Is This a Valid Derivation of Kinetic Energy from Work?

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Discussion Overview

The discussion revolves around the derivation of kinetic energy from the work-energy principle, specifically examining a participant's approach to the derivation and comparing it with textbook methods. The scope includes mathematical reasoning and conceptual clarification related to physics education.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents a derivation of kinetic energy using the work integral, questioning its validity.
  • Another participant notes that their derivation is shorter than the textbook version, expressing confusion over why textbooks often take longer approaches.
  • Some participants suggest that brevity in derivations may overlook important steps necessary for understanding, especially for students new to calculus.
  • There is a recognition that personal taste influences how derivations are presented, with some valuing conciseness while others prioritize thoroughness.
  • A later reply acknowledges the validity of the original derivation while also emphasizing the importance of clarity in educational contexts.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of brevity in mathematical derivations. While some appreciate the concise approach, others argue that it may lead to confusion for learners who are not yet familiar with the concepts involved. The discussion remains unresolved regarding the best method for presenting such derivations.

Contextual Notes

Participants highlight the potential limitations of skipping steps in derivations, particularly for students who may not have a strong background in calculus. There is also mention of the importance of considering the audience's familiarity with the material when presenting mathematical arguments.

Arman777
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I am trying to derive the kinetic energy from the work and can I derive it like this ?

$$W=\int Fdr$$
$$W=\int \frac {dp} {dt}dr=\int (dp) \frac {dr} {dt}=\int (mdv)v=1/2m[v_f^2-v_i^2]$$
 
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My book does this in 7 lines mine took 4. I don't know why books sometimes does things in long way.
 
Arman777 said:
My book does this in 7 lines mine took 4. I don't know why books sometimes does things in long way.

Perhaps they prefer at least a modicum of mathematical rigour!
 
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Arman777 said:
My book does this in 7 lines mine took 4. I don't know why books sometimes does things in long way.
There's a bit of personal taste involved here; what one person considers admirable terseness another may consider skipping important steps.

With textbooks an additional consideration is that skipping steps can be a problem for a student who isn't already familiar with the concept. For example, someone taking intro physics concurrently with their first calculus course may have seen their first integral just a few weeks back - it's easy to imagine that your cavalier treatment of the bounds of integration would confuse them.

For what it's worth... I understand your derivation just fine but it's not what I'd be writing on a chalkboard in front of a class.
 
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Nugatory said:
There's a bit of personal taste involved here; what one person considers admirable terseness another may consider skipping important steps.

With textbooks an additional consideration is that skipping steps can be a problem for a student who isn't already familiar with the concept. For example, someone taking intro physics concurrently with their first calculus course may have seen their first integral just a few weeks back - it's easy to imagine that your cavalier treatment of the bounds of integration would confuse them.

For what it's worth... I understand your derivation just fine but it's not what I'd be writing on a chalkboard in front of a class.
Well yes you are right. Thanks for your reply
 

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