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0

When the rocket sits on the launch pad and the engine fires up, an enormous amount of energy is expended moving the exhaust gases and heating the air. No work is done to move the rocket, but that doesn't mean no energy is expended. But the kinetic / potential energy of the rocket are not changing (actually the potential energy of the rocket becomes less ...


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Useful work=0, power DRIVING the engine =0, ALL useful power lost in the form of wasteful energy, example: exhaust gases, heat, sound and all that..


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In your first example, a rocket on a launchpad that has just started its engines, the equation is exactly correct. The motor is generating huge force, but it has not yet moved the vehicle, so no work or power is being generated. In reality, that moment of no work or power lasts only a tiny fraction of a second. The power moves the rocket and then you can ...


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Yes, the molecular picture has been actually extensively researched theoretically, simulated, and experimentally investigated! The term you might be looking for is nanotribology. There have been several atomic-level studies of friction, there a nice Scientific American article available as well. The latter article explains several experimental approaches ...


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I used to wonder how spacetime curvature could cause you to start moving when you weren't moving already. I got the idea that the path you followed could be bent by the curvature of spacetime, but what if you start out standing still? (Standing still relative to a particular massive body, that is.) If you understand all the other answers that have been ...


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There may tension come into the picture first if the block shear. otherwise friction will come into the picture before tension because there must be some displacement of the block for tension to act. if block get sheared then even without relative displacement between block and surface tension will act.


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Yes, the force points along the vector of the relative velocity between the object and the air. Quadratic drag is an interesting phenomenon. You have to calculate the net velocity vector (which includes a horizontal and vertical component) and compute the force along that axis; when you then decompose it into horizontal and vertical components you will find ...


0

If you open a door by gripping it near the hinge, you apply GREATER energy for a SHORTER time than when you grip it near the outside edge, which requires LESSER energy for a LONGER time. As the friction of the hinge and the weight of the door are equal in both cases, the total energy applied is the same. It's only the time applied and the intensity of ...


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Your force has a component along the slope, so yes, the object will move along the slope. It will not leave the surface though, if that's what you mean by "lift" If you find it counterintuitive why the object has a vertical acceleration component despite your applied force being horizontal, you must think about the normal force. This is always, as the name ...


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The key idea here is that Static friction is self adjusting.... It will oppose and nullify effect of your applied force F , provided maximum friction value is not exceeded. If you imagine or draw a Free body diagram of forces on the body , you'll see that friction due to ground on Body acts in opposite direction to F. So, Tension in string is not developed ...


2

Answer to the first question: This depends to some extent on the 'models' used for the forces of friction and tension. A typical model for string tension is as a restoring force that obeys Hooke's law: $$T = - kx$$ at least for a small positive extension $x$ in the length of the string, or equivalently, displacement of the block along the length of the ...


0

V1 : Velocity in front of panel V2 : Velocity behind the panel Force acting on the plate (Newton second) : F=m*a The accelleration is found by assuming a decrease of velocity from V1 to V2 over a distance s. The acceleration is thus found as: 2*a*s=v1^2-v2^2 . The mass transport over the distance s is m=rho*A *s By insertion: F=rho* A * s* (v1^2-v2^2)/ ...


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See: The string will not develop any tension until force is applied beyond limiting friction, that is, both string and the frictional force b/w the surface will start opposing the motion only when the friction is kinetic. Hence, Static friction acts first, then comes both kinetic frictional force and the force opposing string's elongation. Thanks


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When the force you apply is less than $\mu mg$, the tension is not affected and does not affect the motion of the block. But if you apply a force larger than $\mu mg$, then the tension in the string increases. Obviously, the block will start its motion when the string breaks (if it is a real string). Finally, the friction direction is always perpendicular ...


0

In case the question concerned the case $r \rightarrow 0$, you would reach the situation where the charge (represented by a charged particle like electron, proton, positron) approaches the Coulomb field of the other particle and they would have a tendency to create a kind of a planetary system - but - quantum effects start to play a role here and those two ...


3

If r = 0 then you have a single charge, so the problem reduces to the electromagnetic self-force problem. A charge will interact with the electric field it is in, and that includes the field due to its own charge.As long as the charge is not accelerating, one can pretend as if there is no self-force, but for accelerating charges, the self-force will lead to ...


1

Your target may not stop a very dense, compact bullet: It may exit, departing with some (possibly large) fraction of its initial momentum. If your bullet does not penetrate a target as well, as a less dense rubber bullet might (depening on circumstances including its speed and properties of the target), it might be stopped completely (either stuck in the ...


