How is Mechanical Energy Conserved in a Gravitational System?

[decamij]

I am in grade 12 physics, and i have to practice equation proofs. I am currently studying work, kinetic energy, springs, and potential energy (gravity and elastic).

Does anyone have a good proof?

 

[arildno]

Proof of what?
Your question is much too vague..

 

[decamij]

Sorry, i mean proving an energy-related equation.

 

[arildno]

Which of them?
Specifics, please.

 

[decamij]

i know its a simple equation, but i suck at proofs.

how about the general equation for gravitational potential energy?

 

[arildno]

Ok, I'll derive the conservation of mechanical energy for you, in the case of a point particle under the influence of the force of gravity.
I write Newton's 2.law in vector form:
-mg\vec{j}=m\vec{a}
where \vec{a}=\frac{d^{2}\vec{x}}{dt^{2}} is the acceleration, and
\vec{x}(t)=x(t)\vec{i}+y(t)\vec{j}
and the velocity is given by:
\vec{v}=\frac{d\vec{x}}{dt}
We also have:
\vec{a}=\frac{d\vec{v}}{dt}
1. Form the dot product
between \vec{v} and Newton's 2.law:
-mg\frac{dy}{dt}=m\frac{d\vec{v}}{dt}\cdot\vec{v}
2. Integrate this equation between 2 arbtriray points of time:
\int_{t_{0}}^{t_{1}}-mg\frac{dy}{dt}dt=\int_{t_{0}}^{t_{1}}m\frac{d\vec{v}}{dt}\cdot\vec{v}dt
3. The left-hand side is easy to compute:
\int_{t_{0}}^{t_{1}}-mg\frac{dy}{dt}dt=-mgy(t_{1})+mgy(t_{0})
4. We note the identity:
\frac{d\vec{v}}{dt}\cdot\vec{v}=\frac{d}{dt}(\frac{\vec{v}^{2}}{2})
where \vec{v}^{2}\equiv\vec{v}\cdot\vec{v}
5. Hence, the right-hand side in 2.) may be computed:
\int_{t_{0}}^{t_{1}}m\frac{d\vec{v}}{dt}\cdot\vec{v}dt=\frac{m}{2}\vec{v}^{2}(t_{1})-\frac{m}{2}\vec{v}^{2}(t_{0})
6. Collecting insights from 3. and 5., 2. may be rewritten as:
-mgy(t_{1})+mgy(t_{0})=\frac{m}{2}\vec{v}^{2}(t_{1})-\frac{m}{2}\vec{v}^{2}(t_{0})
7. Or, rearranging 6., we gain:
mgy(t_{1})+\frac{m}{2}\vec{v}^{2}(t_{1})=mgy(t_{0})+\frac{m}{2}\vec{v}^{2}(t_{0})
8. Or, noting that t_{1},t_{0} were ARBITRARY, every mechanical energy amount must remain the same at all times, so we get, by eliminating the specific time parameter:
mgy+\frac{m}{2}\vec{v}^{2}=K
where K is some constant for the whole motion.
That is, the mechanical energy is conserved for the particle