What is the kinetic every, potential energy and acceleration of gravity in this experiment?

Question

If you drop a ball from a certain height, were is the kinetic energy and potential energy? And how about the acceleration energy?

Answer 1

Total energy of anything, any time, anywhere can be written down as TE = KE + PE + WE; where KE is kinetic energy, PE is potential energy, and WE is work energy. Thus TE = constant for a given system and what changes are KE, PE, and WE so that the sum of the three terms always remains TE = constant. All this comes from the conservation of energy law, which says energy is not created or destroyed, but can be converted from one type to another type of energy.

Let's look at your ball in this light. TE(h) = total energy of the ball at height h before dropping it = KE + PE + WE. But KE = 0 because the ball is not moving and WE = 0 because it is neither doing work or being worked on. So that leaves us with TE(h) = PE(h) = mgh; where m is the ball's mass, g ~ 10 m/sec^2, and h is the height of the ball before release.

Now you open your hand and the ball drops (you don't throw it). During the fall from h to h = 0 (on the ground), we still have TE(dh/dt) = KE + PE + WE. But now work is being done; so WE > 0. The work...it's the force of gravity pulling the ball downward as in WE = F(dh) = mg(dh); where dh is a change in height. WE is just the so called work function. And this may be what you're calling "acceleration energy" because the ball is accelerating at the rate g.

OK, then, what's all this work doing to the other two types of energy...KE and PE? Well, as the ball is accelerating from WE's effort, KE, which is based on velocity, is clearly greater than zero and getting bigger. That is, the work is giving the ball kinetic energy. But where is that kinetic energy coming from...we can't create or destroy energy.

It's coming from the PE, the potential energy. PE = mgh to start with. But as h ---> 0, PE ---> 0 also; so as the PE diminishes it is converted into KE by the work WE. And total energy TE remains constant as that ball falls.

Now, for the grand finale...what's TE(h = 0). Well, PE = 0 clearly because PE = mgh and h = 0. Further, there is no more dh, change in height; so WE = mg dh = 0 too. Whoa, where's all the energy then? It's all in TE(h = 0) = KE = 1/2 mv^2; where v is the velocity of the ball m at height h = 0 (actually just before h = 0).

So here's the really cool part. Because of the conservation of energy, we can write TE(h) = PE(h) = mgh = 1/2 mv^2 = KE(h = 0) = TE(h = 0). Then gh = 1/2 v^2 and v^2 = 2gh or v = sqrt(2gh) which is a variation of the famous SUVAT equation v^2 = u^2 + 2gh; where u is some initial velocity, which we assumed u = 0 because you dropped, not threw, the ball.

Answer 2

When you throw a ball up into the air its kinetic energy gradually decreases with time as gravity pulls it down to earth until all of its kinetic energy has been converted to potential energy. This means the kinetic energy of the ball is at a maximum at the instant the ball is thrown and also at the instant before it hits the earth (after it has fallen). The potential energy of the ball is proportional to its height. This means that when the ball is at its maximum height above the earth, it also has its maximum potential energy.