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0

Make the y-axis height and the x- velocity. This will give you a parabolic shape indicating that it is a quadratic variation.


0

Generically, if I have a particle which has potential energy $\phi(x,y,z)$, then the force on that particle will be given by ${\vec F} = - {\vec \nabla}\phi$. So, generically, the motion of particles will "try" to minimze the potential energy. In particular, the only points where the particle will not move will be those points where $\nabla \phi = 0$, or, ...


2

The first one is correct. The problem is that in general the force acting on the particle will no longer be conservative in the moving reference frame and we can no longer associate the potential energy $U$ to it. To see why this is so, realize that a force $\vec{F}$ is conservative if and only if: $$\oint_C\vec{F}\cdot d\vec{r}=0$$ to any closed curve $C$, ...


0

Well there is energy in the magnetic field. (Search for energy density in magnetic field.) It's B^2/(2*mu_sub_zero) in MKS units. (And as long as no magnetic materials are around.) The act of turning on the magnetic field creates a changing B field for a short time. A changing B field creates an electric field, and this can cause current to flow in ...


0

From the answer... ..my question I have learnt that Newton's 3rd law of motion is a direct consequence of law of conservation of energy. When a body moves in a certain direction and an opposing force acts on it , it exerts a reacting force (by Newton's law) ... Hence, the body loses its kinetic energy. Problem ....gravitational force: ...


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.


0

There's a little mistake in the signs, since the potential energy is given by $U=-\dfrac{GMm}{r}$. The problem of finding the time elapsed between two points of the trajectory may be done by realizing that at any point $$\frac{1}{2}m\dot{x}^2-\dfrac{GMm}{x}=E=-\dfrac{GMm}{H},$$ where $H$ is the maximum height of the particle. Hence ...


0

Paily's answer is the correct one. Mechanical energy is defined as the sum of kinetic and potential energy and equal to the work done by non-conservative forces. When these are absent (as when they act perpendicular to displacement) they do no work, therefore ΔK+ΔU = 0 and K+U = constant. In short, the principle of conservation of mechanical energy: any ...


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.


-1

You hold a stone and let it go. What work will the Earths gravity do on the stone to bring it down to the ground? This much: $$W=\Delta K=K_2-K_1=K_2$$ Let's remember this for now and do something different. How much potential energy is lost during this fall? This much (energy conservation): $$E_{\text{total }1}=E_{\text{total }2}$$ $$U_1+K_1=U_2+K_2$$ ...


0

Here are a few points to keep in mind: Potential energy is always described as the potential energy of the system. For example, the gravitational potential energy of the Earth-Moon system, belongs to the system as a whole, not the Earth or the Moon individually. So for your example, if you are for instance throwing a brick upwards, it would be the ...


0

If I understand you correctly, your mistake is in using friction as an analog to gravity. Because friction is a non-conservative force the work done is dependent on the path taken. Furthermore the energy "lost" due to friction is stored in a way that is not spontaneously reversible within the system (e.g. heat, plastic deformations, etc.). Gravity on the ...


0

Nothing is actually stored. (You will not find anything "in" the body :) ) The increase of potential energy means in this case that there is a force (of gravity) acting on a body, and the body's movement away from the source of this force increases the distance the body can p o t e n t i a l l y travel under the influence of this force. So if the body is ...


1

UPDATED: I now think my previous answer was wrong, because the set up would be equivalent to the following question: Is a black body sphere inside a black body shell hotter than the shell? Just change the question to add a carefully crafted lens that focuses all the radiation into the sphere (you could make the shell as large as you want), which of course ...


4

The heat that makes a filament lamp glow is derived from electrons bashing into the lattice of atoms in the filament and transferring energy to them. The kinetic energy of the electrons becomes vibrational energy of the lattice, and this is exactly what heat is. However the interaction of electrons with the lattice is also what resistance is, and ...


3

In short: you have calculated the maximum displacement of the stone, and your friend has calculated the equilibrium displacement. If you place the stone of mass $m$ on the spring (which I assume to be of negligible mass), the stone ends up oscillating around the equilibrium point. So the sum of the potential energy stored in the spring and the kinetic ...


2

Short answer: The distance the rock falls in your thought experiment is not the same as the spring compression described in the actual question. Long answer: In your thought experiment, the rock is released from rest and it falls some distance before turning around. So the question arises: For an object of mass $m$ released from rest just touching a ...


0

Let $v_1$ and $v_2$ be the particle's velocities in the center of mass coordinate system after the collision. by conservation of momentum and energy we have \begin{gather}\tag{1}m_1v_1+m_2v_2=0 \\ \tag{2}m_1v_1^2+m_2v_2^2=m_1(v-v_c)^2+m_2v_c^2 \end{gather} Isolating $v_1$ in $(1)$ and substituting in $(2)$: ...


0

$m_2$ will leave with the same magnitude of momentum but opposite direction. Now the assertion is made that in an elastic collision, $m_1$ and $m_2$ have the same speeds leaving the collision as entering it. In other words, the speed of $m_1$ is $v-v_c$ and the speed of $m_2$ is $v_c$ after the collision. In order to simplify things and to ...


0

For a satellite in a nonperturbed orbit, the change in gravitational potential energy does equal the change in kinetic energy. Increasing the size of a circular orbit requires energy. Half of the energy is used to lift the satellite, while the other half is used to speed up the satellite. For a satellite in a circular orbit perturbed by atmospheric drag, ...


1

The initial potential energy is zero because the ball starts off at essentially ground level, and potential energy is being defined as being zero at ground level. The initial velocity is a vector of magnitude v that points up at an angle $\theta$ from the ground. The components of that initial velocity are $v_x(0)=v \cos\theta$ in the horizontal direction, ...


