# Getting to Work

Alright, Nerds, it's time to learn some physics!

Well, everything I write here involves learning physics, but it's time to go over some fundamental material that will serve as a foundation for upcoming posts I have in mind and provide much-needed background for one of my earliest articles. Plus, this topic is going to come up in my lectures, soon, and I want to have some material that I can refer my students to.

So, let's get to work— literally.

In physics, "work" has a very specific meaning. Say you have an object, you apply a force to that object, and in response, that object moves. Congratulations, you have just done some work!

Figure $$(1)$$: Applying a force $$\mathbf{F}$$ to an object which moves through a displacement $$\Delta \mathbf{x}$$, whose direction is parallel to that of the applied force.

In this case, the amount of work you have done $$W$$ is equal to the magnitude (strength) of the force you have applied $$F$$ times the magnitude (length) of the object's displacement $$\Delta x$$: $$W = F \Delta x. \label{170917work}$$

So there you have it—a basic (albeit incomplete) description of physical work. Nevertheless, it is useful. Why? Well, for reasons that are difficult to appreciate unless you have some experience solving problems in Newtonian mechanics, which I shall attempt to summarize: In Newton's scheme, one must first identify all of the forces acting upon an object. Then, those forces are broken down into components pointing along each perpendicular axis of your coordinate system. Components of different forces along each axis are combined into net forces before Newton's second law is invoked to derive equations of motion for each dimension of space. These equations are then solved to figure out how the object's position and motion change over time.

Sounds like a lot of "work," right? It is, in the linguistic sense, but by invoking the concept of physical work, a lot of the hard "work" is automatically done for us! To see how, let us examine our expression for work in equation \eqref{170917work} above and combine it with Newton's second law of motion, $$\mathbf{F} = m \mathbf{a}. \label{170917n2l}$$

Equation \eqref{170917n2l} tells us that when an object of mass $$m$$ is acted upon by some force $$\mathbf{F}$$, it accelerates in the direction of $$\mathbf{F}$$ and the magnitude of that acceleration is $$a = F/m. \label{170917acc}$$ These facts allow us to replace $$F$$ in equation \eqref{170917work} with the product $$m a$$: $$W = m a \Delta x. \label{170917workma}$$

Now, let us assume that our applied force, and hence our acceleration are constant as the object moves through $$\Delta \mathbf{x}$$, meaning neither quantities magnitude or direction change over time. However, the object's speed will change since it is accelerating! By definition, an object is said to accelerate when its speed changes over some interval of time $$\Delta t$$. If the amount of change in the object's speed throughout that time interval is $$\Delta v$$, then the magnitude of the object's acceleration is $$a = \frac{\Delta v}{\Delta t}. \label{170917accdef}$$ Substituting this expression into equation \eqref{170917workma} produces the following: $$W = m \frac{\Delta v}{\Delta t} \Delta x = m \Delta v \frac{\Delta x}{\Delta t}. \label{070917Wmv}$$

In the last part of equation \eqref{070917Wmv}, I took the liberty to rearrange factors, giving us the quantity $$\Delta x / \Delta t$$. If $$\Delta x$$ is the displacement through which an object moves within the time interval $$\Delta t$$ (which is indeed the case in this example), then we call this ratio the object's average speed, $$v_{avg}$$: $$v_{avg} = \frac{\Delta x}{\Delta t}. \label{170917vavg}$$

Therefore, the work done on our object of mass $$m$$ by the constant force $$\mathbf{F}$$ as it moves through the displacement $$\Delta \mathbf{x}$$ during the time interval $$\Delta t$$ is the product of its mass $$m$$, its average speed $$v_{avg}$$, and its change in speed $$\Delta v$$: $$W = m v_{avg} \Delta v. \label{170917Wmvdv}$$

Whew, worn out yet? Don't worry, we're almost to the finish line!

If the speed of our object is changing throughout the time interval $$\Delta t$$, then that means it has an initial speed $$v_i$$ at the beginning of that period and a final speed $$v_f$$ at the end. The difference in these two quantities is our change in speed $$\Delta v$$: $$\Delta v = v_f - v_i. \label{170917dv}$$

We can also find their mathematical average, $$v_{avg} = \frac{v_i + v_f}{2}, \label{070907meanv}$$which is, in fact, the same thing as the $$v_{avg}$$ defined in equation \eqref{170917vavg} above! Inserting both of these expressions into equation \eqref{170917Wmvdv} produces \begin{aligned}
W &= \frac{m}{2} \left( v_i + v_f \right) \left( v_f - v_i \right)
\\[1.5ex]
&= \frac{m}{2} \left( v^2_f - v^2_i \right)
\end{aligned} \label{170917WE1}

(Foiling out that binomial product is left as an exercise to the reader!)

Alright, let's examine equation \eqref{170917WE1} carefully. It says that the work done on some object of mass $$m$$ by the constant force $$\mathbf{F}$$ as it moves through the displacement $$\Delta \mathbf{x}$$ during the time interval $$\Delta t$$ is the difference in the squares of the object's final and initial speeds times half of the object's mass. At this point, it becomes useful to define a new quantity in terms of these things called kinetic energy: $$KE = E_k = T = \frac{1}{2} m v^2. \label{170917KE}$$

Physicists seem to be incapable of coming to a consensus on which symbol should be used to represent this quantity, as $$KE$$, $$E_k$$, and $$T$$ are all used extensively. But regardless of how it's denoted, an object's kinetic energy is the square of its speed times half of its mass. If the object's speed changes, then its kinetic energy changes: \begin{aligned} \Delta v = v_f - v_i \Longrightarrow \Delta T &= T_f - T_i \\[1.5ex] &= \frac{1}{2} m v^2_f - \frac{1}{2} m v^2_i. \end{aligned} \label{170917WE2}

From this, it is easy to see that the work done on an object of mass $$m$$ by the constant force $$\mathbf{F}$$ as it moves through the displacement $$\Delta \mathbf{x}$$ during the time interval $$\Delta t$$ is equal to the change in its kinetic energy: $$W = \Delta T. \label{170917WE3}$$

Equation \eqref{170917WE3} is a specific case of what we will eventually call the work-energy theorem, and it will prove to be mighty useful in days to come.

So, stay tuned, my fellow Nerds, and as always, thank you for joining me on this adventure!

Aaron