What is the Force and Speed at Different Points on a Rollercoaster?

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Homework Statement


A block of mass m = 1.62 kg slides down a frictionless incline as in the figure(link at the bottom). The block is released at a height of h = 3.91 m above the bottom of the loop.
(a) What is the force of the inclined track on the block at the bottom (point A)?
(b) What is the force of the track on the block at point B?
(c) At what speed does the block leave the track?
(d) How far away from point A does the block land on level ground?
(e) Sketch the potential energy U(x) of the block. Indicate the total energy on the sketch.

Homework Equations


E = K + U
a = m \frac{v^2}{R}

The Attempt at a Solution



So, since the track is frictionless, total energy is constant. Therefore, we can calculate the energy at the top of the track (and this has to be the same at every part throughout the trip).

E = K_{top} + U_{top}
E = 0 + U_{top}
E = mgh Which I can find (62.14 J).

Now, this must be the energy everywhere during the trip. Therefore, at the bottom (point A):
E = K_{A} + U_{A}
mgh = \frac{1}{2}mv_{A}^2 + mgh_{A}
mgh = \frac{1}{2}mv_{A}^2 + 0 Since h_{A} = 0
mgh = \frac{1}{2}mv_{A}^2
v_{A} = \sqrt{2gh} Which I find to be ( 8.76 m/s ).

Now, to answer part (a), at point A:
N - mg = ma = m\frac{v_{A}^2}{R}
N = mg + m \frac{v_{A}^2}{R} Which is (\frac{124.31}{R} + 15.89 Joules)

Is there any way to find what R is?

Part (b):
When the block is at point B,
N - mgcos(45^{\circ}) = m \frac{v_{B}^2}{R}
N = mgcos(45^{\circ}) + m \frac{v_{B}^2}{R}

The problem here is, I don't know v_{B}, and I think I can only figure it out if I know h_{B}, which I cannot see how to find.

Any help would be appreciated! (Then I can continue working the other parts)

http://img27.imageshack.us/img27/3923/physicsg.jpg"

Note: In the figure, \alpha = 45^{\circ}.
 
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Since R was not given, your answers will be in terms of R.
 
Okay, I see. All of my answers will be in terms of R.

Can anyone give me any hints on how to find v_B and h_B?

Thanks!
 
You know the angle and the radius R. So you can find the tangential velocity and the height using trigonometric relations. Again, these answers will be in terms of R.
 
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