The Work–Energy Theorem

53 The Work–Energy Theorem

Going back to our introduction, we figured out that the way to add kinetic energy is to push an object in the direction of motion. Since dot products pick out the parallel components of a two vectors, it must be that work has something to do with the product $\vec F \cdot \vec v.$ We’ll now put all of this together to derive the work–energy theorem. Before we do so, however, it’ll be useful to go through a little bit of general math.

Exercise 53.1: Dot Products and Derivatives

Consider two vectors $\vec A$ and $\vec B$ . To make your algebra easier, assume these are two dimensional. Use the components formulation of the dot product to show that

$\frac{d}{dt}(\vec A \cdot \vec B) = \frac{d \vec A}{dt} \cdot \vec B + \vec A \cdot \frac{d\vec B}{dt}$ .

Notice this is just the ordinary product rule, so again, dot products behave like ordinary multiplication.

The above equation also holds for differentials. That is, if you multiply both sides by $dt$ , you end up with

$d(\vec A\cdot \vec B) = (d\vec A)\cdot d\vec B + \vec A\cdot (d\vec B)$ .

In particular, notice that since $A^2=\vec A\cdot \vec A$ , one has

$dA^2 = 2\vec A\cdot d\vec A$ .

This can be useful from time to time (e.g. the last problem in this chapter).

Exercise 53.2: The Work–Energy Theorem

We start with Newton’s law: $\vec F_{\rm tot}=m\vec a$ .

A. Take the dot product of both sides of the equation with the velocity $\vec v$ , and use the result from the previous exercise to express the right hand side of the equation as a derivative of $v^2$ .

B. At this point, the right hand side of the equation is an exact derivative. That means that we can multiply both sides of the equations by $dt$ and integrate. Do so now and simplify everything you can.

Our final result is

$\frac{1}{2}mv_{\rm f}^2 - \frac{1}{2}mv_{\rm i}^2 = \int_{\rm i}^{\rm f} \vec F_{\rm tot} \cdot d\vec x.$

To make this equation look nicer, we define the kinetic energy via $K\equiv \frac{1}{2}mv^2$ . Thus, the left hand side is simply $K_{\rm f}-K_{\rm i}$ , i.e. the change in kinetic energy. The term on the right hand side is called the total work, $W_{\rm tot}=\int_{\rm i}^{\rm f} \vec F_{\rm tot}\cdot d\vec x$ . With these definitions, the above equation starts to look like the conservation of energy law,

$K_{\rm f} = K_{\rm i} + W_{\rm tot}.$

This equation is called the work–energy theorem. It is clear from this equation that energy has the same units as “force times distance.” In SI units, the unit of energy is called the Joule, corresponding to $1\ {\rm N}\cdot{\rm m}$ .

I will not ask you to memorize the work–energy theorem, since this equation is really just an intermediate step in developing the idea of conservation of energy. To get there, however, we need to introduce potential energy, which in turn comes out from thinking deeply about work.

Exercise 53.3: Work

We have defined work as $W_{\rm tot}=\int_{\rm i}^{\rm f} \vec F_{\rm tot}\cdot d\vec x$ . In differential notation, we can also write this as the much nicer looking equation

$dW = \vec F_{\rm tot}\cdot \vec dx.$

Thus, infinitesimally, work is force times (infinitesimal) displacement.

Let’s make sure this intuition has really sunk in. Consider the diagrams below. Each diagram shows a box of mass $M$ which is acted on by a force $\vec F$ as shown. All boxes move a distance to the right as shown (either $s$ or $2s$ ). The surface on which the boxes move is frictionless.

Explain why the work done by the force $F$ is larger in diagram A than in diagram B.

Consider the ranking you did above. Would the ranking change if the surface is not frictionless. Remember you are ranking by the work done by the force $F$ in the diagram.

Warning: If you have studied physics before, you might have learned that “work is force times distance.” This is not true. The correct expression for work is $dW=\vec F\cdot d\vec x$ , or if you integrate, $W=\int_{\rm i}^{\rm f} \vec F\cdot d\vec x$ . If the force is constant, then you can take the force out of the integral, and recover $W=\vec F\cdot \Delta \vec x$ , or “work is force times distance.” However, this is only valid for a constant force.

Bottom line: Work is NOT forces times distance.

Key Takeaways

53 The Work–Energy Theorem

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