Conservation of Energy: Free Fall, Springs, and Pendulums 

Yo this is literally primary school math drop out of high school if you don't understand any further...

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@@stop2759 Sorry I was a bit rude; I was just trying to sound blunt to give this guy a reality check. If you can't understand the basic maths, then you're going to have trouble with all physics in later years.

@@pubgplayer1720 sometimes it's the teacher, the way it's explained, or it just isn't their passion. give it time and they'll be alright.

@@pubgplayer1720 physics needs to be explained with theory not just the use of equations, as that will help students broaden or lighten the topic they're dealing with.

And, of course, you amply have the great understanding, inasmuch as you know a lot about the science stuff.

omg, I have the exact same feeling when I get the results of a physics test! Dang!

@@Matt-sc6gg same

;_;

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You in college?

@@05bahaafarah11 Nope I was in 9th grade when I made this comment I’m a senior now

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I guess the gravitational force is contributing to the required centrifugal force.Feel free to correct me if I am wrong. :)

@@adithyank3773 Centrifugal force does not exist.

Lynxica is correct. It would be centripetal force acting on the moon' gravitational pull.

The moon is in free fall around the Earth. However, due to the fact that the moon has sufficient tangential speed, the moon is continuing to overshoot its target, which is why it doesn't crash in to the Earth. If the orbit were a perfect circle, the force of gravity would directly match the net force needed to cause centripetal acceleration. The force of gravity would be entirely perpendicular to the moon's velocity, which means there is zero work done by the force of gravity on the moon, and zero change in speed. The orbit is not a perfect circle, so to a lesser extent, the moon does speed up slightly and slow down slightly. The path of the orbit is elliptical, with the Earth-Moon center of mass (which is inside of Earth) at one of the focus points. You can look up the apogee (farthest distance from Earth) and perigee (closest distance from Earth) of the moon to see how far it varies from a perfect circle. There is the parameter called eccentricity that is also a measure of how non-circular an orbital shape is. The moon does speed up as it approaches perigee and slow down as it approaches apogee, so gravity does do a little bit of work on the moon during this path, but not enough to put it on a crash trajectory with Earth.

Because acceleration due to gravity is constant and points in the negative direction, acting on all projectiles and decreasing their vertical velocity all the time.

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ooh good one! i think it would still have zero potential energy, it would only have potential energy if it were rolled off the edge such that it could fall. that may or may not be right!

so if I am on top of a mountain and its highest point is at sea level. does that means I don't have PE until I get rolled off the mountain?

Gravitational potential energy is relative to whatever you choose to be your datum. If you calculate total mechanical energy based on that datum, it will be conserved in a closed system either way. You just can't change the datum at any point during your evaluation.

Was the potential created by removing the circle of dirt?

Good question. It is arbitrary where we define GPE to equal zero. A ball sitting on the ground doesn't necessarily have truly zero GPE, we just define it to have zero GPE by convention. It is mathematically impractical to calculate how much GPE it truly has. We define its initial GPE to equal zero. We then dig a hole 1 meter deep. We let the 1kg ball roll toward the hole. The ball falls to the bottom of the hole, and its GPE becomes -9.8 Joules. The ball lost 9.8 Joules of GPE that became thermal energy, when it falls and comes to rest. When using the universal law of gravitation, we set GPE = 0 at r=infinity, but this is just for mathematical convenience. When we do this, all GPE's end up becoming negative. It is a simple consequence of picking r=infinity as our datum for GPE for mathematical convenience. There is no such thing as absolute GPE, the way we calculate it in Physics.

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good question, i think when it's g and not a, we just use the scalar value rather than the vector so we ignore direction. may want to double check but i think that's correct.

@@ProfessorDaveExplains Thanks for the respond :)

They both are considered mechanical energy. Thermal energy ultimately is kinetic energy, except at the molecular level, instead of macroscopic.

all the arithmetic is there! don't forget to square that initial velocity.

@@ProfessorDaveExplains Could you please explain it briefly as a reply. I'm not able to understand how you got 1640. Would be appreciated.

Is it correlate with projectile motion. I use it and result is the same thanks

@@lorihunter1429 Thanks for that tip, I actually struggled with this problem at first but your comment made me finally solve the equation right.

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Energy can be transferred in or out of a system. That's not a violation of the 1st law. The energy has to come from somewhere or go somewhere. It's conserved.

Professor Dave Explains I was not taking the surrounding into consideration. Thank you

it means the total amount of energy remains constant for a closed system! the amount of energy can be measured, yes, we just have to account for all the different forms of energy. thanks for watching!

Conservation of energy means you cannot create or destroy energy. It only changes form, and changes what object contains it. The sum total energy in an isolated system will equal a constant value. Isolated system means no mass or energy crosses the boundaries of the system.

Well as you say, we are energy in the sense that we are made of matter, and when we die, that matter persists. Our bodies decompose and the atoms go on to be part of other things, like the ground, and a tree, and the air, and so forth.

