Answer
Please see the work below.
Work Step by Step
(a) We know that
$\Sigma F_y=N-mgcos\theta=0$
$\implies N=mgcos\theta$
Now $f_k=-\mu_kN$
$\implies f_k=-(0.23)(0.014)(9.81)cos25^{\circ}$
$f_k=-0.029N$
The negative sign shows that the direction of the force is downward.
(b) We know that
$f_k=\mu_k N$
$\implies f_k=\mu_k mgcos\theta$
We plug in the known values to obtain:
$f_k=(0.23)(0.014)(9.81)cos25^{\circ}$
$f_k=0.029N$
The positive sign shows that the direction of the force is upward.
