Non-Calculus Derivation for the Energy-Equation for a book falling off a Table
Because the acceleration is constant, when we drop the book, it begins to fall (downwards) and its velocity increases (in the negative direction, if up is defined as the positive direction) in a simple linear way with time:
v = - g t
In other words, the velocity's magnitude increases proportionally with time t, and the proportionality constant is simply "g", the acceleration of gravity, by definition. For example, given that the acceleration g is about 10 meters per second squared, then after 1 second, the velocity is 10 meters per second, after 2 seconds its 20 meters per second, after 3 seconds its 30 meter per second, and so on.
Thus you can see that this simple formula gives the instantaneous, or momentary value of the book's velocity at any given time t you choose.
Now, most people are familiar with the notion that "distance equals velocity times time":
Distance = Velocity x Time
So, you might suppose that we could write:
h = hinitial + v t
Here, we are using "h" for distance, which in this case refers to the "height", or distance of the book above the ground. The symbol "hinitial" represents the initial height of the book (which is a constant).
This is almost correct, but strictly speaking, this formula only holds if the velocity in question is either constant (which is not true in our case), or alternatively, its true if we use the average velocity vavg that the book had during any particular time interval from t=0 till time t=tfinal. Hence we should really write:
h = hinitial + vavg tfinal,
where vavg is the average velocity that corresponds to the particular value of tfinal used.
So how do we calculate vavg? This is the crucial step! (Its exactly here that we really avoid the tools of calculus).
Because the instantaneous velocity is given by v = g t (from above), a nice linear relationship, it follows that the average velocity vavg for the interval ending at t=tfinal must equal to the velocity at the mid-point of the time interval (at t = 1/2 tfinal ):
vavg = - g x ( 1/2 tfinal ) = - 1/2 g tfinal
Putting this back into our formula for h above, we obtain:
h = hinitial + vavg tfinal = hinitial + (- 1/2 g tfinal ) x tfinal
h = hinitial - 1/2 g (tfinal)2
So now we have figured out some really significant - that the books height decreases according to the square of the time. This is exactly what always occurs to a position variable for something under constant acceleration.
Now you can actually begin to understand the energy equation above, that is, Total Energy = m g h + 1/2 m v2 = constant. Specifically, we know from the equation above that h is related to time squared. But from our earlier equation, we know that v2 is also related to time squared (one's sees this by simply squaring the equation v = g t: v2 = g2 t2). So, it must be that h is related to v2, as the energy equation suggests.
Now, to actually get the energy equation, we just rearrange our new formula (h = hinitial - 1/2 g (tfinal)2), and also a formula relating tfinal and v, (that is, v = g tfinal ) to eliminate the explicit appearance of tfinal in the equation.
First, we move the term containing tfinal to the left side (for those who want rigor, this is accomplished by adding 1/2 g (tfinal)2 to both sides). We obtain:
h + 1/2 g (tfinal)2 = hinitial
Now note that we have
v = - g tfinal
(This is just the formula for the instantaneous velocity, as above, at the particular time tfinal .)
We can rearrange this to read:
tfinal = - vavg / g.
Substitute this expression into h + 1/2 g (tfinal)2 = hinitial, we obtain:
h + 1/2 g (- vavg / g)2 = hinitial
We're almost there!
The second term simplifies:
1/2 g ( - vavg / g)2 = 1/2 (vavg)2 / g
(Note that the -1 dissappears because it gets squared)
Now our equation now reads:
h + 1 / 2 (vavg)2 / g = hinitial
Now we simply multiply both sides through by mg (just the mass times acceleration - a constant), to get
m g h + 1/2 m v2 = m g hinitial
This is our hard-earned energy equation! Note that the right hand side is a constant - its the initial potential energy (which equals the total potential energy).