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Static friction does not produce or consume work in most of the times. For example for a solid body that rolls without sliding the velocity of the base point $A$ is $\vec v_a = \vec v_{cm} + \vec v_{tangential} \Rightarrow v_a = v_{cm} - \omega R = \omega R - \omega R =0$ which implies that $x_a = 0$. The static friction is a force that acts on $A$ so $W_T = ...


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It's the net resultant that a force would react with on application of a certain change in momentum in the body, on which the force is applied.


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The physics 101 answer is no: it takes more force, but it is compensated by the smaller displacement so the energy stays the same. If we start with a static door, and we end up with a door rotating at some speed, the energy into the door is the work done by the force and it must be the same independently from the point where the force was applied But let's ...


0

Torque=(R) x (F) Energy req. to rotate=(T).(Theta), Ta=Tb, ONLY Fa is less than Fb. & T at Hinge=0, it'll not rotate there. The force required will increase from A to hinge point. Energy needed hence is const from A to hinge(except hinge point). [Ta means torque applied at A]


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It doesn't. The work is done by the active force(ex. A human trying to pull a bull.). This work is converted into frictional energy(ex. Heat generated b/w surfaces)


1

As @Gautham said, W = Torque * angular displacement which is also equal to the energy needed. here the angular displacement is same as we know and torque will be also same(Large distance implies less force needed,less distance implies large force needed but torque will be same) So energy needed will be same in both cases.


3

The answer is NO. Change of energy is work i.e $W = \Delta E$ and here the work done is $$W = \text{Torque} \cdot \text{angular displacement}$$ which is equal in both the cases. The only change is one needs to apply more force to achieve the same amount of torque at a smaller radius. $$\text{Force at "b"} > \text{Force at "a"}$$ but not the work done or ...


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You are right. To open the door during the same time interval (for pushing at $a$ and $b$), you should induce the same angular acceleration. Since rotation in both cases is about the same axis, this means you need the same torque, this gives $$ \frac{F_a}{F_b} = \frac{r_b}{r_a} $$ where $r$ is the distance from the point of contact to the axis. However the ...


2

The most straightforward answer is no, you cannot consider a component to be a vector. A vector is something which has an associated direction, but a component has no direction. It's just a number. For example, if $\vec{F}$ is a vector, the component of $\vec{F}$ in the $\hat{x}$ direction is $\vec{F}\cdot\hat{x}$, and hopefully you know that the value of a ...


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If you only take the component along 45 degrees, you are neglecting the component perpendicular to it. This force counteracts the first component in the x direction. This then results in no net force in the x direction, as expected.


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I'm glad that my head is spheroid instead of pyramidal: then the circle is favored. The eggs have an extremely strong shape: then the circle is favored. The insects are extremely strong and they favor the round shape. The engineering/construction by humans are easier with rectilinear elements and the triangular shape are favored. If I were an ...


2

The short answer is, if you are making a bridge, triangles are, because the way they distribute weight when they are in a group makes them stronger. A single arch is stronger, but when you use lots of triangles when building a bridge it becomes stronger than using one arch. That is why we use triangles for most of our construction. There are tons of places ...


1

Triangular support(or triangle here) is unparalleled in terms of strength they provide to support load because all the hold mass is properly distributed across the support. You may disagree when g=0 :)


1

... smith may use what is called a "dead blow" hammer to reduce the rebound. Thus, measuring the force is not just a question of the instantaneous force of the hammer, but how much it "presses" after that first instant, its impetus so to speak. Now, one idea I had was to use a teeter-totter They are not good solutions. The dead-blow hammer in ...


-1

When the figure is symmetric about y-axis, $x_{cm}=0$. As to $y_{cm}$, I don't know an easy way without integration. Actually the integration is not so hard and I just checked that it is indeed $-r/\pi$. For a half circle with radius $r$, $y_0=\frac{4r}{3\pi}$. There are three half-circles with mass: 4, -1, 1 and corresponding $y_{cm}$: $-y_0$, $-y_0/2$, ...


1

If you cut something by pushing a blade directly into it, here's what happens: On first contact of blade with material, only the very thin edge of the blade is touching the material, the force per unit area is very high, and the blade cleaves the material very easily. That's why it's almost trivially easy to make score marks in things like aluminum using a ...


