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)