The Work-Energy Theorem Along a Trajectory

Work and a Particle’s Motion

A force can change how fast a particle moves, but the force’s effect depends on its direction relative to the motion. A force pointing along the motion tends to increase kinetic energy; a force pointing against the motion tends to decrease it. The work-energy theorem makes this relationship precise: for a particle of constant mass, the net work done over a time interval equals the change in kinetic energy over that interval.

The theorem applies to motion along a straight or curved trajectory. The calculation uses the velocity at each instant, so it keeps track of how the direction of motion changes as well as how the speed changes.

Definition: Let a particle of constant mass $m>0$ have a twice continuously differentiable position $\mathbf{r}(t)$ for $a\le t\le b$. Its velocity is $\mathbf{v}(t)=\mathbf{r}'(t)$, its acceleration is $\mathbf{a}(t)=\mathbf{v}'(t)$, and its kinetic energy is $K(t)=\frac12 m\,\mathbf{v}(t)\cdot\mathbf{v}(t)=\frac12m\lVert\mathbf{v}(t)\rVert^2$. If $\mathbf{F}(t)$ is the net force on the particle, the work done from $a$ to $b$ is $W=\int_a^b\mathbf{F}(t)\cdot\mathbf{r}'(t)\,dt$.

Here “net force” means the resultant of all forces acting on the particle. If several forces act, their individual works add to give the work of the net force, because the dot product and integral are linear. Newton’s second law for constant mass gives $\mathbf{F}(t)=m\mathbf{a}(t)$. Thus the theorem connects the vector work integral to the particle’s kinetic energy at the endpoints.

The Work-Energy Theorem

Theorem: Suppose a particle has constant mass $m$, a twice continuously differentiable trajectory $\mathbf{r}(t)$ on $[a,b]$, and net force $\mathbf{F}(t)=m\mathbf{r}''(t)$. Then the work done by the net force along the trajectory is the change in kinetic energy: $$ \int_a^b \mathbf{F}(t)\cdot\mathbf{r}'(t)\,dt = K(b)-K(a) = \frac12m\lVert\mathbf{v}(b)\rVert^2-\frac12m\lVert\mathbf{v}(a)\rVert^2. $$

The assumptions matter. The force in this statement must be the net force, and it must satisfy Newton’s second law for the particle. The result is stated for constant mass and a sufficiently smooth trajectory so that the derivatives and integral are defined. It does not assert that the work is determined by the endpoints of the path alone: the force and the motion along the trajectory determine the work integral.

Why the Theorem Holds

The key is to differentiate kinetic energy with respect to time. Since the velocity is a vector, the chain rule for its squared magnitude involves a dot product. In coordinates, if $\mathbf{v}=(v_1,\ldots,v_n)$, then $\lVert\mathbf{v}\rVert^2=\sum_{i=1}^n v_i^2$, so its derivative is $2\sum_{i=1}^n v_i v_i'$. Consequently,

\frac{dK}{dt} = \frac{d}{dt}\left(\frac12m\,\mathbf{v}\cdot\mathbf{v}\right) = m\,\mathbf{v}\cdot\mathbf{v}' = m\,\mathbf{v}\cdot\mathbf{a} = \mathbf{F}\cdot\mathbf{v}.

In the last equality, Newton’s second law $\mathbf{F}=m\mathbf{a}$ was used. Since $\mathbf{v}=\mathbf{r}'(t)$, the right side is the instantaneous rate at which work accumulates along the trajectory.

Differentiate kinetic energy.
The chain rule gives $\frac{dK}{dt}=m\mathbf{a}\cdot\mathbf{v}$.

Use the net-force law.
Substitute $\mathbf{F}=m\mathbf{a}$ to obtain $\frac{dK}{dt}=\mathbf{F}\cdot\mathbf{v}$.

Integrate over time.
Integrating from $a$ to $b$, the fundamental theorem of calculus gives $K(b)-K(a)=\int_a^b\mathbf{F}(t)\cdot\mathbf{v}(t)\,dt$.

