Work by a Variable Force: The Integral Definition
- Compute work as the integral of force over displacement
- Interpret work as the area under a force–position graph
- Derive elastic potential energy from the work done by a spring
W = Fd is the special case
The formula W = Fd cos θ holds only when the force is constant. The general definition is the integral W = ∫F · dx, and the constant-force formula is what that integral collapses to when F can be pulled outside it. Whenever a force varies with position — a spring, gravity over astronomical distances, a rope being wound onto a drum — you must integrate. Recognizing the trigger is most of the skill: if the problem gives you a force as a function rather than a number, reach for the integral.
The area under a force–position graph
Because work is an integral of force with respect to position, it is the signed area under an F versus x graph. Area below the axis counts negative, representing a force opposing the displacement. This is the graphical counterpart of the calculus, and it lets you evaluate work for shapes that would be awkward to integrate symbolically — triangles, rectangles and trapezoids read straight off the plot. It is also how the exam most often tests the concept without requiring antiderivatives.
Where ½kx² comes from
The force you must apply to stretch a spring is +kx, growing with extension. The work done stretching it from 0 to x is therefore ∫₀ˣ kx dx = ½kx², and that stored work is the elastic potential energy. The factor of one half is not arbitrary: it appears because the force grows linearly from zero, so the average force over the stretch is ½kx. Every ½ in this course — ½kx², ½mv², ½Iω² — traces back to integrating something that grows linearly.
A force F(x) = 3x² N acts on a particle moving from x = 0 to x = 2.0 m. Find the work done. Then find the work needed to stretch a spring with k = 400 N/m by 0.15 m.
- 1.The force varies with position, so W = ∫₀² 3x² dx.
- 2.Antidifferentiate: ∫3x² dx = x³, so W = (2.0)³ − 0³ = 8.0 J.
- 3.For the spring, W = ∫₀^x kx dx = ½kx².
- 4.W = ½(400)(0.15)² = ½(400)(0.0225) = 4.5 J.
A satellite moves in a perfectly circular orbit. The work done by gravity over one complete orbit is —
A force perpendicular to displacement does zero work. That single fact disposes of the normal force on a level surface, the tension in a conical pendulum, the magnetic force on a charge, and the centripetal force in any circular orbit — all common exam targets.
On a graph of force versus position, the work done by the force between two positions is represented by —
Answer the 2 checkpoints as you read.
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