# Wood block coeffiecent of friction

I also need help with this problem, it involves energy and i just cant crack the problem by using what i know, i can't even start it, so here it is:
A wooden Block whose intitial speed is 6 m/s starts to slide up an incline plane at 25 degrees above the horizontal.
a)if the coeffiecent of friction is .3 how far up the plane does the block go?
b) what will be the blocks speed after it has slid back down the plane to its starting point?

a)the energy way is this but i think that assumes no friction
$$\frac{v^2}{2g} = h$$

u have to add up all the forces on the block $$|ma| = |F_g + F_f + N|$$
which if u know vectors i think gives...
$$a = 7.15 \frac{m}{s^2}$$ down the slope
so the block makes it v = v_0 + at, 0 = 6 - 7.15 t, t = 0.84s and therefore
x = 6(.84s) - 1/2 (7.15) (.84s)^2 = 2.52 m which is the hypotenus
h = 2.52m sin 25 = 1.065 m so ya i put a lot of round off error so check it out
(the energy equation yeilds h = 1.83 m so with out the friction its off 58%!!!)

b) just make sure u account for the friction on the way down

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thanks man, that helps alot