# If an object falls - regarding air resistance - does the Potential or Kinetic energy get converted into thermal energy as it is falling?

I read a paragraph on the transfer of potential energy to kinetic energy and heat from this website:

Even if air resistance slows down the ball, the potential energy is the same (Mb x g x H). But if air resistance is in the way, not all of the potential energy can be converted to kinetic energy. Some of the energy has to be used to push the air molecules out of the way. When that happens, the energy of the air molecules is increased. The air is actually "heated" up by the falling ball.

This text indirectly mentions that the Potential Energy gets converted into heat. However my common sense (for lack of a better term) makes me think that the Potential Energy gets converted into Kinetic Energy which in turn gets converted into Thermal energy. Would someone be able to enlighten me on this please?

Also, as a side question, it isn't called Heat Energy right? A lot of websites seem to be saying that, but heat is just the transfer of Thermal Energy...

• Heat Energy, Thermal Energy - the same thing. Energy manifested in the form of particle motion. The higher the mean particle velocity, the more heat you have and thus a higher temperature. In the end we still don't know what energy 'is'. We only know it's something 'conserved' and we give it many names according to how it manifests itself in space and time. Jan 11 '15 at 21:12
• It isn't only the air that heats up. the ball gets hotter too. So some of the potential energy is going to heating of the ball. Feb 9 '16 at 20:32
• To give some intuition and to answer the question, it is kinetic (u can tell since the drag term always is a function of the velocity explicitly) microscopically see it as the velocity of the shear air is causing the molecules on the side of the object to vibrate vigorously Mar 11 '17 at 2:51

The friction with the air indeed slows down the falling ball. The friction is minimal if the ball stays in place (i.e. there may be only some friction due to air currents.) But if the ball moves the friction is bigger because as the ball moves, it pushes away the air molecules to make room. The friction opposes the ball movement. So, indeed, the heating takes from the kinetic energy through friction, not from the potential energy. The meaning of the text, as I understand, it that, bottom line, the kinetic energy acquired by the ball during its fall is not equal to the total potential energy $M_b gH$.

• So i guess the paragraph was just a bit vague hey :) Thank you for clarifying! Jan 11 '15 at 21:20
• @AdminVoter : yes, the paragraph wanted to say that not all the potential energy goes to the increasing of the ball's kinetic energy. In fact part of the kinetic energy is stolen by the surrounding air molecules through friction. Jan 11 '15 at 21:30
• So saying that just the object itself follows the conservation of energy would be wrong? Because it is neglecting that the energy was taken away from the object and into the air around it, correct? Jan 11 '15 at 21:38

Heat energy and thermal energy are pretty much the same thing. My science teacher taught me, that because potential energy is related to height, it would be the kinetic energy that is converted into th

• *into thermal energy. Feb 9 '16 at 20:16
• 1) Welcome to SE! 2) you can edit your answer (the edit button at the bottom of your post) to fix the unfinished sentence 3) you say that KE is converted into heat, and not PE. Could you elaborate on this, please? I mean, can you explain the reasoning behind your statement? Feb 9 '16 at 20:26

As the ball moves it is in interaction with the local gravitational field, and with the air. The interaction with the air includes both viscosity, and also an adjustment of the regular flow of the air (moving the flow lines), and, related to this, turbulence behind the ball. The ball accelerates at each moment in proportion to the resultant of all these forces. As it moves through any given change in height its gravitational potential energy changes by precisely the associated amount, $$m g (h_f - h_i)$$. Its kinetic energy does not increase by this amount, however, since the ball is doing work on the air. This work gives kinetic energy to the air molecules, some of it in a regular form (so not heat) and some of it in an irregular form, which can be called heat or thermal energy (both terms are ok; but if you want to be strict then you use the word heat for the energy in transit from one system to another, and internal energy for the energy when it arrives in some other system, in this example the air).

The regular part of the air's motion eventually dies way, and then one may say that all the energy acquired by the air is just as if it had absorbed that amount of heat. The ball itself will also acquire some heat and consequently warm up. (And there will then be another process of heat transfer between air and ball).

If an author says that some of the gravitational potential energy has been converted into heat, then they are referring to the overall result; it is a perfectly correct summary, but it is also correct to say that the conversion takes places via the movement of the ball. So your intuition that the ball first picks up kinetic energy, and then this energy is partly passed on to the air, is sound, as long as you realise that the whole process is continuous.

Note that during a fall of a ball, the air friction is negligible, and we call it a free fall, only $$\vec{P}$$ is applied, the work of this force is given by : $$W(\vec{P})=mgl$$.

In this case, the air slows the ball falling, so there absolutely is two forces, the weight and the air contact, let it be $$\vec{f}$$, its work will be : $$W(\vec{f})=\vec{f}.\vec{l}$$

The speed of this ball isn't constant so the $$\sum W \neq 0 \Leftrightarrow \Delta E_k\neq0$$ and the variation of potential energy $$\Delta E_p=-W(\vec{P})$$

Back into the free fall; $$\Delta E_k=\sum W \Leftrightarrow \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2=W(\vec{P})$$

$$\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2=-mg(z_f-z_i)$$

$$\frac{1}{2}mv_f^2+mgz_f=\frac{1}{2}mv_i^2+mgz_i$$

Adding the Constant of the potential energy, we get: $$\frac{1}{2}mv_f^2+mgz_f+C=\frac{1}{2}mv_i^2+mgz_i+C$$

Note that : $$E_p=mgz+C$$ we will have : $$E_{k_f}+E_{p_f}=E_{k_i}+E_{p_i} \Leftrightarrow E_{m_f}=E_{m_i} \Leftrightarrow \Delta E_m=0$$

Therefore during a free fall the mechanical energy (Kinetic and potential) is conserved, but what about this case ?

I mentioned that the ball's speed isn't constant and there's a negative friction force (The work is negative because of the angle note that $$W(\vec{f})=f.l.cos(\pi)=-f.l$$), obviously the mechanical energy isn't conserved and that's why:

$$\Delta E_k= \sum W$$ In this case there's two forces, thus there's two works $$\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2=W(\vec{P})+W(\vec{f})$$ We're doing the same thing here, but there's a new factor which is $$W(\vec{f})$$, We'll have : $$\Delta E_m=W(\vec{f})$$ Now how this friction is transformed to a heat ?

Well to answer this question, you'll need the first law of thermodynamics $$\Delta U=Q+W$$ but I see that it's obvious without any mathematical proof, by the way $$\Delta E_m=-Q$$ so we can conclude that $$W(\vec{f})=-Q$$, so the friction is transformed to a heat (thermal energy) the sign minus is conventional. and you can conclude that when there's any friction force (external force) its work is converted into a thermal energy.