# Tag Info

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While you do spend some body energy to keep the book lifted, it's important to differentiate it from physical effort. They are connected but are not the same. Physical effort depends not only on how much energy is spent, but also on how energy is spent. Holding a book in a stretched arm requires a lot of physical effort, but it doesn't take that much ...

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If you're pushing a 10-ton truck and it's not moving, you are not doing any work on the truck because the distance $ds=0$ and the nonzero force $F$ isn't enough for the product $F\cdot ds$ to be nonzero. Your muscles may get tired so you feel that you're "doing something" and "spending energy" but it's not the work done on truck. You're just burning the ...

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It is all about the loss of energy during each stride - the tendons store some energy, but not a lot. A kangaroo and a greyhound, for example, have far more efficient elastic storage in their legs / tendons, allowing them to achieve (and maintain) greater speeds with less effort. Key phrase from the abstract in that reference: elastic storage of energy ...

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In introductory problems about work you're normally taught that it's force times distance: $$W = F \times x$$ and you treat the force as constant. If you look at the problem this way then you're quite correct that if the force is $F = mg$ then the box can't accelerate so it can't move. However a more complete way to define the work is: $$W = ... 24 By Noether's theorem, there is a conserved quantity (a number) associated with every continuous symmetry of a physical system. Energy is by definition the conserved quantity associated with time translation invariance (i.e. that it doesn't matter whether we perform an experiment today or tomorrow, given that all circumstances are the same). In this sense, it ... 21 The Lorentz force \textbf{F}=q\textbf{v}\times\textbf{B} never does work on the particle with charge q. This is not the same thing as saying that the magnetic field never does work. The issue is that not every system can be correctly described as a single isolated point charge For example, a magnetic field does work on a dipole when the dipole's ... 21 This is about how your muscles work -- the're an ensemble of small elements that, triggered by a signal from nerves, use chemical energy to go from less energetical long state to more energetical short one. Yet, this obviously is not permanent and there is spontaneous come back, that must be compensated by another trigger. This way there are numerous ... 16 Your calculation is incorrect. \text{Work} = \text{Force} \cdot \text{displacement} = F \cdot s The above product is a "dot" or "scalar" product, which means we only consider the displacement that occurs in the direction of the Force, which in the case of gravity is downwards. Can we set this vertical displacement to 0? No we cannot, and here is why: ... 14 This does not contradict newton, because the error is in the calculation: You are calculating in discrete time steps. In that case the calculated speed will increase, as if energy is not preserved. However, Newton's laws apply to continuous time. In the mathematical world in which newton laws are described, speed and acceleration are NOT defined by ... 13 If the cage is completely closed, it doesn't make a difference if the bird is hovering inside it or if it sits on the ground. When flying, the bird pushes air to the ground which will exert a downward force on the cage exactly equal to the weight of the bird. This is a direct consequence of the conservation of momentum and Newton's second & third law. ... 12 The half in the non-relativistic kinetic energy can be traced back to the Work-Energy Theorem^1. Of course, if one is only interested in solving an elastic collision problem for an isolated system of point particles using momentum and kinetic energy conservation, no harm is done by multiplying the energy conservation equation with a factor 2 on both ... 12 The factor of 1/2 is required from Galilean invariance, so that the energy mixes up with the momentum without a factor. This was understood before relativity, but it is largely conventional before relativity, since you could make the energy mix up with the momentum using some coefficient. Once you have relativity, the 1/2 is no longer optional. I'll start ... 11 To prove that experimental Physics is alive and well, I used my kitchen scales to measure the force needed to click the button on my mouse, and it turned out to be 100g i.e. 1 N plus or minus about 10%. The distance the button moves is about a millimeter i.e. 0.001m, plus or minus 20% (OK - you try measuring it without a micrometer to hand) so the work per ... 11 Yes, of course that if a field - magnetic field - is able to make a bar magnet rotate or move, it is doing work. The statement that magnetic fields don't do any work only applies to point-like pure electric charges. Magnetic moments may be visualized as objects with a forced motion of charges (solenoids have the same magnetic field as bar magnets), and if ... 