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physicshelpppp
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Homework Statement
A block with a mass of 1.137 kg is placed at the top of an inclined plane. The plane is d = 2.592 m long. The angle θ the plane makes with the horizontal can be adjusted, changing the height of the top of the plane.
The lower half of the plane has a rough surface with μs = 0.417, μk = 0.375, while the top half of the plane is smooth with μs' = 0.135, μk' = 0.087. These coefficients are given between the plane and the block.
The block is placed at rest at the very top of the slope. What minimum angle should the slope be set for the block to just start sliding downwards?
I calculated the angle to be 7.688 degrees (which is been accepted as correct)
Now I need to calculate - The slope is held at this angle and the block slowly accelerates from rest down the slope. What is the speed of the block halfway down the slope?
The Attempt at a Solution
Vf^2 - Vi^2 = 2as
Vf^2 = 2 x (9.8m/s x sin(7.688) x 1.296m
= 3.398
sqrroot = 1.84m/s
But this is incorrect. Can anyone point me in the right direction or let me know what I am not understanding? Thank you in advance! :)