Integration of a quantity when calculating Work

[Gurasees]

While integrating work we write dw = INT. F.ds. But why can't we write dw = INT. dF.s ?

 

[jedishrfu]

The definition of work is $work = force \times distance$ and while the force can change over position and over time as we move along some path a distance s, we can generalize it as an integral over the path with ds being a very small distance to handle non-linear paths.

However dF isn't a force its a change in force and if used breaks the definition of work which is $work = force \times distance$ not $work = change in force \times distance$.

http://hyperphysics.phy-astr.gsu.edu/hbase/wint.html

 

[Niladri Dan]

First what do you mean by dF multiplied by s...give the physical interpretation...

 

[Buffu]

Isn't by definition $\displaystyle w = \int_C f\cdot ds$ ?

 

[Niladri Dan]

Yes, it is so...

 

[Diegor]

While integrating work we write dw = INT. F.ds. But why can't we write dw = INT. dF.s ?

Check integration by parts. You will see that f.ds is not the same as df.s but there is a relation between them. That is math. then you will have to see if that has physical sense.

 

[nasu]

While integrating work we write dw = INT. F.ds. But why can't we write dw = INT. dF.s ?

The same question was recently asked in the General Physics section.
https://www.physicsforums.com/threads/derivation-of-work-done.914895/
That thread was closed yesterday. Is this an attempt to re-open the same discussion?

 

[Mark44]

While integrating work we write dw = INT. F.ds. But why can't we write dw = INT. dF.s ?

The same question was recently asked in the General Physics section.
https://www.physicsforums.com/threads/derivation-of-work-done.914895/
That thread was closed yesterday. Is this an attempt to re-open the same discussion?

Thread closed. @Gurasees, please take a look at the thread in the link that @nasu posted.