0

Let's take a closer look at the equation: $$\frac{mv^2}{2} = mgh_\text{max} + \frac{m(v\cos\theta)^2}{2}$$ The term on the left is the initial kinetic energy of the cannonball as it leaves the cannon. This is equal to the horizontal kinetic energy plus the vertical kinetic or potential energy. At the maximum height, there is no vertical kinetic energy ...


0

There is no force on the x direction, so the acceleration is zero and the x-component velocity is constant which is know in the initial condition. Plus the conservation of energy at the beginning and at the highest point, you will get that equation


0

Both sides of your equation are already dimensionless, so now it is just a matter of choice. For example, you can define your $\gamma_{...}\equiv V_{...}^2/U^2$ terms as dimensionless variables of your model and $\lambda \equiv L \cdot D_{eq}/D_d^2$ as a dimensionless parameter. Taking $\alpha \equiv \rho_p/\rho_d$, $\beta = (D_p/D_d)^3$ your equation reads ...


0

Kinetic energy of the billiard ball will be maximum when the ball will hit the billiard ball in straight line with a given velocity.$$mv=mu+MU...........1$$And,$$mv^2=mu^2+MU^2............2$$Now the kinetic energy $\frac{1}{2}MU^2$ is a fraction of the kinetic energy $\frac{1}{2}mv^2$$$\frac{1}{2}MU^2=k\frac{1}{2}mv^2............3$$From 2 and 3 find $k$ in ...


0

The way I would approach this problem is through the impact parameter, $b$. You can find the definition on Wikipedia. The first thing to note is that for $b > r + R$, there is no collision. Here $r$ and $R$ are the radii of the two balls. For $b$ less than this limit, you can use geometry to determine the direction of $\mathbf{U}$. This is because the ...


3

Here's a general overview of how to approach this: Since the only external forces are vertical (gravity pulling the balls down, normal force of the surface holding the balls up), we can use conservation of momentum in the plane. Similarly, there is no external torque rotating things in the plane, so that component of the angular momentum is conserved. And ...


1

If you impact the second body its axis of percussion it will purely rotate. By carefully choosing the inertial properties of the two objects you can make the first object stop translating in the process. See this post for more details on a particle to rod impact.


2

To calculate for a situation like this, consider the Law of Conservation of Momentum: Pi = Pf In the case of the billiards: KEi = 1/2 mu1^2 + 1/2 Mu2 (u1, u2 = initial velocity) KEf = 1/2 mv1^2 + 1/2 Mv2^2 (v1, v2 = final velocity) Based on this law, initial kinetic energy and final kinetic energy in an elastic collision are equal. KEi = KEf Hope ...


0

This is not an answer, but is there any contradicción with Noether theorem to say that Kinetic energy gives diferent value according to system of reference?


3

We don't know! If we were to observe a situation where energy conservation did not appear to work, that would be a major puzzle. As you say, either we would have to discover some alternative contribution to the energy that we had been neglecting, or we would have to give up on energy conservation. A priori it is not obvious which one of those two resolutions ...


7

Noether's theorem states that to every continuous symmetry of a physical system there is an associated, conserved quantity. The conserved quantity associated with time translation invariance (i.e. it doesn't matter if you perform an experiment now or tomorrow, provided you set it up the same way) is what we call energy. Therefore, somewhat tautologically, ...


1

A "collision course" is a very fuzzy concept: if you are "barely going to hit" you are on a collision course but don't need a lot of deflection. However, let's assume for a moment a stationary earth, a meteorite of mass $m$ at distance $D$, heading for earth of radius $R$ with velocity $v$. The equations you need are conservation of angular momentum and ...


1

For what it is worth, your formulation of the spherical wave business is flawed. See here for example. These have the form $$u(r, t) = \frac{A}{r} e^{i(\omega t \pm k r)},$$ where the symbols have their usual meaning. So, intensity, which goes as the amplitude squared, will go as $$I = \left(\frac{A}{r}\right)^2$$ falling off as the square of the distance, ...


1

There is a confusion between the terminology "perpetual" , which means "continuously", devices that almost move forever, and a machine that can produce energy. As the other answers point out energy is conserved and if it looks as if energy is provided from nothing a closer analysis shows the mistake, as in the drinking bird perpetual setup. In the case of ...


3

You might be able to get it to work for quite a while, depending on your skill as an engineer. But there is a critical difference between a well-engineered machine that runs for a while, and a perpetual motion machine that runs forever without input. The latter is impossible.


4

The key here is the antimagnetic strip, quite aside from whether or not such a device can be built. When you insert the anti-magnetic strip, you must change the shape of the magnetic field. You must force the magnetic field to "leave" the high permeability ball. The same magnetic induction $|\vec{B}|$ in a high permeability $\mu$ material represents a lower ...


4

If the magnet is strong enough to pull the ball up the bottom slope, it will be too strong to let the ball fall. Even worse, as you have drawn the diagram the magnet is pulling the ball down the lower slope when it is toward the left end. Anywhere to the left of where the perpendicular from the magnet to the ramp, the magnet is pulling more right than up. ...


-1

Too many abrupt changes in direction. Also, is this spring powered? If so, I would think it to be simpler by using an electromagnet to apply and cut voltage. More moving parts mean more opportunities of failure.


1

I interpret your question as What type of equation is the equation $$\int_{t_0}^{t} \vec{F}(\vec{x}(t')) \cdot d\vec{x}(t') = \frac{1}{2} m ((\dot{\vec x}(t))^2 - (\dot{\vec x}_{0})^2) \; \; \;\; (1)$$ where $t$ is variable and $\dot{\vec x}_{0}$ fixed? The answer is, it is a first order integro-differential equation, not a mere ordinary differential ...



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