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Work done over a distance

  1. Mar 10, 2008 #1
    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?
  2. jcsd
  3. Mar 10, 2008 #2
    What is the definition of work?
  4. Mar 10, 2008 #3
    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?
  5. Mar 10, 2008 #4


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    Homework Helper

    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?
  6. Mar 10, 2008 #5
    I would look up the definition of work in your physics text book, or on google, before making assumptions on the definition of work.

  7. Mar 10, 2008 #6
    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 criticise 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.
  8. Mar 10, 2008 #7
    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.
  9. Mar 10, 2008 #8


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    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".
  10. Mar 10, 2008 #9
    Why are they needed? Isn't the magnitide of x vector sufficient enough?
  11. Mar 10, 2008 #10
    I know the link I gave you, and probably in your book, there is a cosine term the definition of work. I really dont 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.
    Last edited: Mar 11, 2008
  12. Mar 11, 2008 #11


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    Staff: Mentor

    Work isn't a vector, it's a scalar. Where on that Wikipedia page did you see that work is a vector?
  13. Mar 11, 2008 #12
    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".
  14. Mar 13, 2008 #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...
  15. Mar 13, 2008 #14


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    Science Advisor

    It means you did zero net work.

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