Problem with the Definition of work

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The discussion centers on the relationship between force, displacement, and work, specifically how they are inversely related when work is held constant. Participants seek clarification on this concept, noting that while force and displacement seem to increase together, they actually exhibit an inverse relationship under constant work conditions. Examples provided include the use of levers and variable brakes, illustrating how less force can be applied over a greater distance to achieve the same amount of work. The conversation also touches on the distinction between work and impulse, emphasizing the importance of understanding these concepts in physics. Overall, the thread highlights the complexities of relating force and displacement in practical scenarios.
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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.
 
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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
 
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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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