Finding work done by friction on a box

[99gray]

1.
A box of mass
m = 59.0 kg
(initially at rest) is pushed a distance
d = 83.0 m
across a rough warehouse floor by an applied force of
FA = 204 N
directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the work done by the force of friction

Homework Equations

W = fdcosΘ
Fƒ=μN

The Attempt at a Solution

I used the two above equations and I found that the applied force of friction would be 57.82 because that is .1*59*9.8 and then if you put that into the first equation, the work would be 57.82*83*cos30 which equals 4156.108 and then since it is opposite the applied force, it would be -4156.108. At least I thought. That's wrong and I don't know why. Please help!

 

[PietKuip]

But in this problem, the normal force is larger than the weight.

 

[azizlwl]

The vertical component of F add to weight.
Remember when resolving vector, here force, noted all the components. Here the X and Y components.