[tex]dane:\\m = 1300 \ kg\\v_{o} = 0\\t = 3 \ s\\v = 15\frac{m}{s}\\szukane:\\F = ?\\\\Rozwiazanie\\\\a = \frac{\Delta v}{t} = \frac{v-v_{o}}{t} = \frac{v}{t} = \frac{15\frac{m}{s}}{3 \ s} = 5\frac{m}{s^{2}}\\\\Z \ II \ zasady \ dynamiki\\\\a = \frac{F}{m} \ \ /\cdot m\\\\F = m\cdot a\\\\F = 1300 \ kg\cdot5\frac{m}{s^{2}}\\\\F = 6 \ 500 \ N = 6,5 \ kN[/tex]
Odp. Musiałaby działać siła o wartości 6,5 kN.