0

Isn't it just a change in the reference frame and therefore the forces are equal? Update: Suppose your initial conditions: the ball ($m_1$) hits the man ($m_2$). Its velocity is equal to $u_1$. As you've said: $\m_1(v_1-u_1)/t$=$\-m_2(v_2-u_2)/t$ Now consider that an observer is moving at the velocity $u_1$. The ball looks static and the man looks like he's ...


0

Here is an intuitive / qualitative answer. Maybe someone else will add some math. I wonder if it's instructive to look at diamond cleaving. As you know, diamond is extremely hard, and conventional machining is very difficult. But if you can find the right fracture plane ((111) and its symmetrical cousins), it's possible to cleave the diamond along that ...


0

The answer is no. Dark energy does not exert a force between objects, but causes space to expand. Objects floating in space are carried along, but remain still in space. That means that if you place two objects in a universe with dark energy, sufficiently far apart that their mutual graviational attraction can be neglected, the speed with which they recede ...


1

For mass $2m$, let $x$ be its displacement to the right from its initial location. For mass $m$, displacement to the right from initial location is $y$. This means the extension of the spring, $e$, equals $y-x$. So look at the forces acting on each mass, and apply Newton's 2nd Law to get the follow equations: $$3F - ke = m\ddot y$$ $$ke - F = 2m\ddot x$$ ...


1

At the instant of maximum extension of the spring both the blocks are moving with same accelerations. Hence the acceleration of (two blocks + spring) system can be found simply by $a=\frac{F^{ext}_{net}}{m+2m} = \frac{2F}{3m}$. Now since the $2m$ mass block is accelerated by only spring force, the spring force is given by, $F_s = F + 2m \times ...


3

I suppose you read this passage in the famous Feynman Lectures. I am fairly certain that what Feynman is referring to (and what you are looking for) is a proof that an electrostatic field is conservative. There are a number of equivalent ways of stating that a vector field is conservative, each of which can be taken as a definition. Let $\vec{F}(x)$ be a ...


0

I was told forces can depend on time, location and velocity, but never on acceleration. That depends on what one means by "force". The concepts of "force" as used in Newton's second law and in Newton's third law are distinct but related concepts. That force and acceleration are intimately connected with one another entire the point of Newton's second ...


1

So let's say you were to consider the forces act on one of the points where the string connects to the painting. Three forces act: The force due to half the load's weight, 2.5N vertically downwards. The tension of the string, direction is at a shallow angle above the horizontal. A sideways reaction force acting horizontally in a manner opposing the ...


2

$F = ma$ when mass is constant: I think that's a common misconception. It doesn't really represent the law correctly (unless mass is constant). Newton's second law of motion states, $F = d/dt(mv)$ the external forces acting on an object in an inertial frame is equal to the change in linear momentum (the measure of motion) I can see why he made that ...


2

An example of a force that depends on position (of the particle) is the force due to a spring: $$F_x = -kx $$ An example of a force that depends on velocity (of the particle) is the force due to a dashpot $$F_v = -c\dot x$$ Now, consider a hypothetical force that depended only on the acceleration of a particle: $$F_a = -d \ddot x$$ The differential ...


2

In terms of the Young's modulus the Hook's law (up to the overall sign) is written as $$F=ES \frac{\Delta L}{L_0}=kx, $$ the corresponding elastic energy (or the work that has be done to stretch a wire) is $$W=ES \frac{\Delta L^2}{2L_0}, $$ here you know the Young's modulus $E=1.3 \times 10^{10}$, the cross area $S=1.7cm^2$ and the initial length $L_0=0.89m$ ...


0

To clarify the law should be stated $$\sum F(t,x,v) = m\,a $$ which is used to solve problems by re-arranging into $$a(t,x,v)=\frac{1}{m} \sum F(t,x,v)$$ The point is that various forces depending on time, position and velocity are used to define accelerations. So yes, you are correct. A less obvious case is in problems of nverse dynamics. Here all ...


1

Unless you are talking only about classical mechanics, I think that it is not true. For example, the Abraham-Lorentz force depends on jerk (time derivative of acceleration).


1

Dark energy is a concept devised to help explain the expansion of the universe. It is presumed to comprise about 68% of the universe (on a mass-equivalence basis), but it is spread so uniformly throughout the universe that its density is on the order of only 10 to the minus 27 kilogram per cubic meter. Dark energy is not presumed to clump in matter, but ...


0

As a commenter noted, dark energy, whatever it turns out to be, is not thought of as a property of any known object. From the Wikipedia article "Dark energy": In physical cosmology and astronomy, dark energy is an unknown form of energy which is thought by some physicists to permeate all of space, tending to accelerate the expansion of the universe. ...



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