Recognize the work integral.
Because $\mathbf{v}(t)=\mathbf{r}'(t)$, the integral is the work along the trajectory.

This proves the theorem under its stated assumptions. It also explains why the direction of force matters: only the component of force in the direction of velocity contributes to the instantaneous change in kinetic energy. A perpendicular force has zero dot product with velocity at that instant, so it does no instantaneous work then.

Worked Calculations Along Trajectories

Worked Example: A Curved Trajectory with Varying Force

A particle of mass $m=2\text{ kg}$ follows $\mathbf{r}(t)=(t^2,t^3)\text{ m}$ for $0\le t\le1$. Find the work done by the net force, and compare it with the change in kinetic energy.

Differentiate position to obtain velocity and acceleration:

\mathbf{v}(t)=(2t,3t^2)\text{ m/s}, \qquad \mathbf{a}(t)=(2,6t)\text{ m/s}^2.

Newton’s second law gives the net force:

\mathbf{F}(t)=m\mathbf{a}(t) =2(2,6t) =(4,12t)\text{ N}.

Take the dot product with velocity and integrate over the given interval:

\begin{aligned} W &=\int_0^1(4,12t)\cdot(2t,3t^2)\,dt\\ &=\int_0^1(8t+36t^3)\,dt\\ &=\left[4t^2+9t^4\right]_0^1\\ &=(4+9)-(0+0)=13\text{ J}. \end{aligned}

Now evaluate kinetic energy at the endpoints. At $t=0$, $\mathbf{v}(0)=(0,0)$, so $K(0)=\frac12(2)(0)=0\text{ J}$. At $t=1$, $\mathbf{v}(1)=(2,3)\text{ m/s}$, so

K(1)=\frac12(2)(2^2+3^2)=1(4+9)=13\text{ J}.

Therefore $K(1)-K(0)=13-0=13\text{ J}=W$, as required. The changing direction and magnitude of the force have both been included in the dot product.

Worked Example: Constant Force Along a Curved Path

A particle of mass $1\text{ kg}$ follows $\mathbf{r}(t)=(t,t^2)\text{ m}$ for $0\le t\le2$. Find the net work and check it using kinetic energy.

The velocity and acceleration are

\mathbf{v}(t)=(1,2t)\text{ m/s}, \qquad \mathbf{a}(t)=(0,2)\text{ m/s}^2.

Thus the net force is constant:

\mathbf{F}(t)=1(0,2)=(0,2)\text{ N}.

The work integral is

\begin{aligned} W &=\int_0^2(0,2)\cdot(1,2t)\,dt\\ &=\int_0^2 4t\,dt\\ &=\left[2t^2\right]_0^2\\ &=8-0=8\text{ J}. \end{aligned}

At the initial time, $\mathbf{v}(0)=(1,0)\text{ m/s}$, giving $K(0)=\frac12(1)(1^2+0^2)=\frac12\text{ J}$. At the final time, $\mathbf{v}(2)=(1,4)\text{ m/s}$, giving $K(2)=\frac12(1)(1^2+4^2)=\frac{17}{2}\text{ J}$. Hence

K(2)-K(0)=\frac{17}{2}-\frac12=\frac{16}{2}=8\text{ J}=W.

The force points in the positive second-coordinate direction. The particle’s second-coordinate velocity grows from $0$ to $4\text{ m/s}$, so the force does positive work even though the path is curved.

Worked Example: Negative Work and a Decrease in Kinetic Energy

A particle of mass $2\text{ kg}$ moves along $\mathbf{r}(t)=(2t-t^2,0)\text{ m}$ for $0\le t\le1$. Find the net work and verify the result from the endpoint speeds.

Differentiate to find

\mathbf{v}(t)=(2-2t,0)\text{ m/s}, \qquad \mathbf{a}(t)=(-2,0)\text{ m/s}^2.

