Launching Blocks up Ramps- Velocity and Height

[Phoenixtears]

Homework Statement

A 3.2 kg wood block is launched up a wooden ramp that is inclined at a 20° angle. The block's initial speed is 8 m/s. (Use µk = 0.20 for the coefficient of kinetic friction for wood on wood.)
(a) What vertical height does the block reach above its starting point?
_____m

(b) What speed does it have when it slides back down to its starting point?
_______m/s down the ramp

Homework Equations

Delta-x= V0*t + .5a(t^2)
Vf^2= V0^2 + 2ax
Vf= V0 + at
F=Ma

The Attempt at a Solution

I don't really understand one aspect of this problem, which I think is the thing that is throwing everything else off. I began by drawing a force diagram and then a horizontal vs. vertical table with initial and final veloctiy, delta-x, acceleretion, and time. We can find the vertical and horizontal inital velocity because we are given the angle and the inital velocity.

Initial Vertical- 8cos20= 2.7362
Inital Horizontal- 8sin20=7.5175

Now here's the part that's throwing me off. I know that to find a) we can say that the final velocity is 0, but wouldn't the acceleration be 9.8? I thought so, but that answer is not correct. I can't figure out what I am doing wrong. Any suggestions?

Also, for b), I don't understand why 8 is not the answer. Is it not a law that the velocities will be the same at the same location?

Thank you,

Phoenix

 

[rl.bhat]

Instead of taking the components of velocity, take the component of weight of the body along and perpendicular to the ramp. Find the total forces acting on the wooden block.Then apply the conservation of energy to it.

 

[Phoenixtears]

We have not learned conservation of energy yet. Is there another path to take to solve this problem?