How to show cons. of Energy from F=change of potential

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The discussion centers on the relationship between force and potential energy, specifically how to derive kinetic energy from the equation mr'' = F(r) = -δV. The user seeks clarification on integrating the left-hand side (LHS) of the equation to show the kinetic energy difference, noting the necessity of a 1/2 term. The response highlights that the derivative of r'² results in the factor of 2, which explains the appearance of the 1/2 term in the integration process.

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trelek2
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Force is equal to change in potential (r is a point in space suppose integrating between points r1 to r2 in time t1 to t2, i did not indicate this but it is not relevant for my quenstion:
[tex]mr'' =F(r) = -\delta V[/tex]
after multiplying both sides by dr it is obvious RHS will give potential difference, but LHS needs to be transformed to show corresponding kinetic energy difference:
[tex]mr'' \cdot dr = m \int_{}^{} r'' \cdot r' \cdot dt = m \int_{}^{} \frac{d}{dt}r' ^{2}dt[/tex]
:(
The problem is there should be a 1/2 term there as well. I don't know how to integrate this LHS. Can you please help- show me clearly step by stem what is done and where this 1/2 comes in...
 
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trelek2 said:
[tex]mr'' \cdot dr = m \int_{}^{} r'' \cdot r' \cdot dt = m \int_{}^{} \frac{d}{dt}r' ^{2}dt[/tex]
:(
The problem is there should be a 1/2 term there as well. I don't know how to integrate this LHS. Can you please help- show me clearly step by stem what is done and where this 1/2 comes in...

Hi trelek2! :smile:

It's because d/dt (r'2) = 2 r' r'' :wink:
 
Yes, I am a retard:P
 

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