Work by a Variable Force

A spring stretches harder the further you pull. Gravity weakens with altitude. Real forces often vary along the path. The total work must then be calculated as a line integral, summing infinitesimal contributions.

Concept

For a force $\vec{F}(\vec{r})$ that varies with position, the work done as a particle moves along a path $C$ from point $A$ to point $B$ is

$W = \int_C \vec{F}\cdot d\vec{r}$

In one dimension with $F = F(x)$,

$W = \int_{x_A}^{x_B} F(x)\,dx$

Geometrically, this is the area under the $F$–$x$ curve between $x_A$ and $x_B$ (with sign).

Example: Hooke's spring. A spring of stiffness $k$ pulled from its natural length to extension $x$ exerts force $F = -kx$ (restoring). The work done by the spring as it extends from $0$ to $x$ is

$W_{\text{spring}} = \int_0^x (-kx')\,dx' = -\tfrac{1}{2}kx^2$

The work against the spring (your hand's work) is $+\tfrac{1}{2}kx^2$.

Derivation

Divide the path into many tiny segments $d\vec{r}_i$. On each segment, the force $\vec{F}_i$ can be considered constant and the elementary work is

$dW_i = \vec{F}_i \cdot d\vec{r}_i$

Summing over all segments and taking the limit of infinitesimal $d\vec{r}$,

$W = \lim_{n\to\infty} \sum_{i=1}^n \vec{F}_i \cdot d\vec{r}_i = \int_A^B \vec{F}\cdot d\vec{r}$

For a 1D force,

$W = \int_{x_A}^{x_B} F(x)\,dx$

which is precisely the area under the $F$–$x$ graph.

Worked Example

A force $F(x) = 3x^2$ N acts on a particle along the $x$-axis. Find the work done as the particle moves from $x = 0$ to $x = 2 \, \text{m}$.

Solution:

$W = \int_0^2 3x^2\,dx = \left[x^3\right]_0^2 = 8 \, \text{J}$

Spring example: A spring of stiffness $k = 200 \, \text{N/m}$ is stretched by 10 cm. Work done by the agent:

$W = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.10)^2 = 1 \, \text{J}$

Common Confusions

• Not $F\cdot s$ for variable $F$. Plug in an "average" force only if you're careful — for a linear force like a spring, average is $F_{\text{avg}} = \tfrac{1}{2}(F_i + F_f)$, which gives $\tfrac{1}{2}kx^2$.
• Sign matters in the integral. Reversing the direction of motion flips the sign of $dx$, so work flips sign too. Going back to the start gives zero net work for a conservative force.
• Path independence is special. For conservative forces (gravity, spring, electrostatic), work depends only on endpoints, not the path. For non-conservative (friction), it depends on path.
• Area under $F$–$x$: sign matters. Below the axis means negative work.

Key Takeaways

• $W = \int \vec{F}\cdot d\vec{r}$ generalizes work to variable forces.
• In 1D, $W$ is the (signed) area under the $F$–$x$ curve.
• Spring work: $\tfrac{1}{2}kx^2$ (against the spring).
• For conservative forces, work depends only on endpoints.
• Always integrate force times displacement, not just multiply.