Talk:Work (physics)/Archives/2013/October

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Mathematical calculation section

Is there any particular reason why the differential of work is introduced as $\delta W=\mathbf {F} \cdot \mathbf {v} dt$ ? Is it possible to just start with $\delta W=\mathbf {F} \cdot d\mathbf {x}$ ? JCMPC (talk) 14:17, 5 October 2013 (UTC)

Good suggestion. I have added

\delta W=\mathbf {F} \cdot d\mathbf {x}

as an extra step in the opening equation. See the diff. Dolphin (t) 00:09, 8 October 2013 (UTC)