Tag Info

New answers tagged

1

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: Energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an only ...


2

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 ...


1

If you realize that the actual Hamiltonian is a time-dependent the whole confusion would evaporate.


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}\\ &= ...


1

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 doesnt matter if he return 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 coordinate, and not in the direction of motion ...


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. ...


0

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 is $$E=\frac Fq=\frac Vd$$ with $V$=voltage, $d$=distance between charged plates \begin{align} \frac Fq&=\frac Vd \\ Fd&=qV \end{align} but $Fd$=energy $$\therefore {\rm energy}=qV$$


-2

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 $K_f = V_i (>0).$


1

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 ...


0

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 ...


2

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 ...


0

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


1

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 ...


1

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 ...


1

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.


0

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 ...


2

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 ...


0

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.


3

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 ...


2

The logic is simple: a body loses energy because opposing forces cancels themselves out. if you want to move in a direction +x against an opposing force -x the net balance must be in the direction +x. The logic is the one that rules the composition of forces or vector addition If $F_2 > F_1 = -18 N$ then the body will move, accelerate in the -x ...


1

When two bodies interact, there is a force between them. Positive work is done on the object for which the dot product of force and velocity is positive. It follows that negative work is done on the object for which the dot product is negative. By Newton's law (for each action there is an equal and opposite reaction), the force on one object is the reverse ...


1

If multiple forces act on a body at the same time, you should compute the net force before considering whether it is increasing or decreasing the kinetic energy of the object, and thus whether the work done by the force is positive or negative. So, using your example, the net force is $F = F_1 + F_2$, and since $F_1 > 0$, $F_2 < 0$ and $|F_2|< F_1$ ...


0

If the mass moment of inertia about the center of mass is $I_C$ then the MMOI about the pivot is $I=I_C +m \ell^2$. The kinetic energy of the rigid body is $$ T = \frac{1}{2} m |\vec{v}_C|^2 + \frac{1}{2} I_C |\vec{\omega}|^2 = \frac{1}{2} m (\ell \dot\theta)^2 + \frac{1}{2} I_C \dot{\theta}^2 \\ T =\frac{1}{2} I \dot{\theta}^2 $$ where $\vec{v}_C$ the ...


12

Yes the effect is real, potentially at least, but no it's not measurable. As an aside, the redshift of light by its gravitational interaction with (homogeneous and isotropic) dust is exactly what the FLRW metric predicts, but this clearly isn't what the question means. The gravitational field of a beam of light is calculated in the paper On the ...


0

If you're concerned about Ohm's law, you might be getting a little muddled in your thoughts (don't worry, I only know this because I've had my turn at being muddled before on this point!) Ohm's law doesn't describe a transformer: Ohm's law may apply to whatever load you connect to the transformer; if so, then Ohm's law combines with the transformer law ...


0

Looking at real life jumpers, Kangaroos, the key element seems to be the efficiency with which the jump energy can be recovered upon landing. This is usually done in all animals, including Humans, by storing energy in the connective tendons.


0

What is the kinetic friction between the bigger block and a surface? Well, you can just use the equation $$ F_k = u_kN $$ where u is the coefficient of kinetic friction between block and table, and N is the normal force. Find the frictional force acting on smaller block. Assuming that the smaller block stays on top of the larger block, then it is ...


0

Kinetic friction is when two objects are in relative motion to each other and you have them rubbing against each other; e.g. a rolling ball on the ground slows down due to kinetic friction. Static friction is when you have two objects that are not moving relative to each other; e.g. an object on a inclined plane - the reason that it does not slide down the ...


3

Thus, work is done on the car, right? No, the car does (positive) work on whatever is stopping it. Alternatively, you could say that negative work is done on the car, but still, the meaning is the same: the car loses energy and something else gains that energy. What that something else is, and what type of energy it gains, depends entirely on how the ...


2

If I eliminate the associated Kinetic Energy of the car, where does this energy go? While conservation of energy dictates that the energy due to the car's motion must be conserved, it does not say how. An old-style braking system converts that kinetic energy into heat. A more modern regenerative braking system converts that kinetic energy into a ...


2

If there was no friction, your breaks couldn't clamp down on your rotors to slow the car, and the car's tires couldn't "stick" to the pavement. Your engine is generating energy, and it does cause pressure in the braking system. That braking system, triggered by you and your foot (and assisted by the car) converts that motion to heat through friction between ...


2

To escape Earth, you'd need to be alive for (see this) $$ \frac{1}{365 E} \left(\frac{GMm}{R}\right) \ \mbox{years} \approx 1 \ \mbox{year}, $$ where $E \approx 2000 \times 1000 \times 4 \ J = 8 \times 10^6 \ J$ is the average daily energy intake (couldn't find a good source for a worldwide average), $G \approx 7 \times 10^{-11} \ m^3 kg^{-1} s^{-2}$ is the ...


3

Some quick scribbling suggests that caloric intake for a $70\;{\rm kg}$ human over the course of $50\;{\rm yr}$ at $2500\;{\rm Cal}$ per day is enough to accelerate that person to almost $75,000\;{\rm m}/{\rm s}$. Escape velocity from Earth is about $11,000\;{\rm m}/{\rm s}$, so even with a reasonable hit for inefficiency (about $2\%$ energy efficiency is ...


0

I will try to give you a different perspective. Hopefully that will be of some help to you. There is nothing abstract about an atomic bomb, the light produced by a light bulb, or your ability to move your arms. These examples are tangible proof that energy is real (it exists). The abstraction is in the minds of people, so that they can model its ...


0

Your definition is a bit funky. I'll explain. Explanation As much as I like that you explicitly express internal energy $U$ as a function of state, for clarity I'm going to drop $\boldsymbol{R}$ for now. We can't directly measure the absolute internal energy of a system, we can only infer its change by understanding some defined process. Consider the ...


-1

Depending on how the energy is stored, it is certainly possible to transfer a lot in a short time. Here are several examples of delivering 1000 J in a short time: A bank of capacitors adding up to 200 mF charged to 100 V holds 1000 J. The right type of capacitor is capable of discharging in 1 ms. Since it may take too long to discharge a capcitor all the ...



Top 50 recent answers are included