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The earth also gains the positive work the same way the rough floor does, only it's not apparent at first. To understand how, let's go to the microscopic level to see what actually happens when you rub an object against a rough surface. You have the molecules and atoms of the object above, and those of the surface below as in the image shown. The ...

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(Summary: In this post I argue that you need at least an energy of $m v_1^2(1-\cos(\theta))$ in the idealized kinematics situation to deflect an asteroid of mass $m$ and velocity $v_1$ by an angle $\theta$ using rockets, without changing the magnitude of $v_1$.) Energy isn't the be-all and end-all of motion. The problem is that momentum also has to be ...

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Newton's 3rd law is nothing but the conservation of momentum. Recall that force can be expressed as the time rate change of momentum, $$\vec{F} = \frac{d\vec{p}}{dt}$$ In order for the conservation of energy to hold, when Object 1 pushes against Object 2, $$m_1d\vec{v}_1 = -m_2d\vec{v}_2$$ Since the mass does not change with time (conservation of mass), ...

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Why do we say the centripetal force does not make any work on a circular uniform motion? In your case, the rocket does not do any work...on the asteroid. As others have noted, no such guarantee is provided for the accelerated propellant spewing out of the back end of the rocket. It is possible to apply a centripetal force to an object without ...

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My Interpretation of your question : Why don't agents of the conservative force gain internal energy (i.e. heat) as in the case of friction, instead of gaining just potential energy. In any Conservative force, Mechanical Energy is conserved. $$\therefore~~ E_{k~(\text{initial})} + E_{p~(\text{initial})} = E_{k~(\text{final})} + ... 0 As long as the asteroid stays in a circle and has a constant speed, the centripetal forces does not do any work. But if you increase the centripetal force, the asteroid will no longer stays in the circle, in fact it will fall. This is equivalent to say that the asteroid develops a velocity in the direction of the centripetal force. So in this case, you do do ... 3 Fat can't just completely be changed to energy, there need to be chemical reactions in the body as the body metabolizes the fat. Some energy in released by the metabolism. By 1933, Tainter and Cutting discovered that dinitrophenol causes cells in the body to waste energy. So, yes, the energy of fat can be extracted without mechanical excercise. ... 1 Your finite square well potential looks like: where V_0 is the potential energy outside the well and V_1 is the potential inside the well. The depth of the well is \Delta V = V_0 - V_1. We normally take V_0 to be zero, in which case V_1 is negative (like your -10eV) and \Delta V = V_1. However you can add any constant value to the potential ... -2 The reason you feel 'stressed' when you grip hard on something solid like say a metal ball is because your hand exerts a lot of force (compared to when you crush a paper ball) to try and crush the metal ball. However by Newton's third law, the metal ball also exerts the same force on your skin (and hand) as you do on the ball. This large force is what causes ... 1 No, gripping something with your hands will make you feel tired because that is how muscles work: blood is pumping, cells are working... But imagin the gripped object on a table, and on it a pile of heavy books. The effect on the object is the same (more or less) than when you grip it with your hand, but the books never get tired. You could harvers energy ... 3 Have a look at the answers to Why does holding something up cost energy while no work is being done?. Gripping things takes energy not because a constant, stationary force does any work but because of the way muscles work. A stationary force doesn't do any work so no energy can be harvested from it. The best you could do is capture the heat given off from ... 0 Fluids is hard. There are actually 2 processes slowing you down: drag and direct friction. In drag, when we push air out of the way, the surface of the parachute applies a positive pressure on the air below, and a negative pressure on the air above. These two pressures create a net upward force, but they also create air currents as the air moves out of the ... 1 I reminded you elsewhere that you know very well when 3rd law is applicable, and that you even teach it to others for example here Newton's third law of motion is not applied on a single body. - user36790 but then, inexplicably, you forget to use it properly in your own posts: Let a body move and a conservative force oppose its motion. ... 1 The problem with your solution is that the inelastic collision and assumption that kinetic energy is conserved are mutually exclusive. You can see that in your math when you try to solve for v_2. Rewriting equation (1) gives v_1=\left(m+M\right)v_2/m which inserted into (2) yields$$ m \left(\frac{m+M}mv_2\right)^2 = \left(m+M\right)v_2^2.$$This ... 0 though momentum as well as energy is conserved but definitely the sum of individual momentum of particles is not equal to sum of individual K.E. of the particles. also there may be different value of K.E. for same momentum. So can not make any result by manipulating the equations. 0 Both energy and momentum are conserved as always. But to understand why this statement is true you have to look at the system as you described it a little more closely: In order for blocks A and B to stick together after the collision, the force between them should be zero when the velocity difference is zero -otherwise that force would continue to ... 5 To add to HDE 226868's correct answer "heat": Even the ram drag component, arising when the parachute losslessly exchanges momentum with the relatively moving air and thus feels a Ram Pressure (see Wikipedia article of this name, and also my footnote), ends up as heat because the air eddies and currents airising from ram effect (see my footnote) then ... 5 Heat Drag is the same thing as air resistance. It's a form of friction. Friction turns some of the kinetic energy of a moving object into heat; drag does the same thing, thus slowing a falling object down. When an object slows down due to friction, it heats up (and some of the heat dissipates to the surroundings). The same principle applies here: There will ... 