Problem with the Definition of work

  • Context: Undergrad 
  • Thread starter Thread starter parshyaa
  • Start date Start date
  • Tags Tags
    Definition Work
Click For Summary
SUMMARY

The discussion centers on the inverse relationship between force and displacement when work is held constant, as described by the equation W = F·d. Participants clarify that while force and displacement typically increase together, they become inversely related when the total work remains unchanged. Examples provided include the use of levers and variable brakes, illustrating how a reduction in force can lead to an increase in displacement, thereby maintaining constant work. The distinction between work and impulse is also emphasized, highlighting the importance of understanding these concepts in physics.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with the work-energy principle (W = F·d)
  • Basic knowledge of scalar and vector quantities in physics
  • Concept of constant velocity versus acceleration in motion
NEXT STEPS
  • Research the mechanics of levers and their impact on force and displacement
  • Explore the concept of work done against friction and variable resistance
  • Study the differences between work and impulse in physics
  • Investigate real-world applications of the work-energy principle in engineering
USEFUL FOR

Students of physics, educators explaining mechanics, and engineers applying the work-energy principle in practical scenarios will benefit from this discussion.

parshyaa
Messages
307
Reaction score
19
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.
 
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: parshyaa
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.
 
  • Like
Likes   Reactions: parshyaa
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.
 
  • Like
Likes   Reactions: russ_watters and sophiecentaur
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:
 
  • Like
Likes   Reactions: parshyaa
  • #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.
 
  • Like
Likes   Reactions: parshyaa

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 48 ·
2
Replies
48
Views
6K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 41 ·
2
Replies
41
Views
4K
  • · Replies 25 ·
Replies
25
Views
2K