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

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SUMMARY

The discussion centers on the derivation of the Work-Energy Theorem, specifically addressing why the mass \( m \) is divided by 2 in the equation \( \int \vec{F} \cdot d\vec{s} = m \int \frac{d\vec{v}}{dt} \cdot \vec{v} dt = \frac{m}{2} \int \frac{d}{dt}(v^{2}) dt \). The key point is that the derivative \( \frac{d}{dt}(v^{2}) \) simplifies to \( 2\vec{v} \cdot \frac{d\vec{v}}{dt} \), leading to the factor of \( \frac{1}{2} \) when integrating. This derivation is crucial for understanding the relationship between force, mass, and kinetic energy.

PREREQUISITES
  • Understanding of calculus, specifically integration and differentiation.
  • Familiarity with vector notation and operations in physics.
  • Knowledge of the Work-Energy Theorem in classical mechanics.
  • Basic concepts of kinematics, including velocity and acceleration.
NEXT STEPS
  • Study the derivation of the Work-Energy Theorem in detail.
  • Learn about the implications of kinetic energy and its relationship to work done.
  • Explore advanced topics in classical mechanics, such as conservation of energy.
  • Investigate the role of mass in different physical systems and its effect on motion.
USEFUL FOR

Students of physics, educators teaching classical mechanics, and anyone interested in the mathematical foundations of the Work-Energy Theorem.

Scalise
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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##
 
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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}##
 

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