Solving for the New Speed of a Brick on a Waxed Plank

[meh6287]

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.

 

[arildno]

But you haven't taken into account the change in potential energy here!

 

[HallsofIvy]

Here's how I would do this, NOT using energy.

The force normal to the plank is -mg cos(30) and the force in the direction of the plank is -mg sin(30). The friction force along the plank is fmg cos(30) (f is the coefficient of friction) so the total force on the brick along the plank is
f mg cos(30)- mg sin(30)= -mg(sin(30)- fcos(30)) and of course, the acceleration is
a= -g(sin(30)- fcos(30)). After a time T, the velocity of the brick will will be aT and the distance traveled is (a/2)T².
Let T be the time required for the brick to reach the end of the plank. Then we are told that aT= 2.5 and (a/2)T²= 2. Since T= 2.5/a, the second equation is
(a/2)(2.5²/a²)= 2 so 4a= 2.5² or a= 2.5²/4.

You can now solve for f. Once you know that, halve it to find the new friction coefficient, find the total force along the plank just as before and the acceleration from that.