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i learn about work and most book define force x distance. The definition given not very informative and the book don't describe much about it. All they do is give above definition and then a lot of example. Later i learn that this concept are related to energy. So basically the are some fact regarding conservation of energy. From this point of view, i know that energy is an abstract quantity store in system/object and those energy can be use to do change on other object/system by transfer its energy. In order to do change on other system the first object must do work on them [what i can recall the only thing can cause change to a system are force and heat. is there anything else?]. And i believe some might define work as transfer of energy. From here i got two definition where subjectively work is transfer of energy and objectively work is Force x distance. So can i said that work is when a system cause change to other static system (subjectively/qualitatively) which value of change can be quantify and measured by Force x distance (objectively/quantitatively). Plus work are done when a change had been made to the system which result either loss or gain in energy. If this really the case can i make a statement where the energy and work concept is actually how physicist trying to discuss about interaction between two system which, when change occur, the interaction will cause one to gain some quantity while other loss in the process of interaction and such quantity are constant when we add them together. In addition, do this concept trying to explain to us that to cause some change to other system one need loss something in order to add something to them. Am i in the right direction?

Another question is why distance? In the formula the work is define by Force X distance. I accept that force are used due to they are the cause of change or i can say they are way/agent of energy transfer. But why we use distance as a unit to quantify work. And most of explanation using a example to explain distance. Do the formula cannot be explain but it just make sense when think logically?

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marked as duplicate by sammy gerbil, Jon Custer, stafusa, John Rennie energy Oct 27 '17 at 5:56

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Force can make a change in a system, yes. You can lift a box up from the ground.

But how much energy must you spend to make a change? Do you spend the same amount of energy to lift the box 1 metre as to lift it 2 meters?

No, you spend double the amount of energy. The distance is needed in the energy transfer consideration as well as the force. Think of it like this:

  • If you push on a wall, nothing happens. It doesn't move. Noone will tell you that you did some good work pushing that wall.
  • Similarly, if you push a balloon it might move very far. But it doesn't require any significant force from you. Again, noone will clap your back and say that you did any proper work to move this ballon.

Before calling something work, we intuitively require both an effort (force) and a change (displacement). Otherwise your table would constantly be doing work just by holding up a book.

what i can recall the only thing can cause change to a system are force and heat. is there anything else?

You are correct here if you just said work and heat instead if force and heat. Work is the energy transfer that does a mechanical change, heat is an energy transfer that does a thermodynamic change.

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Ok, well here is a prove that for a simple object changing velocity work is change in kinetic energy. This should provide some level of foundation for $Work=Fd$

$\Delta E = \frac{mv_f^2}{2}- \frac{mv_i^2}{2}$

$v_f $ is final velocity, $v_i$ is initial velocity.

$\Delta E = \frac{m}{2}(v_f^2-v_i^2)=m\frac{(v_f+v_i)}{2}(v_f-v_i)$

$\Delta E = m\Delta v\bar v$

$\bar v$ is the average velocity and $\Delta v$ is the change in velocity

where $\Delta v= at$
...so

$\Delta E=ma t\bar v $

where $t\bar v=d$
...so

$\Delta E = mad$
... looks familiar, and $F=ma$

$\Delta E = Fd$

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  • $\begingroup$ Sir. I like to ask in history of physic do we found kinetic energy formula first (by experiment) before we found work formula (W=F.d)? Then later the work formula (W=F.d) are derive from the calculation above? $\endgroup$ – Sir.Plz Oct 27 '17 at 15:29
  • $\begingroup$ I don't know that I'm afraid; but there are lots of ways to derrive these, so most likely it was a different way. But it's a possibility. $\endgroup$ – JMLCarter Oct 27 '17 at 20:11

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