Possible explanation for kinetic energy

[physdoc]

why does kinetic energy depend on half the square?
if we add up all the velocities does it not produce half a square?

 

[drvrm]

why does kinetic energy depend on half the square?
if we add up all the velocities does it not produce half a square?

imagine you are pushing a cart and produce a rate of change of displacement i.e. velocity and somebody says you have given K.E. to the cart by pushing through a force F.
now the work being done by you is F.dx if dx is displacement and F is related to change in velocity/momentum - so the energy input
you provided is m.dv/dt . dx for a small time dt involving displacement.
this can be written as m.dv/dx. dx/dt .dx so summing up for all such displacement = m.v.dv integrated from v=0 to v=v and that's the energy .with a factor of 1/2.m. v^2

 

[physdoc]

imagine you are pushing a cart and produce a rate of change of displacement i.e. velocity and somebody says you have given K.E. to the cart by pushing through a force F.
now the work being done by you is F.dx if dx is displacement and F is related to change in velocity/momentum - so the energy input
you provided is m.dv/dt . dx for a small time dt involving displacement.
this can be written as m.dv/dx. dx/dt .dx so summing up for all such displacement = m.v.dv integrated from v=0 to v=v and that's the energy .with a factor of 1/2.m. v^2

Please explain further ...I am not familiar with the mechanics of calculus notation.

 

[russ_watters]

If you accelerate something at 1m/s/s for two seconds, it goes 4x as far as if you accelerate it for one second. 2²=4

 

[physdoc]

Thanks... That is another way to derive it.

 

[drvrm]

Please explain further ...I am not familiar with the mechanics of calculus notation.

when one integrates expression like integral of v.dv the result is v^2/2. people not knowing the integration will sum the values from 0 to v and w

 

[physdoc]

So this is the same as what I said?

 

[physdoc]

what do you mean when you say w

 

[shrey012]

why does kinetic energy depend on half the square?

The work done on an object to move it a distance x in the direction of the force applied is Fx. F = ma and a = (v-u)/t ; we also know that x = 1/2 (u+v) t since this is just the average velocity multiplied by time.

Work done = energy transferred = kinetic energy gained. This is m(v-u)/t * 1/2 (u+v)t = 1/2 m (v^2 - u^2). The kinetic energy is defined as the energy due to motion compared to when the object is at rest, so u = 0.

Therefore, KE = 1/2mv^2

 

[drvrm]

w i referred to the total energy imparted to the cart -which is K.E.-work done by the 'push force' which is responsible for the 'motion' the work done is a check on K.E. value. i.e. 1/2. m. v^2