AcademyKinematics
Academy
Work Done
Level 1 - Math II (Physics) topic page in Kinematics.
Principle
Work measures the energy transferred by a force through a displacement. Only the component of force along the displacement contributes to work.
For a constant force, work is a dot product. For a force that changes along a path, work is a line integral of force along the path.
Notation
\(W\)
work done by a force
\(\mathrm{J}\)
\(\mathbf F\)
force
\(\mathrm{N}\)
\(\Delta\mathbf r\)
displacement
\(\mathrm{m}\)
\(C\)
path followed by the particle
\(d\mathbf r\)
infinitesimal displacement along the path
\(\mathrm{m}\)
\(K\)
kinetic energy
\(\mathrm{J}\)
Method
Step 1: Use the dot product for constant force
When \(\mathbf F\) is constant over the displacement, compute:
Constant-force work
\[W=\mathbf F\cdot\Delta\mathbf r\]
The dot product selects the force component parallel to the displacement.
Resolve force along displacement
\[F_{\parallel}=|\mathbf F|\cos\theta\]
Multiply by displacement length
\[W=F_{\parallel}|\Delta\mathbf r|\]
Substitute the parallel component
\[W=|\mathbf F||\Delta\mathbf r|\cos\theta\]
Recognize the dot product
\[W=\mathbf F\cdot\Delta\mathbf r\]
Step 2: Integrate for variable force
When force changes along a path, add the small contributions \(\mathbf F\cdot d\mathbf r\):
Variable-force work
\[W=\int_C\mathbf F\cdot d\mathbf r\]
Step 3: Relate net work to kinetic energy
The net work done on a particle changes its kinetic energy:
Work-energy theorem
\[W_{\text{net}}=\Delta K\]
Rules
Positive work condition
\[\mathbf F\cdot\Delta\mathbf r\gt0\]
Negative work condition
\[\mathbf F\cdot\Delta\mathbf r\lt0\]
Joule from force and distance
\[1\,J=1\,N\,m\]
- Positive work adds kinetic energy when considering net work.
- Negative work removes kinetic energy when considering net work.
- A force perpendicular to displacement does zero work over that displacement.
- Work depends on the path when the force is not conservative.
Examples
Question
A constant force
\[\mathbf F=3\mathbf i+4\mathbf j\]
N moves a particle by \[\Delta\mathbf r=2\mathbf i\]
m. Find the work.Answer
Use the dot product:
\[W=(3\mathbf i+4\mathbf j)\cdot(2\mathbf i)=6\]
J. The \[\mathbf j\]
component does not contribute because the displacement has no \[\mathbf j\]
component.Checks
- Use the component of force along displacement, not necessarily the full force magnitude.
- Check the sign of the dot product before interpreting energy transfer.
- Use a line integral when the force varies along the path.
- Work and energy have the same unit, the joule.