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Work and Energy

We have seen how much fun it is to multiply the SECOND LAW by a scalar (dt) and integrate the result. What if we try multiplying through by a vector? As we have seen in the chapter on VECTORS, there are two ways to do this: the scalar or ``dot'' product $\mbox{\boldmath$\vec{A}$\unboldmath } \cdot \vec{\bf B}$, so named for the symbol $\cdot$ between the two vectors, which yields a scalar result, and the vector or ``cross'' product $\mbox{\boldmath$\vec{A}$\unboldmath }\times\vec{\bf B}$, whose name also reflects the appearance of the symbol $\times$ between the two vectors, which yields a vector result. The former is easier, so let's try it first.

In anticipation of situations where the applied force $\mbox{\boldmath$\vec{F}$\unboldmath }$ is an explicit function of the position^11.5 $\mbox{\boldmath$\vec{x}$\unboldmath }$ -- i.e.   $\mbox{\boldmath$\vec{F}$\unboldmath } (\mbox{\boldmath$\vec{x}$\unboldmath })$  - let's try using a differential change in $\mbox{\boldmath$\vec{x}$\unboldmath }$ as our multiplier:

  $$\begin{eqnarray*}\mbox{\boldmath$\vec{F}$\unboldmath } \cdot d\mbox{\boldmath$ . . . . . . {v}$\unboldmath } \cdot d\mbox{\boldmath$\vec{v}$\unboldmath } \end{eqnarray*}$$

where we have used the definitions of $\mbox{\boldmath$\vec{a}$\unboldmath }$ and $\mbox{\boldmath$\vec{v}$\unboldmath }$ with a little shifting about of the differential dt and a reordering of the dot product [which we may always do] to get the right-hand side [RHS] of the equation in the desired form. A delightful consequence of this form is that it allows us to convert the RHS into an explicitly scalar form: $\mbox{\boldmath$\vec{v}$\unboldmath } \cdot d\mbox{\boldmath$\vec{v}$\unboldmath }$ is zero if $d\mbox{\boldmath$\vec{v}$\unboldmath } \perp \mbox{\boldmath$\vec{v}$\unboldmath }$ -- i.e. if the change in velocity is perpendicular to the velocity itself, so that the magnitude of the velocity does not change, only the direction. [Recall the case of circular motion!] If, on the other hand, $d\mbox{\boldmath$\vec{v}$\unboldmath } \parallel \mbox{\boldmath$\vec{v}$\unboldmath }$, then the whole effect of $d\mbox{\boldmath$\vec{v}$\unboldmath }$ is to change the magnitude of $\mbox{\boldmath$\vec{v}$\unboldmath }$, not its direction. Thus $\mbox{\boldmath$\vec{v}$\unboldmath } \cdot d\mbox{\boldmath$\vec{v}$\unboldmath }$ is precisely a measure of the speed v times the differential change in speed, dv:

$$\mbox{\boldmath$\vec{v}$\unboldmath } \cdot d\mbox{\boldmath$\vec{v}$\unboldmath } = v \, dv$$ (11.6)

so that our equation can now be written

$$\mbox{\boldmath $\vec{F}$\unboldmath } \cdot d\mbox{\boldmath $\vec{x}$\unboldmath } = m \, v \, dv$$

and therefore

$$\int_{\mbox{\small\boldmath$\vec{x}$\unboldmath }_0}^{\mbox{\ . . . . . . v v \, dv = m \left( {1\over2} v^2 - {1\over2} v_0^2 \right)$$ (11.7)

(Recall the earlier discussion of an equivalent antiderivative.)

Just to establish the connection to the mathematical identity   $a \, dx = v \, dv$,  we multiply that equation through by m and get   $m a \, dx = m v \, dv$. Now, in one dimension (no vectors needed) we know to set  ma = F  which gives us   $F \, dx = m v \, dv$  or, integrating both sides,

$$\int_{x_0}^x F \, dx = {1\over2} m v^2 - {1\over2} m v_0^2$$

which is the same equation in one dimension.

OK, so what? Well, again this formula kept showing up over and over when people set out to solve certain types of Mechanics problems, and again they finally decided to recast the LAW in this form, giving new names to the left and right sides of the equation. We call   $\mbox{\boldmath$\vec{F}$\unboldmath } \cdot d\mbox{\boldmath$\vec{x}$\unboldmath }$  the work  dW  done by exerting a force $\mbox{\boldmath$\vec{F}$\unboldmath }$ through a distance $d\mbox{\boldmath$\vec{x}$\unboldmath }$ [work is something we do] and we call   ${1\over2} m v^2$  the kinetic energy  T. [kinetic energy is an attribute of a moving mass] Let's emphasize these definitions:

$$\int_{\mbox{\small\boldmath$\vec{x}$\unboldmath }_0}^{\mbox{\ . . . . . . d\mbox{\boldmath$\vec{x}$\unboldmath } \equiv \Delta W \, ,$$ (11.8)

the WORK done by $\mbox{\boldmath$\vec{F}$\unboldmath } (\mbox{\boldmath$\vec{x}$\unboldmath })$ over a path from $\mbox{\boldmath$\vec{x}$\unboldmath }_0$ to $\mbox{\boldmath$\vec{x}$\unboldmath }$, and

$${1\over2} m v^2 \equiv T \, ,$$ (11.9)

the KINETIC ENERGY of mass m at speed v.
 
Our equation can then be read as a sentence:

$\parbox{8cm}{\noindent \lq\lq When a force acts on a body, the {\sl kine . . . . . . t equal to the {\sl work\/} done by the force exerted through a distance.'' }$

One nice thing about this ``paradigm transformation'' is that we have replaced a vector equation   $\mbox{\boldmath$\vec{F}$\unboldmath } = m \, \mbox{\boldmath$\vec{a}$\unboldmath }$  by a scalar equation   $\Delta W = \Delta T$. There are many situations in which the work done is easily calculated and the direction of the final velocity is obvious; one can then obtain the complete ``final state'' from the ``initial state'' in one quick step without having to go through the details of what happens in between. Another class of ``before & after'' problems solved!