Problem with the Definition of work

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Discussion Overview

The discussion revolves around the definition of work in physics, particularly the relationship between force and displacement when work is held constant. Participants explore the implications of Newton's second law and seek clarification on the inverse relationship between force and displacement in the context of work.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants express confusion regarding the relationship between force and displacement, questioning how they can be inversely related when work is constant.
  • Examples are provided to illustrate the inverse relationship, such as using a lever or a variable brake, where less force is required for greater displacement.
  • Participants discuss the need for constant work to observe the inverse relationship and provide various scenarios to clarify this concept.
  • One participant mentions the difference between work and impulse, suggesting that the mathematical treatment of these concepts may not be the same.
  • Another participant raises a point about the implications of vector versus scalar treatment in the context of Newton's second law and work.

Areas of Agreement / Disagreement

There is no consensus on the clarity of the inverse relationship between force and displacement, as participants express differing levels of understanding and seek examples to illustrate the concept. Multiple viewpoints and examples are presented without a definitive resolution.

Contextual Notes

Participants highlight the importance of holding work constant to observe the inverse relationship, but some express difficulty in relating this to real-life examples. The discussion includes various scenarios and examples, but no unified conclusion is reached.

parshyaa
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From Newton's 2nd law F = ma and a = F/m(acceleration and mass are inversely related when force is constant)
But in w = F.d , F =w/d(but d and F are not inversly related just as above)
I think there's something wrong in my question, please point it to me or please answer it.
 
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parshyaa said:
But in w = F.d , F =w/d(but d and F are not inversly related just as above)
This is more of a statement than a question. For constant amount of work, F and d are inversely related.
 
lewando said:
This is more of a statement than a question. For constant amount of work, F and d are inversely related.
So they are inversely proportional, give me examples, i can not imagine their inverse relation, as force increases,displacement also increases how it will decrease, but direction is also important
 
Work must be held constant in order for the inverse relationship to be seen. An example:
1 J = (10 N)(0.1 m)
1 J = (0.1 N)(10 m)
 
lewando said:
Work must be held constant in order for the inverse relationship to be seen. An example:
1 J = (10 N)(0.1 m)
1 J = (0.1 N)(10 m)
Could you explain it with a physics example(not mathematically)
Just like i could say as mass increases obviously its hard for a body to accelarate and we could relate this with real life very easily but I am not able to relate inverse relation of force and displacement when work is constant, i.e i am not able to give an example on this, so please help me
 
parshyaa said:
Could you explain it with a physics example(not mathematically)

How about a lever? It reduces the force by increasing the desplacement.
 
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parshyaa said:
Could you explain it with a physics example(not mathematically)
A lever would be one possibility. Another one would be with a variable brake, resisting motion. The same amount of energy could be expended pushing a long distance against a lightly applied brake and a short distance against a heavily applied brake.
Using a situation that includes acceleration just makes things harder; it's much easier when the velocity is constant. Cases where there is acceleration are pretty common (rockets and boy racers) but best to deal with them after the constant speed situation is sorted out. The difference between Work (change of KE, for instance) and Impulse (change of Momentum) is important.
 
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parshyaa said:
Could you explain it with a physics example(not mathematically)
Just like i could say as mass increases obviously its hard for a body to accelarate and we could relate this with real life very easily but I am not able to relate inverse relation of force and displacement when work is constant, i.e i am not able to give an example on this, so please help me

Drop 10 identical apples from a height of 1.0 meter.
Drop one of those apples from a height of 10.0 meters.

In both cases the work done by gravity is the same.
 
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Carry a box of books up one flight of stairs. Same work as carrying one-half the box up two flights. Or two boxes one half flight.

Keep in mind, I am talking about the work done on the books.
 
  • #10
I may be wrong, but on the treatment of the 2nd law, (at least for non-relativistic systems) we have a vector as the product of a vector by a scalar F=ma. In the treatment of work, we have the scalar product of two vectors, resulting in a scalar F.d = W. I don't know if such consideration has implications to the problem present, but we have at least to suppose that the mathematical treatment is not the same.
 
  • #11
sophiecentaur said:
A lever would be one possibility. Another one would be with a variable brake, resisting motion. The same amount of energy could be expended pushing a long distance against a lightly applied brake and a short distance against a heavily applied brake.
Using a situation that includes acceleration just makes things harder; it's much easier when the velocity is constant. Cases where there is acceleration are pretty common (rockets and boy racers) but best to deal with them after the constant speed situation is sorted out. The difference between Work (change of KE, for instance) and Impulse (change of Momentum) is important.
Sir why i am not able to make a private conversation with you on physics forum
 
  • #12
parshyaa said:
Sir why i am not able to make a private conversation with you on physics forum
Sorry but I just don't do them. I prefer to talk on open forum. :smile:
 
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  • #13
parshyaa said:
So they are inversely proportional, give me examples, i can not imagine their inverse relation, as force increases,displacement also increases how it will decrease, but direction is also important
The more drag a car has the shorter the range on a tank of gas or charge.
 
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