​@@dixztube Sure, I suppose that's one way of putting it.

shut up kendal

www.physicsforums.com/threads/law-of-the-lever-conservation-of-energy-or-angular-momentum.833317/

" Then where is the energy of another 5KG?" The 5kg doesn't cease to exist, it just changes form. When 10 kg of U235 decay, 5 kg become something else, and 5 kg are still remaining as the original isotope of U235. There are biproducts when U-235 decays, and in its natural decay cycle, the next atom is Thorium-231. That means only about 4 amu of mass are annihilated and scattered as energy, while the rest remains as matter in the biproduct of Thorium. The natural decay chain of U235 eventually settles on lead-207, which is stable. When U235 decays to Pb207, there are 28 amu of mass that get emitted as gamma rays, alpha particles, and free electrons, the gamma rays of which get absorbed and converted to thermal energy. The alpha particles become natural sources of helium on our planet. If you look at the energy balance of Earth, based on the incoming solar radiation, outgoing thermal radiation in the form of infrared, you find that you need to add energy generated in the Earth's interior to make up the balance. This energy comes from the radioactive decay of naturally existing radioactive material

"If we use the Sun which is doing fusion every moment. The energy should be transformed into heat energy. then during the travel to the earth, where is the lost energy? " The energy is conserved on its journey from the sun to the Earth. It is just that most of it, continues to travel to the other planets and to deep space. Only a small fraction based on the angular size that the Earth would be in a hypothetical sky from the sun, actually makes it to Earth. But everything emitted in that angular fraction of the sun's surface, does indeed make it to Earth. You can calculate the extraterrestrial Watts/meter^2 that you expect at Earth's orbit, based on the inverse square law, and comparing it to measured data, it is indeed consistent with the sun's total luminosity and rate of nuclear fusion in its core.

Your pretense that it started with zero energy when it was infinitely far away from the planet, is incorrect. We assign infinitely far away to have zero potential energy, but that is nothing more than a mathematical convenience. We don't know where true zero gravitational potential energy really would occur, and it is impractical to calculate, so we pick infinitely far away because that is what makes the math as simple as can be. It comes from integrating G*M*m/r^2 relative to r, which produces a +C arbitrary constant of integration. Since most of the time, we are only interested in differences in potential energy, rather than absolute potential energy (wich doesn't really exist), we keep it simple and assign C=0 when defining potential energy as it derives from N's law of gravitation.

@@carultch Nice Logical Answer but .. Both Situation ( Whether object are inside of the planet's finite Gravity field or Field with Infinite range from highest to lowest strength . ) Seems Violate the Conservation . Take Hypothetical example for Gravity as finite strength - Suppose Sun has Gravitational Field of 10AU unit and its very sharp boundary of it's G - field and there is an object with 1 kg mass , just meter away from sun's G- Field and it just need very little energy to push that object in side that sun's Gravity region ( 10AD assumed ) but because of Sun's gravitational acceleration that little triggering energy will end up millions of times stronger when that 1KG object will hit the sun . It seem Funny thought experiment but was just for saying - it doesn't matter if Gravitational field slowly fade out over distance or never ends towards infinity .

@@omsingharjit There are planets that are farther than 10 AU from the sun in the solar system, so your premise is already false. 10 AU is about the orbital position of Saturn. You have to go far beyond the edge of our galaxy to find a hard-stop to the sun's gravitational field. It does exist eventually, just not at the scale of 10 AU. Eventually, the expansion of the universe gets in the way, and causes space to be created between us and distant galaxies. Where enough space is being created, the gravitational waves cannot propagate fast enough to get to the distant galaxies, and such distant galaxies are no longer in gravitational contact with our own sun and our own galaxy. It is lucky that we are alive at a time when we can discover this, because other than the select few local galaxies like Andromeda and the Magellanic Clouds, most galaxies are far enough away from us that they are out of gravitational contact, and will perpetually move away. Eventually, they will move farther away than the observable universe, and no receding galaxies will be visible to us as evidence of the universal law of gravitation. How we account for the energy of universe expansion is still a mystery, and is a concept that cosmologists call dark energy.

The reason its sign changes, is that you assign a sign convention and coordinate system as you see fit to solve the problem. In any event, the gravitational field g acts downward, and your assignment of signs needs to account for this, as the downward direction fits your coordinate system. Sometimes it is more convenient to assign positive to the downward direction, other times it is more convenient to assign positive to the upward direction.

I guess you have to assume that the initial velocity is 0

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why?

@@hajimajinja4941 Because energy is conserved. There is no energy added to the swinging pendulum, when it fell from the person's face, to the opposite end of the room, and back to the person's face. The pendulum can go no higher, than the point at which it started. Time would need to run backwards, for the air drag and friction to do positive work on the swinging pendulum, which would speed it up. As crazy as this sounds, it is perfectly OK with the first law of thermodynamics. It is the second law of thermodynamics, that makes forces like friction and air drag slow objects down instead of speed them up.

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9.8 meters per second per second is the acceleration due to gravity on earth

Free fall pendulum is an oxymoron. The bob is still supported by the constraint force in the rod and pivot. There is no closed form equation for high amplitude pendulums, where you can't approximate sin(theta) as theta. You can solve it with an infinite series if you need to model it, but you cannot get an exact equation in closed form for the period. It also no longer undergoes simple harmonic motion, when you can no longer take credit for the approximation. Hyperphysics has a large amplitude pendulum period calculator, that enables you to see this effect.

ooohhh you mean like will the metal ball have its rate of free fall affected by the magnetic field of the earth? very interesting question. i would be confident that any impact would be negligible and we could not detect any difference. however the more fundamental question of whether there would be any difference whatsoever, even if not measurable, is pretty interesting, and i'm not sure of the answer.

my guess is that in order to stop the current from inducing there would be a decrease in its speed, so that energy remains conserved. im not that sure though since current itself gets induced in accordance with energy conservation (lenz's law).

Yes