Proof that energy of a free body is constant, using the derivate

Question

Ok, what I'm trying to prove is the law of conservation of energy for a free fall. Let the downward direction be positive. We want to prove that:

$$mgh+\frac{mv^2}{2}=constant$$

For this, we try to prove that the derivate of the function is constant:

$$\frac{d}{dt}(mgh+\frac{mv^2}{2})=0 $$ $$mg\frac{dh}{dt}+Fv=0 $$

Applying Newton's law we get $$F=mg$$

Now, $\frac{dh}{dt}=v$

$$mgv+Fv=0 $$ $$Fv+Fv=0 $$

And it doesn't cancel out..

I understand the principle, and I have been able to prove it without using the derivate..but now it seems that I can't do it. I've been applying in a lot of problems. What am I getting wrong?

Answer 1

Note that $v=-\frac{dh}{dt}$. It increases as you come down and decrease $h$

Answer 2

Applying Newton's 2nd law and projecting on z, we get : $$-m\frac{dv}{dt}=\parallel \vec{W}\parallel=mg $$ Then you multiply by v $$-mv\frac{dv}{dt}=mgv\\ \frac{-d(\frac{mv²}{2})}{dt}=mg\frac{dh}{dt}\\ \frac{d(\frac{mv²}{2}+mgh)}{dt}=0\\ QED$$