Kinetic energy's dependence on velocity.

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Kinetic energy is dependent on the square of velocity due to the mathematical relationship derived from the work-energy principle, which states that work done on an object results in a change in kinetic energy. This relationship implies that as velocity increases, the energy required to accelerate an object also increases significantly, making it more energy-intensive to raise the speed from higher velocities compared to lower ones. Real-life examples include vehicles needing more power to accelerate from 100 to 200 m/s than from 0 to 100 m/s, illustrating the quadratic nature of kinetic energy. Doubling the velocity results in quadrupling the kinetic energy, reinforcing the concept that energy increases with the square of velocity. Understanding these principles is crucial for applications in physics and engineering.
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This is not homework but rather a personal question.

Why is kinetic energy dependent on the square of the velocity. Is there a physical answer or is it simply mathematically derived. If there are some real life examples that can explain this then it would be greatly appreciated.

Similarly, why does it take more energy to accelerate an object at high velocities than low. For example, it takes more energy to raise an object's velocity from 100 - 200 m/s than it does to raise the velocity of the same object from 0 - 100 m/s.
 
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This is the way I remember explained - using the product of force over distance, which becomes
W = m \int{v dv}, with an equivalence between work and energy.
http://scienceworld.wolfram.com/physics/KineticEnergy.html

So since kinetic energy is proportional to v2, and the change would be related to v dv, which changes linearly with v.

Also, if one doubles v, e.g. vf = 2 vi, then Ef is proportional to vf2 = (2 vi)2 = 4 vi2 , or Ef = 4 Ei.
 
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