# Roller Coaster Problem with KE and PE

1. Apr 9, 2013

### chris61986

1. The problem statement, all variables and given/known data

https://dl.dropbox.com/u/92857534/physicsproblem.png [Broken]

Note: Solving for part D and E

2. Relevant equations

Unsure. Language in the problem seems ambiguous.
Perhaps (KE + PE)b = (KE + PE)a

3. The attempt at a solution

So I think it's done this way.

From A to B and B to C, the equation would look this way:
PEb + KEb = KEa + PEa
PEa = 0, move KEb to right side, left with:
PEb = KEa - KEb
so WGH = 1/2mv'^2 - 1/2mv^2

Am I going the right way about this? If so, where do I go from here when I don't know both velocities?

Also, for parts A, B, and C, it's simply W = mgΔh, correct? I'm not used to things being easy :P

Last edited by a moderator: May 6, 2017
2. Apr 9, 2013

### mukundpa

Work energy rule says that the increase in kinetic energy of a body is equal to work done on it.

3. Apr 9, 2013

### chris61986

Do you think you could be a little less cryptic?

4. Apr 9, 2013

### mukundpa

Work done by gravity is equal to the increase in kinetic energy of the car. The same is the loss in potential energy. W = mgΔh is correct.
In second part I think potential energy is to be calculated at different points.