AcademyWork, Energy, and Power
Academy
Work by a Force
Level 1 - Physics topic page in Work, Energy, and Power.
Principle
Work transfers energy through the force component along displacement.
Notation
\(\vec{F}\)
force doing the work
\(\mathrm{N}\)
\(\Delta\vec{r}\)
displacement
\(\mathrm{m}\)
\(W\)
work done by the force
\(\mathrm{J}\)
\(\theta\)
angle between force and displacement
rad or deg
\(F_{\parallel}\)
force component along displacement
\(\mathrm{N}\)
Method
The diagram isolates the part of the force that points along the displacement; the perpendicular part does not transfer energy along this path.
Projection turns the vector force into the signed component that acts along the displacement.
Displacement
\[\Delta\vec{r}=\vec{r}_f-\vec{r}_i\]
Project force
\[F_{\parallel}=F\cos\theta\]
Define work
\[W=F_{\parallel}\Delta r\]
Dot product
\[W=\vec{F}\cdot\Delta\vec{r}\]
Net work
\[W_{\mathrm{net}}=\sum_i W_i\]
Rules
Constant-force work
\[W=\vec{F}\cdot\Delta\vec{r}\]
Angle form
\[W=F\Delta r\cos\theta\]
Component form
\[W=F_x\Delta x+F_y\Delta y+F_z\Delta z\]
Net work
\[W_{\mathrm{net}}=\sum_i W_i\]
Examples
Question
A
\[30\,\mathrm{N}\]
force pulls \[4\,\mathrm{m}\]
at \[40^\circ\]
to the displacement. Find the work.Answer
\[W=F\Delta r\cos\theta=30(4)\cos40^\circ=91.9\,\mathrm{J}\]
Checks
- Joules are newton metres.
- Perpendicular force gives zero work.
- Opposing force gives negative work.
- Work is scalar, not vector.