forces down an incline?

Question

A 23.0 kg box is released on a 37.0° incline and accelerates down the incline at 0.272 m/s2. Find the friction force impeding its motion.
____N
How large is the coefficient of friction?

Answer 1

F=Fd-f=ma
also
Fd= mg sin(37)
f= umg cos(37)

we have
f= mgsin(37) - ma=
f= m(gsin(37) - a)
f= 23.0 (9.81 sin(37) - 0.272)
f=130N

since f= umg cos(37)
u=f/(mg cos(37))
u= 130/(23.0 x 9.81 x cos(37))
u=0.721