Prove energy of a falling body is constant

[DorelXD]

Homework Statement

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

Homework Equations

We need to prove that:

[tex] \frac{mv^2}{2} + mgh [/tex] 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:

[tex] mg = F = ma [/tex]

Now I differentiated the expression and I got:

[tex] Fv + \frac{dh}{dt}mg [/tex]

If I replace mg with F, I get:

[tex] F(v+\frac{dh}{dt}) [/tex]

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.

 

[paranormal]

To prove that the energy of a falling body is constant, we can use the law of conservation of energy. This law states that energy can neither be created nor destroyed, but it can only be transferred or converted from one form to another. In the case of a falling body, the initial potential energy (mgh) is converted into kinetic energy (mv^2/2) as the body falls. The total energy remains constant throughout the fall.

To prove this mathematically, we can use the derivative of the total energy with respect to time. Since energy is constant, its derivative must be equal to 0. Let's start with the expression for total energy:

E = \frac{mv^2}{2} + mgh

Taking the derivative with respect to time, we get:

\frac{dE}{dt} = m\frac{dv}{dt}v + mg\frac{dh}{dt}

Using Newton's second law, we can substitute ma for mg:

\frac{dE}{dt} = ma\frac{dv}{dt} + ma\frac{dh}{dt}

We know that acceleration (a) is equal to the rate of change of velocity (dv/dt). So we can rewrite the equation as:

\frac{dE}{dt} = ma\frac{dv}{dt} + ma\frac{dh}{dt} = ma\frac{dv}{dt} + ma\frac{dv}{dt}

Since the body is falling, its velocity (v) is increasing and its position (h) is decreasing. This means that both derivatives (dv/dt and dh/dt) are negative. Substituting this in the equation, we get:

\frac{dE}{dt} = ma\frac{dv}{dt} + ma\frac{dv}{dt} = ma(-) + ma(-) = -ma^2

Since acceleration (a) is always negative in this case, the derivative of energy is also always negative. This means that the energy is decreasing over time. However, we know that the total energy is constant, so the derivative must be equal to 0. Therefore, we can conclude that the energy of a falling body is constant.

In summary, we can prove that the energy of a falling body is constant by using the law of conservation of energy and taking the derivative of the total energy with respect to time. The resulting derivative