Work-Energy Theorem and Conservative Forces
Sneakily deriving the Work-Energy Theorem from Newton's second law via the chain rule, then defining potential energy and unpacking what conservative forces actually are!
When I first saw this chapter I was like “oh nice, easy win” — then I sneeeakily peeked inside, didn’t look easy at all, so I noped right out.
Like — what could possibly be in Newtonian mechanics that goes past high school????
First of all, do you even learn the chain rule in high school?????????????????????
(Ha… am I old? Why can’t I remember being a student…)
$$\overrightarrow{F} = m\overrightarrow{a} = m\overrightarrow{\ddot{x}}$$Rewrite this with the chain rule:
$$F = m\ddot{x} = m\frac{d^2x}{dt^2} = m\frac{d}{dt}\frac{dx}{dt} \\ = m\frac{d}{dt}\frac{dx}{dx}\frac{dx}{dt} \\ = m\frac{dx}{dt}\frac{d}{dx}\frac{dx}{dt} \\ = mv\frac{d}{dx}v \\ = mv\frac{dv}{dx}$$Boom — everything time-related got rewritten in terms of $x$.!
The time variable? Sneeeakily tucked away.
Or if you want it even more blatant:
$$F = \frac{m}{2}\frac{d}{dx}\left(v^2\right)$$This is the form you use when you want to yank out info about position instead of time!!!!!
And one more thing!!!
$F(x)$ is literally what you get when you differentiate $\frac{1}{2}mv^2$ with respect to position $x$!!!!
So the definition of energy is:
$$\frac{d}{dx}\left(\frac{1}{2}mv^2\right) = F(x)$$We’re defining this one-half-m-v-squared thing as the physical quantity called Energy!!!
So the concept of “force” is: “by how much did the energy change as distance $x$ changed?? (did it pump some in? or yank some out?)”
And flipping it: “energy” is —
how much force got applied per little $dx$, aaaall added up!!!!
right?
$$\int d\!\left(\frac{1}{2}mv^2\right) = \int F(x)\,dx = T - T_0 \\ = \int dT$$Hold on. $\int F\,dx$… wait — that’s work. Like literal ‘work’ from high school!?!?!?!!
YES. This is where it comes from! The “Work-Energy Theorem!!!”
$$-\frac{dV}{dx} = F(x)$$Now let’s define a function $V(x)$ that satisfies this.
Call $V(x)$ potential energy!!!
Integrating $F(x)$ with respect to $dx$ once more using the above:
$$\int -dV = -V(x) + V(x_0) = \int F(x)\,dx \\ \therefore\; T - T_0 = -V(x) + V(x_0) \\ T + V(x_0) = T(x) + V(x)$$How clean is that?!
Apparently this is called the “energy equation” haha
Looks obvious — but one thing to flag:
$$-\frac{dV}{dx} = F(x)$$we got here assuming a $V(x)$ satisfying this exists….!!!
But! There are forces that play along with this assumption!!!! Namely, the forces
that depend only on position $x$. People call those conservative forces!!
Now — someone’s gonna come at me:
“Bro, seriously? A force that depends only on position? In what universe, man?! -_-”
Hmm… forces that depend only on position include
gravity, electric force, elastic (spring) force…
$$\text{gravity} = \frac{GMm}{\text{position}^2} \quad \text{electric force} = \frac{1}{4\pi\epsilon_0}\frac{Qq}{\text{position}^2} \quad \text{elastic force} = -k(\text{position})$$We’ll learn about these conservative forces~~~~ properly later, but
$$\nabla \times F = 0$$i.e., the curl is zero~~~~
(Forces whose curl is zero are called “conservative forces.”)
Looking at it this way, it kinda seems like a lot of forces in the world depend only on position…..
But… nope. Not really.
Like, just off the top of my head:
$$\text{magnetic force} = q\!\left(\overrightarrow{v} \times \overrightarrow{B}\right) \\ \text{friction force} = -\mu N$$the magnetic force has a $v$ in it,,,,
friction has zero dependence on position,,,,
Non-conservative forces are probably way more dominant than the conservative ones, honestly~~~~~~~~!!
So why’d we learn all this~~~? Looking at it,
it was so we could start handling forces that don’t depend on position from here on — air resistance, for example.
You’re riding in a car, roll the window down, stick your hand out. When is the resistance on your hand big, and when is it small?
If you ask that, you’d say “big when the car’s fast, small when it’s slow~ ^^” right????
“Ohhh — so drag is proportional to velocity, got it~”
Valid conclusion???
BZZT. Wrong.
How would you know if it’s proportional to $v$, to $v^2$, or to $\sqrt{v}$? T_T
Oh… T_T this is getting hard….. T_T
OK OK OK OK. Fractional powers? Power series expansion — boom, they become integer powers,
and we’re left with: a term proportional to $v^2$, a term proportional to $v$.
Let’s say F is a linear combination of these. (Yes, we’re making an assumption.)
Alright then:
$$F(v) = c_1 v + c_2 v^2 = c_1 v + c_2 v|v|$$Let’s write it like this~~~!!! (Why the $|v|$? Apparently — to keep track of velocity direction. That’s what the book said anyway.)
$F(v)$ can only be pinned down experimentally, apparently.
Some scientist or engineer ran the experiments and got:
$$c_1 = 1.55 \times 10^{-4} D \\ c_2 = 0.22\; D^2 \quad (D : \text{diameter of sphere}\; [m])$$Now we can use this to figure out “which term dominates~~~~~”:
$$\frac{c_2 v|v|}{c_1 v} = 1.4 \times 10^{-3} |v| D$$whether this is greater or less than 1 tells us which term shows up more dominantly in $F(v)$, apparently !!! heh
Let’s try a problem ;; heh
1. (Assume linear air resistance dominates.)
2. (Assume nonlinear ($v^2$) air resistance dominates.)
Originally written in Korean on my Naver blog (2015-06). Translated to English for gdpark.blog.