New answers tagged

3

The way I understand the setup: The block is initially at rest at some height above the spring. The spring is initially at rest oriented vertically with one end on the ground and at it's natural length (though if its not a massless spring it would compress somewhat due to its own weight). Then, the block is dropped, it lands on the spring compressing it ...


0

The Gravitaional Potential Energy is calculated taking the potential at Infinty as zero, ie , if you come from infinity to the point at which you are calculating gravitational potential (say point P), Energy will continuously decrease because $$E \varpropto -\frac{1}{r}$$ (r is distance from the center of object to P) as r decreases, the fraction increases, ...


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An orbiting observer is a bit problematic because there are no stable orbits for $r \le 3r_s$, so let's instead consider an observer hovering at some distance $r$ from the black hole. In that case as $r \rightarrow r_s$ the blue shift does indeed $\rightarrow\infty$ and the observer would indeed be roasted. But this shouldn't be surprising. The acceleration ...


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If you are wondering if $F=\frac{-\partial U}{\partial x}$ holds, you may check the equivalent condition that you posted, $\nabla\times\vec{F}=0$. In this case the curl will come out to be zero, so you may then proceed to write the force as the negative gradient of a potential. This appears to work because the del operator is not taking time into account, ...


0

No work is required to rotate the rod about its center by 90 degrees. The center of the rod is at height $h/2$. The PE of a horizontal rod at a height $h/2$ is $m g h/2$. Another way to see this: If all of the mass were concentrated at the top of the rod, the PE would be $m g h$. If all of the mass were concentrated at the bottom of the rod, the PE would be ...


1

The state of equilibrium is characterised by a minimum in free energy $F=U-TS$ (Helmholtz for simplicity), not a minimum in potential energy. What this means is that while the system is indeed attempting to minimise the potential energy $U$, it is simultaneously trying to maximise the entropy $S$. The balance, i.e. which term dominates, is determined by the ...


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Negative sign just indicates the attractive potential. The factor of $1/2$ is to average the double counting of the same term. It is the average of two terms for every value of $i$ and $j$.


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My guess is that it would be impractical. Your reasoning is basically correct: a longer right arm will grant you a larger $\Delta h$, increasing the speed of the projectile. But there would be some issues: The structure would have to be higher, otherwise the weight would smash into the ground. Hence, it will be heavier and more difficult to carry. The ...


3

If I understand your notation right, then by $h_L$ you mean the height of the left weight? In that case your formula doesn't make sense: To account for energy conservation, you would have to take the heights before and afterwards of the right weight. Then you get \begin{align} m_R*g*h_{before} = m_R*g*h_{after} + \frac{1}{2} m_L * v^2 \end{align} You ...


1

This is a homework-like question, so I will not provide a full answer. Here are a couple of good things to think about on your way to the answer: What is special about the velocities (or momenta or kinetic energies) of the particles at the instant of minimum separation? What quantities are conserved throughout the interaction?


4

Actually, as someone pointed out in the comments, potential/kinetic classification is the only meaningful classification in physics. Potential energy is the energy which comes from interaction, and kinetic energy is the energy which comes from motion. Maybe you stumbled upon terms like chemical energy, thermal energy and so on. But chemical energy is just ...


0

The basic fact for telling this is that central forces are conservative in nature. This means the work done by a conservative force between any two points is independent of the path, but depends only on the initial and final points only. In such a case, the work done by a conservative force in traversing from point $a$ to point $b$ is given by: ...


1

If the force is conservative then it is the gradient of the potential: $$ \mathbf F = -\nabla \mathbf V $$ If we write this out in component form (in polar coordinates) we get: $$\begin{align} F_r &= \frac{d\mathbf V}{dr} \\ F_\theta &= \frac{d\mathbf V}{d\theta} \\ F_\phi &= \frac{d\mathbf V}{d\phi} \end{align}$$ Since we are told that the ...


10

Note that this is an incomplete answer. Imagine an object of mass $m$ at a distance $r$ from the centre of a black hole of mass $M$. The gravitational potential energy is $$ U(r)=-G\frac{Mm}{r} $$ This has its highest value when $r=\infty$ and its lowest value when $r$ is at the event horizon of the black hole, i.e. the Schwarzschild radius $$ R = ...


0

Electric potential energy is stored in your system of a large mass charged body $A$ and a small mass charged body $B$. When an external force is applied to body $B$ that external force will do work on the system (body $A$ and body $B$) and change the electric potential energy of the system and possibly change the kinetic energy of body $B$. However it is ...


1

Force is a vector. Potential energy is a scaler. Forces which have associated potential energy functions as called conservative forces. Conservative forces act in such a direction that, if released from rest, the potential energy function associated with that force will decrease (and the kinetic energy will thus increase) with the velocity increasing, until ...


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Perhaps the confusion comes from the often belief that if a force field is only a function of position then the force is conservative, and that non-conservative forces are those that include friction dependence on speed, and other variations. But this is incorrect, as alfred centauri commented this field is not conservative. But this is not just because ...


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Apparently you are not considering the work done by the electric field, and you are confusing yourself with the "energy storage" concept. After traveling a $\Delta x$ distance into the electric field $\vec{E}$ (avoid using $dx$ because $d$ usually stands for infinitesimal changes), the particle B with charge $q$ will have suffered change in its "horizontal" ...


1

Yes you are. If a force is conservative, its work does not depend of any path between any points $A$ and $B$. Since the work integral can depend only on the initial and final points themselves, we define $$W_{A\rightarrow B}=\int_A^B\vec F\cdot d\vec r\equiv U(A)-U(B).$$ Now define the mechanical energy as $E=K+U$ so that $$dE=dK+dU.$$ Suppose there are two ...


2

If voltage is all that you know, then the answer is No. If you know how much charge $Q$ in Coulombs is added, you only have to divide by the charge $e$ on each electron (in Coulombs). Otherwise, if you have a parallel plate capacitor and you know the capacitance $C$, you can work it out from $Q = CV$. Your suggestion that the extra electrons 'push more' ...


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Try to consider the changes in KE, and how it alternates with EPE and GPE. You should also try and think about what happens to the velocity and acceleration of the mass when it reaches equilibrium, and when it is at maximum amplitude above and below the equilibrium point (consider the energies involved in those 3 specific points). This should give you some ...


1

The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g). Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg. If you want to break it down further, I guess mg = Q/sec. I'm not sure if this answers your question, but it at least gets ...



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