Energy Proof

1. Oct 16, 2004

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?

2. Oct 16, 2004

arildno

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

3. Oct 16, 2004

decamij

Sorry, i mean proving an energy-related equation.

4. Oct 16, 2004

arildno

Which of them?

5. Oct 16, 2004

decamij

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

how about the general equation for gravitational potential energy?

6. Oct 16, 2004

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