Work done by forces pulling a sled at constant speed

[Jac8897]

A boy pulls a 14.6kg sled 12.5m at a constant speed across a horizontal floor. if the rope makes a 32.4degrees?
angle with the horizontal and the coefficient of friction is .489, what is the work done by each of the forces acting on the sled?
2 days ago - 2 days left to answer.
Additional Details
my work,
from the free body diagram I get

EFy= n+Fsin@-mg=0 "@ = theta"
EFx= Fcos@-f=0 "constant speed"

solve Fy= n=mg-Fsin@
plug it into Fx

friction=un
Fcos@-u(mg-Fsin@)=0
Fcos@-umg+uFsin@=0 >>
Fcos@+uFsin@=umg >>
F(cos@+usin@)=umg>>
F=(umg)/(cos@+usin@)
F=63.3N >>

work of the force
WF=63.3N(12.5)cos0>> WF= 791J

no solve by friction
Fcos@-f=0 >>> f=Fcos@
f=(63.3N)cos(32.5) >> f=53.4N

work of friction
Wf=(53.4N)(12.5M)cos180 >> Wf= -668J

I ignore the N force and the gravity and Fsin@ because they are perpendicular.

is my physics process good?
is my answer right?
thanks

 

[willem2]

All the forces you calculated are OK. The total work done on the sled isn't 0. This isn't possible since the sled gains no kinetic or potential energy. Are you sure that both of the forces involved are pulling in the same direction as the motion of the sled?

 

[Jac8897]

so the total work is W=WF+Wf

W=791J-668J
W=123J
the total work done on the sled is 123J