Maximizing Height and Final Velocity on a Tilted Rough Surface

[hastings]

Homework Statement

An object passes through point P0 in the instant t=0, going up with a velocity v0=3m/s along a tilted rough surface (dynamic friction coefficient µd=0.3) which forms an angle α=30° with the "ground". The length L of the slanted surface is practically endless.
Find
a)h, the maximum height the object can reach
b)vf, the final velocity when the object re-passes through point P0 assuming that the static friction coefficient, µs, is such that the object doesn't stop when it reaches the maximum height.
Note: This problem can be solved either by applying the 2nd dynamics principle or through energetic deductions.
[Answers: a) h=30 cm ; b) vf=1.69 m/s]

Homework Equations

K(B)+U(B)= K(A)+ U(A) (kinetic energy and potential energy)
F=µ∙N (friction force, a non-conservative force)
W=⌠F∙ds (work as integral of force and movement)

The Attempt at a Solution

I tried to solve the first question, find h , through the conservation of mechanical energy. I called B the point of max height, A=P0 the starting point.
E(B)=E(A) - Work(A→B)
½m∙v²(B)+mgh=½m∙v²(A) - F∙L
where F is the friction force, a non-conservative force: F=µd∙N=µd∙(mg∙cos(30°)).
But now how do I continue? L is ∞, and I don't know v(B). should I assume v(B)=0 in order to solve the 1st question? Is what I have done till this point, right?
the picture of this problem is a bit confusing so I'm sending you a copy just as it is on the paper.

 

[skeeter]

a) at max height, kinetic energy = 0.
displacement along the incline is d = h/sin(30). so, d = 2h

initial kinetic energy - work done by friction = final potential energy
(1/2)m(vo)^2 - u*mg*cos(30)*2h = mgh ... solve for h.

b) once you determine h from part (a) ...

initial potential energy - work done by friction = final kinetic energy
mgh - u*mgcos(30)*2h = (1/2)m(vf)^2 ... solve for vf.

 

[hastings]

yeah, I figured it out. Thank you very much for replying!