Mechanical Work as a Vector Dot Product

Why It Matters

Mechanical work is often introduced in one dimension as force times displacement. That shortcut is useful, but the stricter definition is vector-based: work depends on the component of force along the displacement. This distinction prevents common errors in inclined planes, circular motion, field forces, and force-displacement graphs.

Definition

For a constant force acting during a straight displacement, the work done by the force is:

$$W=\vec{F}\cdot \Delta \vec{r}$$

Equivalently,

$$W=F\Delta r\cos \theta$$

where $\theta$ is the angle between the force vector $\vec{F}$ and the displacement vector $\Delta \vec{r}$.

This can be interpreted in two equivalent ways:

$$W=F_{\parallel }\Delta r$$

where $F_{\parallel }=F\cos \theta$ is the component of force along the displacement, or:

$$W=F\Delta r_{\parallel }$$

where $\Delta r_{\parallel }=\Delta r\cos \theta$ is the component of displacement along the force.

Work is a scalar, even though force and displacement are vectors.

Key Representations

If the force is parallel to the displacement:

$$W=F\Delta r$$

If the force is opposite to the displacement:

$$W=-F\Delta r$$

If the force is perpendicular to the displacement:

$$W=0$$

For a force that changes during the motion, work is the line integral:

$$W={\int }_C\vec{F}\cdot d\vec{r}$$

where $C$ represents the path followed by the object.

In one-dimensional motion, after choosing a positive direction, this becomes:

$$W={\int }_{x_i}^{x_f}{F}_x(x)dx$$

Here $F_x$ is the signed component of the force along the chosen $x$-axis. The signed area under an $F_x$-$x$ graph gives the work done by that force.

Common Exam Points

• Use displacement, not distance, in the definition of work.
• Use the angle between force and displacement, not necessarily the angle to the horizontal.
• A force perpendicular to displacement does no work at that instant.
• The normal contact force does no work for horizontal motion on a flat surface, but it can do work if the contact point moves along the normal direction.
• In circular motion, a purely radial centripetal force does no work because it is perpendicular to the instantaneous displacement.
• For variable force questions, check whether the graph gives the force component along displacement. If it does, work is the signed area under the graph.

Links

• Prerequisite: vectors
• Prerequisite: forces
• Prerequisite: kinematics
• Related: work energy and power
• Related: work and energy transfer
• Related: kinetic energy and work energy theorem
• Related: potential energy and conservative forces