Now the case is not like this what you are thinking the displacement will be zero and then net workdone will also be zero let us find out how,
First of all as you know that
$W = |F||S|cosθ$
So, for simplicity in this case the overall displacement is zero and it's fine but if we break this case in...
Yes, that's what I'm trying to say because we've fixed the origin at the point A (from where the body actually started moving) and displacement is the position vector so if we go behind this origin then the displacement vector will come out to be negative (considering motion in 1 dimension), although I've edited the post and explained it very thoroughly than before you can check that out as well.
No. Displacement is change in position, ie final position - initial position. During the first segment displacement is A to B (B - A) and the force is also in the direction from A to B so the work done is positive. During the second segment the displacement is from B to A (A - B) and the work done is also from B to A, so the work is again positive. If the force is constant then zero displacement results in zero work, but the force is not constant - its direction reverses.
@Peter the statement you've said "If the force is constant then zero displacement results in zero work, but the force is not constant - its direction reverses" is completely wrong and misleading just consider the case of SHM and calculate the workdone for starting from position zero to amplitude and from its amplitude back to mean position you will get the net work done which will come out to be zero.
(Our case) The displacement is still positive but the direction of force is reversed so the workdone will come out to be negative.
If once we've fixed the directions then you can't manipulate it and consider it as a new case.
@TejasDahake Firstly, I said if, not only if. A car accelerating around a circular track will have changed energy back at the start. Secondly, displacement during an interval is not the position vector but the change in the position vector (ie $\Delta x$). I know the word displacement is sometimes used to mean the displacement from the origin (ie position vector), but that is clearly not what is meant here. An example meeting the description of the problem would be a car accelerating from A to B, performing a U-turn without changing speed, then accelerating further from B back to A.
@Peter first of all I've taken the position at which the object was initially at rest (i.e A) as origin you are very correct that displacement is change in position but here it is a simple case so considering the point A as origin will not be an absurd thing. second of all, please be clear to your statements $if$ and $only$ $if$ means the same here because once you have made the statement and the readers will interpret the same. Third of all....
If once we've considered the direction form A to B as positive then you can't do anything you have to solve as it is, the case you've considered of a car going from A to B and then from B to A means that displacement is zero, are here we are calculating NET WORKDONE ON THE OBJECT, we are not calculating the total workdone by all the forces during this whole journey......
They are both the different things I think I'm repeating myself hope it's clear to you now what I'm trying to say.
The workdone you are calculating here as 200J is only true when you will manipulate the directions you are considering that from B to A its positive that's why because of this the direction of the force will come out to be positive and the angle between them will be zero and hence the workdone will come out to be positive so by adding them you will get 200J.
Yes. OP's problem appears to be that he thought getting 200J was not compatible with going back to the starting point. For a constant force situation (like gravity) that would be a problem. When the force varies anything could happen - you need to calculate it. My car suggestion is the only half-realistic example I could think of. BTW I suspect some moderator is likely to delete this whole conversation, especially as the question has been closed.
Yes that's what the case, I've fixed the directions and considered only origin (point A) as the objects initial point and you've considered the objects initial point as point A and as well as B that's why this thing happend. And also you've manipulated the directions.
Yes the example seems reasonable because in this case the forces were exerted by a person or something and in this case (of car) the same role of the force will be done by the friction!
I suppose another example might be a book pushed by a hand on a long rough bench where friction would stop the book (almost) instantly if the force ceased. In both directions the hand force would do positive work, while friction would do negative work. Total work of all forces would be zero with 200J done by the hand and -200J done by friction.
Anyway, it's after midnight for me so I need to go to bed. Thanks for your contributions. I hope OP learned something.
Discussion between Tejas Dahake and P…
Discussion between Tejas Dahake and P…