Find the coefficient of kinetic friction between the crate and the surface

[hannsparks]

Coefficient of kinetic friction

I have a mojor physics test thursday (two days) and here is a pretest question that I cannot figure out if anyone can figure it out I'd greatly appreciate it. I actually know what the answer is uk=0.253 I just don't know how to get that answer.
" An apple crate with a weight of 49 N accelerates at a rate of 0.5 m/s2 along a horizontal surface as the crate is pulled with a force of 14.5 N as shown in Figure 1. The angle at which the crate is being pulled is 19 degrees.
Find the coefficient of kinetic friction between the crate and the surface."

heres my attempt: N=49+Fsin19 y direction
Fcos19-uk=(5)(0.5) x direction
I then tried to combine the two equations to figure out N and F but I am stuck.

 

[daniel_i_l]

First of all is the angle 19, the angle over the horizontal or under? If it is over then you have to write that N=49-Fsin19 . Also, uk isn't a force, N*uk is the force of friction.

 

[mrjeffy321]

Draw a picture and label the forces.

If the create is accelerating there must be a net force (F = m*a).
You know that a 14.5 N force is being exerted on the box at 19 degrees. What part of this 14.5 N is going to accelerate the box horizontally? What part tries to accelerate it vertically?

Remember,
The force of friction = coefficient of friction * normal force.
The force of friction points in the opposite direction of motion.

 

[hannsparks]

im so confused I have no idea what to do next.. do I know have to combine the two equations to figure out N and F?

 

[hannsparks]

hey guess what..I figured it out! Thanks for all your help :)