Can't figure it out.(Force of tension)

[stealthking]

A sledge loaded with bricks has a total mass of 18kg and is pulled at a constant speed by the rope.The rope is inclined at 20 degrees above the horizontal, and the sledge moves a distance of 20m on a horizontal surface.The coefficient of kinetic friction between the sledge and surface is .500.
(a)What is the tension of the rope?
(b)How much work is done by the rope on the sledge?
(c)What is the mechanical energy lost due to friction?
I keep getting the wrong answer for (a).
Havn't gotten to (b) and (c).
I tried to get the normal force of the sledge by this formula...
m(9.8)-(m(9.8sin20))
I just can't get anywhere with this problem.
My teacher can't teach right either.

 

[Leong]

I will consider the sledge together with its bricks as a particle. There are 4 forces acting on this particle; its weight(mg), the rope tension(T), the normal force(N), the kinetic frictional force(F). Since the particle moves with constant speed, the sum of all forces along the incline is zero. so does the sum of all forces which are perpendicular to the incline. i will choose the positve x-axis along the incline pointing upward and the positve y-axis perpendicular to the incline pointing upward.
y component forces :
N+(-mgcos20)=0
x component forces :
T+(-mgsin20)+(-F)=0 with
$$F=\mu_{k}*N$$

 

[stealthking]

the answer is 79.4N
I don't see how they got it.

 

[Leong]

I don't know neither.

 

[stealthking]

lol
i give up on it,.