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

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)