Calculating Friction Force on 18.8kg Box on 38° Incline

[p0ink]

An 18.8kg box is released on a 38.0o incline and accelerates down the incline at 0.281m/s2. What is the magnitude of the friction force impeding its motion.

F = m*g*cos(38)
sum forces parallel to the plane
m*g*sin(38) - (mu)*(F) = m*a, or
m*g*sin(38) - (mu)*m*g*cos(38) = m*a
masses cancel out
[9.81*sin(38)-(.281)]/[9.81*cos(38)] = .75 = mu

but it says I'm wrong. i can't be.

i only have one try left, and that's it.

 

[Doc Al]

All you are asked to find is the friction force. No need to compute the normal force (what you call F, for some reason) or find mu. Rewrite your equation for forces parallel to the plane using "F" to represent the friction force.