
#1
Jun2309, 09:28 AM

P: 76

Hi all,
I was doing an Alevel 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)(e^{2})/(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 rewritten 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 1/2mv^{2} 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! 



#2
Jun2309, 09:47 AM

HW Helper
P: 4,442

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



#3
Jun2309, 09:55 AM

P: 76

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




#4
Jun2309, 10:09 AM

HW Helper
P: 4,442

Derivation of Kinetic Energy
Yes.



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