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2

The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles. Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is $$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p ...


2

There is the (non-genetal) relation between the free energy of interacting of two currents $J^{a}, J^{b}$ and the propagator: $$ U = -\frac{1}{2} \int d^{4}xd^{4}y J^{a}(x) D_{ab}(x - y)J^{b}(y). $$ It's not general, but it realizes the simple example which can help you to understand how to get the expression for force. The structure of field which causes ...


0

The astronauts do have a potentially deleterious affect on the vibrational environment on the International Space Station. They have to exercise multiple hours a day to keep bone and muscle loss down to a tolerable minimum. The unmitigated vibrations from all that exercising would be harmful to very sensitive micro-g experiments. Before I get to the ...


0

Here's the crux of the problem: break it up. As an example, let's find the forces on charge 1. First, we have to find the force on charge 1 ($q_1$) by charge 2 ($q_2$). We then use Coulomb's law, $$F=k\frac{q_1q_2}{r^2}$$ We now have to find $r$. Using the Pythagorean theorem, $r^2=r_x^2+r_y^2$, so $r=\sqrt{r_x^2+r_y^2}$. If we know $x_1$, $x_2$, $y_1$, and ...


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: ...


1

Think of it like this and you will never be confused: Friction always tries to keep the two objects together. This regards both static and kinetic friction ($f_s$ and $f_k$) . If something is sliding to the right over asphalt (kinetic friction), friction will try to stop this relative motion, and hold the object and the asphalt together. So $f_k$ will ...


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 ...


1

A kinetic friction force that acts on a body always has an opposite direction to that in which the body is moving. So, if you have a system Surface+Body A+Body B, where B is on top of A and A is moving to the right, both the surface and B will exert a friction force on A, pointing to the left. Moreover, according to Newton's third law, A will exert a ...


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

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 ...


1

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 ...


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 ...


0

I'm not sure what you're referring to by 'straight' acceleration. If that angle is the angle at which a force, $F$, is being applied, then the horizontal force is: $$ F_x=F\cdot cos(\theta)\\ F_y=F\cdot sin(\theta) $$ These will get the forces along the X and Y axis respectively. Assuming width, $w$, length, $l$, height, $h$, mass, $m$ and $\text{'some ...


1

You need some kind of a force, either friction, or gravity (if the road is slanted for instance) to account for the "forces" side of the 2nd Newton law: $$ \sum\vec{F}=m\vec{a} $$ Whereas the Right Hand Side (RHS) should contain the centripetal acceleration, i.e.: $$ m\vec{a}=-m\frac{v^2}{r}\hat{r} $$ And the minus means it is directed towards the center of ...


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

Lets assume there are no non-conservative forces like air drag and you are dropping the ball from rest. While the ball gains kinetic energy, it loses potential energy. This means that $$ K_\text{in} + P_\text{in} = K_\text{f} + P_\text{f} $$ $$ 0 + mgh = \frac{1}{2} mv^2 + 0 $$ which means $v= \sqrt{2gh}$ and there is no mass term here. Hence the (final) ...


-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 ...


3

In principle there is an effect, but firstly it's tiny and secondly it averages to zero. The mass of the ISS is about 420 tonnes, or about 5000 times the mass of an astronaut. That means if an astronaut pushes themselves off a wall at 1 m/sec the ISS moves in the other direction at about 0.0002 m/sec. But the ISS isn't very large so after only a couple of ...


0

When an object is spinning, there are indeed stresses developing that keep the object from ripping apart. But every material has a finite strength (the yield stress), and when you exceed that it will deform plastically (it will not go back to its old form after you remove the stress), and may even break (when you reach the ultimate tensile strength). If you ...


1

This page has a helpful summary of the history--it seems he initially accepted the Aristotelian idea that objects could only continue to move if some "force" inside them was moving them (keep in mind this is before his technical definition of 'force'), and it took him a while to switch to the idea that bodies naturally tend to keep moving unless acted on by ...


1

Yes, alternatively you can use that the potential energy of the spring is transformed into kinetic energy of the object. This is simpler than considering the work $L$ done by the spring. The result is however the same: $$ U=E_{kin}\\ \Rightarrow \frac{1}{2} k x^2=\frac{1}{2}m v^2\\ \Rightarrow v=x\sqrt\frac{k}{m} $$


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

How much is the force of gravity on the $3m$ mass, and how much mass must it accelerate? Similar question for the $2m$ mass? Then carry on as you have...


0

Mass cannot just cease to exist. It can detach from the main body or annihilate with anti-matter to produce EM radiation. The effect on the velocity in such cases will depend on the details of the process (for example, whether the mass is lost in all directions like for evaporating water droplet or in one preferred direction like for rocket).


