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If you suppose that the scale works like a spring, which seems reasonable, then during standard use, the displacement $x$ of the scale is proportional to the mass $m$. The equilibrium relation is $$mg=kx,\tag{1}$$ where $k$ is the stiffness of the spring. Assuming that, when you dropped a mass $M$ from a height $h$, all kinetic energy (which is equal to ...

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As the comments to the questions state, this is a question on the research state about the generation of the universe, and the first moment is modeled in the Big Bang model. The real beginning point is not yet known even in this model since gravity has not been consistently quantized within the model, only effective theory is used. Nevertheless existing ...

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This is actually the paradox that led Einstein to the equivalence of mass and energy, and to General Relativity. Consider a special case: An electron and positron are at the Earth's surface. Bring them together and they annihilate, creating gamma rays (which is very energetic light). The gamma rays travel up to the Space Station, where they are converted ...

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Your guess at the solution to this paradox is correct. "Pumping energy up" to the space station, regardless of the method you choose, would require an input of at least the amount of energy you would gain in kinetic energy on the way down. This is just a variation on the impossible perpetual motion machine concept. In practice, you would not only not gain ...

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You are right about how in the first scenario the energy you put into the magnets results in kinetic energy, but the second scenario is incorrect. Saying that the magnets are using energy to repel each other is like saying that the chair I'm sitting in is using energy to keep me from sinking into the ground. I hope this clarifies your understanding.

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The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...

