Work-Energy Theorem from Newton's Second Law

1. Work and Variable Forces

You are likely familiar with the definition of work for a constant force: $W=Fd\cos \theta$. But what happens when the force is not constant, such as the force exerted by a stretching spring or the gravitational pull on a rocket leaving Earth?

In such cases, we can't simply multiply force by distance. Instead, we must sum up the work done over many infinitesimally small displacements, $d\vec{r}$, over which the force $\vec{F}$ is approximately constant. This summation is, by definition, an integral.

For motion along a straight line (say, the x-axis), the work done by a variable force $F(x)$ from position $x_i$ to $x_f$ is:

Graphically, this integral represents the area under the force-vs-position curve.

To familiarize yourself with this integral formulation, please read the short introductory section from the following resource.

2. Deriving the Work-Energy Theorem in One Dimension

Now, let's connect this concept of work directly to Newton's second law. Our goal is to transform $F=ma$ into a statement about work and energy. The key is a clever application of the chain rule from calculus.

Consider an object of mass $m$ moving in one dimension, with a net force $F(x)$ acting on it.

1. Start with Newton's second law: $F(x)=ma=m\frac{dv}{dt}$.
2. The force is a function of position $x$, but acceleration is a derivative with respect to time $t$. To relate them, we use the chain rule to rewrite acceleration: $$a=\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=\frac{dv}{dx}\cdot v$$
3. Substitute this back into Newton's law: $$F(x)=m\left(v\frac{dv}{dx}\right)$$
4. Now, we can separate variables. Move the $dx$ term to the left side: $$F(x)dx=mvdv$$
5. Finally, we integrate both sides from an initial state (position $x_i$, velocity $v_i$) to a final state ($x_f$, velocity $v_f$): $${\int }_{x_i}^{x_f}F(x)dx={\int }_{v_i}^{v_f}mvdv$$
6. Evaluating the integrals gives: $$W_{net}={\left[\frac{1}{2}m{v}^2\right]}_{v_i}^{v_f}=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$$

The left side is the net work done on the object. We define the quantity $\frac{1}{2}m{v}^2$ as the kinetic energy (K) of the object. Thus, we have derived the Work-Energy Theorem:

The following video provides a clear, step-by-step walkthrough of this exact derivation.

The remainder of the video (after 5:47) discusses how the Work-Energy theorem is universally true, unlike some other energy principles that have specific conditions. This universality is what makes it so powerful.

3. Generalization to Three Dimensions: The Line Integral

The 1D derivation is clear, but how does it work for a particle moving along a curved path in 3D space? We need a more general definition of work. The infinitesimal work $dW$ done by a force $\vec{F}$ over an infinitesimal displacement $d\vec{r}$ is given by the dot product:

The total work is the integral of this quantity along the particle's path, from an initial position ${\vec{r}}_i$ to a final position ${\vec{r}}_f$. This is known as a line integral:

Now let's derive the work-energy theorem in this general vector form. This derivation uses a different, equally elegant mathematical technique.

1. Start with Newton's second law in vector form: $\vec{F}=m\frac{d\vec{v}}{dt}$.
2. Take the dot product of both sides with the velocity vector $\vec{v}$: $$\vec{F}\cdot \vec{v}=m\left(\frac{d\vec{v}}{dt}\cdot \vec{v}\right)$$
3. Recall the product rule for the derivative of a dot product: $\frac{d}{dt}(\vec{v}\cdot \vec{v})=2\vec{v}\cdot \frac{d\vec{v}}{dt}$. Since $\vec{v}\cdot \vec{v}=|\vec{v}{|}^2=v^2$, we can write: $$\vec{v}\cdot \frac{d\vec{v}}{dt}=\frac{1}{2}\frac{d}{dt}(v^2)$$
4. Substitute this into our equation: $$\vec{F}\cdot \vec{v}=m\left(\frac{1}{2}\frac{d}{dt}(v^2)\right)=\frac{d}{dt}\left(\frac{1}{2}m{v}^2\right)=\frac{dK}{dt}$$ This is a remarkable result in itself. The quantity $\vec{F}\cdot \vec{v}$ is the instantaneous power delivered by the force, and it equals the rate of change of kinetic energy.
5. To get the work-energy theorem, integrate both sides with respect to time from $t_i$ to $t_f$: $${\int }_{t_i}^{t_f}(\vec{F}\cdot \vec{v})dt={\int }_{t_i}^{t_f}\frac{dK}{dt}dt$$
6. Using the definition of velocity, $\vec{v}=\frac{d\vec{r}}{dt}$, we have $\vec{v}dt=d\vec{r}$. The left-hand integral becomes the line integral for work: ${\int }_{{\vec{r}}_i}^{{\vec{r}}_f}\vec{F}\cdot d\vec{r}$.
7. The right-hand integral, by the Fundamental Theorem of Calculus, is simply $K_f-K_i$.

This gives us the general work-energy theorem, valid in three dimensions for any path:

The following video presents this exact vector-based derivation concisely and clearly.

For a more detailed textbook presentation of both the 1D and 3D derivations, you can refer to the following resource. It also provides worked examples that apply the theorem.

Conclusion

In this lesson, you've derived one of the most fundamental principles in mechanics. By integrating Newton's second law with respect to position instead of time, we have developed a new and powerful tool for analyzing motion.

Key Takeaways:

• Work done by a variable force is calculated by integrating the force over the path of displacement: $W=\int \vec{F}\cdot d\vec{r}$.
• Kinetic energy is the energy of motion, defined as $K=\frac{1}{2}m{v}^2$.
• The Work-Energy Theorem states that the net work done on an object by all forces is equal to the change in its kinetic energy: $W_{net}=\Delta K$.
• This theorem is a direct mathematical consequence of Newton's second law and holds true universally.

Preview of the Next Lesson:
We have now seen that integrating $F=ma$ with respect to position gives us the work-energy theorem. What happens if we integrate $F=ma$ with respect to time? This leads to another powerful conservation principle related to impulse and momentum. In the next lesson, we will derive this relationship and see how it is particularly useful for analyzing collisions and systems where mass changes, such as rockets.