zero work
Positive work: the force has a part along the motion (it speeds things up). Negative work: it has a part against the motion (friction, braking).
Zero work: the force is perpendicular to the motion, like the normal force on a level floor or the tension in a string twirling a ball in a circle.
For a spring, the work depends on x², so stretching from 0.1 m to 0.2 m takes three times the work of stretching from 0 to 0.1 m.
3. Example, then try it
A constant force F = (4 î − 2 ĵ) N acts while an object moves Δr = (3 î + 5 ĵ) m. Find the work.
W = FxΔx + FyΔy = (4)(3) + (−2)(5) = 12 − 10 = 2 J.
A force F(x) = 6x − 1.5x² (N) acts on an object (graphed below). Find the work it does from x = 0 to 4 m, and from x = 4 m to 5 m.
Antiderivative: ∫(6x − 1.5x²) dx = 3x² − 0.5x³.
0 to 4 m: (48 − 32) − 0 = 16 J.
4 to 5 m: (75 − 62.5) − 16 = 12.5 − 16 = −3.5 J. The force has turned backward, so it does negative work.
Answer: +16 J, then −3.5 J.
4. Common mistakes
A ball on a string moves in a horizontal circle at constant speed. How much work does the string’s tension do on the ball in one revolution?
The tension points toward the center, perpendicular to the velocity at every instant, so it does no work. That’s why the speed stays constant.
Ready to practice?
5 questions on Work, each with a hint and an explanation for every wrong answer.