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Why can't the work done by a non-conservative force be zero? The displacement along a closed path is always zero. So, whatever be the type of force, variable or constant, the work has to be zero. Why do we need to calculate the work done for individual paths?enter image description here

This is a non-conservative force that starts from $A$ moves via Path 1 to $B$ and then back to $A$ via Path 2. Since the displacement is anyways going to be zero, why can't work done be zero?

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    $\begingroup$ The work done can be zero is particular cases, but the definition of "non-conservative" is such that it is not zero in general. So you have to compute it each time rather than relying on a handy rule like you can for conservative forces. That is real physical forces get classified as "non-conservative" because there is no handy rule that depends only on the endpoints of the path. $\endgroup$ – dmckee Jun 29 '14 at 18:42
  • $\begingroup$ Doesn't that violate the definition of work itself? Wouldn't that mean that W= F.d is valid only for conservative forces, because for the cases of non-conservative forces, displacement seems to have least significance in calculating the work done? $\endgroup$ – Swami Jun 30 '14 at 3:54
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    $\begingroup$ The $d$ is not displacement, it's distance traveled. In the infinitesimal form those are the same, but not in the finite form. Many books choose a different symbol (often $s$) to emphasize this. $W = \int_s \vec{F} \cdot \mathrm{d}\vec{s}$ or similar. $\endgroup$ – dmckee Jun 30 '14 at 13:27
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For forces that change along the way, displacement is not the thing to calculate work with. Let $\gamma : [0,1] \rightarrow \mathbb{R}^3$ be the (closed or open) path that the particle the force is exerted on follows. Then, the work done along that path is

$$ W[\gamma,F] = \oint_\gamma \vec{F}(\vec{x})\cdot \mathrm{d}\vec{x}$$

which is a line integral. If $\vec{F}$ is conservative, there is a function $V(\vec{x})$ such that $\nabla V(\vec{x}) = \vec{F}(\vec{x})$, then we can apply Stokes' theorem (or, less fancy, the fundamental theorem of calculus) to calculate the work by

$$\oint_\gamma \vec{F}(\vec{x})\cdot \mathrm{d}\vec{x} = \int_{\partial\gamma} V(\vec{x}) = V(\gamma(1)) - V(\gamma(0))$$

For closed paths, $\gamma(1) = \gamma(0)$, so this is zero. If there is no potential with $\nabla V = \vec{F}$, we cannot apply this argument and have to actually calculate the line integral, which may be anything, especially not zero.

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  • $\begingroup$ Spring force is a variable force, but is conservative. I am trying to understand what inspires us to calculate the work done using integration? Is it force being variable or non-conservative? $\endgroup$ – Swami Jun 30 '14 at 3:56
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    $\begingroup$ @Swami: The first formula is the definition of work for all forces. For constant forces, $\vec{F}\cdot\vec{x}$ (with displacement $\vec{x}$) is the way to calculate the work, and for constant forces, the integral is just this. But for forces that vary along the way, we need to do the integral - it represents applying $\vec{F}\cdot\vec{x}$ for every infinitesimal part of the path and then summing over all these infinitesimal parts, hence the suggestive notation $\vec{F}(\vec{x})\cdot\mathrm{d}\vec{x}$. $\endgroup$ – ACuriousMind Jun 30 '14 at 11:12
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Well, we can do a simple counter-example. Let $$ \vec{F}(\vec{x}) = F_0 \cdot \varrho(\vec x) $$ where $\varrho$ is the function that rotates vectors by 90° counter-clockwise (in matrix form $(\begin{smallmatrix}0 & -1\\1 & 0\end{smallmatrix})$ if you prefer that). Clearly, for the closed path $$ \vec{\gamma}\colon\quad [0, 2\pi]\ \to\ \mathbb{R}^2 , \qquad t\ \mapsto \begin{pmatrix}\cos(t) \\ \sin(t)\end{pmatrix} $$ we have always $\dot{\vec\gamma} \cdot \vec{F}(\vec\gamma) = F_0$. So $$ \oint_\gamma \mathrm{d}{\vec x}\cdot \vec{F}(\vec{x}) = 2\pi\cdot F_0. $$ That's basically the answer to your question, although you've worded it wrongly:

Why can't the work... be zero

It can actually be zero. Consider $$ \vec{\gamma}_2\colon\quad [0, 2\pi]\ \to\ \mathbb{R}^2 ,\qquad t \mapsto \begin{cases}\vec\gamma(t) & \text{if $t<\pi$} \\ \vec\gamma(2\pi - t) & \text{otherwise}\end{cases} $$ This simply takes half the path of $\gamma$, but then turns around and goes back the same way it came, thereby integrating the opposite force scalar product, so the result comes out as $0$ here although the force, as I proved above, is not conservative. Only, in a non-conservative field not all closed paths have zero total work. (More obviously: in any field the closed "path" that simply stays at the same point forever has zero work.)

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While the total displacement as shown in your figure is zero, this does not mean that the work is zero! Work is force scalar displacement, $W = \vec F\cdot \vec s$. Diving the path $A \to B \to A$ into infinitesimal steps we are led to $$dW = \vec F \cdot d\vec s$$ and for the total work, adding up the contribution $$W = \int \vec F\cdot d\vec s.$$ I think that from here, you want to distribute the integral over the product, $$W \overset{?}{=} \int F \cdot \int d\vec s$$ and since the total displacement is $0$, the product should vanish. But this equality does not hold in general, which is why you can't conclude that total work done vanishes just because total displacement does.

Of course as other answers have remarked it could happen by accident that the total work is still zero, it is just that for a non-conservative force we are not guaranteed that it does.

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  • $\begingroup$ Your expression suggests that F has to be variable. Would that mean that every non-conservative force has to be a variable force only? $\endgroup$ – Swami Jun 30 '14 at 3:47
  • $\begingroup$ Why do I need to divide the path to calculate the work done? Why can't I simply do the dot product of force (a constant force, say friction, which is non-conservative) and displacement (which is zero for a loop)? $\endgroup$ – Swami Jun 30 '14 at 3:50
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    $\begingroup$ My expression allows $\vec F$ to vary but does not require it. For a loop, it is not possible for friction to be constant: it is always in the direction opposite to the motion, so it must vary in direction, or we can't come back to where we started. $\endgroup$ – Robin Ekman Jun 30 '14 at 3:59
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As non conservative forces are those forces whose work done depend upon path followed by a particle. So around a closed path it's value will be positive because some path have been followed by particle in moving Around a closed path. But in case of conservative forces work depend only upon initial and final points not on path. So around a closed path initial and final points coincide so work done is zero.

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