physics.. check ans for me!?

Question

A block is projected up an inclined plane that makes and angle z to the horizontal. it returns to its intial postion with half its initial spped. what is the coeff of friction between the block and the plane in terms of the angle of incline?(hint: do any forces acting on the block change direction during its travels?)

this is what i got:
SUM OF Fx=Nx+wx+(Fk)x
= -mgsinz-Fk
=MAx
sum of fY=Ny+Wy+(Fk)y
= n-mgcos z
=MAy=0
max=-mgsinz-(miuk)(mgcosz)
miuk=(-Ax-sinzcosz)/gcosz
=(2ax-sin2z)/2gcosz

Answer 1

The block returns to it's original position with 1/2 of it's original kinetic energy. Since the PE is the same at the start and at the end, then the energy lost is due to friction, which is
.5*m*v1^2-.5*m*v1^2/4=2*d*m*g*cos(z)*u

(the left side is the energy lost and the right side is the normal force times the total distance traveled)

d is the distance up the incline the block traveled before reversing direction.

You could also write
.5*m*v1^2=m*g*d*sin(z)-
d*m*g*cos(z)*u

the left side is the starting kinetic energy and the right side is the gain in potential energy less the loss due to friction for the trip up the incline to the point where it reverses direction)

from above
.5*m*v1^2=8*d*m*g*cos(z)*u/3
so
8*cos(z)*u/3=sin(z)-cos(z)*u
and
u=3*sin(z)/(11*cos(z))
or
u=3*tan(z)/11

j