Crate being pulled

[Gentec]

Good evening.

I would appreciate if someone could verify the following please.

The question:

A 100kg crate is pulled across a horizontal floor by a force P that makes an angle of 30 degrees above the horizontal. The coefficient of kinetic friction between the crate and floor is 0.200. What is P if the net work done is zero.

My solution:

First determine Ff = umg. This would give me 196N. My next step is to determine P. I use 196N and put that over cos30. My answer is 226.3N. Does this seem reasonable.

thanks for your time

[Doc Al]

First determine Ff = umg.
No, F_f = \mu N, where N is the normal force. To solve this problem, recognize that the crate is in equilibrium. Set up equations for the vertical and horizontal forces. (Net force in each direction equals zero.)

[Gentec]

Thanks for the direction.

Is this what you meant?
Horizontal:
mgsin30 = f + Fcos30 = uN + Fcos30
Vertical:
N = Fsin30 = mgcos30

To resolve = Add N into N in equation for Horizontal

[Doc Al]

I don't understand how you got your equations. Here's what I get:
Horizontal: P cos(30) = \mu N
Vertical: N + P sin(30) = mg