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

[Ghost803]

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.

 

[robphy]

Work Energy Theorem

 

[Pengwuino]

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

 

[lalligagger]

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=$$\int$$F$$\bullet$$dr
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.

 

[Fredrik]

$$W=\int_{x_1}^{x_2} F dx=\int_{t_1}^{t_2} m\ddot x \dot x dt=\int_{t_1}^{t_2}\frac{d}{dt}(\frac 1 2 m\dot x^2)dt=\frac 1 2 mv_2^2-\frac 1 2 mv_1^2$$

 

[arildno]

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.

 

[Ghost803]

Thx for the replies, everyone. l'll check them all out.