[tex]Dane:\\m = 1,5 \ kg\\F_{c} = 10 \ N\\T = 5 \ N\\Szukane:\\a = ?\\\\Rozwiazanie\\\\Z \ II \ zasady \ dynamiki \ Newtona\\\\a = \frac{F_{w}}{m}\\\\ale \ \ F_{w} = F_{c} - T = 10 \ N - 5 \ N = 5 \ N\\\\a = \frac{5 \ N}{1,5 \ kg}\\\\a \approx3,3\frac{N}{kg} \approx3,3\frac{kg\cdot\frac{m}{s^{2}}}{kg}\approx\boxed{3,3\frac{m}{s^{2}}}[/tex]
