[latex]dane:\m = 1000 kg\v_1 = 36frac{km}{h} = 36cdotfrac{1000m}{3600s} = 10frac{m}{s}}\v_2 =72frac{km}{h} =72cdotfrac{1000m}{3600s}= 20frac{m}{s}\s = 150 m\F_{o} = 25\%F_{g} = frac{1}{4}F_{g}=frac{1}{4}cdot mcdot g\g = 10frac{N}{kg}\szukane:\P = ?\\Rozwiazanie\Moc, to iloraz pracy i czasu:\P = frac{W}{t}\\Liczymy prace:\W = E_{k} + F_{o}cdot s\\E_{k} = frac{1}{2}(mv^{2}_2}-mv^{2}_1)=frac{1}{2}m(v^{2}_2-v^{2}_1)\F_{o}cdot s=frac{1}{4}mgs[/latex]
[latex]W = frac{1}{2}m(v^{2}_2 -v^{2}_1)+frac{1}{4}mgs\\W = frac{1}{2}m(v^{2}_2 - v^{2}_1 + frac{1}{2}gs)[/latex]
[latex]Liczymy czas:\s = v_{o}t+frac{1}{2}at^{2}\\v_{o} = v_1\a = frac{Delta v}{t}=frac{v_2-v_1}{t}\\s = v_1t+frac{1}{2}at^{2}\\s = v_1t = frac{1}{2}frac{(v_2-v_1)}{t}cdot t^{2}\\s = v_1t + frac{1}{2}(v_2-v_1)t |cdot2\\2v_1t+v_2t-v_1t = 2s\\v_1t + v_2t = 2s\\t(v_1+v_2) = 2s /:(v_1+v_2)\\t = frac{2s}{v_1+v_2}[/latex]
[latex]Moc:\\P = frac{frac{1}{2}m(v^{2}_2-v^{2}_2+frac{1}{2}gs)}{frac{2s}{v_1+v_2}}\\P = frac{frac{1}{2}cdot1000kg((20m/s)^{2}-(10m/s)^{2}+frac{1}{2}cdot10m/s cdot150m))}{frac{2cdot150m}{10m/s+20m/s}}=frac{500(400-100+750)}{10} W\\P = 52 500 W = 52,5 kW[/latex]
Odp. Moc silnika wynosi 52 500 W (52,5 kW).
