Section II, Chapter Nine: Kinetic Energy, Potential Energy, and Work

Can our notion of work tell us anything interesting about classical physics that Newton’s laws of motion cannot?

Note—this is a draft of Chapter Twelve of a book I’m writing. The book is going to cover humanity’s deepest ideas from philosophy, physics, epistemology, and economics. Each chapter is meant to be short and digestible. Most of the chapters will explain just one or two ideas.

‘Chapter []’ indicates a future chapter that I’ve not yet written.

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Consider again a rock falling vertically towards the Earth’s surface. Our commonsensical understanding of work might tell us that Earth’s gravity is doing work on the rock, and this indeed coincides with the notion of work as understood by physicists. But what about a pair of clashing winds that smack against the rock from opposite sides? Do they perform work as well? If the winds are equal in magnitude, then, despite each one exerting a force on the rock, they do not move the stone at all and so perform no work on it.

In Newtonian physics, the work W that a force does on a system along its trajectory from initial point A to final point B is defined as the path integral ∫_A^B F *dr*cos(θ):

At every point along the system’s path, the force acts on it with a magnitude and direction. The system’s infinitesimal displacement, dr, also has a direction—it tells us the direction in which the system is moving. θ (read ‘theta’) is the angle between the force and the infinitesimal displacement, and cos(θ) is a function of θ that runs from -1 to 1.

When F and dr point in the same direction, cos(θ) is 1, and when they are at right angles to each other, cos(θ) is 0. This is why gravity performs zero work on, say, a sled being pulled across a flat plane of ice—the sled’s direction of displacement is perfectly horizontal, while gravity is perfectly vertical . This is also why gravity does perform work on or falling rock—the force of gravity and the rock’s direction of displacement point in the same direction (downward).

In many circumstances, we don’t need the general path integral formula for work. For example, if the force acting on the system and the angle between the force and the system’s displacement are both constant, we may simply write W = F*x*cos(θ), where x is the total displacement from initial point A to final point B and θ is the angle between the force and the system’s displacement.

Since force is measured in Newtons and displacement in meters, and since cos(θ) is unitless, work is measured in Newton-meters, which we call Joules by SI convention (again, the International System of Units).

The Work-Energy Theorem

We’ve defined work in such a way that some forces acting on a system perform a nonzero amount of work on a system (such as gravity on our falling rock), while others perform zero work (such as clashing horizontal winds on our falling rock). But this doesn’t seem very insightful. Can our notion of work tell us anything interesting about classical physics that Newton’s laws of motion cannot? It can: work helps us understand the energetics of a system.

There is a deep connection between work and a quantity called kinetic energy, or (½)m*v²: the work-energy theorem states that the work done by a force F on a system between initial point A and final point B is equal to the system’s change in kinetic energy from A to B. Mathematically, the work-energy theorem says that W = (½)m*v_B² - (½)m*v_A², where V_A is the system’s velocity at point A and V_B is the system’s velocity at point B. Unlike velocity, which is a vector and thus has both a magnitude and a direction, kinetic energy is a scalar and therefore only has a magnitude.

Consider again dropping a 1 kg stone from 100 meters above the Earth’s surface. How much work does gravity do to it from the moment we drop it to the moment it reaches the ground?

Its initial velocity, v_A, is zero, and so (½)m*v_A² = (½)m*(0)² = 0. In other words, the rock’s initial kinetic energy is zero.

What about its final kinetic energy, (½)m*v_B²? We know its mass m already, so we just need to determine its velocity the instant it reaches the ground. Rearranging the kinematic equation (see Chapter Ten) for displacement as a function of time, x(t) = (½)g*t² + 100, we have t = sqrt((x(t) - 100) / ((½)g)). We know that x = 100 m at the time at which the stone reaches the ground, and g is a constant equal to 9.8 m/s². Plugging in, we find that t = 4.52 s.

Then we may use the kinematic equation for the stone’s velocity as a function of time to find its final velocity: v(t = 4.52) = g*4.52 = 9.8*4.52 = 44.3 m/s.

The total work done on the stone by gravity is therefore W = (½)m*v_B² - (½)m*v_A² = (½)*1*(44.3)² - 0 = 981.25 Joules.

Because the work done on a system depends only on its initial kinetic energy and final kinetic energy, we do not need to know the system’s intermediary values for kinetic energy to calculate the total work done on it. Whether the object accelerates at a constant rate during its trajectory, or it ‘winds’ this way and that, whether it takes a long time or a short time to travel from A to B, the total work performed on it depends only on its first and final kinetic energies.

This useful feature of work is called path independence (see Chapter []), and it only holds when the forces acting on the system are conservative. Path independence does not hold when non-conservative forces act on a system. In that case, the path that the system takes does affect the total work done on it. For example, if we took air resistance into account in our example, then the total work done on the stone would depend on the stone’s particular path.

Potential Energy and the Conservation of Energy

In our example of a falling rock, clearly kinetic energy is not conserved: it begins at zero and increases monotonically until the rock reaches the ground. However, given the particularities of our system, we know that its maximum kinetic energy is 981.25 Joules regardless of where it is along its trajectory. So, even though its kinetic energy changes in time and therefore is not conserved, its maximum kinetic energy is a constant.

We can make explicit the rock’s capacity to increase its kinetic energy by introducing the concept of potential energy into the picture such that the sum of the system’s potential energy and kinetic energy is a constant. In our example, potential energy is a maximum of 981.25 Joules and kinetic energy is at a minimum of 0 Joules just as the rock is dropped from the rooftop, and potential energy is at a minimum of 0 Joules and kinetic energy is at a maximum of 981.25 Joules just as the rock reaches the ground.

In our example, potential energy must be a function of displacement, since its value must change in concert with kinetic energy, which itself changes with displacement. In general, the potential energy function, V, of a given (conservative) force is given by V = ∫F(x)*dx. In our example, then, V = m*g*x, where x is the distance from the ground to the rock’s vertical position.

If the potential energy of an object in the Earth’s gravitational field increases with its distance from the Earth’s surface, then we may get more work out of objects that are higher up than those that are nearer to the ground. Similarly, we may get more work out of a coiled spring than a flaccid one (the former has more potential energy than the latter). This idea generalizes to all sorts of systems beyond those studied in Newtonian mechanics—chemical systems, electrostatic systems, magnetic systems, and so on. We’ll discuss work and energy in greater detail in Section III—thermodynamics.

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Thanks to Dennis Hackethal for early feedback.

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