# Tag Info

## Hot answers tagged work

39

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 ...

35

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 ...

23

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 = ... 20 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 ... 19 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 ... 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: ... 13 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 ... 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 ... 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 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 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$$You could equally well define the potentials f=kr^2, or f=kr^5, etc., which will produce vector fields that are respectively linear and quartic with r. 9 Recall that a force perpendicular to the direction of motion does no work but simply changes the direction of the velocity vector. The same thing happens here: Initially the ball's motion is perpendicular to the force of gravity and hence at this very moment, gravity does no work but slightly "rotates" this velocity vector towards the downward direction; as ... 8 So, you require a simple answer... Let us consider a body of mass m at rest. Initial velocity u=0. Now, The body moves with a velocity v. Force on the accelerating body is F=ma,$$\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$$Hence, the total work ... 7 Officially, I completely agree with the other answer given. I would like to offer this answer as a simplistic, intuitive answer to the question. No math involved. I understand where your question comes from. In fact, depending on your current level of education, this question could indicate a high potential for future scientific success. We all know that ... 7 In the context of classical mechanics as you describe, negative work is performed by a force on an object roughly whenever the motion of the object is in the opposite direction as the force. This "opposition" is what causes the negative sign in the work. Such a negative work indicates that the force is tending to slow the object down i.e. decrease its ... 7 Ayush: Isn't the question telling that the bullet always loses 1/n th of its velocity no matter which plank? Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing 1/n^\text{th} of its velocity per plank (however, the fact that the question does not ... 7 If the displacement of the object is zero, then one can calculate the work done by each individual force, the work done by each force is zero. Why? Work is not defined in terms of what would have happened to the object in the absence of other forces; it is defined in terms of the motion that actually occurred. More concretely, if from time t_a to time ... 6 Well, you simply need to accept that work is given by Force time Distance, and it doesn't matter how long it takes. For example, the work done on a mass m lifted a distance h against gravity with an acceleration g is given by:$$W=F\times h=mgh$$If you are told that someone is going to drop a 1 kilogram mass on your head from a height of 10 ... 6 Magnetic fields never do work directly. This is because the magnetic force on any charged particle,$$\mathbf{F}=q\mathbf{v}\times \mathbf{B,}$$is always orthogonal to the velocity, and therefore the power transferred, \mathbf{F}\cdot\mathbf{v}, is zero. On the other hand, this seems to contradict much of our intuition about how magnets behave. If you ... 6 How is it proved to be always true? It's a fundamental principle in Physics, that is based on all of our currents observations of multiple systems in the universe, is it always true to all systems? Because we haven't tested or observed them all. Could it possible that we discover/create a system that could lead to a different result? A physical theory, ... 6 When it comes to modeling the human body and its movements it becomes a lot more complex, because you actually have to model the different components of your body and how they attribute to the system. Plus you are not traveling at a perfectly constant velocity when walking at a steady pace. If you were to model the work done by the human body while ... 6 It is a matter of convention. The sign convention of Clausius and the sign convention of IUPAC are the two prevailing sign conventions. Both of these assign a sign to the work done differently. The former, used primarily in physics assign a positive sign to the work done by the system while the latter assigns positive sign to the work done on the system. ... 6 In addition to the other answers, bear in mind that  E = \frac 1 2 m v^2  is a quadratic form. Quadratic forms arise whenever you integrate an f(x) = k\cdot x function. Examples:$$ E_{kin} = \int m v\; \mathrm d v = \tfrac 1 2 m v ^ 2 \\ A_\circ = \int \tau r \; \mathrm d r = \tfrac 1 2 \tau r^2 \quad \tau = 2\pi \\ E_{spring} = \int k x ...

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The reason is because of the Work-Energy theorem, which states that if a force is applied on an object over a distance, that the integral of the force along the distance, or the work applied on the object, must equal the change in the kinetic energy, as the energy transferred through work must be conserved. So for an object at rest, if a force $F$ is ...

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