A: The Concept of Net Work in Physics

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Discussion Overview

The discussion centers around the concept of work in physics, exploring its definition, implications, and applications. Participants engage in clarifying the mathematical formulation of work, its scalar nature, and the conditions under which it is considered useful or relevant in various contexts.

Discussion Character

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

Main Points Raised

  • Some participants express confusion about how walking a distance forward and back results in zero work, questioning the definition of work.
  • There is a discussion about the definition of work as force times distance, with some participants suggesting that this idealized definition may not always be practical.
  • One participant proposes an alternative definition based on adding distances, arguing it may better reflect actual energy expenditure.
  • Another participant clarifies that work is the scalar product of force and displacement, emphasizing the importance of direction in calculating work.
  • Concerns are raised about the terminology used in physics, with one participant noting the potential confusion caused by borrowing common words for scientific concepts.
  • Some participants discuss the scalar nature of work, with one asserting that work is not a vector and questioning the source of any claims that it is.
  • There is mention of conservative forces and how work can be path-independent, with examples provided to illustrate this point.
  • One participant suggests that the usefulness of work may be better understood through its relation to changes in mechanical energy.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the definition and usefulness of work, with multiple competing views and ongoing questions about its application and implications.

Contextual Notes

Some participants express uncertainty regarding the definitions and implications of work, particularly in relation to vector and scalar quantities, and the conditions under which work is calculated. There are also references to specific mathematical formulations that may not be fully resolved in the discussion.

Who May Find This Useful

This discussion may be of interest to students and individuals seeking to understand the concept of work in physics, particularly those grappling with its definitions and applications in various scenarios.

Surreal Ike
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I don't understand how walking 10 feet forward and then 10 feet back means you did zero work. If I pull a 1 pound weight 4 feet forward and 3 feet right, have I only done 5 foot-pounds of work?
 
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What is the definition of work?
 
Force times distance, I suppose. But how can this idealized definition where you assume that you took the most efficient path ever be useful? My adding-the-distances definition would be more useful when you wanted to calculate the actual energy it would take to move something. When is the idealized definition ever useful?
 
Work done is when a force moves its point of application in the direction of the force.

If something moves in a specified direction, is it still a scalar?
 
OK, so I'm beginning to get it now. Work equals force times distance. Force equals mass times acceleration, and acceleration equals velocity per second. Since velocity is a vector, that means that when you multiply it by a bunch of other numbers, it's still a vector.

That still doesn't answer my question about the usefulness of work. Also, speed refers to the magnitude of the velocity vector, right? So why don't we have words that refer to the magnitude of the acceleration vector, the force vector, and the work vector?

Also, isn't it terribly counterintuitive that hitting someone with a ball will exert a negative force on them? The ball starts decelerating as soon as it leaves your hand, right?

This is going off topic a little, but maybe the fundamental problem isn't the usefulness or integrity of these terms, but rather the very human tendency to borrow words from the common parlance to describe something precise and scientific. For instance, according to the scientific definition of "berry", blueberries, raspberries, blackberris, and strawberries are not "berries", while tomatos, avocados, eggplants, and chili peppers are. Makes you wonder why they didn't just come up with a new word. I think it's because making up new words like "blorf" makes you sound silly.

To Cyrus: I read about work in my math textbook, so I've got a good reason to believe what I said about work. If you want to criticize me for asking newbie physics questions in this forum full of geniuses, that's a far more valid criticism :-). Hope I'm not bothering you guys too much.
 
Alright, thanks for the Wikipedia link. I now know that mechanical work refers to the magnitude of the work vector. What about acceleration and force? Does putting absolute value signs around a vector make it clear that you're referring to the scalar?

Sorry for asking so many questions. I learn better that way. Maybe I'll go find a physics IRC channel. I hate to be a pain.
 
Surreal Ike said:
Force times distance, I suppose. But how can this idealized definition where you assume that you took the most efficient path ever be useful? My adding-the-distances definition would be more useful when you wanted to calculate the actual energy it would take to move something. When is the idealized definition ever useful?

