Work Along a Path showed a force whose work changed when the path changed. Gravity is different: lift a book straight up or carry it up the stairs, and gravity does the same work either way. That difference sorts all forces into two kinds.
Key idea
A force is conservative when the work it does between two points is the same along every path; equivalently, its work around every closed loop is zero. Gravity, ideal springs and the electric force between charges are conservative. Friction and drag are not: their work depends on the length of the path, and around a loop it is negative.
Why it works
The two definitions are the same statement. Go from A to B along path 1 and come back along path 2: the loop's work is $W_1 - W_2$, since reversing a path flips the sign of each $\vec{F}\cdot d\vec{r}$. That is zero for every loop exactly when $W_1 = W_2$ for every pair of paths. For gravity near Earth, $\vec{F} = \langle 0, -mg \rangle$, so $W = \int -mg\,dy = -mg\,\Delta y$, which depends only on the heights of the ends. Friction always points against the motion, so every step of every path adds negative work, and longer paths add more.
Worked example: Gravity on two paths
A $2\ \text{kg}$ ball moves from a height of $5\ \text{m}$ to a height of $1\ \text{m}$, once by falling straight down and once along a winding slide. Find gravity's work each way.
- $W = -mg\,\Delta y = -2(9.8)(1 - 5)$.
- $W = +78.4\ \text{J}$ on both paths: only the change in height matters.
Answer: $78.4\ \text{J}$ either way.
Worked example: Friction on two paths
A $5\ \text{kg}$ box is slid across a floor ($\mu_k = 0.2$) between opposite corners of a $3\ \text{m}$ by $4\ \text{m}$ rectangle, once along two sides and once along the diagonal. Find friction's work each way.
- Friction's size is $\mu_kmg = 0.2(5)(9.8) = 9.8\ \text{N}$, always against the motion, so $W = -9.8\ \text{N} \times (\text{path length})$.
- Along two sides: $-9.8(7) = -68.6\ \text{J}$. Along the diagonal: $-9.8(5) = -49\ \text{J}$.
- Different paths, different work: friction is not conservative.
Answer: $-68.6\ \text{J}$ along the sides; $-49\ \text{J}$ along the diagonal.
Common mistake
Treating friction like gravity, as if it did the same work on any route between two points. Friction's work grows with the distance traveled, not with the displacement; drag a box around a room and back to its spot, and friction has done plenty of negative work while gravity has done none. Check: ask whether the force does negative work on every piece of a round trip. If so, it can't be conservative.
A test without drawing paths: in two dimensions, a force $\langle F_x, F_y \rangle$ defined on a region without holes is conservative exactly when $\dfrac{\partial F_x}{\partial y} = \dfrac{\partial F_y}{\partial x}$. For $\langle -y, x \rangle$ these are $-1$ and $1$, so it fails. For a conservative force you can then find a potential energy, as in Potential Energy from Force.