# Law of conservation of energy

1. Feb 24, 2014

### LagrangeEuler

1. The problem statement, all variables and given/known data
Particle goes to free fall from point $A$ to point $B$. $\upsilon_{A}=0$. How to define law of conservation of energy. Coordinate system is selected in such way that $y$ coordinate od point $A$ is zero. (See figure energy)

2. Relevant equations
For free fall $\upsilon_{B}=\sqrt{2gy}$.

3. The attempt at a solution
My problem is with formulation of law of conservation of energy. Energy of point $A$ is in this case $0$? Right? Because potential energy is zero ($y$ cordinate of point $A$ is zero) and in case of free fall $\upsilon_{A}=0$.
And energy of point $B$ is also zero $-mgy+m\frac{\upsilon^2}{2}=0$.
Am I right? I'm confused how the energy of the particle could be zero? Can you help me? Give me some other view?

#### Attached Files:

• ###### energy.JPG
File size:
6.8 KB
Views:
70
2. Feb 24, 2014

### Staff: Mentor

What you take to be the zero of energy is arbitrary. At point B, the particle as $-mgy < 0$ potential energy (it has lost some potential energy) and $mv^2/2 > 0$ of kinetic energy (it has gained kinetic energy). (Note: I take it that the y axis is pointing downwards.)

If you choose point B to be the zero of potential energy, then at point A, you have $-mgy + 0 > 0$ (since $y < 0$) and at point B $0 + mv^2/2$, with $mv^2/2 = -mgy$, which is exactly the same as before.

In both cases, you see potential energy converted into kinetic energy.