Roller Coaster Problem with KE and PE

[chris61986]

Homework Statement

https://dl.dropbox.com/u/92857534/physicsproblem.png

Note: Solving for part D and E

Homework Equations

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

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

 

[mukundpa]

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

 

[chris61986]

Do you think you could be a little less cryptic?

 

[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.

 

[Nickie]

Yes, you are on the right track for solving parts D and E. The equation you have written is correct, and you can use it to solve for the final velocity, v'. To do this, you will need to know the initial velocity, v, and the values for the potential energy and kinetic energy at point B.

For parts A, B, and C, yes, you are correct. The work done by gravity is simply the change in potential energy, which is given by the equation W = mgh. Keep in mind that the change in height, Δh, may be different for each section of the roller coaster.