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When someone moves in a field then the work generated by field forces is independent of motion and related to the starting and ending positions. The quantity $U$ $$\int_{r_1}^{r_2} \mathbf F( \mathbf r) \cdot d \mathbf r = [U ( \mathbf r)]^{r_2}_{r_1}$$ is called potential energy and is associated with a potential $V$.By work energy thm we get $$dK = W = ... 0 mechanical energy conservation is sum of all conservatives energy such that there is no external forces like frictional or an explosion. basically we mainly deal with Blockquote GRAVITATIONAL energy and KINETIC energy Blockquote . mgh + 1/2m sq.(v) = constant when no other external forces like i perviously mentioned do not act. if they do so ... 0 Let's look at the example of roller coaster. At a given moment, 2 forces are acting on the train: Gravity: towards -z direction. Normal force: direction normal to the rail. The normal force does no work, since work is given by dW=\vec{F}\cdot d\vec{x}, and the normal force is always normal to the direction of the rail, thus normal to the moving ... 2 For example, given the initial peak height of the roller coaster, I can predict the velocity at any point, despite the fact that there are various loops and curves It is very simple: if you know the height h of a body you know its potential energy which is mgh. This energy is given to the body (transformed into Kinetic energy) at every lower ... 0 One thing that might help you address your question is that conservation of energy is deeply connected with the idea of path independence. If you have a force field--say, gravity--it can be "conservative" (conserves energy) or "non-conservative" (...guess). An example of a force that is not conservative is friction, because a body experiencing friction loses ... 0 The so called Pulse and Glide is a driving technique that is reported to save quite much fuel. It consists of moderate accelerations (that better exploit the petrol engine) interleaved with moderate decelerations (that save fuel). This is even more effective with hybrid vehicles which can completely turn off the petrol engine without issues like the absence ... 0 In principle, neglecting rolling friction and air friction, and neglecting the fact that the hilly road is longer, it should make no difference. On the other hand, if on the hilly road you ever have to either hit the brakes, or use engine braking, because of too much speed, that will cost a lot of energy. Also on the hilly road, if you have to downshift ... 2 A nice answer to this question may be found in the 2009 MIT Course on the Physics of Energy (Lecture 3) by R. Jaffe and W. Taylor. They work out the energy balance of car transport as an example of mechanical energy and its conservation. Let me briefly recap their findings in my own words. The two "towns" they consider are Boston and New York, both assumed ... 9 ...two roads of the same length. One road is flat, the other road goes up and down some hills. Will an automobile always get the best mileage driving between the two towns on the flat road versus the hilly one..? The question two roads (between two towns A, B) with a different profile cannot be of same length The flat road (1) is shorter, ... 13 I believe, the answer is a small but quantifiable, yes, there is a non flat road configuration that would lead to better gas mileage between any two points at the same height. I have numerically solved for such an optimal path. I believe I can give a nice explanation of why that is, but it will take some work, so bear with me. Granted, you can only expect ... 1 You actually don't need to take any time derivatives here. Since the energy, E, is a constant, you only need to know it at one moment in time (say at t=0 as an initial condition), and you know it at all other times. Thus, just solve for \dot{x}(t) in terms of the other quantities (x,E,...) in the second equation that you have to get the equation of ... 1 From a very fundamental point of view one can see this with respect to Noether's theorem. Every symmetry is related to a conservation law, e.g. the symmetry of space with respect to rotational symmetry (space does not change if we rotate our coordinate system) results in conservation of angular momentum. In this framework energy is related to time symmetry. ... 4 First, it's not true that energies are generally in the form \frac{a_1a_2^2}{2}. Take the gravitational potential energy U = \frac{GMm}{r} as an example. However, it is generally true that kinetic energy takes that quadratic form. Why? Kinetic energy is the energy traded when some agent applies a force on some system that causes it (the system) to ... 7 Even if the laser had perfectly reflecting, i.e. lossless, mirrors at either end of the cavity, and both ends were sealed so no light could escape it would still require a continual power input. That's because excited atoms/molecules can decay by mechanisms that don't involve a photon e.g. collisional de-excitation. The lost energy goes into heating up the ... 0 Is there any fundamental reasons seemingly many physical phenomenon has the energy in that form? Yes. Think about a 10kg cannonball in space. If it’s travelling at 10m/s relative to you, you say it's “got” kinetic energy KE=½mv², and you say it's "got" momentum p=mv. Now brace yourself, then apply some constant braking force with your spacesuit jets, and ... -3 These terms of energy are defined so because the expression keeps on popping up in various scenario. Its similar to the reason why Moment of Inertia is defined. 0 Answer #11: You need 1. downhill slope to get the car to an optimal speed - corresponding the highest gear+optimal rpm. 2. At the optimal performance regime at the point of upslope start, you start the motor and keep adding the energy necessary to get to B with v=0, then stop the motor and get to B. Reasons: I assume there is lower efficiency at lower ... 0 When the sloth starts falling, it's potential energy starts getting converted to kinetic energy, and just before it hits the ground, all of its potential energy has been converted to kinetic energy. Now, the problem you're facing is because of the fact that you're considering just the sloth, which is wrong, because the sloth is interacting with the ... 0 There is an external agent that removes mechanical energy from the sloth, namely the normal force exerted by the ground. You are right that no net work is done, but remember it is the work-kinetic-energy theorem: The net work equals the change in kinetic energy of the sloth. 3 Conservation of energy, as you note, holds for "the system." For instance, if you push on a ball, that ball gains energy, but the energy of the ball is not conserved--only the energy of you and the ball. In this case, the system needs to include more than just "the sloth" because the sloth is not an isolated system--there are external forces at work. Here, ... 0 Like you say, it would be a very difficult problem if we had to resort to using Newton's laws to calculate the forces and accelerations at every point along the trajectories. Thankfully, the universe has conservation laws which make this problem immensely easier. Try using the conservation of mechanical energy and the conservation of total momentum. 0 The way I would start the problem is by dividing it up in subproblems: Express the speed v(\theta) of the ball at angle \alpha Calculate the angle \beta (between the rope at angle \alpha and the part of the rope with the ball attached) that the ball will travel in circular motion upwards until the tension drops to 0 and the motion becomes ... 0 I will choose in function of the wind factor. Wind is a positive source of energy if it blows from the rear and... The road thru the hills has the potential to minimize the exposure to the wind if it comes from a frontal direction. In the valley the exposure is minimal. The plain road has the potential to maximize the exposure to the wind if it comes from ... 0 After the elongation of x units in the spring, the spring develops energy in the form of spring compression or elongation energy(the work done in elongation is converted into spring energy), spring tries to get into its mean position. Now, if the system is isolated the spring will perform SHM otherwise friction and other forces will make it stop its motion. ... 1 In an elastic collision involving two objects both momentum and energy are conserved. You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore one object will have the opposite momentum of the other. Since energy is the square of the momentum divided by twice the mass, this means ... 1 Look at the original, Heinrich Hertz, 1889! If E=B=0 on a plane, S=0 => no energy Transport through that plane. I agree with Annix: "The pictures may be depicting a static wave, but certainly, not a propagating wave." 0 Answer to the title of your question : "NO!"... First of all,in general, in most of the mechanics questions, you should make it clear that what is your system and what conservation laws are applicable on it. Like if you have a system on which some constant net external force is acting throughout the time duration you are analyzing the system, you can not ... 0 Ok... I see the confusion... See when you bring a ball from infinity to a position at a distance r from the center of the earth with uniform velocity, you need to apply some external force on it ; otherwise naturally the ball would be accelerated (because of gravitational force of earth ) and the decreased potential energy would appear as the kinetic energy ... 1 I think what may be referred to in the question is the determination of gravitational potential energy (GPE). The definition for GPE is the work that was done in bringing the two masses to a distance r apart from an initial separation that was infinite. (Tsokos, 396) What is important here is that this happens at a very slow uniform speed so the Kinetic ... 1 When a ball of mass m is brought with uniform velocity from infinity into the g field of the earth at a distance r from it, the potential energy of the ball earth system decreases from 0 to -GMm/r. What does this lost energy appear as? Kinetic energy, which is typically radiated away into space. Imagine your ball starts off a long long way from Earth, and ... 0 In an electronic car, as you say, or any other device that uses electrical energy to produce some mechanical work, [the term "uses" means transforms one kind of energy-one kind of microscopical interactions(here electromagnetic)- to another], what happens? You have an electrical energy source that, through a mechanism, gives you mechanical work, so the car ... 1 Your question is similar to the common question of putting a wind turbine on an electric car to generate electricity while driving and provide power, and is based on the same conceptual misunderstanding. To understand why it is not possible you should read about why perpetual motion machines are not practically possible. Specifically relating to your ... 0 First of all, the work needed depends on the way in which the final velocity is reached. If you write the expression for the work done by the drag force you obtain$$W=-\int F ds = -b \int v^3 dt$$and it's easy to see that if the object is carried to its final speed in a very short time W can be arbitrarily small because v in the integrand is always ... 0 Drag force:$$\mathbf{F_{\text{drag}}} = \dfrac{1}{2}\cdot \text{density} \cdot \text{cross sectional area} \cdot \text{drag coefficient} \cdot \text{velocity}^2$$. Now,$$\text{energy lost due to drag} = \mathbf{F_{\text{drag}}} \cdot \text{displacement}$$. It might give the exact or quite straight approximation. :) 4 .... We are told that all the subsequent collisions involving the balls and floor are elastic. We are asked to determine the maximum height to which the small sphere will rise on the rebound. The problem does not mention any radii, but if we did know the radius of each of the spheres, would it be valid to bypass conservation of linear ... 0 Yes - but it needs the right sort of hill. A family member, now sadly passed away, used to conserve fuel during 2nd world war by going into town down a hill with a very long and very gentle incline downwards and turning the engine off - the road is also pretty straight for about 3 or 4 miles. This is hinted at in the answer of Edward. Sorry, this has ... 2 You need to use the equations of motion in the expression for dE/dt. Lets take a simple example. For the Hamiltonian$$\mathcal{H} = \frac{1}{2}\dot{x}^2 + V(x)$$(which is equivalent to the Lagrangian \mathcal{L} = \frac{1}{2}\dot{x}^2 - V(x)) we have that the time-derivative of the energy is$$\frac{d\mathcal{H}}{dt} = \dot{x}\ddot{x} + ...

