Work done by a father pushing a sled up a hill

[endeavor]

A father pushes horizontally on his daughter's sled to move it up a snowy incline, as illustrated in Fg. 5.23b. If the sled goes up the hill with a constant acceleration of 0.25m/s², how much work is done by the father in getting it from the bottom to the top of the hill? (The total mass of the sled and the girl is 35kg and the coefficient of kinetic friction between the sled runners and the snow is 0.25)

the slope must be 3.6/sin 15º.
the Normal Force is N=mg cos 15º = 331.31N
Therefore the kinetic friction is f_k=u_k = 82.83N

Now here's the part I think I'm doing wrong, because I get the wrong answer:
F_net=ma = 8.75N
F_net=F cos 15º - f_k - mg sin 15º

solving for F, I get F=186.7N

to find the amount of work done by the father, I use F cos 15º (because it is the force in the same direction as the displacement).

W = F cos 15º * 3.6/sin 15º = 2508 J

However, the answer is 2700 J. What have I done wrong?

 

[Doc Al]

the Normal Force is N=mg cos 15º = 331.31N
Therefore the kinetic friction is f_k=u_k = 82.83N

Realize that the father's horizontal force affects the normal force; the normal force is not simply [itex]mg \cos \theta[/itex].

 

[endeavor]

Realize that the father's horizontal force affects the normal force; the normal force is not simply [itex]mg \cos \theta[/itex].

thanks, i got the right answer now