Work, energy with kinetic friction

[Bcisewski]

Can anyone provide some assistance? I know this comes in two parts, one in locating the velocity, which I belive comes down to v=sq root of 2(9.8)(6.34), however the second part is creating havoc. Any suggestions on part II's formula?

A box slides down a frictionless 6.34 m high hill, starting from rest. At the bottom of the hill, the box slides along a level surface where the coefficient of kinetic friction uk = 0.246. How far from the bottom of the hill does the box come to rest? The final answer will be 25.8m

[Pyrrhus]

The First part

mgh = \frac{1}{2}mv^2

The Second Part:

1)

Use the kinetic constant acceleration formula

v^2 = v_{o}^2 + 2a(x-x_{o})

and Newton's 2nd Law on the box where there's friction to find the acceleration

\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}

2)

Use

\Delta E = W_{f}

\frac{1}{2}mv^2 = \mu mgd