New answers tagged

2

If you accept that no external work was done, then if there is a change in the state of a system through which the kinetic energy changed, there must be a corresponding change in potential energy. The key to understanding the (rather poorly narrated) video is that the lecturer implies (at T=2:30) that $\Delta E=0$ from which it follows that $\Delta KE= - \...


1

A simple pendulum performs simple harmonic motion when it is displaced very slightly. You can say that a simple pendulum performs a periodic motion which can be treated as simple harmonic motion in small oscillation Now lets move forward supposing it to be pure SHM. The time period of a simple pendulum does not depend on how much it is displaced( but it ...


2

Newton's second law, force f is $$f=m\frac{d^2 x}{d t^2}$$ x is position vector of the particle. $$f=-\frac{d v}{dx}$$v is the potential energy. $$m\frac{d^2 x}{d t^2}=-\frac{d v}{dx}$$ Multiply both sides with $\dot x$ $$\frac{m}{2} \frac{d\dot x^2}{dt}=-\frac{dv}{dt}$$ $$ \frac{d}{dt}(\frac{1}{2}m\dot x^2+v)=0$$ i.e., $$\frac{dE}{dt}=0$$Energy is ...


5

Newton's third law tells us that the momentum imparted on one body is equal and opposite to the momentum imparted on another if they interact. We then have $$ \Delta \vec p_1~=~-\Delta\vec p_2. $$ The change in momentum is $\Delta \vec p_i~=~m\vec a_i\Delta t$, $i~=~1,~2$. The change in momentum is with Newton's second law due to a force so that $$ \vec F_1~...


3

The classical interference pattern is explained by the equations governing the behavior of light, and energy there is treated as a collective phenomenon, using the Pointing vector Energy transfer in a light beam can be best understood as an emergent phenomenon from the underlying quantum mechanical level. Innumerable photons create the visible interference ...


0

$h=R(1-\cos\theta)=R(2\sin^2\frac{\theta}{2})$ $v=\sqrt{2gh}=\sqrt{4gR}\sin\frac{\theta}{2}$ $v$ is a maximum at $\theta=\pi$. The tangential acceleration is $\frac{dv}{dt}=\sqrt{4gR}\cos\frac{\theta}{2}.\frac12\frac{d\theta}{dt}=\sqrt{4gR}\cos\frac{\theta}{2}.\frac12.\frac{1}{R}\sqrt{4gR}\sin\frac{\theta}{2}=g\sin\theta$ as expected. This is a maximum at $\...


0

When the bullet hits the plasticine and embedded inside the plasticine, there is friction exerts on the bullet and hence heat is dissipated to the surrounding. Therefore, conservation of mechanical energy cannot be applied when the bullet is embedding the plasticine. But as for the bullet and the plasticine together displaced upward, conservation of ...


0

The height $h$ is given as the vertical height so you do not need to worry about the angle $\theta$ and assuming that there is no friction then use the law of conservation of energy: $$KE_\text{initial} + PE_\text{initial} = KE_\text{final} + E_\text{final}$$ to find the speed after the mass has fallen a height $h$.


0

My guess is using the Hamiltonian which is potential energy + kinetic energy (or is it minus) So assuming you have a gravitational field $h = m \ g + \frac{1}{2} \ m \ v^{2}$ but its at an incline so $m \ g \ \sin{\theta}$?


4

In newtonian mechanics the angle does not matter, but in relativity it does. For example: An object close to the speed of light launched horizontally will orbit circular at a distance of 3GM/c² from the center of mass (the so called photon sphere), but it will escape if launched vertically. When you launch it at a distance just above 2GM/c² (the so called ...


6

Essentially, yes. The derivation of the escape velocity is based only on the energy balance and energy does not depend on the direction. This follows from the property of the gravitational field to be conservative, so work required to move between any 2 points is independent from the path you take. Vague intuition: the vertical speed you estimate is about $...


1

The idea of a "Zero-Energy Universe" is a theory held by a limited number of scientists. There are several stackexchange question that expand on the theory and may help you. Zero energy universe Total energy of the Universe


0

The variables in your experiment were : the mass of water in the calorimeter $m_w$ (independent variable) and the temperature rise in the calorimeter $\Delta T = T_{final} - T_{initial}$ (dependent variable). Your equation can be written as : $(m_w c_w + m_c c_c)\Delta T = m_s c_s (\Delta T' - \Delta T)$ $\frac{m_w}{m_s} + \frac{m_c}{m_w} \frac{c_c}{c_w} = \...


1

The efficiency is being miscalculated because they left out the energy being used to heat up the device to 135 degrees! This is similar to claiming that an amplifier that takes a 1 mW input signal and "converts" it into 1 W output signal, as having an efficiency of 1000%! They are leaving out the energy being supplied to the amplifier "external" to the ...


1

We put the circular orbit of the particle on a straight line and convert the motion to a 1-dimensional rectilinear motion as follows : The arc length, the natural parameter $\:s(t)\:$ is the distance travelled on the straight line till time $\:t\:$. The speed $\:v(t)\:$ on the straight line is the magnitude of the tangent to the circle velocity. Now, on ...


