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There is no friction and no inelastic collision so, yes, energy is conserved. The energy of the compressed spring is $\frac 1 2 kd^2$ so conservation of energy gives you: $\frac 1 2 mv_m^2 + \frac 1 2 Mv_p^2 = \frac 1 2 kd^2$ and conservation of momentum gives you: $mv_m + Mv_p = 0$

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You are assuming, simultaneously, two things that cannot be true at once: Both stars convert the same amount of energy from thermal energy per unit time, and Both stars emit the same amount of energy per unit time as electromagnetic radiation. If you want the same thermal energy conversion rate, there will be different amounts of energy in the ...

2

If you lift one ton one inch the energy you need is $U=mgh$ = 1 ton * $g$ * 1 inch, where of course $g$ is the gravity. This force has to be balanced by the work $L$ done by the force $F$ pushing the tip and that is $L=$F*1260inches. We can "convert" the force in weight by dividng it by $g$, i.e. $W=F/g$ ($W$ being the weight "equivalent" to the force $F$ in ...

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$\sqrt {gR}$ is the speed at the top of the loop, not at the bottom. It is found by setting the centripetal force equal to the weight, $mg$, added to the reaction force, $N$, at the top. For the minimum speed, $N=0$. $$\frac{mv^2}{R}=mg+N=mg$$ This gives $v=\sqrt{gR}$. To find the speed at the bottom of the loop, $u$, use conservation of energy. $$\frac{1}{... 2 In the Lagrangian formalism, the Lagrangian is a function L(x, \dot{x}, t). The notation \frac{\partial L}{\partial t} means nothing but "the partial derivative of L with respect to its third argument". The partial derivative notation always means differentiating a function with respect to a certain "argument slot", regardless of what you put in the slot ... 4 Explicit dependence would mean \partial_tL\ne0. Note that \partial_t\dot{x}=0. 1 If a mirror reflection affected the energy of the photons to a large extent the colors would change, and it would not be a "true mirror" The fact that the colors do not change for a "true" mirror , means that the interaction of the photons is elastic, i.e no energy is lost in our frame, the lab, reference. Elastic scattering keeps the energy of the photon ... 1 Bandgap insulators have colors that are determined mostly by one parameter, the size of the band gap between occupied and unoccupied states. Molecules are different. Pigments have electronic transitions between different levels, and there can be different pairs of levels with strong absorption. There are simple inorganic pigments, like transition metals or ... 1 Your confusion might stem from the way operational amplifiers and the like are depicted in schematics - the power supply is often omitted. And there always is a power supply, unless you're dealing with a transformer, which amplifies the current (or the voltage) at the cost of the voltage (or the current), thus keeping the power constant. 5 The energy comes from the power supply. If you have a guitar amplifier and you feed in a few milli-watts from the pick-up, you'll get a many tens of watts to the speaker. The extra power came from the power cable that plugs into the mains supply. If you want to check, try unplugging it. 2 So will it not make the power at the secondary side more that the power No. A transformer cannot generate power out of thin air and so the power on both sides is (roughly) the same. That means if the voltage on the secondary side is higher then the secondary current is actually lower than the primary. Let's look at simple example. Let's say you have a ... 1 You don't seem to be considering the currents in the primary and secondary. If you connect a resistive 'load', for example an old-fashioned filament lamp, across the secondary, there will be a current through it. There will then also be a (larger) current through the primary in a step-up transformer. To a fair approximation, the mean power supplied to the ... 1 But to have a constant energy over a diminishing mass requires that r diminishes, so I don’t understand how this argument is used to justify an increasing radius. Mechanical energy and angular momentum cannot both be conserved in a variable mass gravitational system. If energy is conserved then \frac{\Delta r}r = \frac{\Delta M}M and angular momentum must ... 3 It is harder to debunk the argument about energy than to explain the phenomenon in terms of forces. As the Sun’s pull (g) decreases, the Earth (in roughly circular orbit at a given radius) find itself going a trifle too fast for its centrifugal acceleration to balance g. The effect on the orbit is the same as if the Sun’s mass had remained constant but the ... 0 Consider the kinetic energy of a system of particles with respect to a given inertial frame of reference, it will be kinetic energy of centre of mass of system with respect to the frame + kinetic energy of system of particles with respect to the centre of mass. I think that given this clue, you can try to prove it yourselves. 0 Hints: (1) For both questions use the work energy theorem which states the net work done on an "object" equals its change in kinetic energy. For the first question the "object" is the two block system. For the second the "object" is the two block-Earth system. (2) Regarding your first thought the work done by gravity is negative when an object is being ... 