51.
[tex]m = 4 \ kg\\F_1 = 6 \ N\\F_2 = 8 \ N\\t = 5 \ s\\F_{w} = ?\\a = ?\\s = ?[/tex]
a)
[tex]F_{w} = F_2 - F_1 = 8 \ N - 6 \ N = 2 \ N\\\\a = \frac{F}{m} = \frac{2 \ N}{4 \ kg} = 0,5\frac{N}{kg} =0,5\frac{m}{s^{2}}\\\\s = \frac{1}{2}at^{2} = \frac{1}{2}\cdot0,5\frac{m}{s^{2}}\cdot(5 s)^{2} =6,25 \ m[/tex]
b)
[tex]F_{w} = F_1+F_2 = 6 \ N = 8 \ N = 14 \ N\\\\a = \frac{F}{m} = \frac{14 \ N}{4 \ kg} = 3,5\frac{m}{s^{2}}\\\\s = \frac{1}{2}at^{2} = \frac{1}{2}\cdot3,5\frac{m}{s^{2}}\cdot(5 \ s)^{2} = 43,75 \ m[/tex]
52.
[tex]m = 3 \ kg\\F_1 = 6 \ N\\a = 12\frac{m}{s^{2}}\\F_2 = ?\\\\Z \ II \ zsady \ dynamiki\\\\F_{w} = m\cdot a\\\\F_{w} = 3 \ kg\cdot12\frac{m}{s^{2}} = 36 \ N\\\\F_2 = F_{w} - F_1\\\\F_2 = 36 \ N - 6 \ N = 30 \ N[/tex]
Odp. Siła F₂ a wartość 30 N, kierunek i zwrot zgodny z kierunkiem i zwrotem siły F₁.
