[tex]dane:\\m = 1 \ t = 1000 \ kg\\t = 5 \ s\\\Delta v = 20\frac{m}{s}\\szukane:\\F_{w}} = ?\\\\Rozwiazanie\\\\a = \frac{\Delta v}{t} = \frac{20\frac{m}{s}}{5 \ s} = 4\frac{m}{s^{2}}\\\\Z \ II \ zasady \ dynamiki:\\\\a = \frac{F_{w}}{m} \ \ /\cdot m\\\\F_{w} = m\cdot a\\\\F_{w} = 1000 \ kg\cdot4\frac{m}{s^{2}}\\\\F_{w} = 4 \ 000 \ N = 4 \ kN\\\\(1 \ kN = 1000 \ N)[/tex]
Odp. Siła wypadkowa ma wartość 4 kN.
