Problem with the Definition of work

In summary, force and acceleration are inversely related when mass is constant according to Newton's second law. However, in the equation for work, force and displacement are not inversely related. To see the inverse relationship between force and displacement, work must be held constant. This can be seen in examples such as using a lever or a variable brake. It is important to note that when dealing with acceleration, it is easier to understand the inverse relationship when velocity is constant. Additionally, the mathematical treatment for force and work is different, with force being a vector and work being the scalar product of two vectors. Private conversations on the physics forum are not preferred by the expert summarizer.
  • #1
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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  • #2
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.
 
  • #3
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
 
  • #4
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)
 
  • #5
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
 
  • #6
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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  • #7
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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  • #8
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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  • #9
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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1. What is the definition of work in science?

In science, work is defined as the transfer of energy that occurs when a force is applied to an object and the object moves in the direction of the force.

2. Why is there a problem with the definition of work?

The problem with the definition of work is that it only applies to situations where the force and the movement of the object are in the same direction. This does not account for situations where the force and the movement are not in the same direction, such as when an object is being lifted or pushed at an angle.

3. How can the definition of work be improved?

To improve the definition of work, it can be expanded to include situations where the force and the movement are not in the same direction. This can be done by using vector quantities, such as displacement and force, to calculate work done on an object.

4. Why is it important to have an accurate definition of work?

An accurate definition of work is important because it allows scientists to accurately measure and calculate the amount of work done on an object. This is essential in many fields of science, including physics and engineering, where understanding the transfer of energy is crucial.

5. Are there any other factors besides force and movement that affect work?

Yes, there are other factors that can affect work, such as the angle at which the force is applied, the distance over which the force is applied, and any external forces acting on the object. These factors must be taken into account for a more accurate calculation of work.

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