I'm working through an 11th grade physics course and am trying to grasp the concept of kinetic potential energy. According to an equation I'm given, potential energy can be expressed as $$m * h * g.$$ Thus, as an object falls, it's potential energy decreases, converting into kinetic energy. Now according to this equation, if an object (like a ball) is at the edge of a cliff, it has lots of potential energy. However, if that same ball is at ground level, it's potential energy is zero.
But really, isn't gravity pulling on both objects with virtually the same strength? And aren't they both sitting on a solid surface? Thus, the tension, or "stored energy" in each object should be the same, right?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
Welcome to physics stack exchange ! Actually there a couple of concepts which have to be cleared up for answering your question: Essentially, kinetic energy and potential energy: First of all: a body, a ball, for instance, only has kinetic energy when it moves. So the ball at the edge of the cliff does not have kinetic energy, not yet, only when it starts falling, then it gets kinetic energy. On the other hand, it has at the edge of the cliff "potential energy", this is what in simple words saying is "stored energy", energy is can be potentially be converted into kinetic energy. So the formula you cite: $m\cdot g\cdot h$ is actually the potential energy : $E_{pot}=m\cdot g\cdot h$, it turns into kinetic energy if the ball starts falling from the cliff if the cliff is of height $h$ with respect to the ground at the foot of the cliff.
A very important thing to know is that the observer has to fix a reference level for the potential energy, that means the level where the potential energy is supposed to be zero. In the experiment you propose the reasonable level for this is the ground at the foot of the cliff. This makes sense, because the ball on the ground is at height zero, so the potential energy $E_{pot}=m\cdot g\cdot h =0$. Actually, the choice of the reference where to define zero potential energy is free. One could choose another level for defining zero potential energy, but it would unnecessarily complicate the reasoning if we chose another reference level in the "ball at the cliff experiment".
Finally I come to your last point: You are right: All objects fall with the same strength, or to be more clear: All object fall with the same acceleration: On earth, the acceleration of gravity is called $g$ and you use it the cited formula. This fact can be used to compute the speed the ball achieves when it reaches the ground. When this happens, all potential energy has been converted into kinetic energy $E_{kin}$ and we can set up the following equation:
$E_{kin}=E_{pot}=mgh$
And finally knowing that kinetic energy can be computed from the velocity and the inertial mass of the object: $E_{kin}=\frac{1}{2}mv^2$, we can say:
$\frac{1}{2}mv^2=E_{kin}= E_{pot}=mgh$
And now we see that on both sides of the equation we can cancel out the mass $m$:
$\frac{1}{2}v^2=gh$
and we can compute the velocity of the ball when it reaches the ground: $v = \sqrt{2gh}$. This velocity is independent of the mass! Amazing, isn't it ?
Actually, this already a lot of explanation, may be even far too much, so if you don't understand it, send a comment. Then I'll try to explain more. Nevertheless I hope it helps you understand the concepts of kinetic and potential energy better.
answered Sep 12, 2017 at 14:27
-
$\begingroup$ Thanks for the answer! I actually have no problem using the equations, and I'm pretty sure I'll ace this test, it's just the actual concept of potential energy that's bothering me. I suppose my real question is: is potential energy just a theoretical expression, or is there really some sort of "energy" stored in an object on a cliff as opposed to an object on the ground? $\endgroup$ Commented Sep 12, 2017 at 15:22
-
$\begingroup$ When we say potential energy is stored in the object, one has always to keep in mind that this type of energy depends on the reference point. What counts is the potential energy difference between the object at the edge of the cliff and the object on the ground. This concept is also the basis of the energy conversation. Actually, you can store energy by putting a massive object on a cliff and release it when necessary(well better not a bit dangerous) the same as using some chemical compound like fuel that stores (this time chemical) energy which can be burnt for driving a motor when needed. $\endgroup$ Commented Sep 12, 2017 at 17:40
-
$\begingroup$ @CalebBertrand: I've removed the phrase about the gravitational mass. After thinking about it I realized that in both types of energy $E_{kin}$ and $E_{pot}$ the mass $m$ is inertial mass. It's the independence of $g$ on the mass of the falling object which expresses the behaviour found by Galilei. $\endgroup$ Commented Sep 14, 2017 at 9:17
$\begingroup$
$\endgroup$
The $h$ in the formula $U=mgh$ is a height. So you can ask: a height from where? You need a reference.
- The ball on the high shelf causes potential energy to be present when compared to the ground.
- The ball in the ground causes no potential energy to be present compared to the ground - but it does when compared to the bottom of a deep hole.
The point is that potential energies of any form requires a reference. Their values are nothing more than a difference to the reference level.
