[tex]dane:\\m = 60 \ kg\\F_1 = 400 \ N\\F_2 = 600 \ N\\F_3 =200 \ N\\szukane:\\F_{w} = ?\\a = ?\\\\Rozwiazanie\\\\F_{w} = F_1 + F_2 + F_3\\\\F_{w} = 400 \ N + 600 \ N + 200 \ N\\\\F_{w} = 1200 \ N[/tex]
[tex]a = \frac{F_{w}}{m}\\\\a = \frac{1200 \ N}{60 \ kg}\\\\a=20\frac{m}{s^{2}}[/tex]