To be exact, work is the scalar product of the force times the displacement vector [tex]W = \vec{F} \cdot \vec{\Delta r}[/tex] So to calculate the work done by a force you have to specify in what direction the force is acting and in what direction the motion is taking place. If you hold a weight in your hand and you walk around (without raising or lowering the weight) at constant velocity it does not matter what distance you travel and in what direction (left/right/forward/backward) you do zero work. There is nothing about a "most efficient path".
 
Surreal Ike said:
That still doesn't answer my question about the usefulness of work. Also, speed refers to the magnitude of the velocity vector, right? So why don't we have words that refer to the magnitude of the acceleration vector, the force vector, and the work vector?

Why are they needed? Isn't the magnitide of x vector sufficient enough?
 
  • #10
Surreal Ike said:
OK, so I'm beginning to get it now. Work equals force times distance. Force equals mass times acceleration, and acceleration equals velocity per second. Since velocity is a vector, that means that when you multiply it by a bunch of other numbers, it's still a vector.

That still doesn't answer my question about the usefulness of work. Also, speed refers to the magnitude of the velocity vector, right? So why don't we have words that refer to the magnitude of the acceleration vector, the force vector, and the work vector?

Also, isn't it terribly counterintuitive that hitting someone with a ball will exert a negative force on them? The ball starts decelerating as soon as it leaves your hand, right?

This is going off topic a little, but maybe the fundamental problem isn't the usefulness or integrity of these terms, but rather the very human tendency to borrow words from the common parlance to describe something precise and scientific. For instance, according to the scientific definition of "berry", blueberries, raspberries, blackberris, and strawberries are not "berries", while tomatos, avocados, eggplants, and chili peppers are. Makes you wonder why they didn't just come up with a new word. I think it's because making up new words like "blorf" makes you sound silly.

To Cyrus: I read about work in my math textbook, so I've got a good reason to believe what I said about work. If you want to criticize me for asking newbie physics questions in this forum full of geniuses, that's a far more valid criticism :-). Hope I'm not bothering you guys too much.

I know the link I gave you, and probably in your book, there is a cosine term the definition of work. I really don't know what your going on about in terms of usage of words. You have way too many basic questions, which leads me to believe that you need to read a physics book and then come back with any questions. You're asking basic questions that are answered in any quality undergrad physics text.
 
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  • #11
Surreal Ike said:
Alright, thanks for the Wikipedia link. I now know that mechanical work refers to the magnitude of the work vector.

Work isn't a vector, it's a scalar. Where on that Wikipedia page did you see that work is a vector?
 
  • #12
Surreal Ike said:
Force times distance, I suppose. But how can this idealized definition where you assume that you took the most efficient path ever be useful? My adding-the-distances definition would be more useful when you wanted to calculate the actual energy it would take to move something. When is the idealized definition ever useful?

Firstly, work is defined as the dot product of force dot displacement. Is it defined as a scalar, because that is the point of a dot product! There is not work vector.

We usually use work when dealing with conservative forces. Or forces that are path independent and conserve mechanical energy. Gravity is such a force because the energy of the ball will be exactly the same if you hold a ball at a height of 1 m then move it to 0 m then back to 1m. This is where we can drop the assumption of a "most efficient path" since, in a sense, all paths to the same place are equally efficient. When we're dealing with work we're generally interested in the beginning and end result, not what is happening in the middle of the process since it is not important. Work can be derived as a change in mechanical energy (kinetic of potential). Since it can be derived as such it is effectively accomplishing what you said you can do by "adding-the-distances".
 
  • #13
Hi Pals,

It sounds like you all need to work your mechanics lecture notes...

[tex]W_{A \to B} = \int_{t_A}^{t_B} \textbf{F} \cdot \textbf{v}\ dt[/tex]

where [tex]\textbf{F}[/tex] is the external applied force and [tex]\textbf{v}[/tex] the velocity

[tex]\delta W = \textbf{F} \cdot \textbf{v}\ \delta t[/tex] is the elementary work,

and [tex]\textbf{F} \cdot \textbf{v}[/tex] is power. Hope it helps...
 
  • #14
Surreal Ike said:
I don't understand how walking 10 feet forward and then 10 feet back means you did zero work.

It means you did zero net work.

CS
 

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