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Derivation of Kinetic Energy

by RoryP
Tags: derivation, energy, kinetic
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RoryP
#1
Jun23-09, 09:28 AM
P: 76
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 realised 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!
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rl.bhat
#2
Jun23-09, 09:47 AM
HW Helper
P: 4,435
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
#3
Jun23-09, 09:55 AM
P: 76
ok brilliant thanks! what is dW? respect to work done?

rl.bhat
#4
Jun23-09, 10:09 AM
HW Helper
P: 4,435
Derivation of Kinetic Energy

Yes.


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