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According to my text book Total Work = Delta Kinetic Energy = KEf - KEi

But then work is defined to be dot product of Force (vector) and Displacement (vector).

Also to my knowledge work is positional.

So, if we assume an object running in circle, and it completes one cycle,

Is it correct to say net work = 0? or is net work = total work?

I have one more question,

if gravitational force is only thing acting on the system, in which the object is moving downward vertically, do we say Work is Kinetic Energy, and Gravitational Force is Potential Energy? or the opposite of what I think it is?

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So, if we assume an object running in circle, and it completes one cycle, Is it correct to say net work = 0?

No. It depends the nature of Force field against which you are doing work. I say force field because it is a technical term used to identify the direction and magnitude of the Force a body will experience in the given region of space. For e.g the gravitational force field.

Now to prove you wrong, I will let you work out a counter example. Consider you are sliding along the circumference inside a friction less torus loop. Also consider there are no gravitational or viscous force of any kind. enter image description here

Once you are set to motion inside the torus, you will keep on moving inside it. Now consider a stream of water made to run in your opposite direction inside the torus. If you didn't apply any effort (force) against the flow, you will eventually stop losing energy on colliding with the incoming water molecules and continue motion along the direction of the water stream. This water stream can be visualized as a force field $V = v(r)\hat\theta$ (try to find what the terms mean by yourself). Consider also that you have a motor of some kind that will help you to steer forward against the flow. If you set it on, you are working against the water stream or the force field. In other words you are spending energy. Now think of what happens when the velocity of water flow is different in different $\theta$. i.e. $V = v(r,\theta)\hat{\theta}$. Hint: Consider a simple function and find the line integral. In either case you are either spending energy (positive work) or gaining energy (negative work).

Image courtesy : http://pages.vassar.edu/magnes/advanced-em/derek/

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Work is defined as the line integral $\int \mathbf{F} \cdot \mathbf{d\ell}$. The force on an object can be a function of position or time, and could represent external forces placed on the system. Net and total work refer to the same concept, the sum of all work done on an object.

For your example, you cannot simply say work is 0 because the object returns to its starting location. Say your object is a block, initially at rest, which I push around the entire circle. Assuming, I don't apply a force to stop the block, it starts with 0 Kinetic Energy and ends with some kinetic energy $K$. As $W=\Delta K$, I have clearly done work on the block.

There is a case where the work done would be 0, which is if the force on the object was conservative and exclusively position dependent, like a gravitational field.

Concerning the gravitational force, we say the gravitational does work on the object, giving it kinetic energy. The work the gravitational field does is, by conservation, exactly equal to the amount of potential energy it loses.

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Work is equal to force multiplied by displacement. Despite this seemingly simple explanation, there are several caveats to keep in mind:

1) Only the displacement that is parallel to the "resisting" force that is involved, contributes to work. Thus, if I carry a hole puncher across my class room at constant velocity, and ignore the acceleration that was involved in getting it to constant velocity, I am not doing any work on it because the resisting force is gravity, which acts downward, and I am only moving the hole puncher horizontally.

2) If I am sliding the hole puncher horizontally across my desk, work is involved, because the resisting force is friction, which acts horizontally, and I am displacing the hole puncher horizontally, which is parallel to the resisting force.

3) If I am pushing the hole puncher across my desk with a force that is equal to the force of friction, there is no net force on the hole puncher, which will move at a constant velocity. I am doing positive work (pushing in the same direction as the displacement) and friction is doing negative work. This leads to the concept of "net work", which is equal to the net force on the object multiplied by its displacement. If the net force is zero, the net work is zero.

4) If I can find a frictionless desk and push the hole puncher, there will be no dissipative forces trying to stop me. In that case, the work/kinetic-energy theorem definitely applies, and the work that I put into the hole puncher will indeed equal its change in kinetic energy. This means that your text book used an implicit assumption of no dissipative forces (i.e., friction) when work was applied to an object.

5) If you push an object in a circle across a frictionless horizontal surface, there will be no dissipative forces involved, and when you end back up at your starting point, displacement will be zero and work will be zero.

6) If you push an object in a circle, at constant speed, across a horizontal surface which is "rough" (friction is involved), there will be work involved all the way around the circle as friction tries to stop you. In this case, the positive work that you do will be matched by the negative work that friction does. Net work will be zero, and all of the work that you put into this experiment will heat the surface of the desk and the object that you pushed.

7) If you lift an object straight up, you are doing work against gravity. If you then slowly lower the object, gravity is doing work against you. If the object ends up at its starting point, the positive work and negative work are equal, so no net work was done.

The "normal" notion of work is often subtly and substantially different from the physics definition. Positive work, negative work, net work, and zero work, require a very careful specification of the conditions under which the work was done. This naturally means that you are not likely to be able to read a problem that involves forces and displacement, and immediately plug numbers into an equation to arrive at a correct answer. Only by working a variety of problems can you get the intuition to know what hidden assumptions are contained in the problem statement.

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I am trying to go to a bit basic level. The formula work=Force*Displacement works only if the force is constant and not changing its direction or magnitude. When an object moves in circle,the force continuously changes its direction. So to calculate it we have to use integral of F with dl,assuming that force remains constant for a very short displacement dl. And net work and total work are same,just two different words of english.Also if there is a conservative force in space,the work done by that force does not depends upon in what path is object moving. It just depends upon final displacement in direction of force.

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  • $\begingroup$ The formula work=Force x Displacement works only if the force is constant and not changing its direction or magnitude. This statement is absolutely wrong. Work done around any loop in a conservative field $F$($\bigtriangledown \times F = 0$) is 0. The constant field is merely a special case. $\endgroup$ – Sathyam Dec 15 '15 at 9:55
  • $\begingroup$ Would you be kind to provide correct version of my statement? $\endgroup$ – user247833 Dec 21 '15 at 12:47
  • $\begingroup$ Please edit your answer to include mathematical formulas. $\endgroup$ – Sathyam Dec 21 '15 at 13:46

protected by Qmechanic Oct 3 '15 at 19:11

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