What type of of force is applied force? Conservative ,Nonconservative

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Discussion Overview

The discussion centers around the nature of applied force, specifically whether it is classified as a conservative or non-conservative force. Participants explore the implications of work done by applied forces in various scenarios, including closed loops and curved paths, and the role of displacement versus distance in calculating work.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that in a closed loop, the net work done by an applied force is zero due to zero displacement, questioning why applied force is considered non-conservative.
  • Others clarify that when the force is not constant, work must be calculated using integration, suggesting a deeper understanding of calculus is necessary.
  • One participant explains that work done along a curved path depends on the path taken, stating that even if the displacement is zero, the work done can be non-zero if the force is applied along a distance.
  • Another participant emphasizes that the work done is path-dependent for non-conservative forces, contrasting this with conservative forces where work is independent of the path taken.
  • There is a discussion about the distinction between distance and displacement, with some participants noting that displacement has direction, which affects the calculation of work done when forces are applied in opposite directions.
  • One participant points out that the elementary work can be expressed as a dot product of force and displacement vectors, highlighting the vector nature of work calculations.

Areas of Agreement / Disagreement

Participants express differing views on the classification of applied force and the implications of work done in various scenarios. The discussion remains unresolved, with multiple competing perspectives on the nature of work and force.

Contextual Notes

Participants highlight the importance of understanding the definitions of distance and displacement in the context of work calculations, as well as the need for clarity on the conditions under which work is considered zero or non-zero.

Miraj Kayastha
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In a closed loop when we apply an applied force on an object the object starts at point A and stops at point A.

Since the displacement is 0, Work done by the applied force on the object is = F x s x cosθ
= 0

So the net work done by the applied force is 0 but why is applied force a non-conservative force?
 
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When the force is not constant during the motion the work is calculated by integration. Have you studied calculus?

ehild
 
Yes I have.
 
The work done along a curved path is ##W=\int F_s d_s ## where Fs is the component of the force tangent to the segment of the curve and ds is the length of the line segment. If you apply a force of constant magnitude, always tangent to the curve, ##W=F \int d_s =F L## where L is the length of the curve. The work is not zero along a closed loop.

If you push a crate along a rough horizontal surface with force F you do F*D work during D distance. When you push it back with force of the same magnitude, you do F*D work again. The displacement is zero, but the net work is not.

In general, the work depends on the way it is done. The conservative forces are exceptions, their work does not depend on the path taken between the initial and final points.

ehild
 
Last edited:
ehild said:
If you push a crate along a rough horizontal surface with force F you do F*D work during D distance. When you push it back with force of the same magnitude, you do F*D work again. The displacement is zero, but the net work is not.
ehild

Here when you said "when we push it back with the same magnitude, you do F x D work again" I think we should also account the direction of the force because the force is constant both in magnitude and direction throughout the motion. So shouldn't the work done by the force on the crate when crate is coming back be - F x D ?

And then the net work is zero.
 
Miraj Kayastha said:
Here when you said "when we push it back with the same magnitude, you do F x D work again" I think we should also account the direction of the force because the force is constant both in magnitude and direction throughout the motion. So shouldn't the work done by the force on the crate when crate is coming back be - F x D ?

The D in the work formula is not a "distance". It is a "displacement". The distinction is that a distance is always positive and has no direction. A displacement has a direction.

When you push the crate back toward the starting point, both force and direction are reversed. So it's -F x -D.
 
The elementary work can be written either as dot product of the force vector with the displacement vector, ##dW=\vec F \cdot \vec{dr}## or as the product of two scalars, Fs the component of force along the path taken and the elementary path length, ds dW=Fsds.

ehild
 

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