Introduction
A force can act on an object without producing the same effect in every direction. To calculate the work done by a constant force, we need both the force and the object's displacement. Their dot product combines the force's size with the part of its direction that aligns with the displacement.
This makes work a useful application of the dot product: the result is a single number, not a vector. The calculation also explains why using force magnitude alone is not enough. The direction of the force relative to the displacement matters.
Definition of Work
Suppose a constant force \(\mathbf{F}\) acts on an object while the object moves from an initial position to a final position. The displacement \(\mathbf{d}\) is the final position minus the initial position. The work \(W\) done by the force is the dot product of \(\mathbf{F}\) and \(\mathbf{d}\).
In three-dimensional Cartesian coordinates, write the force and displacement as \(\mathbf{F}=(F_x,F_y,F_z)\) and \(\mathbf{d}=(d_x,d_y,d_z)\). The component formula for their dot product gives
In two dimensions, omit the \(z\)-components. When force is measured in newtons and displacement in metres, work is measured in newton-metres, which are called joules: \(1\text{ J}=1\text{ N}\cdot\text{m}\). Each force component must be multiplied by the displacement component along the same coordinate axis.
Using the Angle Between Force and Displacement
The dot product can also be calculated from vector magnitudes and the angle between the vectors. If \(\theta\) is the angle between \(\mathbf{F}\) and \(\mathbf{d}\), then
This form is useful when the magnitudes and angle are known. The angle must be the angle between the force direction and the displacement direction, not an angle measured from an unrelated axis. If the vectors are given in components instead, the component formula is usually more direct.
Worked Example: Force at a Known Angle
A constant force of magnitude \(12\text{ N}\) acts at an angle of \(60^\circ\) to a displacement of \(5\text{ m}\). Find the work done.
Use the angle form of the dot product. Substitute the given magnitudes and angle:
Since \(\cos 60^\circ=\frac12\),
To check, the component of the force along the displacement is \(12\cos60^\circ=6\text{ N}\). Multiplying this component by the \(5\text{ m}\) displacement gives \(6\cdot5=30\text{ N}\cdot\text{m}=30\text{ J}\), as before.
Why the Dot Product Gives Work
The angle formula follows from the force component in the direction of motion. For a nonzero displacement, define its unit direction vector by \(\mathbf{u}=\mathbf{d}/\|\mathbf{d}\|\). The signed scalar component of the force along that direction is \(\mathbf{F}\cdot\mathbf{u}\). The force part perpendicular to the displacement contributes no work for that displacement; only the parallel component is used.
The vector component of \(\mathbf{F}\) parallel to \(\mathbf{d}\) is \(\mathbf{F}_{\parallel}=(\mathbf{F}\cdot\mathbf{u})\mathbf{u}\), where \(\mathbf{u}\) is the unit vector in the displacement direction.
The work is the signed force component along the displacement multiplied by the displacement length: \(W=(\mathbf{F}\cdot\mathbf{u})\|\mathbf{d}\|\).
Since \(\mathbf{d}=\|\mathbf{d}\|\mathbf{u}\), the expression in Step 2 equals \(\mathbf{F}\cdot\mathbf{d}\).
In full, this is the short derivation:
The argument assumes \(\mathbf{d}\ne\mathbf{0}\) when defining \(\mathbf{u}\). If the displacement is zero, the dot-product definition gives \(W=\mathbf{F}\cdot\mathbf{0}=0\) directly, without needing a unit direction vector. The dot-product formula applies to a constant force; it is not, by itself, a rule for a force that changes during the motion.
Calculating Work from Components
When force and displacement are specified by components, multiply matching components and add the products. Keep the signs of the components throughout the calculation. A negative component contributes a negative product when paired with a positive component, and the reverse is also true.
Worked Example: Three-Dimensional Force and Displacement
A constant force is \(\mathbf{F}=(4,-2,5)\text{ N}\), and the displacement is \(\mathbf{d}=(2,3,-1)\text{ m}\). Find the work.
