Solving Friction Force and Coefficient on a 17 kg Box on a 35 Degree Incline

[shrtweez13]

i've been trying to figure this question out for about half an hour now and I'm still where I started. if some one could help me i'd be much obliged.

a 17 kg box is released on a 35 degree incline and accelerates down the incline at .270 m/s^2. Find the friction force impeding its motion. How large is the coefficient of friction?

 

[UrbanXrisis]

F-Ffriction=ma
(17kg*9.8m/s^2*sin35)-Ff=17kg*.27m/s^2
Ff=91N

 

[shrtweez13]

i kind of understand what you did but its not the answer. I'm using the webassign program that let's you check your homework and i tried that answer and it didn't turn out right.

 

[UrbanXrisis]

what answer did you plug in? Try 91N, 92N, or 93N

I'm leaving the coefficient of friction for you to solve

 

[primarygun]

92.9N.
I got this