Finding the Height of a Block Released from a Frictionless Track

[Elbobo]

Homework Statement

A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track (as shown in: http://img392.imageshack.us/img392/5711/25li4.jpg ).
It leaves the track horizontally, flies through
the air, and subsequently strikes the ground.
The acceleration of gravity is 9.81 m/s^2.

At what height h above the ground is the block released? Answer in units of m.



I tried a bajillion different methods, and my latest brought me to figuring out the speed of the block is right before the rough area in order to find the unknown height. However, since I can't find that speed, I can't find that height either. Soemone help please :D

 

[Doc Al]

Why can't you find the speed of the block as it leaves the track in terms of h?

 

[Elbobo]

Err, is this what you meant...:

2.4 = (gt^2)/2
t = sqrt (4.8 / g)
vx = x / t = 3.69/ (sqrt (4.8/g))

vx = v

v = sqrt(2g (h-2.4)) = 3.69 / (sqrt (4.8/g))
h = 3.81834375 m

It's wrong though ><

 

[Elbobo]

Oh and I'm pretty sure that I need to incorporate the coefficent of kinetic friction, but I can't figure out a way to do it with this problem using Fnet = Fa - Ffr

 

[Doc Al]

To find the speed of the block, use energy conservation:
Energy(initial) + Work done by friction (which is negative) = Energy(final)

Where Energy means mechanical energy: KE + PE.

 

[Elbobo]

OH! I can't believe I didn't think of that. Thank you, that worked.