1 It is the gravitational potential energy of the water and of the dominos. When siphoning water, once started, it flows from a higher point to a lower point. The weight of the water in the tube, below the water level, will pull water up over the edge of the cup. In the case of the dominos, when you stand them up you are adding gravitational potential to them. ... 5 The shape of the track is identical to the graph of the marble's potential energy vs. distance. Flipping the graph upside down shows you a graph of the marble's kinetic energy by distance, because a loss of PE means a gain in KE. If you put the two kinetic energy graphs on top of one another, you can see that graph Y has higher kinetic energy than graph X at ... -1 You are right in remarking that they will have the same velocity at the end of the track. But Y will start losing potential energy, and converting it into kinetic energy, much earlier than X. So it will soon start covering a lot of ground, while X moves much more slowly. So it will reach the end of the track much earlier. Sure, towards the end, it will not ... 0 Conservation of energy law applies to a system, not to one component of it. So in this case it applies to hammer+nail system, which can be expressed in terms of momentum too :$$ p_{h0} + p_{n0} = p_{h1} + p_{n1} $$where subscript 0,1 indicates body momentum before and after an event has occurred (hammer impact this time) Or to be expressed in general, ... 0 Sound is just pressure waves propagating back and forth, so kinetic energy, when it gets absorbed by an object it just jiggles its constituent particles, so basically it stays kinetic energy; macroscopically you could say it turns into thermal energy since the thermal energy of an object is the total kinetic energy of its constituent atoms/molecules. 1 if we don't accelerate the body how can we raise the body to height of ℎ in the example? You can either assume that you accelerate it from rest by a very small amount (so small we can ignore it), or you can assume that it magically begins and ends with some constant speed v, so no acceleration is required. but this time we have a problem with W(byus) =... 0 Let the potential energy if the system be zero at infinity then gravitational potential energy of the system is given as$$E=\frac{-GMm}{r}$$You have missed the negative sign in your answer which accounts for the error.When the system is released from rest when it is at a separation of 2r then kinetic energy of the system increases which comes at the cost ... 3 You're missing a negative sign. The gravitational potential energy U of two masses m and M separated by a distance r is given by$$U=-G\frac{mM}{r}$$As the objects get closer together, the gravitational potential energy becomes more negative, which is another way to say that it decreases. As the potential energy decreases, the kinetic energy ... 1 You have a sign error. The potential energy for gravitation is lower the closer two objects are to each other, not higher. The potential energy of two objects separated by r is$$U = U_0 -\frac{GM_1 M_2}{r}$$where U_0 is the potential energy at infinite separation. U_0 is conventionally taken to be zero, but it doesn't really matter what value it ... 0 If a force is provided to lift a system (let's say a box) vertically upwards and there is no resistive force, from my understanding (though I may be misguided),it is said that the work done on the box is equal to the gravitational potential energy that it has gained. That is correct provided the velocity of the box is zero at a given height so that it ... 1 You are getting mixed up between work done by a single force and the net work done. The net (total) work done on an object is equal to it's change in kinetic energy$$W_\text{net}=\Delta K$$However, it is most likely that in your education of the box lifting example they are talking about the work just done by the lifting force as well as applying the ... 3 The conservation of the four-momentum by itself implies the conservation of energy, the conservation of 3 momentum, and the conservation of invariant mass. 0 The equations for plastic collision with a coefficient of restitution \epsilon are$$\begin{aligned} v_1^f & = \frac{m_1 - \epsilon m_2}{m_1+m_2}v_1 + \frac{\epsilon m_2+m_2}{m_1+m_2} v_2 \\ v_2^f & = \frac{\epsilon m_1+m_1}{m_1+m_2} v_1 + \frac{m_2 - \epsilon m_1}{m_1+m_2} v_2 \end{aligned} $$1 The frequency of the reflected light will be a tiny bit smaller than the incident light. So the energy (and also the momentum) gained by the recoiling mirror is taken from the light. This can be understood by considering the reflection as a collision between the photons and the mirror. During reflection the photon's energy E=h\nu gets a little bit ... 1 I suspect the discrepancy arises because the collision was not elastic. 2 No, this is not always true. According to conservation of energy, the change in total mechanical energy of a system is determined by the work done by forces external to the system:$$\Delta E=\Delta K+\Delta U=W_\text{ext} What you are proposing is true if $W_\text{ext}=0$, because then as $\Delta K\geq0$ it must be that $\Delta U\leq0$. Since we started ...

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If only internal forces are doing work (no work done by external forces), then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. In these situations, the sum of the kinetic and potential energy is everywhere the same.

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The recoil energy is extremely small, but you are correct the in principle it is not exactly zero. The energy of the photon (eVs) is many orders of magnitude smaller than that of the slit system ($10^{26}$ eVs) .

