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In order for the work-energy theorem to have meaning, work (F·d) and kinetic energy (0.5 .M.v^2 should have the same units. Show that they do

 Sep 20, 2020
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In order for the work-energy theorem to have meaning, work (F·d) and kinetic energy (0.5 .M.v^2 should have the same units. Show that they do.

 

Hello Guest!

 

\(W=F\cdot d\\ W=1N\cdot 1m=1Nm\\ W=1Nm\cdot \frac{kg\cdot m}{N\cdot s^2}\\ \color{blue}W=1\ \frac{kg\cdot m^2}{s^2}\ = 1\ Joule\)    

 

\(E=\frac{1}{2}\cdot m\cdot v^2\\ E=\frac{1}{2}\cdot 1kg\cdot (1.414\ \frac{m}{s})^2\\ \color{blue}E=1\ \frac{kg\cdot m^2}{s^2}=1\ Joule\)

 

laugh  !

 Sep 20, 2020
edited by asinus  Sep 21, 2020

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