1

The ideal shape for the water to just 'hang' is actually the cube that you've proposed. In this case all of the forces on the water are uniform across the interface. For one region to slip down, another needs to move up (as you've indicated in your diagram). But since every location is experiencing the same forces, there's no reason any spot to start ...


0

You appear to understand the situation quite well. For a conservative force, the amount of work done going from A to B is independent of the path taken. A slightly more formal definition is that the curl of the force field is zero. This results in any closed path having zero net work done (by Stokes's theorem). As the route from A to B and back again by a ...


1

The localized vector is a vector which we know its magnitude and direction and its point of application , which is the point where the force acts , if the force were free vector , how you will calculate for example the moment ? that won't make any sense ok the second principle i found it in Vector mechanics book called PRINCIPLE OF TRANSMISSIBILITY. check ...


0

For a force, you need the following bits of knowledge: The direction and magnitude of the force. The point it acts on. Some people call that pair of specifications a "localized vector", because it consists of both a location (#2) and a vector (#1).


0

Is a nice design, although it seem rather complex it only has one degree of freedom related to the expansion. So it is a ingenious ensemble of pieces that are connected to each other in such a way that the system is only stable in 2 configurations. Is composed of two types of pieces: disc-like piece: there are 4 of them visible for each color, they are ...


24

No. The answer is clearly no. This building is 800 meter high. Some comparison: Skydivers are falling more kilometers in free fall. They experience absolutely no damage from the pressure increase. Scuba divers moving fast upwardly or downwardly also don't get any wounds, although 10 meter deep water has the same pressure as there is between the sea level ...


0

Originally I wrote this as a comment to ask for clarification but I think it might qualify as an answer (if I interpret your question correctly). I see two magnetic effects: electrons moving in the region of strong magnetic field feel the vertical force; and to the left / right you have an increase / decrease in magnetic field (as you enter / leave the ...


1

At 1:02 in that video, you see the "people" at the end of the swing do something like a half revolution with a radius of 15 meter in 0.3 seconds (rough estimate of dimensions and time - feel free to come up with your own estimates by counting frames etc - then substitute those numbers in the equations below). The centrifugal acceleration is given by $$a = ...


0

I am curious about the following line in your derivation: Our quadcopter measures 47cm motor-to-motor, so r = 0.235m. Let’s calculate the torque. Presumably to pitch the copter without changing the over all lift, you need one pair of rotors to slow down and the other pair to speed up. It was not clear to me that the factor of two that this results in ...


1

It appears to be rolling friction. Even on the Moon, a wheeled vehicle will eventually come to rest because of friction at the axle and friction between the wheel and the ground. The wheels on that vehicle are deformable. That means that the reaction force by the ground is not purely vertical. It has a horizontal component that resists the rolling motion of ...


2

Any force greater than zero can stop the car. Only it will take longer and the distance moved by it by the time it stops also will be greater. If the force is larger these parameters (time to stop and distance traveled before stopping) will decreasing. Theoretically, infinite force is required to stop it instantaneously.


5

The tension is not the same for all parts of the rope. If the tension is $T_1$ between the hands and $T_2$ between the top hand and the ceiling then. $$ T_2 = T_1 + F_2 \\ T_1 = F_1 $$ where $F_1+F_2=W$ are the forces acted upon the arms. You arrive at this if you make two free body diagrams, one at each hand. The result is that $T_2 = W$ and ...


0

Hydrogen is a special kind of intermolecular interactions that involve the electrostatic attraction between an electronegative atom, H which is bonded to N, O or F. Similarly, van der Waals force is a kind of intermolecular interactions between permanent dipoles or induced dipoles. For example, the interactions between HCl is called van der Waals(permanent ...


3

Newton law of gravitation is given by: $$F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2$$ The gravitational constant, $G$, the weight of Earth, $m_1$, and the radius are constants, so: $$G \frac {m_1}{r^2}=(6.6742 \times 10^{-11}) \frac{5.9736 \times 10^{24}}{(6.37101 \times 10^6)^2}=9.822$$ Hence, the equation simplifies to $$F =(9.822) ...


1

See the wiki page . There is no difference between the two . Weight of a body of mass $M$ is $M.g$ which is equal to gravitational force on the body.


0

You have a typo in the above; "9.8*5.2" should read "9.8*52", in two places. But that's apparently just a typo that doesn't reflect your actual calculation; your answer of $3.07 m/s^2$ looks correct. It does look like a mistake in the book.


3

The perpendicular force is no longer the weight of the box, you need to consider the force by the two workers in the perpendicular direction as well.



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