1. The problem statement, all variables and given/known data A 2.0kg wood block is launched up a wooden ramp that is inclined at a 25° angle. The block's initial speed is 9.0m/s . The coefficient of kinetic friction of wood on wood is μk = 0.200. a) What vertical height does the block reach above its starting point? b) What speed does it have when it slides back down to its starting point? 2. Relevant equations (solve this without using the conservation of energy equation -- the 1/2*m*v^2 one because we haven't gone over that in class yet) Vf2 = Vi2+as 3. The attempt at a solution I tried a lot of different things, but my numbers just wouldn't work out. I put my Y axis along the ramp, and my X axis perpendicular to the ramp. For part a: I know that I have to: 1) Find the acceleration (mgCos[itex]\theta[/itex] for X direction and mgSin[itex]\theta[/itex] for Y direction) 2) Somehow incorporate the friction into the acceleration (I have no idea how to do this... We haven't gone over anything like this in class yet, that's why I'm so confused on this extra credit problem!) 3) Plug the acceleration I found into the Vf2 equation to find s (height). For part b: I'd have to find the acceleration going downward incorporating the friction. But, again, I have absolutely no idea how to use the friction... My work so far: I already have my axis set up and, for the most part, figured out. I'm still getting the wrong answer though. Given: 2kg = 2kg(9.8) = 19.6N weight Angle = 25° Vi = 9m/s μk = .2 X-axis is parallel to the ramp, Y axis is perpendicular to the ramp. Find the height (which is in the X direction). Find ax: Fgx + Fk = max mgSin∅ + mMkCos∅ = max gSin∅ + MkgCos∅ = a g(Sin∅ + MkCos∅) = a 9.8(Sin(25) + (9.8)(.2)(Cos(25)) = a a = 5.918 Plug into Vf2 = Vi2 + ah 0 = 92 + (5.918)h 13.68 = h But it said this answer for height was wrong... I'm not sure how to fix what I've done so far since to me it seems like it should work.