Chapter 11 - Rolling, Torque, and Angular Momentum - Problems - Page 326: 71c

$F_f = (4.0~N)~\hat{i}$

In part (a), we found that the magnitude of the acceleration of the center of mass is $a = 1.6~m/s^2$ We can use Newton's Second Law to find $F_f$: $F_{app}+F_f = Ma$ $F_f = Ma-F_{app}$ $F_f = (10~kg)(1.6~m/s^2)-12~N$ $F_f = 4.0~N$ $F_f = (4.0~N)~\hat{i}$