KAHIBARO
Table of Contents
When the Force Changes with Position
In many real situations, a force is not constant. A spring pulls harder the more it is stretched. Gravity changes with distance from a planet. Air resistance can change with speed and direction. In such cases, we cannot use the simple constant-force expression $W = F d \cos\theta$ over the whole motion unless the force stays the same throughout.
The basic idea of work is still the same. We add up the tiny amounts of work done over tiny pieces of motion. This leads to an integral.
Small Work Over a Small Displacement
Imagine an object moves through a very small displacement $d\vec r$ while a force $\vec F$ acts on it. Over that tiny step, the small amount of work is
$$dW = \vec F \cdot d\vec r$$
This says that only the component of the force along the displacement contributes to work.
If the object moves from point $A$ to point $B$, the total work done by the variable force is
$$W = \int_A^B \vec F \cdot d\vec r$$
For a variable force, the general formula for work is
$$W = \int_A^B \vec F \cdot d\vec r$$
Do not use $W = F d \cos\theta$ for the whole motion unless the force is constant over that motion.
One-Dimensional Motion
The simplest case is motion along the $x$ axis, where the force depends on position, $F(x)$. Then the displacement is just $dx$, and the work becomes
$$W = \int_{x_i}^{x_f} F(x)\,dx$$
This is the most common formula for work done by a variable force in one dimension.
If $F(x)$ is positive while the object moves in the positive $x$ direction, the work is positive. If $F(x)$ is negative while the object moves in the positive $x$ direction, the work is negative.
Interpreting the Integral Geometrically
In one dimension, work can be understood from a graph of force versus position. The work done from $x_i$ to $x_f$ is the signed area under the curve $F(x)$.
Area above the $x$ axis gives positive work. Area below the $x$ axis gives negative work.
In one-dimensional motion,
$$W = \int_{x_i}^{x_f} F(x)\,dx$$
This equals the signed area under the $F$ versus $x$ graph.
Example, A Force That Increases with Position
Suppose a force depends on position as
$$F(x) = 3x^2$$
and an object moves from $x=1$ m to $x=3$ m. The work is
$$W = \int_1^3 3x^2\,dx$$
Now integrate:
$$W = \left[x^3\right]_1^3 = 27 - 1 = 26\ \text{J}$$
So the variable force does $26$ joules of work.
Example, A Force That Changes Sign
Consider
$$F(x) = 4 - x$$
from $x=0$ to $x=6$ m. Then
$$W = \int_0^6 (4-x)\,dx
= \left[4x - \frac{x^2}{2}\right]_0^6
= 24 - 18 = 6\ \text{J}$$
Even though the force becomes negative for $x>4$, the total work is still positive because the positive area is larger than the negative area.
This example shows that total work is the net signed area, not just the total geometric area.
A Very Important Case, The Spring Force
A classic variable force is the spring force. For an ideal spring,
$$F(x) = -kx$$
where $k$ is the spring constant and $x$ is displacement from equilibrium. The minus sign means the spring force points toward equilibrium.
If an external agent slowly stretches the spring from $x=0$ to $x=x_f$, the external force needed has magnitude $kx$ in the direction of stretching. The work done by the external agent is
$$W_{\text{ext}} = \int_0^{x_f} kx\,dx = \frac{1}{2}kx_f^2$$
The work done by the spring itself during that same stretching is negative:
$$W_{\text{spring}} = \int_0^{x_f} (-kx)\,dx = -\frac{1}{2}kx_f^2$$
For an ideal spring, the work done by an external force in stretching or compressing it from $0$ to $x$ is
$$W_{\text{ext}} = \frac{1}{2}kx^2$$
The work done by the spring over the same change is
$$W_{\text{spring}} = -\frac{1}{2}kx^2$$
Work Along a Path in More Than One Dimension
If the force and motion are in two or three dimensions, work is found by integrating along the actual path taken:
$$W = \int_A^B \vec F \cdot d\vec r$$
This matters because both the force and the direction of motion may change from point to point.
For example, if
$$\vec F = F_x \hat i + F_y \hat j + F_z \hat k$$
and
$$d\vec r = dx\,\hat i + dy\,\hat j + dz\,\hat k$$
then
$$dW = F_x\,dx + F_y\,dy + F_z\,dz$$
so
$$W = \int_A^B \left(F_x\,dx + F_y\,dy + F_z\,dz\right)$$
Constant Force Versus Variable Force
The difference between constant and variable force can be summarized clearly.
| Case | Formula for work |
|---|---|
| Constant force, straight displacement | $W = \vec F \cdot \vec d$ |
| Variable force in one dimension | $W = \int_{x_i}^{x_f} F(x)\,dx$ |
| Variable force in general | $W = \int_A^B \vec F \cdot d\vec r$ |
The integral formula includes the constant-force case as a special case.
Units
Work done by a variable force still has the same unit as any work, the joule:
$$1\ \text{J} = 1\ \text{N}\cdot\text{m}$$
This is consistent with the integral $\int F(x)\,dx$, since force is measured in newtons and position in meters.
Common Mistakes
A common mistake is to multiply the final value of the force by the total displacement. That only works if the force is constant. Another common mistake is to ignore the sign of the force. Negative work is physically important, because it means the force removes energy from the object.
A third mistake is to confuse area under an $F$ versus $x$ graph with area under a graph of some other quantity. Only the area under a force-position graph gives work directly.
Important checks for variable-force work:
$$W = \int F(x)\,dx$$
only for a force plotted against position.
Positive area gives positive work, negative area gives negative work.
Use the actual force function over the interval, not just one force value.
Final Idea
Work done by a variable force is found by breaking the motion into tiny pieces and adding the tiny contributions. This is why integration is needed. In one dimension, the result is the area under the force-position curve. In more general motion, the work is the line integral of force along the path. This idea is essential for understanding springs, changing gravitational forces, and many other physical systems.