2 Let's make a concrete example with numbers: Suppose that v_a = 6m/s and v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2 . According to the conservation of energy and momentum: Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an ... 0 The thing to keep in mind is that for inelastic collisions, energy is not conserved within the system. Some energy is lost as thermal energy, and after all, it takes some energy to get the blocks to stick together. This doesn't mean that the Law of Conservation of Energy is false, though. It just means that energy has left the system that you're studying ... 1 In the 1d particle in the box the energy of the particle should be completely determined by the momentum of the particle that you observe correct? The Hamiltonian in the position basis is$$\hat H = \begin{cases}-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}, & 0\lt x \lt a\\\infty, & \text{otherwise} \end{cases}and the energy ... 9 The source of gravity is not mass, but stress-energy-momentum, so you are correct that the energy converted in this process already has gravity and that that gravity is only rearranged The change in the gravitational field needs time to propagate, though, and this does indeed happen at the speed of light. 0 Why does torque, which is an analogy of force have the same units as energy, but force does not? The magnitude of (net) force is, in a sense, work done per unit displacement thus we can think of it as, e.g., Joules per meter. In a similar sense, the magnitude of (net) torque is work done per unit angular displacement, e.g., Joules per radian. But ... 1 This is a side-effect of treating angles as dimension-less. For translational systems, we have \begin{align*} [\text{linear momentum}] &= [\text{action}][\text{length}]^{-1} \\ [\text{force}] &= [\text{linear momentum}][\text{time}]^{-1} \\&= [\text{energy}][\text{length}]^{-1} \end{align*} Correspondingly, for rotational systems, we have ... 0 Torque is a cross product, and work is a dot product. So one big difference is that torque is a vector and work is a scalar. Another way to think about it is that work is a force being applied over a length interval, where only the force applied in the direction parallel to the displacement counts toward the work performed. On the other hand, torque is ... 6 The momentum operator P in the infinite well can be defined as a self-adjoint operator by infinitely many ways with respect to the boundary conditions by:P_\theta=-i\hbar\frac{d}{dx}\\ \mathcal{D}(P_\theta)=\left\{\psi\in \mathcal{H}^1[0,a]:\psi(a)=e^{i\theta}\psi(0)\right\},where \mathcal{H}^1[0,a] is the Sobolev space, on the interval [0,a]. In ... 1 Is it possible for electrons to carry more than one charge? If by, one charge, you mean more electric charge than (the negative of) the elementary charge, the answer is no. More specifically, an 'electron' would not be an electron if its charge were not -e. However, electric charge is not the only type of charge electrons 'carry'. But this is a ... 2 Each electron has a fixed charge of -1.602\times10^{-19}\,\mbox{C} (where (C stands for coulombs). If you gather 6.24\times10^{18} electrons, the total charge will be \begin{align*} \mbox{Total charge} &= \mbox{Charge per electron}\times\mbox{Number of electrons}\\ &= ... 0 A conservative force only returns the energy back when the object moves in a closed path, that is, it returns to the initial position (it doesn't matter if he returns due to other forces). This can be demonstrated as a theorem, but the intuitive explanation is that a conservative force depends only on the spatial coordinates, and not in the direction of ... 1 But in the real world,instead of storing that work, the agent of the conservative force returns it to the body itself which will be stored as potential energy. I don't believe this is the case. For example, consider the simple mass-spring system. When the spring is compressed and the mass is motionless, all of the energy of the mass-spring system ... 0 Almost all electrical machines can be run in both ways (generator or motor). If you're talking about the direction of rotation, it will work too. However, your message is so vague, we can't help you : we don't have any clue about the kind of machine, about the frequency, voltage, use... 1 There are several (equivalent) ways to look at this. One is to say that for any conservative force \mathbf{F}, one can define the potential energy Ep as an associated potential field such as \mathbf{F}=-\frac{\partial Ep}{\partial r}, or maybe more formally \mathbf{F}=-\nabla(Ep). That's no more than a definition of the potential energy. ... 1 I think there is some confusion about the terms and the relationship between conservative forces and energy conserving systems. I will make a few general statements: 1) A conservative force can be described by a potential function \Phi(x) with \vec F=-\nabla{\Phi}, a non-conservative force can not be reduced to such a function. 2) Friction is a ... 0 Regarding scenario 1: we can simply send another truck with the same mass and velocity in the opposite direction to collide with the first one, and both of them will stop because of the conservation of momentum. If one assumes a totally inelastic collision then, by conservation of momentum, it is true that both trucks (objects) stop. This requires ... 0 The motorcicle will reduce the speed of the truck a little bit, because of the work made by the tension of the rope. The motorcicle will also slow down and actually reverse direction until it is the same than that of the truck. At that point they all will keep moving at the same speed without interacting any longer (the rope tension is zero). Momentum is ... 0 A momentum-based analysis is the way to go for the motorcycle-rope-truck scenario. In your kinetic energy argument, you are assuming that kinetic energies add like vectors. This is not the case. If you want to properly apply a kinetic-energy-work argument, you need to think about the force F that the rope exerts on the truck and the distance d over ... 0 The truck keeps going. Assuming the rope does not break, then the kinetic energy ends up in elastic potential energy in the rope. The rope will stretch. If you ask what happens if you assume a rope that cannot stretch, bzzt. No such thing. The rope will either stretch or break. -2 electric field strength isE=\frac Fq=\frac Vdwith V=voltage, d=distance between charged plates \begin{align} \frac Fq&=\frac Vd \\ Fd&=qV \end{align} but Fd=energy\therefore {\rm energy}=qV

