Why Is Mass Halved in the Work-Energy Theorem Derivation?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
Scalise
Messages
1
Reaction score
0
Hello,

Someone could explain me why in the derivation below the mass m is divided by 2 in the last step?:

##\int\vec{F}\cdot d\vec{s}=m\int\frac{d\vec{v}}{dt}\cdot\vec{v}dt=\frac{m}{2}\int \frac{d}{dt}(v^{2})dt##
 
Physics news on Phys.org
Scalise said:
Hello,

Someone could explain me why in the derivation below the mass m is divided by 2 in the last step?:

##\int\vec{F}\cdot d\vec{s}=m\int\frac{d\vec{v}}{dt}\cdot\vec{v}dt=\frac{m}{2}\int \frac{d}{dt}(v^{2})dt##

The derivative ##\frac{d}{dt}(v^{2})## is equal to ##2\vec{v}\cdot\frac{d\vec{v}}{dt}##