Work Done by a Variable Force

The problem with variable forces

The formula $W = Fd\cos\theta$ works only when force is constant. In many real situations, force changes as the body moves — a spring, gravity near a planet, an electric force.

For a variable force, we cannot simply multiply force by displacement. We need calculus.

Building the integral

Divide the total displacement from $x_1$ to $x_2$ into many tiny intervals $dx$. Over each tiny interval, the force $F(x)$ is approximately constant (since $dx$ is infinitesimally small).

Work done in each tiny interval:

$dW = F(x) \, dx$

Total work = sum of all tiny pieces:

$W = \int_{x_1}^{x_2} F(x) \, dx$

Geometric meaning

The integral $\int_{x_1}^{x_2} F(x) \, dx$ is the area under the force-displacement graph between $x_1$ and $x_2$.

• Area above the $x$-axis (force in direction of motion): positive work
• Area below the $x$-axis (force opposing motion): negative work
• Net work = algebraic sum of all areas

This geometric interpretation is extremely useful — many problems can be solved by finding areas of simple shapes (rectangles, triangles, trapezoids) without explicit integration.

Example: Spring force

For a spring with spring constant $k$, force $F = -kx$ (restoring, opposing displacement).

Work done by spring as body moves from $x = 0$ to $x = x_0$:

$W_{spring} = \int_0^{x_0} (-kx) \, dx = -k\frac{x^2}{2}\Big|_0^{x_0} = -\frac{1}{2}kx_0^2$

Negative — the spring does negative work as it is stretched (it resists the stretching).

Example: Gravity near Earth's surface

$F = -mg$ (downward, taking upward as positive).

Work done by gravity as body moves from height $0$ to height $h$:

$W_{gravity} = \int_0^h (-mg) \, dy = -mgh$

Negative — gravity does negative work when the body moves upward.

Work done by gravity as body moves downward by $h$ (from $h$ to $0$):

$W_{gravity} = \int_h^0 (-mg) \, dy = mgh$

Positive — gravity does positive work on a falling body.

Reducing to the constant force case

If $F$ is constant (does not depend on $x$):

$W = \int_{x_1}^{x_2} F \, dx = F(x_2 - x_1) = F \cdot d$

Consistent with the constant-force formula for $\theta = 0$.