Let’s start our investigation of energy with the simplest possible situation: One particle of mass m is acted on by one constant force $\vec{F}$ that acts parallel to the direction of motion, pushing or pulling on the particle as it undergoes a displacement $\text{Δ}s$. We define the particle to be the system—a one-particle system—while the agent of the force is in the environment. FIGURE 9.2 shows both an interaction diagram and a new kind of pictorial representation, a before-and-after representation, in which we show an object before and after an interaction and, as usual, establish a coordinate system and define appropriate symbols.

FIGURE 9.2 The interaction diagram and before-and-after representation for a one-particle system.

You know what’s going to happen. If the force is in the direction of motion—the situation shown in the figure—the particle will speed up and its “energy of motion” will increase. Conversely, if the force opposes the motion, the particle will slow down and lose energy. Our goal is to make this idea precise by discovering exactly how the changing energy is related to the applied force.

We’ll start by writing Newton’s second law for the particle:

Newton’s second law tells us how the particle’s velocity changes with time. But suppose we want to know how the velocity changes with position. To answer that question, we can use the chain rule that you’ve learned in calculus:

where in the last step we used $ds/dt=v_s$. With this, we can write Newton’s second law as

Now we have an alternative version of Newton’s second law in terms of the rate of change of velocity with position.

To use Equation 9.4, we first rewrite it as

Now we can integrate. This is going to be a definite integral over just the motion shown in the before-and-after representation. That is, the right side will be an integral over position s from the initial position $s_{\text{i}}$ to the final position $s_{\text{f}}$. The left side will be an integral over velocity $v_s$, and its limits have to match the limits of the right-hand integral: from $v_{\text{i}s}$ at $s_{\text{i}}$ to $v_{\text{f}s}$ at $s_{\text{f}}$. Thus we have

We have two integrals to examine, and we’ll do them one by one. We can start with the integral on the left, which is of the form$\int xdx$. Factoring out m, which is a constant, we find

You’ll notice that we dropped the subscript s in the next-to-last step. $v_s$ is a vector component, with a sign to indicate direction, but the sign makes no difference after $v_s$ is squared. All that matters is the particle’s speed v.

The last step in Equation 9.7 introduces a new quantity

which is called the kinetic energy of the particle. Kinetic energy is energy of motion. It depends on the particle’s mass and speed but not on its position. Furthermore, kinetic energy is a property or characteristic of the system. So what we’ve calculated with the left-hand integral is $\text{Δ}K=K_{\text{f}}-K_{\text{i}}$, the change in the system’s kinetic energy as the force pushes the particle through the displacement $\text{Δ}s\text{.}\text{Δ}K$ is positive if the particle speeds up (gain of kinetic energy), negative if it slows down (loss of kinetic energy).

The unit of kinetic energy is mass multiplied by speed squared. In SI units, this is $\text{kg}{\text{m}}^2{\text{/s}}^2$. Because energy is so important, the unit of energy is given its own name, the joule. We define

All other forms of energy are also measured in joules.

To give you an idea about the size of a joule, consider a 0.5 kg mass ($\approx 1\text{ lb}$ on earth) moving at $4\text{m}\text{/}\text{s }(\approx 10\text{ mph})$. Its kinetic energy is

This suggests that everyday objects moving at ordinary speeds will have energies from a fraction of a joule up to, perhaps, a few thousand joules. A running person has $K\approx 1000\text{J}$, while a high-speed truck might have $K\approx {10}^6\text{J}$.