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Apply the work energy theorem. viz. $\Delta K +\Delta V = \Delta W$ where $\Delta K = K_f - K_i$ etc. Now we have $\Delta W = 0 ,V_f = 0, V_i >0, K_i = 0$. Hence it follows that $K_f = V_i (>0).$

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An initially ice-skater pushes away from a railing and then slides over the ice. Her kinetic energy increases because of an external force F⃗ on her from the rail . However, that force does not transfer energy from the rail to her. Thus, the force does no work on her. This is simply confused: This example is nothing but an elastic collision ...

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I think this is a good example, and should be studied and understood carefully. But I don't know where the book goes after making it. Whether or not it is poorly stated depends on the surrounding context. It focuses attention on two things: 1.) the skater is a deformable body. The center of mass is not fixed in the body. 2.) work is defined as a ...

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I agree with CuriousOne that the example is more confusing than helpful, but this is the way I would explain it. Suppose you take a spring, place it on the ground then compress it. If you now suddenly let go of the spring it will rebound and bounce upwards off the ground: The spring clearly has work done on it because its kinetic energy increases and ...

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Momentum: p=mv Kinetic energy: E(k) = 1/2mv^2 So the answer is both. Kinetic energy and momentum are both functions of mass and velocity, and so functions of each other. One goes up, the other goes up: from momentum: v = p/m plugging back into Kinetic energy: E(k) = 1/2m[p/m]^2

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Now if they are both shot at ballistic gelatin, which one is expected to cause more damage if both are stopped by the gelatin ? By damage here I mean more penetration, bigger cavity, heat and any other deformation of the gelatin. In other words, if the velocity of a projectile is doubled, will the amount of damage it causes when it collides with ...

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At this point, we simply need to remember Newton's 1st law. For every force, there is an equal and opposite force. First off, make no mistake, the railing did provide a force for her. The railing pushed her at exactly the same amount that she pushed it. However, since it is cemented into the ground, the force was not enough to move it whatsoever. Work is ...

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I think what you are missing is that these energies are eigenvalues of the time-independent Hamiltonian. i.e. They correspond to stationary states that do not change in time. The scenario you describe is not time-independent - therefore the difference between the energy levels will carry some uncertainty corresponding to the lifetime of the excited state.

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There was a time, when trains worked on coal. What sort of energy, do you think, it was? - Thermal. There's also solar energy, which is, if you think thoroughly, thermal energy. You turn heat (received by a solar panel) into an usable energy source. However, as I know, solar energy produces so little electricity, it's not worth all the effort. The way, I ...

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To turn thermal energy into useful work completely one would need a thermal bath at the temperature of absolute zero. This is explicitly forbidden by the third law of thermodynamics. The best one can do is given by the efficiency of the (theoretical) Carnot cycle: http://en.wikipedia.org/wiki/Carnot_cycle. Th efficiency of the Carnot cycle only depends on ...

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It is possible up to the Carnot limit, which is never 100%. In particular, the Carnot limit is higher if the temperature before extracting is higher. But you could only hit 100% if the temperature could reach infinity, which it can't.

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The total energy of the universe is a vexed issue since different commentators have different views about what the concept means. See the question Total energy of the Universe for a sampling of the various viewpoints. If you Google for zero energy universe you'll find several papers purporting to show that the total energy is zero. However since their ...

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