Tag Info

5

As a general rule (regardless of the definition of V) we have: $$\frac{\mathrm{d}}{\mathrm{d}t}(\bf{V}\cdot\bf{V})=\frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\cdot \bf{V}+\bf{V} \cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}$$ $$\frac{\mathrm{d}}{\mathrm{d}t}V^2=2\left(\bf{V}\cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\right)$$ ...

4

The first method is giving the correct answer. In writing the work done by the force, you are assuming that the force $F$ itself is constant throughout the extension. However, this is not true. While extending the spring in a quasi-static way, the force $F$ must always match exactly the spring force at that time. This is needed so that at the end of the ...

4

If you use a constant force along the path, the spring will move past the position where $F=kx$, because it will reach that point at some speed. Thus it is incorrect to use the force method in the way you used it, because at maximal extension $v=0$ but $a\neq0$. The energy method as you used it will give the correct answer. If, instead, the force is used to ...

4

The Hamiltonian $H(\theta,p_\theta)$ needs to be formulated in terms of the coordinate $\theta$ and its canonically conjugate momentum $p_\theta = \frac{\partial L}{\partial \dot{\theta}} = R^2 \dot\theta$. The correct expression for the Hamiltonian is \begin{align} H(\theta,p_\theta) & = p_\theta \dot{\theta}(\theta,p_\theta) - ...

3

You can do so. The energy put in is the integral of $VI$, the product of the voltage and current. It may be hard to calculate $I$ from $V$ because of the back emf and changing circuit resistance. The energy absorbed is the increasing magnetic field energy caused by the expansion of the current loop and the increasing kinetic energy of the wire and ...

3

An incompressible liquid is never completely incompressible, more like quasi-incompressible. So when you apply considerable force $F$ on the piston, pressure will wise by say $\Delta p$:: $$\Delta p=\frac{F}{A},$$ where $A$ is the cross-section of the piston (and assuming constant $F$). But the liquid will have decreased slightly in volume by $\Delta V ... 2 The total energy stored in the magnetic field goes down - that's where the energy to move the needle comes from. The "internal energy" of the atoms inside the needle has nothing to do with it. 2 The total energy at the top is $$T = \frac{1}{2} m v^2 + m g L$$ The total energy at some other point is $$B = \frac{1}{2} m f^2 + m g L \cos\theta$$ Energy is conserved so $$f = \sqrt{v^2 + 2 L g (1-\cos\theta) }$$ 2 A physical system in GR is never isolated, in general, as it interacts with the curved metric, i.e., the gravitational background. (However a notion of isolated system can be given in the particular case of an asymptotically flat spacetime as discussed in auxsvr's answer.) Apparently this fact prevents the existence of conserved quantities because the ... 1 Your calculations are wrong, Hint:$\frac{1}{2}I\omega^2 + \frac{1}{2}Mv^2 = Mgd\sin(t)wr=v$1 There are different forms of energy. Energy can be converted from one form to another but cannot be destroyed. In this case the kinetic energy of the hammer is driving the nail into the wood which is breaking the molecular bonds in the wood fiber. The energy is converted to heat energy as a result of the breaking of the bonds and the friction of the nail in ... 1 Imaging the balls on a string. You are launching N balls per second, at a velocity$u$. This means the distance between the balls is$u/N$. And$N$balls per second will pass a certain point in space. Now if the car is moving at a velocity$v$(same direction as$u$), fewer balls per second can hit it - because subsequent balls on the string have further to ... 1 The initial kinetic energy$E_k$gets partly dissipated as friction,$E_f$, and partly converted to gravitational potential energy,$E_g$. The sum of these two must equal the original energy input, so $$E_k = E_f + E_g$$ 1 In the first part you wrote$E_{k}=E_{g}$because kinetic energy is fully converted into potential energy. But in the second part, some of the initial kinetic energy$(E_{f})$lost due to friction and part of energy left is$E_{k}-E_{f}$. Only this part is converted to potential energy$E_{g}$. Thus,$E_{g}=E_{k}-E_{f}$and this simplified as ... 1 No. Look up something called the Carnot efficiency, among many others. There is no extractable heat energy without a temperature difference. For your apparent level of physics, it's probably best to take this as a fundamental principle that just is. 1 The city is not a perpetual motion machine, so it will not run forever. For example, all light will be absorbed by buildings, pavements, citizens, etc, and the resulting heat will not be recoverable. The same applies for any heat source including the metabolisms of occupants and pets. The final result will be a city which becomes slowly warmer, while usable ... 1 I prefer to look at it based on certain Laws and observations. The First and second Law of Thermodynamics are true Energy is conserved and the universe is moving towards increasing entropy. The Hubble constant is accurate and the universe is expanding at an exponential rate given by$a(t) = e^{Ht}$, where the constant$H$is the Hubble expansion rate and ... 1 Hint: Use $$m\ddot{x}=-kx-x^3 \\\ddot{x}=v\frac{dv}{dx} \\-\frac{kx^2}{2}-\frac{ax^4}{4}=\frac{m}{2}\left(\frac{dx}{dt}\right)^2$$ It will reduce to a form $$\frac{dx}{dt}=ix\sqrt{c^2+x^2}$$ This is a standard integral, and can be solved, then use $$U=-\int f(x) dx \\T=\frac{1}{2}m\dot{x}^2$$ Total energy$E=T+U\; .\$

1

For these kinds of system we often define a pair of quantities, one which is characteristic of objects or systems and one which is characteristic of interactions. Examples of these pairs are work (interaction) and energy (system) or impulse (interaction) and momentum (system). There is no commonly applied name for the interaction quantity that pairs with ...

1

Let me discuss a simpler version of your rocket-question: one where there is no gravity, so that we don't have to worry about gravitational potential energy. Consider a rocket in free space (vacuum), and consider that the rocket is at rest. Now the rocket fires it's engine for a short time. The engine accelerates the rocket. The rocket now has kinetic ...

Only top voted, non community-wiki answers of a minimum length are eligible