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It is all quantum mechanic, i.e the Heisneberg uncertainty principle (HUP) is within the framework of quantum mechanics, not of classical particles running along on straight tracks. What does it mean for a single photon? It means if one measures the $x$ coordinate very accurately, the $p$ component uncertainty is constrained by the HUP. $ΔxΔp>h/{2π}$ ...

3

Momentum is a vector. That vector is changed, not its magnitude. Otherwise we'd see a slit changing the wavelength of the photons. The change does cause a force on the barrier structure, but the barrier is not moving. If the barrier moved, there would be an energy transfer between photon and barrier, and the photon wavelength would change. BTW, an ...

1

Both approaches are actually the same, if you do them correctly. I will address your second case first. You are correct to use conservation of energy and say that the potential energy stored in the spring at the lowest point is equal to the sum of the kinetic energy and the potential energy due to gravity at the equilibrium point. So you were correct with ...

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Neither energy nor forces are something we know to exist. And neither of them are really "fundamental". We often want to predict what will happen when we release a roller-coaster cart down a track, or throw a satellite at Jupiter, or any of a number of other things we might do. Presumably we cannot ever know how Mother Nature truly functions (if that even ...

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Let me discuss it for classical mechanics While Work-Energy mechanics is very powerful, it also has an achilles heel. A necessary condition for Work-Energy mechanics is that the system is such that an unambiguous potential energy can be defined. You'd think that you would have to invent a very contrived case to run into that problem. But here is the ...

3

This is called a super-elastic collision. It is not physically impossible - in a sense, it's just the reverse of an inelastic collision, and by following that reasoning, you can then easily come to understand how it's possible. Here's how it goes. In an inelastic collision, the energy after collision is lower than the energy before. You still have a ...

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Intuitively, is it right to think of Energy as a more fundamental quantity than Force? The “fundamental-ness” of something is difficult to quantify so it is hard to say whether something is more or less fundamental. However, I would tend to agree with you, but for different reasons. There are two basic approaches to classical mechanics. The Newtonian ...

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No, it is false. You have this impression just because you are considering one-dimensional problems where the forces are conservative or they do not produce work. Newton's equation of motion, in this situation, is essentially equivalent to the theorem of mechanical energy conservation. If there are non-conservative forces and/or the motion involves more ...

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I'm not sure why answers here only discuss changing reference frames. You can operate in the same reference frame and still have an increase in kinetic energy. For example, if one object has a compressed spring attached to it that is set to release upon collision. Then the extra energy comes from what was the potential energy. All you need to do is apply ...

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I have written this answer to point out that the increase in kinetic energy has nothing to do with changing the reference frame and answers which hint at this being true are wrong. Although individual particles might appear to have a different kinetic energy in another reference frame the total kinetic energy of a system under consideration does not change. ...

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Yes final kinetic energy may be greater/less than the initial if you view from a different frame of reference as it is a frame dependent physical quantity.In this problem it is clearly mentioned that it is asking for relative velocity between particles.

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Macroscopic kinetic energy is reference frame dependent. For example a car with velocity $v$ with respect to the reference frame of the road has kinetic energy of $\frac{mv^2}{2}$ with respect to a person standing on the road, but it has zero kinetic energy in the reference frame of the driver. Note that the problem asks for the RELATIVE final velocity ...

2

According to the Work-Energy Theorem, the net work done to some object is equal to the difference in its kinetic energy, this is, \begin{align} W_{net}=K_{final}-K_{initial}. \end{align} In order to understand what you mean by "independent of velocity direction", suppose we have an object at rest, and you can move it either to the right or to the left. ...

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In general relativity there is (in general) only local conservation of energy, global conservation of energy does not exist the way it does in the rest of the physics. That is due to the difficulties in defining gravitational energy, which cannot really be done "correctly". However you can define gravitational energy "uncorrectly" (f.e. using Landau-...

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FWIW, the total potential energy is $U^{\prime}(r)=U(r)+U_{\rm cf}(r)$, where $U_{\rm cf}(r)=\frac{\ell_0^2}{2mr^2}$ is the centrifugal potential. The region with $E<U^{\prime}(r)$ is classically forbidden as it would require the radial kinetic energy to be negative, which is classically impossible.

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By setting $E=0$ you are saying that the object's speed is equal to escape velocity $v_e(r)=\sqrt{\frac{2GM}{r}}$. By fixing angular momemtum $l_0$ you are saying that the angular speed of the object is $v_{\theta}(r)=\frac{l_0}{mr}$. But if $r \lt \frac{l_0^2}{2GMm^2}$ then $\frac{l_0^2}{m^2r^2} > \frac{2GM}{r}$ so $v_\theta(r) > v_e(r)$. All this is ...

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