[tex]dane:\\m = 4 \ kg\\F_1 = 7 \ N\\F_2 = 3 \ N\\szukane:\\a = ?\\\\Rozwiazanie\\\\F_{w} = F_1 + F_2 = 7 \ N + 3 \ N = 10 \ N\\\\Z \ II \ zasady \ dynamiki:\\\\a = \frac{F_{w}}{m}\\\\a = \frac{10 \ N}{4 \ kg}\\\\a = 2,5\frac{N}{kg} = 2,5\frac{m}{s^{2}}[/tex]