2

@dvij gave the equation $$g\sin \theta =R\frac{d^2\theta}{dt^2}=R\frac{d\omega }{dt}$$ If we multiply this by omega, we obtain: $$g\sin \theta \frac{d\theta}{dt}=R\omega\frac{d\omega }{dt}$$ If we integrate this equation between 0 and t, we obtain: $$g(1-\cos \theta)=\frac{R}{2}\omega^2$$ So we have $$mg\cos\theta-2mg(1-\cos \theta)=N=mg(3cos\theta-2)$$ I ...


0

As @dmckee said, you should use polar coordinates for this problem. Equations of motion are below: ($\omega$ is angular velocity and $\alpha$ is angular acceleration) $$mg\cos \theta-N=mR\omega^2\;\tag 1$$ $$mg\sin \theta=mR\alpha\;\tag 2$$ From $(2)$, we have $\alpha=\large{\frac gR}\sin \theta$ On the other hand, we know: $$\alpha\mathrm d \theta=\omega\...


0

If a 'law' of Physics can be really uncared for and still you can predict the outcome of an experiment completely accurately then it is not a law of Physics. So if energy conservation is a physical fact here then either implicitly or explicitly, we are going to use that fact - otherwise, we must not be able to predict the complete outcome. So I assume your ...


-2

Kinetic energy and photon absorption Particles - when moving into a magnetic field - have gained before kinetic energy. As GRB wrote: Every time a charged particle has to be accelerated, a photon has to be involved. If you want to linearly accelerate a charged particle, you have to shoot photons at its back. To be precise this one of the possibilities ...


4

The underlying reason for OP's flawed argument is that a premature use of EOMs in the stationary action principle $$ S~=~\int\!dt ~L(r,\dot{r};\theta,\dot{\theta}), \qquad L(r,\dot{r};\theta,\dot{\theta})~=~\frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) -V(r),\tag{A}$$ destroys the variational principle. Concretely OP is implicitly assuming that (3) is a ...


2

Effective potential is defined by the formula $E=T_{radial}+V_{eff}(r)$. Your calculation shows that once you make this identification it is not true that $\mathcal L = T_{radial}-V_{eff}$, but that is fine. This happens because centrifugal term (i.e. the one with angular momentum) is really kinetic term and not a true potential. Hence it must enter the ...


0

You are wrong, the orbit of the moon is changing, it is receding from the earth so its potential and kinetic energy is changing : The notional tidal bulges are carried ahead of the Earth–Moon axis by the continents as a result of Earth's rotation. The eccentric mass of each bulge exerts a small amount of gravitational attraction on the Moon, with the ...


0

I know the final kinetic energy is 0.1566J How did you find the final kinetic energy? To find the final state of the system you have to assume that there are no external torques acting on the system and that angular momentum is conserved.


1

When you reading the plot in the paper, when efficiency is above 100%, the temperature is 135 degC (275 F) and the light output power is low. This makes me think, when you heat an steel bar to hot red, it emits light without electricity. You can pretend to inject electricity into the bar, and your efficiency is infinite. Energy is from hot environment. ...


4

Given that the device is extracting heat (vibrational energy) from the lattice that the LED is embedded in, the conservation of energy issue can be understood. The question you really should be worried about is, "does this violate the second law of thermodynamis?" Is the overall free energy increasing? You can use thermoelectric coooling, for example, ...


19

The device is apparently working as a heat pump, for which I give a brief theoretical analysis here. In the example given, the $P_h=69{\rm pW}$ light output comprises the $W=30{\rm pW}$ input by the researchers together with $P_c=39{\rm pW}$ of heat that was formerly in the chip. We can model the process by ideal heat pumping as follows. Heat drawn from ...


1

Feynmann's machines are idealised devices and we don't know or care about how they work internally. All we know is that we can connect the weights to the machines and they will move them according to the rules. The 1kg and 3kg weights start at height zero. We connect them to machine B and it lowers the 1kg weight to $h=-1$ and raises the 3kg weight to $h=Y$....


3

The conceptual problem seems to be however there is no way to define the friction for the rope since it has two parts, with apparently distinct potential energies. Don't worry about the "distinct potential energies". You can just compute the gain in potential energy for each part separately. For the bit already hanging down, as the rope slides by a ...


0

Comments to OP's post (v4): OP is trying to prove via Noether's theorem that no explicit time dependence of the Lagrangian leads to energy conservation. OP's transformation seems to be a pure horizontal infinitesimal time translation $$\tag{A} t^{\prime} - t ~=:~\delta t ~=~-\epsilon, \qquad \text{(horizontal variation)}$$ $$\tag{B} q^{\prime i}(t) - q^i(t)...


0

The easier way of doing this is to just consider a generic transformation, G, such that the canonical co-ordinates of the Hamiltonian are shifted as below: $$ \delta p = \frac{\partial G}{\partial q} \delta \lambda$$ and $$ \delta q = - \frac{\partial G}{\partial p} \delta \lambda\,,$$ where $\lambda$ is the transformation parameter determining how much of ...



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