[tex]Dane:\\a_1 = 5\frac{m}{s^{2}}\\F_1 = 2 \ 500 \ kN\\a_2 = 7\frac{m}{s^{2}}\\Szukane:\\F_2 = ?\\\\Rozwiazanie\\\\Z \ II \ zasady \ dynamiki \ Newtona:\\\\a = \frac{F}{m} \ \ \rightarrow \ \ m = \frac{F}{a}\\\\m = \frac{F_1}{a_1}\\oraz\\m = \frac{F_2}{a_2}\\\\\frac{F_1}{a_1} = \frac{F_2}{a_2}\\\\F_2\cdot a_1 = F_1\cdot a_2 \ \ |:a_1\\\\F_2 = F_1\cdot\frac{a_2}{a_1}\\\\F_2 = 2500 \ kN\cdot\frac{7\frac{m}{s^{2}}}{5\frac{m}{s^{2}}}\\\\F_2 = 3 \ 500 \ kN[/tex]
[tex]\Delta F = F_2 - F_1 = 3500 \ kN - 2500 \ kN\\\\\underline{\Delta F = 1000 \ kN}[/tex]
Odp. Dodatkowa siła wsteczna to 1000 kN.