Why is kinetic energy 1/2mv^2 instead of mv^2?

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SUMMARY

The kinetic energy (K.E.) of an object is defined as K.E. = 1/2 mv^2, where m is mass and v is velocity. This formula is derived from the Work-Energy Theorem, which states that work done (W) is equal to the integral of force (F) over distance (d). The relationship between force and distance leads to the inclusion of the 1/2 factor in the kinetic energy equation, distinguishing it from the simpler expression mv^2, known as "vis viva." Understanding this derivation clarifies why kinetic energy is expressed with the 1/2 coefficient.

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  • Understanding of Newton's Second Law (F = ma)
  • Basic knowledge of calculus, particularly integration
  • Familiarity with the concepts of work and energy in physics
  • Knowledge of kinematic equations and their applications
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Ghost803
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The definition of joules, th unit of energy is the 1 J = 1 kg * m2/v2.

And that 1 joules is the amount of energy needed for the work done by one Newton traveling one meter.

From all this, I got the impression that to get energy, you would have to multiply force times distance. Is work, then just a measurement of energy.

Assuming my first assumption that energy = F x D, let's sat a 5kg object is accelerating towards Earth from 20 feet above at 9.8 m/s2 . To get the force, you would have to do 5 kg x 9m/s2 Then to get energy, that force times distance. So if it traveled 5 meters. The kinetic energy should be F(5kg x 9.8 m/s2) x D(5 meters.)

Which would end up being 5kg x 49m2/s2.

Sooo, can someone explain why ke is 1/2 mv2 instead of just mv2.
 
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Work Energy Theorem
 
I hope you'll notice that nowhere did you determine its actual speed after falling the 5 meters. If you use your kinematic equations and solve for the time it took to reach the 5 meters and subsequently found the velocity at that 5meter point, you'd see that the energy is infact, mv^2/2
 
Assuming my first assumption that energy = F x D

Right, work is the energy added to or taken away from a system when it encounters a force over some distance. The definition of work is actually the integral,
W=\intF\bulletdr
Where dr is the change in position, the D in your equation.
The equation for the kinetic energy of an object, K.E.=1/2mv^2 is derived by making a series of substitutions in the above equation and integrating.
 
Last edited:
W=\int_{x_1}^{x_2} F dx=\int_{t_1}^{t_2} m\ddot x \dot x dt=\int_{t_1}^{t_2}\frac{d}{dt}(\frac 1 2 m\dot x^2)dt=\frac 1 2 mv_2^2-\frac 1 2 mv_1^2
 
Well, there would be no conceptual problems whatsoever to define a quantity mv^2, and use this instead of "kinetic energy".

This has already been done; mv^2 is called "vis viva".

But, the work-"vis viva" theorem is less aesthetic than the equivalent work-energy theorem, due to the explicit inclusion of the factor of 1/2 in the formula.
 
Thx for the replies, everyone. l'll check them all out.
 

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