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Why does the work done on a particle is equal to W=F.dx but not W=x.dF ? Where F is the applied force on the particle
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marked as duplicate by sammy gerbil, Qmechanic♦ Feb 4 '18 at 17:26
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1$\begingroup$ Possible duplicate of Why is small work done always taken as $dW=F \cdot dx$ and not $dW=x \cdot dF$? $\endgroup$ – sammy gerbil Feb 4 '18 at 17:23
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The required explanation is given below, if you have some doubt in this please revert back !!
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$\begingroup$ Sorry but What do you mean by force applied by every dx displacement, and why deltax not x $\endgroup$ – user182794 Feb 4 '18 at 17:28
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$\begingroup$ I said delta x, because x specifies the position of particle at some time. Whereas delta x tells the change in displacement caused by force acting on it. And Force applied for every dx displacement means, for moving some distance delta x, some force is applied, but if we reduce the delta x to infinitesimally small change in displacement we get dx. So Force applied for every dx displacement to occur. $\endgroup$ – Saurav Hiremath Feb 5 '18 at 17:06