Apply the component formula, pairing the \(x\)-, \(y\)-, and \(z\)-components respectively:
Check the result using the projection in the displacement direction. The squared displacement length is \(2^2+3^2+(-1)^2=14\), so \(\|\mathbf{d}\|=\sqrt{14}\text{ m}\). The signed force component along the displacement is \((\mathbf{F}\cdot\mathbf{d})/\|\mathbf{d}\|=-3/\sqrt{14}\text{ N}\). Multiplying by the displacement length gives \((-3/\sqrt{14})(\sqrt{14})=-3\text{ J}\), confirming the component calculation.
Finding Displacement from Positions
Sometimes the problem gives the object's initial and final coordinates instead of giving the displacement directly. Subtract the initial position from the final position, coordinate by coordinate. Reversing the subtraction changes the displacement vector and can change the computed work.
Worked Example: Work from Endpoint Coordinates
An object moves from \(A=(-1,2)\text{ m}\) to \(B=(3,5)\text{ m}\) under the constant force \(\mathbf{F}=(6,8)\text{ N}\). Find the work done.
First calculate the displacement by subtracting the initial coordinates from the final coordinates:
Now take the dot product:
Check using the projection formula. The displacement length is \(\sqrt{4^2+3^2}=5\text{ m}\), and the force component along the displacement is \((\mathbf{F}\cdot\mathbf{d})/\|\mathbf{d}\|=48/5\text{ N}\). Their product is \((48/5)(5)=48\text{ J}\), so the result is consistent.
A Common Mistake: Multiplying Magnitudes Only
A frequent error is to calculate work as the force magnitude times the displacement magnitude and omit the directional factor. That product is correct only when the angle makes \(\cos\theta=1\). In general, the angle or the components must be included.
Worked Example: Correcting a Magnitude-Only Calculation
Let \(\mathbf{F}=(3,4,0)\text{ N}\) and \(\mathbf{d}=(4,-3,0)\text{ m}\). A mistaken calculation might find \(\|\mathbf{F}\|=5\text{ N}\) and \(\|\mathbf{d}\|=5\text{ m}\), then report \(W=5\cdot5=25\text{ J}\).
This omits the angle between the vectors. The correct calculation uses the dot product:
As a check, \(\mathbf{F}\cdot\mathbf{d}=0\), so the vectors are perpendicular by the dot-product criterion. The angle is \(90^\circ\), and the magnitude formula gives \(5\cdot5\cos90^\circ=0\text{ J}\). Thus the magnitude-only answer of \(25\text{ J}\) is incorrect because it ignores direction.
Practical Checks
Before accepting a work calculation, check the setup as well as the arithmetic. The force and displacement should be expressed using the same coordinate axes, and each component product should pair matching directions. If displacement comes from positions, confirm that it is final minus initial. Finally, check that the units reduce to joules.
- Use the constant-force formula only when the force is constant over the displacement being considered.
- Use either the angle formula or the component formula, choosing the one that matches the given information.
- Keep negative component values when multiplying; do not replace components with their magnitudes.
- Include units: force in newtons and displacement in metres produce work in joules.
Why Work Is a Useful Quantity
The dot product turns directional information into a scalar that measures the work associated with a force and a displacement. It makes calculations possible in any coordinate orientation: components can be used directly, or magnitudes and the included angle can be used when those are known. The projection derivation also gives a reliable check on whether a calculation uses the force component along the motion rather than the full force magnitude.
Check Your Understanding
Answer these questions using the definitions and methods in this tutorial.
- A force of \(10\text{ N}\) acts at \(60^\circ\) to a displacement of \(4\text{ m}\). What is the work?
- For \(\mathbf{F}=(2,-1,3)\text{ N}\) and \(\mathbf{d}=(1,4,2)\text{ m}\), calculate the work from components.
- An object moves from \((2,-1)\text{ m}\) to \((5,3)\text{ m}\). What displacement vector should be used in the work calculation?
- Why does the angle form of the work formula include a cosine factor?
- What assumption about the force is needed for \(W=\mathbf{F}\cdot\mathbf{d}\) to apply in this tutorial?