Derivative of Force and Work in Respect to Time

[dextercioby]

Indeed that is what i meant. Don't mind dextercioby with his useless remarks. He is just angry and not willing to see he's just regurgitating self-adapted physics...

let's not get into that...

YOU ARE RIGHT...

marlon

I'm not angry... For the last couple of days,i've been laughing my a$$ out... I ain't got nobody to be angry at...If somebody proves me wrong,i'll accept it,even if it's Marlon... So far,i haven't had sufficient evidence to admit i was/am/will be wrong...

Self-adapted physics u say... I wish i could believe you...

Daniel.

 

[marlon]

I'm not angry... For the last couple of days,i've been laughing my a$$ out... I ain't got nobody to be angry at...If somebody proves me wrong,i'll accept it,even if it's Marlon... So far,i haven't had sufficient evidence to admit i was/am/will be wrong...

Self-adapted physics u say... I wish i could believe you...

Daniel.

good for you...

marlon

 

[marlon]

This is wrong,if 'W' stands for potential energy.If it stands for work,then the force should not depend on time:
$$\frac{dW}{dt}=\frac{d}{dt}(\vec{F}\cdot \vec{r})=\vec{F}\cdot \vec{v}=P$$(1)
,where P is the mechanical power.

this is just plain wrong again...

$$\Delta W = \int mvdv = \Delta E _{kinetic}$$

from this formula you need to start...

marlon