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Jacobpm64
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Homework Statement
A block of mass [tex] m = 1.62 kg [/tex] slides down a frictionless incline as in the figure(link at the bottom). The block is released at a height of [tex] h = 3.91 m [/tex] 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 [tex] U(x) [/tex] of the block. Indicate the total energy on the sketch.
Homework Equations
[tex] E = K + U [/tex]
[tex] a = m \frac{v^2}{R} [/tex]
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).
[tex] E = K_{top} + U_{top} [/tex]
[tex] E = 0 + U_{top} [/tex]
[tex] E = mgh [/tex] Which I can find (62.14 J).
Now, this must be the energy everywhere during the trip. Therefore, at the bottom (point A):
[tex] E = K_{A} + U_{A} [/tex]
[tex] mgh = \frac{1}{2}mv_{A}^2 + mgh_{A} [/tex]
[tex] mgh = \frac{1}{2}mv_{A}^2 + 0 [/tex] Since [tex] h_{A} = 0 [/tex]
[tex] mgh = \frac{1}{2}mv_{A}^2 [/tex]
[tex] v_{A} = \sqrt{2gh} [/tex] Which I find to be ( 8.76 m/s ).
Now, to answer part (a), at point A:
[tex] N - mg = ma = m\frac{v_{A}^2}{R} [/tex]
[tex] N = mg + m \frac{v_{A}^2}{R} [/tex] Which is ([tex] \frac{124.31}{R} + 15.89 [/tex] Joules)
Is there any way to find what [tex] R [/tex] is?
Part (b):
When the block is at point B,
[tex] N - mgcos(45^{\circ}) = m \frac{v_{B}^2}{R} [/tex]
[tex] N = mgcos(45^{\circ}) + m \frac{v_{B}^2}{R} [/tex]
The problem here is, I don't know [tex] v_{B} [/tex], and I think I can only figure it out if I know [tex] h_{B} [/tex], 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, [tex] \alpha = 45^{\circ} [/tex].
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