11 What I've been telling the students in the intro mechanics class I TA is that each force corresponds to some amount of work. The individual contributions of work add up to the net (or total) work, just as the individual forces add up to the net force. So, for instance, when you lift a box up from the floor to a table, there are two forces acting on that ... 11 Perhaps an analogy is in order. Lets hold up the book by using an electromagnet (say we put a piece if steel under it ). If the coils were made of superconducting material it would take no energy input to maintain the position/field strength. But if we use ordinary wire, ohmic loses within the coil must be made up for by externally supplied electrical ... 10 The reason is that you need to spend energy to keep muscle stretched. The first thing you need know is that the work W=F \Delta x is the energy transfer between objects. Hence, there are no work done on the book when it is put on the table because there are no movement. When your arm muscle is stretched, however, it consumes energy continuously to keep ... 10 So, you require a simple answer... Let us consider a body of mass m at rest. Initial velocity u=0. Now, The body moves with a velocity v. Force on the accelerating body is F=ma,$$\implies F=m\frac{dv}{dt}$$The small amount of work done in moving the body over a small distance ds is$$dw=F.dsdw=m\frac{ds}{dt}dv=mvdv$$Hence, the total work ... 10 In polar coordinates, the velocity being tangent to the circle, it is directed along the \hat{e}_{\theta} vector. The centripetal force is directed along the \hat{e}_r vector. So \frac{F}{m}\hat{e}_r = \frac{d\vec{v}}{dt} = \frac{d(|v|\hat{e}_\theta)}{dt} \underbrace{=}_{\text{Chain Rule}} \frac{d|v|}{dt}\hat{e}_{\theta} + ... 10 Not really, no. A field is conservative if it can be written as the gradient of a potential function:$$\mathbf{F}=\nabla f$$You could equally well define the potentials f=kr^2, or f=kr^5, etc., which will produce vector fields that are respectively linear and quartic with r. 9 Recall that a force perpendicular to the direction of motion does no work but simply changes the direction of the velocity vector. The same thing happens here: Initially the ball's motion is perpendicular to the force of gravity and hence at this very moment, gravity does no work but slightly "rotates" this velocity vector towards the downward direction; as ... 8 To add to Dirk Bruere's answer. As I describe in detail in this answer here, the accountant's idea of a budget is a good analogy: the budget seems "abstract" but the law of conservation of energy is experimentally proven and in this way energy is very "real": a system fulfilling conservation of energy behaves in a way that is measurably very different from ... 8 You say: Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out and the book moves with a constant velocity. so I'm guessing your reasoning is that the net force on the book is zero so the amount of work done on the book is zero. And you are absolutely correct - no work is done on the book and ... 7 The answer to your question lies in simple dimensional analysis. Joules are the units of energy, so also the units of work. Work, as we all know, satisfies the relationship$$W=\vec F \cdot \vec s$$Meanwhile,$$\vec F= m \vec a$$Substituting in this relationship gives us$$W=m\vec a \cdot\vec s$$Now, let's look at the units of this equation: ... 7 The kinetic energy is given by E_c=\frac12mv^2, as the factor 1/2 is dimensionless, you can see that \mathrm{[m^2.s^{-2}]=[J.kg^{-1}]}. Dimension analysis remains correct if the velocity v takes the value c, because c is also a velocity. 7 Officially, I completely agree with the other answer given. I would like to offer this answer as a simplistic, intuitive answer to the question. No math involved. I understand where your question comes from. In fact, depending on your current level of education, this question could indicate a high potential for future scientific success. We all know that ... 7 In the context of classical mechanics as you describe, negative work is performed by a force on an object roughly whenever the motion of the object is in the opposite direction as the force. This "opposition" is what causes the negative sign in the work. Such a negative work indicates that the force is tending to slow the object down i.e. decrease its ... 7 The factor \frac12 comes in because we're integrating the equation$$ \frac{\mathrm dE}{\mathrm dv}=mv $$once. Less abstract and only using basic arithmetics, the story goes like this: When accelerating a body by applying a (constant) force F along a distance \Delta s, the body gains energy according to$$ \Delta E=F\Delta s  which is just the ...

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Ayush: Isn't the question telling that the bullet always loses 1/n th of its velocity no matter which plank? Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing $1/n^\text{th}$ of its velocity per plank (however, the fact that the question does not ...

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If the displacement of the object is zero, then one can calculate the work done by each individual force, the work done by each force is zero. Why? Work is not defined in terms of what would have happened to the object in the absence of other forces; it is defined in terms of the motion that actually occurred. More concretely, if from time $t_a$ to time ...

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