What is the momentum of the box at the bottom?

[punjabi_monster]

A 98.0 N box slides 25.0 m along a 35.0° incline. If the force of friction along the incline is 32.0 N, and the box starts from rest at the top, what is the momentum of the box at the bottom?

This is how i attempted to solve this question. The answer in the book is 1.1 * 10^2 Kgm/s. Can you please tell me what error i have made. Thank-you.

Eptop + Ektop = Epbot + Ekbot
mgh + 0 = 1/2mv^2 + 0
(9.81m/s^2)(25.0m*Sin35) = 1/2v^2
v=16.77 m/s

Fg=mg
m=98.0 N/9.81m/s^2
m=9.9898 Kg

p=mv
p=(9.9898 Kg)(16.77 m/s)
p=168 Kgm/s

 

[Tide]

You didn't take into account the energy lost to friction.

 

[phy]

Lol shaani nu gaal kadh

 

[punjabi_monster]

i think i figured it out:

Ep=Ek + Ffd
mgh=1/2mv^s + Ffd

THen solve for v and use that in p=mv