Finding the kinetic friction between the block and surface

[gijungkim]

Homework Statement

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The block shown is pulled across the horizontal surface at a constant speed by the force shown. If M = 5.0 kg, F = 14 N and Theta= 35°, what is the coefficient of kinetic friction between the block and the horizontal surface?

Homework Equations

Fx = Fcos(theta)
Ff = Fn * Uk (Friction equation)
F=ma
Fn = mg

The Attempt at a Solution

I did it by myself and got 0.234, but the answer says it's 0.28. Can you explain this to me?

My work:
a = F / m
a =( Fcos(35) - Fn * Uk ) / m= 0
Fcos(35) / Fn = Uk
14*cos(35) / (5 * 9.8) = 0.234 = Uk

I don't know where I did wrong

 

[gijungkim]

Nevermind! I got it! I forgot the Fsin(theta)
It is
Fcos(theta) - Uk(mg - Fsin(theta)) = 0
14 * cos(35) / (5*9.8 - 14 sin(35)) = 0.28 = Uk