Prove energy of a falling body is constant

[DorelXD]

Homework Statement

Prove that the energy of a falling body remains constat using the derivate of a function

Homework Equations

We need to prove that:

$$\frac{mv^2}{2} + mgh$$ is constant, trhat is its derivate equals 0

The Attempt at a Solution

Let the positive direction be downward. Applying Newton's second law we get:

$$mg = F = ma$$

Now I differentiated the expression and I got:

$$Fv + \frac{dh}{dt}mg$$

If I replace mg with F, I get:

$$F(v+\frac{dh}{dt})$$

Now, how do I show that this equals 0? I don'y get it. I know that probably some elemntary fact is escaping me, but what? Could you please explain me? Thank you very much!

 

[TSny]

Hello.

How is dh/dt related to v?

 

[DorelXD]

I think I've found the answer. If dx is the change in the position of the body, then dx=-dh, so dh/dt=-dx/dt=-v, right? Another way I thought about it would be. Let A be a point such that h=AB, where B it's at ground level. Let O an arbitrary point in space, that it's on the direction of AB. Then we have that OA+ AB is constat, that is x+h is constant, and this means that d(x+h)/dt=0 . I believe that both arguments are solid and correct, but could you please give me your opinion?

 

[TSny]

Yes. That looks good.