Answer
(a) $0.030N$ down the incline.
(b) $0.042N$ up the incline.
Work Step by Step
(a) We know that
$\Sigma F_y=N-mgcos\theta=0$
$\implies N=mgcos\theta$
As $f_k=-\mu_kN$
$\implies f_k=-\mu_kmgcos\theta$
We plug in the known values to obtain:
$f_k=-(0.23)(0.014)(9.81)cos180^{\circ}$
$f_k=-0.030N$
The negative sign shows that the force is directed down the incline.
(b) We know that
$\Sigma F_x=f_s-mgsin\theta=0$
$\implies f_s=mgsin\theta$
We plug in the known values to obtain:
$f_s=(0.014)(9.81)sin18^{\circ}$
$f_s=0.042N$
The positive sign shows that the force is directed up the incline.
