1. Jan 13, 2017

### Kaneki123

Okay...I have an example of work calculation in my book in which a girl has carried a bag upstairs( i.e. to some height)...The force the girl has applied is taken equal to the weight of bag( W )...According to my understanding, The girl will apply a force greater than W when she lifts it up from floor, will apply a force equal to W in transit, and will apply a force less than W when putting it down..(Correct me if wrong)...I have a question here that why we are taking the force equal to W when calculating work when that force is not actually causing displacement?????

2. Jan 13, 2017

### Staff: Mentor

You don't have to take it as equal but both quantities are proportional, so as longs as you don't want to have some exact figures according to a certain system of units, one could as well see the proportion factor to be one. Why do you think there is no displacement? She carried the bag upstairs, didn't she? Do you know the connection between force and work in this context?

3. Jan 13, 2017

### Kaneki123

Work is done when a force displaces a body(in its direction)...The girl applied a force (>W) when picking up the bag which give it some vertical velocity...However during her transition from down to upstairs, she does'not cause any net force on the bag...(and this is time time when she is applying W force)...In other words , the W force did not cause displacement, the >W force did....

4. Jan 13, 2017

### Staff: Mentor

She increased the potential energy of herself and the bag, because she extended the distance between the two masses earth and bag. She applied a force opposite to the gravitational force of the earth.

5. Jan 13, 2017

### A.T.

Because "causation" is irrelevant to the definition of work.

6. Jan 13, 2017

### sophiecentaur

Yes. It's Work Done On and not Work Done By. However the bag gets up there, the work is the same.

7. Jan 13, 2017

### A.T.

No idea what that means. My point was that "the cause of motion" is irrelevant to computing work. In fact "the cause of motion" is ambiguous, because motion (and work) depend on the choice of reference frame, while the force (as the supposed "cause") doesn't.

8. Jan 13, 2017

### gmax137

The OP seems not to understand that the relevant displacement is the elevation change from downstairs to upstairs. The lifting of the bag from the floor to the carrying position is also a displacement normally ignored in such examples. The final act of letting the bag back down to the floor (the upstairs floor) cancels that out anyway. The only things that matter (to the "net" work) are the initial and final elevations.

9. Jan 13, 2017

### Staff: Mentor

Work depends on the frame of reference. From the frame of reference of the earth, there is an upward force exerted on the bag, and it is displacing upwards, so work is done (as reckoned from this frame of reference). From the frame of reference of the girl, there is an upward force exerted on the bag, but it is not displacing upwards, so no work is done (as reckoned from this frame of reference).

10. Jan 13, 2017

### Kaneki123

Which did not cause any kind of displacement in the body...Which just stopped ''retardation'' of bag...(i dont know if stopping retardation is considered as causing displacement)...If that force did not cause displacement, then it should not do work...??

11. Jan 13, 2017

### Kaneki123

Work is when a force displaces a body...In the considered case, the force does not displace a body...(unless stopping retardation is considered a displacement)

12. Jan 13, 2017

### Staff: Mentor

As I said, as reckoned from the girl's frame of reference, no work is done. When one is talking about the amount of work done, one must specify what frame of reference is being used to reckon the work.

13. Jan 13, 2017

### jbriggs444

Work is done when a force is applied to a body which is displaced. Regardless of what other things might be deemed to cause the displacement.

14. Jan 13, 2017

### Khashishi

Kaneki, why are you saying there's no displacement? The bag was moved to a higher elevation. That's a displacement.
Now, if the girl was riding an escalator up while carrying the bag, then she didn't do the work of raising the elevation of the bag. But if she climbed the stairs while carrying the bag, she did the work. She had to counter the weight of the bag by pushing down on each step with her feet.

15. Jan 13, 2017

### Staff: Mentor

That is the work done on the bag, as reckoned from the laboratory frame of reference. As reckoned from the girl's frame of reference, there was no displacement and no work done on the bag (by the girl). There is no inconsistency here.

16. Jan 13, 2017

### jbriggs444

As I understand @Kaneki123's concern, it is not the reference frame which is the problem. It is the question of causation -- the [mistaken] idea that a force only does work when it causes motion.

The lack of a displacement is noted not because @Kaneki123 is considering the girl to be at rest. Rather, it is because the bulk of the displacement takes place during a steady state when the bag ascends the stairs at a steady rate under a net force of zero. During this phase, the support from the girl and the down-force from gravity are in balance. @Kaneki123 sees this as a condition where the movement which she agrees does exist does not count toward work done because the force applied by the girl during this interval does not cause the motion.

However, as @A.T. has pointed out multiple times, the notion of causation is simply irrelevant.

17. Jan 13, 2017

### Staff: Mentor

@Kaneki123 Is that actually what you are saying?

18. Jan 13, 2017

### A.T.

19. Jan 14, 2017

### Kaneki123

Okay I understand now...One last thing, if work is done when displacement is in same direction as force, so where does the concept of ''negative work'' comes from???

20. Jan 14, 2017

### jbriggs444

When displacement is in the direction opposite to the force being considered.

For example, the force of a man's hands on a thrown ball that he is catching.

21. Jan 14, 2017

### sophiecentaur

Not too difficult for you, I'm sure' . Your post says the same thing.
"On and By": However you perform an action 'on' an object, the work done on it is the same. This may involve a lot more work being done 'by' you and this is what confuses people when they describe the Energy they use, just holding a heavy weight without lifting it up. Efficiency is a very understandable reason for the difference. [Edit: The idea of 'dead weight' is also relevant here. The hook and pulleys on a crane can be of similar mass to the object they lift. That can double the work done 'by' the crane, compared with that done 'on' the load. This may appear to be trivial but it is highly relevant to first timers.]
Reference Frames are, of course, relevant but that term introduces a further level of complexity into a simple everyday situation. It's always possible to ramp up a conversation by introducing more Scientific concepts. I try to address the basic concerns of the OP first, when it's possible. I would suggest that the question was posed in a School scenario and I doubt whether that the term 'Reference Frame' would have been mentioned at that stage of the course.

Last edited: Jan 15, 2017
22. Jan 15, 2017

### Kaneki123

''In physics, a force is said to do work if, when acting, there is a displacement of the point of application in the direction of the force''(Wikipedia)....So how there be a negative work when a ''displacement in direction opposite to force'' does not fit te definition...In other words, the work can only be done if the displacement is in direction of force...So how can there be a negative work (in which displacement and force are in opposite directions)???

23. Jan 15, 2017

### Staff: Mentor

Did you consider that there could be two forces acting on the body in opposite directions, and one of these forces causes the body to move in a direction opposite to that of the other force? Or, even if there is only one force, the body may be already moving in the direction opposite to the force, and the force is acting to slow it down.

24. Jan 15, 2017

### jbriggs444

You are looking at the first line on the Wiki page. That line is a rough summary that captures only some of the full definition. The proper way of defining work is as the dot product of the force vector times the displacement vector: $W = \vec{F}\ \cdot\ \vec{d}$

Without using vectors, this may be calculated as force times distance times the cosine of the angle between the two. If force and distance are parallel, the cosine is one and you get the formula that is most often quoted. Work = force times distance. If force and distance are not quite in the same direction you get almost but not quite the full work. If force and distance are at right angles you get no work at all. If force and distance are pointed in opposite directions, the cosine term is negative.

25. Jan 15, 2017

### Staff: Mentor

The definition of work is $W=f\cdot d$ for a constant force or $dW/dt = f \cdot v$ for a varying force. Nothing about "cause" or even "net" shows up in the definition.