The net force is $\mathbf{F}(t)=2(-2,0)=(-4,0)\text{ N}$. Therefore,

\begin{aligned} W &=\int_0^1(-4,0)\cdot(2-2t,0)\,dt\\ &=\int_0^1(-8+8t)\,dt\\ &=\left[-8t+4t^2\right]_0^1\\ &=(-8+4)-0=-4\text{ J}. \end{aligned}

The initial velocity is $\mathbf{v}(0)=(2,0)\text{ m/s}$, so $K(0)=\frac12(2)(2^2)=4\text{ J}$. The final velocity is $\mathbf{v}(1)=(0,0)\text{ m/s}$, so $K(1)=0\text{ J}$. Thus $K(1)-K(0)=0-4=-4\text{ J}=W$. The force opposes the motion during the interval, and the particle’s kinetic energy falls to zero.

Reading the Sign of Work

The theorem lets us interpret the sign of net work without treating work as a measure of distance traveled. If $W>0$, then $K(b)>K(a)$, so the final speed is greater than the initial speed. If $W<0$, then $K(b)<K(a)$, so the final speed is smaller. If $W=0$, the endpoint kinetic energies—and therefore the endpoint speeds for a fixed positive mass—are equal. Zero net work does not mean that no forces acted or that the particle did not move; different contributions can cancel.

Kinetic energy depends on speed, not on the direction of velocity. A particle can change direction while keeping the same speed, and its kinetic energy can remain unchanged. Conversely, the theorem does not say that the force must point along the trajectory throughout the motion. It says that the accumulated dot product of force and velocity determines the net change in kinetic energy.

Takeaway: For a particle of constant mass acted on by a net force satisfying Newton’s second law, calculate work using $\mathbf{F}\cdot\mathbf{v}$, then compare it with the final kinetic energy minus the initial kinetic energy. The two quantities must agree.

A Common Sign Error

A frequent mistake is to replace the force vector by its magnitude in the work integral. In the third example, the force has magnitude $4\text{ N}$, and the particle travels a distance of $1\text{ m}$ in the positive first-coordinate direction. Using magnitudes alone would give $4\text{ J}$, but this loses the fact that the force points in the opposite direction.

The incorrect calculation would be

\int_0^1 \lVert\mathbf{F}(t)\rVert\,\lVert\mathbf{v}(t)\rVert\,dt = \int_0^1 4(2-2t)\,dt = \left[8t-4t^2\right]_0^1 = 4\text{ J}.

This is not the work integral, because it omits the angle between force and velocity. The correct calculation retains the dot product:

\int_0^1\mathbf{F}(t)\cdot\mathbf{v}(t)\,dt = \int_0^1(-4,0)\cdot(2-2t,0)\,dt = -4\text{ J}.

The negative sign is also confirmed by the endpoint kinetic energies: the particle slows from $2\text{ m/s}$ to $0\text{ m/s}$, so its kinetic energy changes by $0-4=-4\text{ J}$. Magnitudes can replace a dot product only when the relevant directions are known to align appropriately; in general, use the vector dot product.

Why the Relation Is Useful

The work-energy theorem provides two equivalent ways to analyze motion. If the force and trajectory are known, the work integral can be evaluated and used to find the change in kinetic energy. If the endpoint speeds are known, their kinetic-energy difference gives the net work without requiring the force to be integrated point by point. In either direction, the comparison is a useful check on a calculation.

The theorem is especially helpful when a path bends or the force varies during motion: the chain-rule argument still applies at each instant, provided the stated assumptions hold. It does not replace the definition of work. Rather, it explains why the work of the net force along a particle’s actual motion has exactly the value needed to account for the change in kinetic energy.

Check Your Understanding

Use the theorem’s assumptions and keep track of the direction of each vector.

For a constant-mass particle, what equation connects net force and acceleration in the work-energy theorem?
Why does differentiating $\frac12m\,\mathbf{v}\cdot\mathbf{v}$ produce the dot product $m\mathbf{a}\cdot\mathbf{v}$?
If the net work over an interval is zero, what can you conclude about the initial and final speeds? What can you not conclude about the forces during the interval?
A force is perpendicular to the particle’s velocity at every instant. What is the instantaneous rate of change of kinetic energy?
Why can $\int \lVert\mathbf{F}\rVert\lVert\mathbf{v}\rVert\,dt$ give the wrong value for work?