Answer
The truck skid $13.5m$ before coming to a stop.
Work Step by Step
Since there is kinetic friction, the energies are not conserved. According to the work-energy theorem, $$E_f-E_0=W_{nc}$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=W_{f_k} (1)$$
We have $W_{f_k}=-f_ks$ (since $f_k$ opposes the motion along the displacement) $=-\mu_kF_Ns$
As shown in the figure below, the component of $mg$ opposing the normal force is $mg\cos15$. Since there is no vertical acceleration, $F_N=mg\cos15$
Therefore, $W_{f_k}=-(\mu_kmg\cos15)s$
We know $\mu_k=0.75$, $g=9.8m/s^2$, $\cos15=0.966$, so $W_{f_k}=-7.1ms$
Plug this back to (1): $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=-7.1ms$$ $$\frac{1}{2}(v_f^2-v_0^2)+g(h_f-h_0)=-7.1s$$
The question asks how far the truck goes, which is the value of $s$. We know $v_0=11.1m/s$, $v_f=0$. As we can see in the figure, $h_f-h_0=-(s\sin15)=-0.259s$ $$\frac{1}{2}(-11.1^2)-9.8(0.259s)=-7.1s$$ $$-61.605-2.538s=-7.1s$$ $$s=13.5m$$
