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A brick slides down a wooden plank 2.0m long tilted so that one end is at a height of 1.0 m. The brick's speed at the bottom is 2.5 m/s. The plank is then sanded smooth and waxed so that the coefficient of friction is half what it was before and the brick is slid down again. What is the new speed of the brick at the bottom?

so far all i could come up with is that theta=30degrees

then i did f x m x g x cos30) x d = .5 x m x v^2 for the first block

and then w=(.5f x m x g cos30) x d = .5 x m x v^2 for the second block

then i canceled the m's, and i pluggedin v for the first equation and solved for f, and then i tried to put that f value in and come up with an answer for v but that answer is wrong because it comes out to less than 2.5 which doesn't make sense.

Thank you for the help in advanced.

so far all i could come up with is that theta=30degrees

then i did f x m x g x cos30) x d = .5 x m x v^2 for the first block

and then w=(.5f x m x g cos30) x d = .5 x m x v^2 for the second block

then i canceled the m's, and i pluggedin v for the first equation and solved for f, and then i tried to put that f value in and come up with an answer for v but that answer is wrong because it comes out to less than 2.5 which doesn't make sense.

Thank you for the help in advanced.

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