In the thermodynamics video I watched, it was stated that objects at higher elevations have more energy. And that objects must speed up as they fall. My question is, "what happens if an object reaches it's terminal velocity? It's still falling but no longer speeds up" Doesn't this make the statement that "objects must speed up as they fall", incorrect?
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$\begingroup$ Also what happens to the potential energy of an object that can no longer speed up, it takes x amount of energy to put an object at 900 feet . However, the object reaches it's terminal velocity at 800 feet. Shouldn't there be energy left over , or there mass, gravity, drag, and other factors that come into play. ? Look, I have a PhD in Religion , humor me please while I'm trying to learn at age 53. $\endgroup$– Jeremy Shayne HarringtonJan 17 at 5:56
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$\begingroup$ Which video? Which minute? $\endgroup$– Qmechanic ♦Jan 17 at 6:14
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$\begingroup$ Related: physics.stackexchange.com/q/669118/247642 $\endgroup$– Roger VadimJan 17 at 8:29
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$\begingroup$ Because it's a thermodynamics video, I assume that it is trying to teach you about energy, and ignoring air resistance $\endgroup$– user253751Jan 17 at 14:50
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$\begingroup$ @JeremyShayneHarrington Air resistance steals kinetic energy from the object and turns it into heat and wind. The energy isn't "left over", it went into the air. In a vacuum this doesn't happen. $\endgroup$– user253751Jan 17 at 14:55
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6 Answers
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Only in a vacuum do falling objects continue to speed up the whole time they are falling. in air, they achieve constant velocity when the kinetic energy gain in descending matches the energy loss due to drag.
answered Jan 17 at 6:12
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$\begingroup$ So why doesn't light speed up in vacuum. Actually there is no vacuum anywhere, even if it's there one can't measure it scientifically, yes can randomly. Try to measure current with galvanometer of different value. Is radiation emitted by a body is loss due to friction or to attain thermal equilibirium. $\endgroup$ Jan 17 at 15:24
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2$\begingroup$ @NeilLibertine: Light doesn't speed up in vacuum because light doesn't speed up, period. It has a fixed velocity of $3 \times 10^8$ m/s in all reference frames. $\endgroup$– KevinJan 17 at 17:04
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$\begingroup$ @Kevin Do you know what vacuum means, no resistance. So technically anything have infinite speed. But as it is said that light from stars comes to us through vacuum, so it has to be inifinitely fast. This is why it slow down in denser medium. $\endgroup$ Jan 18 at 2:12
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$\begingroup$ @NeilLibertine: You should ask a new question if you want to discuss the speed of light. $\endgroup$– KevinJan 18 at 15:54
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$\begingroup$ @Kevin That came from assumption in answer. Why are comment section for then. $\endgroup$ Jan 19 at 3:07
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No, it doesn't make the statement incorrect. It only it makes true for specific situations.
It is always true that in a system where there is no resistance, a falling object hits the ground with kinetic energy that depends on how high it was prior to being let go. Specifically, $$E_k=\frac 12mv^2=mgh$$ is how much kinetic energy it hits the ground with if there is no resistance. Note that this is equal to its potential energy as its dropped.
It is a fair question to ask, "well if an object has a certain amount of gravitational potential energy and is released, but then reaches a point where it can no longer accelerate, then where does the potential energy that should have been converted to kinetic energy, go?"
The answer is of course, it is converted into heat (heating the object and atmosphere) due to the frictional forces of the air$^1$.
$^1$ What's happening is that as the object is falling, it is colliding with the atoms/molecules in the atmosphere, and each time this happens, kinetic energy is transferred between the object and particles. If you had a very advanced machine that could calculate the kinetic energy of the object and particles before and after each collision, the total kinetic energy of the entire system (collisions + kinetic energy on impact with ground) will equal the initial gravitational potential energy.
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$\begingroup$ Moving object doesn't heat nearby air but cooling it down. Have you seen pictures of aircrafts breaking sound barrier or having speed higher than a mach. You have noticed that there form girth of cloud around aircraft, that is condensation of water molecules. So faster you move, colder air becomes. $\endgroup$ Jan 17 at 16:00
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2$\begingroup$ Niel, if this were true, spacecraft would not need heat shields to survive reentry. $\endgroup$– IzzyJan 17 at 17:01
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$\begingroup$ @NeilLibertine "Moving object doesn't heat nearby air but cooling it down." When an object moves through air, both the object and the air are heated. If you knew the frictional force $F_r$ and the distance $d$ it moved, the energy of this heating will be $U=F_r\cdot d$ Although in practice, such calculations are usually complicated to do, it is nevertheless objectively true. The fact that heating occurs in dynamic systems where there are dissipative forces is basic physics - your comment contradicts basic physics. Cheers. $\endgroup$– joseph hJan 18 at 0:36
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$\begingroup$ @Izzy This is true from air-conditioner to fighter jets flying at high speed. Even when hurricane comes, temperature fall down. NASA and academicians also need to learn many things and in past they done it, learn from others. $\endgroup$ Jan 18 at 2:23
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$\begingroup$ @josephh Okay, why do we shiver in winter. Science says to heat up by vibrating muscles, while we do it to cool down. Molecules vibrate when get energy to dispense that energy. So in winter, difference between internal and external (of skin) temperature is more. So internal body vibrate to pass heat to surface. $\endgroup$ Jan 18 at 2:30
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No, given statement is correct, because it assumes an ideal situation - free fall in a vacuum and only under influence of gravity all other forces excluded. All science is pretty much similar,- models abstracts unnecessary details away. Because only in this way we can extrapolate nature laws. Otherwise, if we would put each and every law and/or situation in one basket, we would get a chaos which would not help us to understand how nature operates in the nutshell at all.
