Topic 3.2 Work: define work as the dot product of force and displacement, compute the work done by a variable force as an integral, and interpret work as the area under a force-position graph.

What this topic is asking

The College Board (Topic 3.2) wants you to define work as the dot product of force and displacement, to compute the work done by a variable force as an integral, and to read work off a force-position graph as an area. In AP Physics C the variable-force case is central: a typical exam problem gives a force as a function of position and asks you to integrate it, often combining the result with the work-energy theorem.

Work for a constant force

The dot product means only the component of the force along the displacement does work. A force at $\theta = 0$ (along the motion) does maximum positive work; at $\theta = 90^\circ$ (perpendicular) it does none; at $\theta = 180^\circ$ (opposing the motion, like friction) it does negative work. This is why carrying a box horizontally does no work against the supporting force, and why the centripetal force in circular motion does no work, it is always perpendicular to the velocity. The sign of work tells you whether energy flows into the object (positive) or out of it (negative).

Work done by a variable force

When the force changes with position, you cannot use a single value of $F$. Instead you integrate the force over the displacement:

This is the defining AP Physics C generalization of work. The integral sums up the infinitesimal contributions $dW = F\,dx$ as the object moves. The spring is the classic example: $F = kx$ grows with displacement, so $W = \int_0^x kx'\,dx' = \tfrac{1}{2}kx^2$, recovering the elastic potential energy. In general you are handed an $F(x)$, set up the integral with the correct limits, and evaluate it. If the force is a vector and the path is curved, the full statement is the line integral $W = \int \vec{F}\cdot d\vec{r}$, but most exam problems are one-dimensional.

The work-energy theorem revisited

Integrating Newton's second law over displacement gives the work-energy theorem: the net work done by all forces equals the change in kinetic energy,

This holds whether the force is constant or variable, which is why energy methods are so powerful: you can find a final speed from the net work without solving the detailed motion. When several forces act, you can either sum their works (the net work) or compute each separately; the works of gravity and springs are conveniently repackaged as potential energy in the next topic.

Try this

Q1. A $50$ N force pulls a crate $8.0$ m along the floor at $30^\circ$ above the horizontal. Calculate the work done by the force. [2 points]

• Cue. $W = Fd\cos\theta = (50)(8.0)\cos 30^\circ = 400(0.866) = 346$ J.

Q2. A force-position graph is a straight line from $F = 10$ N at $x = 0$ to $F = 0$ at $x = 4.0$ m. Calculate the work done over this interval. [2 points]

• Cue. Work is the triangular area: $\tfrac{1}{2}(4.0)(10) = 20$ J.