We apply the work-energy theorem, which states that the net work done on an object equals its change in kinetic energy ($W_{net} = \Delta K$). Alternatively, the work done by non-conservative forces equals the change in mechanical energy ($W_{nc} = \Delta E = \Delta K + \Delta U_g$). The object starts with kinetic energy $K$ and ends with zero kinetic energy, so $\Delta K = -K$. The change in gravitational potential energy is $\Delta U_g = mgh = mgd \sin\theta$, where $d$ is the distance traveled along the incline.

[Q1] For the frictionless case, the only non-conservative force is the normal force, which does no work. Mechanical energy is conserved.

$$ \Delta K + \Delta U_g = 0 $$ $$ -K + mgd_1 \sin\theta = 0 $$ $$ d_1 = \frac{K}{mg \sin\theta} $$

[Q2] With friction, the non-conservative work is done by the kinetic friction force, $f_k = \mu_k N$. The normal force is $N = mg \cos\theta$. The work done by friction is $W_f = -f_k d_2 = -\mu_k mg d_2 \cos\theta$.

$$ W_{nc} = \Delta K + \Delta U_g $$ $$ -\mu_k mg d_2 \cos\theta = -K + mgd_2 \sin\theta $$ $$ K = mgd_2 \sin\theta + \mu_k mgd_2 \cos\theta $$ $$ K = mgd_2 (\sin\theta + \mu_k \cos\theta) $$ $$ d_2 = \frac{K}{mg(\sin\theta + \mu_k \cos\theta)} $$