phy question

With reference to the diagram below:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Oct07/Crazymech1.jpg


圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Oct07/Crazymech2.jpg

1.The accleration when it travels up the inclined plane

When it is moving up the incline plane, there is NO force pushing it in the same direction.

However, forces acting on it down the incline plane are:
frictional force f = 0.25N = 0.25 x mgcos35
& mgsin35

thus,
by 2nd Law,

net force = m x a
-0.25mgcos35 + (-mgsin35) = ma
-0.25 x 10 x cos35 - 10 x sin35 = a
a = -7.784 metres per square second (-ve means decelerating)


2.the distance it travels up the plane

When it reaches the highest position, its velocity is ZERO.
given:
u = 32,
v = 0,
a = -7.784
s = ??

applying
v^2 - u^2 = 2as,
0 - (32)^2 = 2 x (-7.784) x s
s = 65.78 metres


3. the acceleartion when it slides down the incline plane

When it slides down the incline plane,

downward force acting on it is
mgsin35

upward force acting on it is
frictional force f = 0.25mgcos35

by 2nd Law,
net F = ma

mgsin35 - 0.25mgcos35 = ma
10 x sin35 - 0.25 x 10 x cos35 = a
a = 3.688 metres per square second (down the incline plane)


4. the speed of the block when it return to its starting point
at highest position, u = 0
a = 3.688,
s = 65.78
v = ??

applying v^2 - u^2 = 2as,
v^2 - 0 = 2 x 3.688 x 65.78
v = 22.03 metres per second


5. the time taken for the block to return it starting point
for moving up the incline plane,
u = 32,
v = 0,
a = -7.784
s = 65.78
t = ??

applying v = u + at,
0 = 32 - 7.784 x t
t = 4.111

for moving down the incline plane,
u = 0
a = 3.688,
s = 65.78
v = 22.03

applying v = u + at
22.03 = 0 + 3.688 x t
t = 5.973

so the time required is 4.111 + 5.973 = 10.08 seconds