What's the "d" in that formula? (work formula)

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Discussion Overview

The discussion centers around the interpretation of the variable "d" in a work formula, exploring its definition and implications in the context of calculus and physics. Participants examine whether the equation assumes constant force and the nature of work done in relation to infinitesimal displacements.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • Some participants propose that "d" represents a basic calculus term indicating a change, likely as a function of time.
  • Others argue that the equation does not assume a constant force, as work must be evaluated by integrating force over the appropriate range of displacement.
  • A participant clarifies that the expression dW = \vec F · d\vec r represents an infinitesimal amount of work done over an infinitesimal displacement, with the force being approximately constant during that displacement.
  • It is suggested that thinking of "d" as differentials may not be appropriate, as work done can depend on the path taken, and this is sometimes denoted differently in thermodynamics literature.

Areas of Agreement / Disagreement

Participants express differing views on whether the equation assumes constant force, indicating that multiple competing interpretations remain unresolved.

Contextual Notes

There are limitations regarding the assumptions made about the nature of force and work, as well as the dependence on the path taken in evaluating work done.

Austin Gibson
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What's the definition of the "d" in that formula?

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Austin Gibson said:
This equation assumes a constant force?

No it doesn't. This is because to find W, you have to integrate F over the appropriate range of dr.

Zz.
 
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[itex]dW=\vec F \cdot d\vec r[/itex] is an infinitesimal amount work done over the infinitesimal displacement [itex]d\vec r[/itex],
where [itex]\vec F[/itex] is approximately constant during that displacement.

When this is evaluated over a path, then [itex]\vec F[/itex] and [itex]d\vec r[/itex] will vary as you progress along the path.

It's probably not a good idea to think of these [itex]d[/itex]'s as differentials (as in [itex]dW[/itex] is a differential of [itex]W[/itex]
since there is generally no such [itex]W[/itex] because the work done generally depends on the path. In thermodynamics books, this is sometimes written as [itex]đW[/itex].)
 
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