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In almost every place the statement regarding non-conservative forces is:

In the case of non-conservative forces, the work done is dependent on path taken between two fixed points. Longer the path, greater is the work done against friction.

In the last (italicised) part it is written that if path taken is longer work done is more, but work done depends on displacement only ,, no matter whatever is the distance covered the work done will be the same if displacement is same. Then, how in the case of non-conservative forces we consider more work when distance is more?

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The trick here is that work is not just the product of the force time the displacement, but rather the integral of the force over the displacement. $$W=\int_{C}\mathbf{F}\cdot\mathbf{ds}$$ For a conservative field, the work will only depend on the starting and ending points of the contour and $\mathbf{F}$ will be a function of position only.

For the example of friction as a non-conservative force, $\mathbf{F}$ will always be opposite in direction to $\mathbf{ds}$ in addition to possibly varying with position. This difference means that the work will grow as the path gets longer, assuming the magnitude of the frictional force remains the same.

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but work done depends on displacement only

That's not correct, as discussed below.

In physics, distance is a scalar quantity that essentially refers to "how much ground an object has covered" during its motion, whereas displacement is a vector quantity that refers to "how far out of place an object is" , i.e., the object's overall change in position. The general definition of work is

$$W=\int_1^2\vec F\cdot d\vec s$$

We see that work depends on the dot product of the force times the differential displacement of the object, not simply the product of the force and the displacement of the object. The dot product is a scalar quantity equal to the distance covered in the direction of the force between the initial and final position of the object. The work done by a non conservative force depends on that distance.

Case in point is work done against kinetic friction between two points. The direction of the kinetic friction force is always opposite to the direction of the motion of the object. When the force applied to a sliding object changes direction, the opposing friction force also changes direction so that the direction of the force is always in line with (and opposite to) the direction the displacement. Since the angle between the applied force and displacement is always zero, the dot product is unity (positive for the work done by the agent pushing the object and negative for the work done by friction) and thus the magnitude of the total work done is the average force times the total distance covered.

The displacement is a vector that points from the initial position to the final position of the object. When the work done depends only on the initial and final positions, and not the path between, the work is done by a conservative force.

Case in point is the work done by gravity when an object is raised to a height $h$ above the ground. Since the force of gravity is always vertically downward, the dot product (and thus work done by gravity) is only non zero for the component of the displacement in the vertical direction, and zero for any horizontal deviations in the path.

Hope this helps.

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  • $\begingroup$ Thanks !! I got to know my misconception. $\endgroup$
    – Learner
    Apr 10 at 5:16
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You are right that the statement is imprecise (or actually wrong). Consider the example of friction a sled experiences when dragged over a flat surface. Drag it in a circle of radius $10R$, then load 20 identical sleds onto it and drag it in a circle of radius $R$. The work in the latter case is greater (assuming that friction is proportional to the normal force caused by gravity).

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  • $\begingroup$ So is there any better way we can define non-conservative forces? $\endgroup$
    – Learner
    Apr 9 at 16:56
  • $\begingroup$ What? This follow-up makes no sense to me. You asked about work, now you want to redefine non-conservative forces (which you can't). $\endgroup$
    – kricheli
    Apr 9 at 17:12
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    $\begingroup$ A Conservative forces closed line integral is zero as is path independant. A non Conservative forces closed line integral doesn't have to be zero. The definition of a non Conservative vector field is a field where $$\nabla × \vec{F} ≠ 0$$ $\endgroup$ Apr 9 at 18:53
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[...] but work done depends on displacement only [...]

This is false. As you say, the work done by non-conservative forces is path-dependent. The total displacement is not enough to calculate the work done by non-conservative forces.

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