Net Work Calculus: Understanding mΔx(Δv/Δt)

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    Calculus Net Work
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Discussion Overview

The discussion revolves around the manipulation of the equation for net work, specifically the transition from one form of the equation to another involving the terms of mass, change in velocity, and change in position over time. Participants explore the mathematical reasoning behind these transformations, focusing on the interpretation of the terms involved.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about how the term involving time (Δt) is moved under the term for position (Δx) in the equation for net work.
  • Another participant attempts to clarify the manipulation by providing a numerical analogy, suggesting that the transformation is similar to basic arithmetic operations.
  • A third participant notes that Δx, Δv, and Δt are numeric approximations to differentials, emphasizing that the discussion involves numerical manipulation rather than differential calculus.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the mathematical manipulation, though there is a shared understanding of the basic operations involved. The discussion remains somewhat unresolved regarding the initial participant's confusion.

Contextual Notes

There are limitations in the understanding of how the terms relate to differentials versus numeric approximations, and the discussion does not fully resolve the initial participant's confusion about the manipulation of the equation.

Nile3
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In the equation:

Wnet=FnetΔX=[m*Δv/Δt]Δx=m*Δv*Δx/Δt

I just don't understand how the 3rd part of the equation is changed into the 4rth (Dt going under the Dx). If I can get some help with this, that would be great.

The way I think about it is [m*Δv/Δt]Δx = mΔx(Δv/Δt)

I have no idea how Dt went under the Dx...
 
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Nile3 said:
In the equation:

Wnet=FnetΔX=[m*Δv/Δt]Δx=m*Δv*Δx/Δt

I just don't understand how the 3rd part of the equation is changed into the 4rth (Dt going under the Dx). If I can get some help with this, that would be great.

The way I think about it is [m*Δv/Δt]Δx = mΔx(Δv/Δt)

I have no idea how Dt went under the Dx...
It's no different from writing
[tex]\frac{3}{4}\cdot 5 = 3\cdot \frac{5}{4}[/tex]
 
Note that [itex]\Delta x[/itex], [itex]\Delta v[/itex], and [itex]\Delta t[/itex] are numeric approximations to the differentials, not the differentials. They are simply numbers and what is going on is simply manipulation of numbers.
[tex]a\frac{b}{c}= \frac{ab}{c}= \frac{ba}{c}= b\frac{a}{c}[/tex]
 
Ah thank you. I couldn't see it for some reason...
 

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