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Suppose we are given a 2D force vector field $F(x,y)$ (position dependent only, no time dependency) and we're given a path C in the plane, and we compute $\int_C F \cdot ds$ and get zero. I was told (without explanation) that this means the force cannot move any object from the start of the path to the end of that path. Question 1: Is this (generically) true? (That is, generally speaking, one can think of it this way, and why, or is it simply false.) Question 2: What if we compute $\int_C F \cdot ds > 0$. Would this mean it "could" push an object from the beginning to the end of the path?

Someone could come up with a better example, but here is an explicit question one could ask: Supposing $C$ was the parametrization of some track, if $\int_C F \cdot ds = 0$, and we put a cart (of any mass) at the beginning of the track, that force $F$ could not move the cart from the start of the track to the end of the track? If $\int_C F \cdot ds > 0$, then there would be a cart we could place at the beginning of the track and the force would move it to the end?

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  • $\begingroup$ Note that you never specified any restrictions on the force vector other than 2D. So could could be position or time dependent. Also, you never specified a velocity at the beginning of the path. so it could be non-zero. $\endgroup$ – Bill N Jan 24 at 22:03
  • $\begingroup$ @BillN I assume F=F(x,y) is position dependent and F is the only Force acting on anything put on the track. I assume we place a cart at the beginning of the track initially at rest. If $\int_C F \cdot ds = 0$, can the force move the cart from start to finish? If $\int_C F \cdot ds > 0$ would there exist a cart for which the Force moves it from start to finish? $\endgroup$ – Curiosity Jan 24 at 22:24
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No, that is not true, considering that you did not state that $\vec{F}$ is constant. The statement is concerning the integral over a specific path, so it is entirely possible for $$\int_C \vec{F}\cdot\mathrm{d}\vec{s} = 0 $$ without $\vec{F}\cdot\mathrm{d}\vec{s}= 0$ at every point.

If $\vec{F}$ is position dependent, such as a spring force, consider a mass attached to an initially compressed spring. It is released with zero kinetic energy. It moves to some extremum where the kinetic energy is again zero. The work integral is zero. But the object has moved.

On the other hand, if $\vec{F}$ is constant, but the initial velocity is not zero (\v_0 > 0), then the force could change the path, as in uniform circular motion, and the integral would be zero while the object moves from the beginning of the path to the end.

If the initial velocity is zero and the force is constant, then no, the work integral can't be zero.

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    $\begingroup$ A practical example: A skateboarder at the top of a half pipe starts skating down, and then up the other side. They stop at the other side. It's trivial to show that work of gravity = 0 (end at the same height and at same speed), but gravity can indeed pull the skater down from the initial position on the half pipe. $\endgroup$ – Cort Ammon Jan 24 at 23:34
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This is not "generally" true. To do work, a force must displace an object, and that displacement must be in the direction of the force or a component of the direction of the force. Examples:

1) A horizontal rope is pulling a box of mass $m$, along a sidewalk at constant velocity in the presence of friction. There is work being done, because the force on the box is horizontal, the box is moving horizontally, and there is a displacement involved.

2) A rope is being pulled upwards at a non-vertical angle $\theta$, and the box is moving horizontally down a sidewalk in the presence of friction. There is work being done, because there is a horizontal component of the force that is displacing the box horizontally down the sidewalk.

3) A rope is tied to a 1 kg sphere, which is being swung in a horizontal circle. There is displacement of the sphere in this case, and the rope is obviously moving the sphere in its path to make it go in a circle, but despite this, no work is being done on the sphere. The instantaneous velocity, and the instantaneous displacement of the sphere, are tangent to the circle that the sphere is swinging in, but the force on the sphere is always directed toward the center of the circle. That centripetal force is always at right angles to the displacement of the sphere, so no component of that force is in the direction of the displacement, meaning that from a physics standpoint, no work can be done on that object by the centripetal force that it is experiencing.

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Good Question!!

$\int_C \vec{F}\cdot\mathrm{d}\vec{s}$ yields you 0 in 2 cases:

1)The displacement of the object is 0.

2)The Force and ds vectors are always perpendicular.

What you were told was only the 1st case. Answer to your 1st question: No

Consider an object tied to a string moving in a circular path(with constant speed). The tension in the string wont do any work as it is always perpendicular to the displacement but still tension was able to move the ball in the circular path, isn't it.

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Yes. I believe all you have described here is true. Just consider gravity as an example. If you define a path over flat ground, then gravity will never ever be the cause of any object moving along that path. The product of force and displacement along this path will always be zero, because no displacement happens, which can be contributed to that force.

This is what the dot product tells us: a dot product equalling zero mathematically means that the two vectors (force and displacement) have no parallel components. And as we know from Newton's laws, a force causes motion along its own direction. A force will always be parallel to the displacement it causes (or at least a component of the force is parallel to the displacement it causes). Surely, if any displacement does happen, it is not due to a force that has no component pointed that way. No work by gravity is done on an object moving over flat ground.

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A simple counter example:

  1. I push on an object, from a standing start, with a force F for L meters. It gets up to some sped and moves.

    1. Later, I push with magnitude F but opposite direction. After a length L, it comes to a stop at the end of its path.

The total Fds is zero by construction but it moved the object.

More generally, any restoring force oscillation is going to have this property. Pull a pendulum aside and let it go. From one side of the swing to the other, the force does zero total work between the start and stop of the swing. Yet the pendulum moves!

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  • $\begingroup$ I wasn't clear, but I want the force vector field F to be F(x,y), position dependent only. (No time dependency like in your example.) $\endgroup$ – Curiosity Jan 24 at 23:06
  • $\begingroup$ Not a problem. See paragraph added at end. $\endgroup$ – Bob Jacobsen Jan 24 at 23:10
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Whenever work was done by the force on the particle(assuming that mass of the particle is constant), then there is a change in the kinetic energy of that particle. Whenever work done is zero around a path length, one of the possibility is that the component of force is zero along that path length. Another possibility is there are components of forces at every point along the path length, but eventually gets canceled throughout the path. Also,whenever the work done on particle is zero,it doesn't imply that the state of particle is unchanged.But the speed of the particle is unchanged,not the velocity.

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