Kinetic energy and the work-energy theorem


Newton's second law 

$$\overrightarrow{a}_A= \frac{\overrightarrow{F}^{net}_A}{m_A}$$

tells us that forces acting on an object tend to change the object's velocity, that it's the sum of all the forces acting on the object (the net force) that does it, and that it's a vector law — that changes in direction require a force just as changes in speed do.

Now suppose that we just want to know about how the object's speed changes — without paying any attention to direction. Could we extract that information out of Newton's second law directly — without having to pay all that attention to vectors? The answer, of course, turns out to be yes (otherwise we wouldn't raise the question here), and furthermore, the answer turns out to be both very interesting and immensely useful.

Changing speed in 1D motion

As we often do in physics, let's start with the simplest case. Since we're only interested in changing speed and not direction, we'll start considering motion in 1D — along a line without worrying about the possibility of changing direction. (We can handle going back and forth in this example — just not going off at an angle.)

Let's open up our acceleration, treating it as a change in velocity over a small time interval. Since we are only working in 1D, we'll work with the x-component of everything — position, velocity, acceleration, and force (but for now we will leave off the little subscript x to simplify our notation). N2 becomes

$$\frac{\Delta v}{\Delta t} = \frac{F^{net}}{m}$$

where our time interval takes us from our initial position and velocity to our final ones. We know that if we exert a force in the same direction as the object is moving we will speed it up.  If we exert a force in the opposite direction from the object's motion, we will slow it down. Thinking of pushing a moving object for a small distance, let's multiply both sides of N2 by $Δx$.

$$\frac{\Delta v}{\Delta t} Δx= \frac{F^{net}}{m} Δx$$

This turns out to do us a lot of good. Since we understand our derivative (acceleration is the time derivative of velocity) as a ratio of small changes, we can slide the $Δt$ under the $Δx$ to make an average velocity.

$$ Δv \frac{\Delta x}{\Delta t}= \frac{F^{net}}{m} Δx$$

$$ Δv \langle v \rangle= \frac{F^{net}}{m} Δx$$

Now for small time intervals, we can treat the force as if it is constant (not changing significantly), so we can write our linear change for average velocity and put in our initial and final values.

$$Δv \langle v \rangle = (v_f - v_i) \bigg(\frac{v_f + v_i}{2}\bigg) = {1 \over 2} \big(v^2_f - v^2_i \big) $$

That collapses nicely.  If we put this back into our N2 equation and multiply by $m$ we get

$${1 \over 2} mv^2_f -{1 \over 2}  mv^2_i  = F^{net} \Delta x $$

Let's see if we can make sense of this.

Interpreting the result: Kinetic energy

What has come out after all our manipulations is a change in a quantity associated with motion, ${1 \over 2}mv^2 $. This is kind of like momentum in that it counts both the mass and the velocity, but it differs in that momentum is proportional to the velocity vector -- so it is very directional. Reversing momentum is a big deal even if the speed doesn't change. For our new quantity, since it is proportional to $v^2$ instead of to $v$, the direction of motion doesn't matter. You get the same$v^2$ whether $v$ is positive or negative. If our general result turns out to only depend on the magnitude of $v$ and not the direction (it will), we will have solved our problem and learned what it is that changes an object's speed (not caring about direction).

The quantity that appeared from our manipulations, ${1 \over 2} mv^2 $, is called the kinetic energy.  It is a measure of "how much an object is moving". What our final equation of the last section tells us is that multiplying the force by the object's displacement gives us a quantity that changes the object's kinetic energy.  This quantity, the force times the displacement, is called the work.  

What if it's not 1D?

If the motion is not in 1D but the forces are changing the direction of the velocity as well as the speed, can we still do something like this?  Sure! And it's pretty obvious what we have to do if we remember how vectors work. We can always replace vectors by other vectors that have the same sum as the one we started with. (Like writing 23 + 17 = (20 +3) + (20 - 3) to make adding easier.) Often we broke up a vector in an arbitrary direction into two vectors — one in the $x$ direction and one in the $y$ direction—- in order to make adding easier.

Here what we want to do is take a force vector acting on an object and pull it into two parts:

  • one that is along the direction of motion of the object
    • either acting in the same direction as the object is moving to speed it up,
    • or acting in the opposite direction as the object is moving to slow it down, and
  • one that is perpendicular to the direction of motion
    • acting only to change its direction. 

For each force the object is feeling, we'll break it up as follows:

$$\overrightarrow{F} = \overrightarrow{F}_{\perp} + \overrightarrow{F}_{\parallel} $$

where the little upside down T indicates a right angle and means the part of the vector perpendicular to the object's motion and the two parallel lines means the part of the vector parallel to the object's motion. We can't get away with an $\hat{i}$ or a $\hat{j}$ since as the object moves it changes direction and therefore what we mean by "parallel" or "perpendicular" changes. We just have to be careful.  (The general extraction of the part of a vector in the same direction as another is done by a vector operation called the dot product.  Go to the link to read about how to do it in general.) 

The result is what's called The Work-Energy Theorem. The Kinetic Energy definition remains the same, but now for the work we will define the work for each force as being

The work a force $\overrightarrow{F}$ does on a moving object as it moves through a displacement $Δ\overrightarrow{r}$  is equal to the component of that force parallel to the displacement $Δ\overrightarrow{r}$ times $Δr$.

We can write this as

$$W = F_{\parallel} \Delta r$$

or, being more explicit:

$$W_{B \rightarrow A} = F^{\parallel}_{B \rightarrow A} \Delta r$$

The second way is more explicit about reminding us that every force acting on an object A is exerted by some other object.

The Work-Energy Theorem

This leaves us our final result that tells us how to look at the forces from N2 and see what part of them cause the speed (only) to change and defines a way of specifying "the quantity of motion an object has" in a way that will generalize spectacularly — into the concept of energy.

$$\Delta \big({1 \over 2} \;  m_A v_A^2 \big) = F^{\parallel}_{net\;on\; A} \Delta r$$

$$\Delta \big({1 \over 2} \;  m_A v_A^2 \big) = \big( F^{\parallel}_{B \rightarrow A} + F^{\parallel}_{C \rightarrow A} + F^{\parallel}_{D \rightarrow A} + ... \big)\Delta r$$


Joe Redish 10/27/11