Dane:
m = 10 kg
F₁ = 3 N (→)
F₂ = 4 N (→)
F₃ = 5 N (→)
F₄ = 6 N (←)
F₅ = 7 N (←)
Szukane:
a = ?
Rozwiązanie
Z II zasady dynamiki Newtona:
[tex]a = \frac{F_{w}}{m}\\\\ale\\\\F_{w} = F_4+F_5 - (F_1+F_2+F_3)=6 N + 7 N - (3 N + 4 N + 5 N) = 13 N - 12 N = 1 \ N\\\\a = \frac{1 \ N}{10 \ kg} = \frac{1 \ kg\cdot\frac{m}{s^{2}}}{10 \ kg}\\\\\boxed{a = 0,1\frac{m}{s^{2}}}[/tex]
