Question: conservation of E and Newton's second law

[huhmattg]

I was wondering if anyone had some input on how, via calculus, to show how the law of conservation of mechanical energy for an object in free fall (ideal) is a direct consequence of the of Newtons second law.

 

[Stingray]

The energy of a particle moving in 1 dimension through a potential field V(x) is
$$E=\frac{1}{2} m ( dx/dt )^{2} + V(x)$$
Then the rate of change of the energy is
$$dE/dt = m ( dx/dt ) ( d^{2} x/dt^{2}) + dV/dt$$

Use Newton's second law on the first term to give
$$dE/dt = F ( dx/dt ) + dV/dt$$

But by definition,
$$F= - dV/dx = - (dV/dt) / (dx/dt)$$

Substituting this into the previous expression gives dE/dt=0, as expected.

 

[huhmattg]

Thanks! but what about 2-dimensional case?

 

[Norman]

Thanks! but what about 2-dimensional case?

Just substitute the correct kinetic energy into the problem. If you have rotation you will have to look the torque. And remember that:

$$\mathbf{F}=-\mathbf{\nabla}V(\mathbf{r})$$

 

[huhmattg]

Sorry but I am still confused, is there a way to express the relationship algebraically?

 

[Stingray]

What do you mean?

 

[huhmattg]

I am having a probelm understanding the formulas

 

[Norman]

Do you know what the gradiant operator is?
if not check out mathworld.com:
http://mathworld.wolfram.com/Gradient.html