You can get terminal speed from drag and weight equilibrium :
$$ {\tfrac {1}{2}}\,\rho \,v^{2}\,C_{D}\,A = mg $$
Upon reaching terminal speed, due to force(s) balancing weight,- body further falls at constant speed. Parachute is based on this idea,- it stops gravitational acceleration at low terminal speeds, because parachute has big cross-section area $A$, and hence produces high drag force.
answered Jan 17 at 8:22
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$\begingroup$ Did you notice that $v^2$ is there for acceleration used in force equation or when acceleration becomes zero. The term like drag coefficient is useless or specific thing. See in given equation its presence affecting speed gained by a body just as frontal area of a body. While when it is isolated, it is being reduced by area. Now main thing is that while you always say that air drag is limiting factor and it increases with increase in speed but it reduces on increasing speed. Don't say that speed increases when drag reduced. $\endgroup$ Jan 17 at 16:23
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$\begingroup$ Seems you need to study drag force more. 1) Drag depends not on frontal area of body, but rather on cross-section of the body, which is $A$ term in equation. 2) And yes, drag force $F_D \propto v^2$ , or in case of very small Reynolds numbers $F_D \propto v$. Hence on speed increase, drag force increases all other factors being equal. $\endgroup$ Jan 17 at 19:10
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$\begingroup$ And cross sectional area is area perpendicular to motion or facing along the motion, frontal. You are unnecessarily dragging, a long body has less drag generally. Second, when you equate coefficient of drag, it is inversely dependent on speed. $\endgroup$ Jan 18 at 2:18
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$\begingroup$ Your logic is doomed. So according to you, flying bullet has zero drag because it's frontal area is almost zero ? Or falling man with a parachute has low drag because of his low frontal area of his foots ? Nope, cross-section of body should be taken for the maximum referenced area perpendicular to body motion. Second- you don't know how coeff. of drag depends on speed, until you know body exact shape (which you haven't specified), so it's just your speculation about some specific case. Besides it's not clear who will win $F_D \propto v$ or $F_D \propto C_D(v)$ (concurrent processes if any) $\endgroup$ Jan 18 at 8:56
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$\begingroup$ Frontal area of a car is not limited to front grill and headlamps but including windshield. So now you got that frontal area of a bullet is not tip but almost cylindrical head, shape is for streamlinig and that includes in drag coefficient. Rest is that when you isolate drag coefficient one side and rest other side in given equation, you will find that drag coefficient depends inversely on speed while generally it increases with speed because drag force increases. $\endgroup$ Jan 18 at 11:57
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The statement "objects must speed up as they fall" is a bit of fast-and-loose casual conversation and not a definitive statement.
Generally speaking, these sorts of gravitational-potential energy problems are stated in Newtonian kinematic terms where "the force of gravity" is acting upon a falling object that has mass. Because "$F_{grav}=ma$" one might say that this gravitational force causes an acceleration of the mass, and thus a "speed up", but this is assuming the absence of all other forces.
In more complex situations with additional forces (like the normal force of a table surface, air friction in freefall, etc.) acting upon the object, it will be the case that $F_{total}\neq F_{grav}$ and thus the overall acceleration of "$F_{total}=ma$" might be positive, zero, or even negative depending upon the relative magnitudes and directions of all the forces involved.
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All physical laws are simplified descriptions of reality. How much simplification is too much? How much error can you tolerate?
If you want to know how long it takes a cobble stone to fall from a second story window to the ground below, you can ignore the air resistance. It won't make enough of a difference for you to measure unless you measure timing, height, and initial velocity with high-precision instruments.
If you want to know how long it takes an inflatable beach toy to fall to Earth from an aircraft, then you can ignore Newton's laws altogether, and just base your answer on the altitude from which the toy was dropped and its terminal velocity.
For situations in between those two extremes, you may need to use a more complicated method that combines both Newtwon's laws and aerodynamic laws.
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Force is spatial rate of energy transference. When it is applied on an object, the object gains momentum or loss depending upon state of motion of an object and direction of applied force. It is never become constant instantaneously. When it becomes constant that means an object reach at equilibirium and rate of gain in momentum equals rate of loss in momentum.
Just like thermal equilibirium in which rate of absorption equals to rate of emission there is mechanical equilibirium in which rate of gain of momentum equals to loss of momentum. At mechanical equilibirium, an object attains constant speed. For mathematical expression visit link below.
