[tex]dane:\\m = 1 \ t = 1000 \ kg\\v_{o} = 0\\t = 10 \ s\\v = 25\frac{m}{s^{2}}\\szukane:\\F_{w} = ?\\\\Rozwiazanie\\\\Z \ II \ zasady \ dynamiki\\\\a = \frac{F_{w}}{m} \ \ |\cdot m\\\\F_{w} = m\cdot a\\\\ale \ \ a = \frac{\Delta v}{t} = \frac{v-v_{o}}{t} = \frac{v}{t} = \frac{25\frac{m}{s}}{10 \ s} = 2,5\frac{m}{s^{2}}\\\\F_{w} = 1000 \ kg\cdot2,5\frac{m}{s^{2}} = 2 \ 500 \ N = 2,5 \ kN[/tex]
Odp. Siła wypadkowa ma wartość 2,5 kN.
