Is Kinetic Energy Derived from an Integral and Work Done?

[RoryP]

Hi all,
I was doing an A-level Mechanics paper the other day and one of the quesitons was to show that, starting with Hookes law T=(lambda)(x)/(l), show that the energy stored in an elastic rope is (lambda)(e²)/(2l).
This was ok, I just said that energy stored would be equal to the sum of the work done stretching the rope a small distance δx and as δ tended to zero it would be dx. Which could be re-written as integral with limits e and 0 dx.
Which leads to the equation for elastic potential energy.
After doing this i realized that kinetic energy is in a similar form, i.e power of 2 and has a multiplying factor of 1/2 which leads me to my question, is 1/2mv² the result of an integral? Has it also got something to do with work done? but with respect to v?
Any help would be greatly appreciated!

 

[rl.bhat]

dW = f*dx = m*a*dx = m*dv/dt*dx = m*v*dv ( since dx/dt = v)
Take the integration between 0 to v.

 

[RoryP]

ok brilliant thanks! what is dW? respect to work done?

 

[rl.bhat]

Yes.