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Any energy principle is not being violated since the speed of the photon is never less than $c$ and hence the momentum is unchanging (in the classical sense). Why light travels slower than $c$ in a medium is because of the photons being absorbed and reradiated by atoms in the material. In a sense you can make the analogy of light traveling a longer path in ...

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I see the viewpoint you're coming from. From the wording of "potential energy", it sounds like the energy might not exist until it becomes kinetic energy. Potential energy is as real as kinetic energy though. The idea is that the energy gets stored in the fields that produce the force, and that the energy is transferred between kinetic energy and field ...

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Well your question was not perfect, but acceptable. the idea of energy may sound easy, but deeply it is a very strange Idea. but the answer to your question: no it doesn't mean it "potentially" gonna gain. I see that you read Feynman's lectures, that's very good, but these speeches actually are for people who accepted the idea of energy blindly, and didn't ...

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The link Qmechanic has suggested is a duplicate and does discuss the question you ask. However there is another point that is worth making here. In general relativity we describe the universe as a manifold equipped with a metric, and the metric is the FLRW metric that desciribes expanding spacetime. However the FLRW metric does not include the point(s) at ...

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This is simply a statement of conservation of energy. Consider an imaginary surface between plates $k$ and $k+1$. The flux through this surface is: $$J=\sigma T_k^4-\sigma T_{k+1}^4,$$ where the second term is negative because the energy flows in the opposite direction. If the system is in a state of equilibrium, then the temperatures are not changing ...

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There is no net energy being transferred. The energy each wave has is cancelled by the other wave! You might be confusing this with the "two slit" experiment. In this case you do have both, destructive and constructive interference. This is the result of two simultaneous waves with a given spatial separation causing the waves to "cancel" each other at ...

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