Work and Work Graphs
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
Work is how a force transfers energy by pushing something through a distance: W = Fd cos θ, where θ is the angle between the force and the displacement. Only the part of the force along the motion does work. On a force–position graph, work is the area under the graph.
1. Positive, negative, or zero
Pull a wagon with a rope at an angle: only the horizontal part of your pull, F cos θ, does work. The upward part doesn’t, because the wagon doesn’t move up.
Friction on a sliding crate points opposite the motion, so it does negative work.
The normal force on level ground and the string in uniform circular motion are perpendicular to the motion, so they do zero work.
Work is a scalar: it has a sign but no direction.
2. Work from a graph
When the force changes, you can’t just multiply F × d. Find the area between the force–position graph and the x-axis instead.
Units check: the vertical axis is newtons and the horizontal is meters, so the area is N·m = joules.
3. Example, then try it
Use the graph below. How much work does the force do from x = 0 to x = 6 m?
Triangle (0 to 4 m): ½(4 m)(8 N) = 16 J.
Rectangle (4 to 6 m): (8 N)(2 m) = 16 J.
Total: 32 J.
A box is dragged 5.0 m across a level floor by a 20 N rope pulling 60° above horizontal. Friction on the box is 6.0 N. Find the work done by each force and the net work. (cos 60° = 0.50)
Rope: W = Fd cos θ = (20)(5.0)(0.50) = +50 J.
Friction: opposite the motion, θ = 180°: W = (6.0)(5.0)(−1) = −30 J.
Gravity and normal force: both perpendicular to the motion: 0 J each.
Net work: 50 − 30 + 0 + 0 = +20 J.
Answer: rope +50 J, friction −30 J, gravity and normal 0 J; net work +20 J, so the box gains 20 J of kinetic energy.
4. Common mistakes
You carry a heavy backpack across a level hallway at constant speed. How much work does your upward force on the backpack do?
Your force on the backpack is straight up, and the backpack moves horizontally. The force is perpendicular to the displacement, so cos 90° = 0 and the work is zero.
Ready to practice?
4 questions on Work and Work Graphs, each with a hint and an explanation for every wrong answer.