Question about net work and displacement

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Discussion Overview

The discussion revolves around the relationship between net work and displacement in physics, particularly focusing on scenarios involving pushing an object, the effects of friction, and the implications of conservative versus non-conservative forces. Participants explore theoretical concepts and practical examples, including circular motion and energy dissipation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that if displacement is zero, then work done is zero, but others challenge this by presenting scenarios where work appears to be done despite zero net displacement.
  • One participant discusses the role of friction as a dissipative force that does work opposite to the work done by the applied force, raising questions about the net work on the object.
  • There is a suggestion that the total work done by friction is equal in magnitude but opposite in sign to the work done by the pushing force, leading to a discussion about the object's kinetic energy.
  • Participants mention that the work done is path-dependent, especially when forces are not constant, referencing the integral formulation of work.
  • Some participants clarify that the definition of zero work for zero displacement applies specifically to conservative forces, while in general cases, work can be non-zero even with zero displacement.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between work, displacement, and the role of friction. There is no clear consensus on the implications of these concepts, and multiple competing interpretations remain present throughout the discussion.

Contextual Notes

Limitations include assumptions about the constancy of forces and the nature of the forces involved (conservative vs. non-conservative). The discussion also highlights the complexity of calculating work in scenarios involving friction and varying forces.

FQVBSina_Jesse
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I know that by physics definition if displacement is zero, work is zero.

However, if I push an object 5 m to the east, and then move to the other side of the object and push it 5 m back to the west. I think in this case I have always done positive work on the object and hence the total work done shouldn't be zero.

If I swing the object around in a circle then my force is perpendicular, so the work done is zero, but what about if I pushed a cart around in a circle from behind?
 
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FQVBSina said:
I think in this case I have always done positive work on the object and hence the total work done shouldn't be zero.
When it stopped moving on each leg, which way was the force that stopped it pointing?
 
Ibix said:
When it stopped moving on each leg, which way was the force that stopped it pointing?
Stopped by friction. As in if I stop pushing then it doesn't move. And friction points in the opposite direction. But friction is a dissipative force
 
FQVBSina said:
Stopped by friction. As in if I stop pushing then it doesn't move. And friction points in the opposite direction. But friction is a dissipative force
So the total work done by friction has the opposite sign to the work you did. What's the magnitude of the work it did compared to the work you did?
 
Ibix said:
So the total work done by friction has the opposite sign to the work you did. What's the magnitude of the work it fid compared to the work you did?
Ok, I still don't know why you are asking about friction since friction would just be calculated into the net force. But the friction's work would be whatever the friction force is times the total distance (not displacement since friction is not conservative), which will be less than the pushing force.
 
FQVBSina said:
But the friction's work would be whatever the friction force is times the total distance (not displacement since friction is not conservative), which
So you are saying that you do work on the object, implying that it gains kinetic energy. And the frictional force does work of the opposite sign and a smaller magnitude. So the object is left with kinetic energy - it hasn't stopped. How are you going to stop it?
 
Ibix said:
So you are saying that you do work on the object, implying that it gains kinetic energy. And the frictional force does work of the opposite sign and a smaller magnitude. So the object is left with kinetic energy - it hasn't stopped. How are you going to stop it?
I am going to stop pushing on it so the friction dissipates the energy and it stops. Then I move to the other side and push the object back.
 
FQVBSina said:
I am going to stop pushing on it so the friction dissipates the energy and it stops. Then I move to the other side and push the object back.
Right. So the frictional force does the same magnitude of work as you do, but with the opposite sign. So the total work on the object is?
 
Ibix said:
Right. So the frictional force does the same magnitude of work as you do, but with the opposite sign. So the total work on the object is?
The total work is (myforce - friction) x 5 + (myforce - friction) x 5
 
  • #10
FQVBSina said:
I know that by physics definition if displacement is zero, work is zero.
This is not a general definition, it is only the definition for the specific case of a constant force. In your description the force is not constant.

The general formula is ##W=\int_S F \cdot ds ##
 
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  • #11
Dale said:
This is not a general definition, it is only the definition for the specific case of a constant force. In your description the force is not constant.

The general formula is ##W=\int_S F \cdot ds ##
There we go! This is what I was looking for. Work integral is path dependent. Thanks!
 
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  • #12
FQVBSina said:
The total work is (myforce - friction) x 5 + (myforce - friction) x 5
And that is? (Hint: it starts with zero kinetic energy, it ends with zero kinetic energy - what's the change in energy?)
 
  • #13
Not only for constant force but for a conservative force, the total work is zero if the total displacement is zero. This is a consequence of the definition of the conservative force and the definition of work and the gradient theorem.

In the general case a force might not be conservative, the gradient theorem cannot be applied, so the work becomes path-dependent,so the total work is not necessarily zero even if the total displacement is zero.
 
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