[tex]\displaystyle\sf\\m=4t=4000kg,~\eta=60\%,~\alpha=30^{\circ},~L=8m~iar~t=10s.\\--------------------------\\a)~\boxed{\sf v=\frac{L}{t}=0,8m/s}.\\b)~conform~teoremei~unghiului~de~30^{\circ}\implies \boxed{\sf h=\frac{1}{2}L=4m}.\\c)~\eta=\frac{Lu}{Lc}=\frac{G_tl}{(G_t+f_f)l}=\frac{mgsin\alpha}{mgsin\alpha+f_f}.\\stim~ca~randamentul~planului~inclinat~este~60\%~adica~0,6 \implies\\\frac{mgsin\alpha}{mgsin\alpha+f_f}=0,6 \implies 0,6(mgsin\alpha+f_f)=mgsin\alpha \Leftrightarrow \\[/tex]
[tex]\displaystyle\sf\\0,6mgsin\alpha+0,6f_f=mgsin\alpha \implies \boxed{\sf f_f=\frac{mgsin\alpha-0,6mgsin\alpha}{0,6} = 1,33333\cdot10^4}.\\d)~\boxed{\sf P=\frac{L_m}{t}=\frac{Fd}{t}=\frac{(mgsin\alpha+f_f)L}{t}\approx 26666,64W}[/tex]
