7.
F₁ = 130 N (→)
F₂ = 30 N (←)
Fw = F₁ - F₂
Fw = 130 N - 30 N
Fw = 100 N (→)
Fw = 0
Fw = 100 N - Fr = 0
Fr = 100 N (←)
Odp. B.
8.
[tex]Dane:\\F = 5 \ N\\a = 2,5\frac{m}{s^{2}}\\Szukane:\\m = ?\\\\Wzor \ (II \ zasada \ dynamiki):\\\\F = m\cdot a \ \ /:a\\\\m = \frac{F}{a}\\\\Obliczenia:\\\\m = \frac{5 \ N}{2,5\frac{m}{s^{2}}} = \frac{5 \ kg\cdot\frac{m}{s^{2}}}{2,5\frac{m}{s^{2}}} = 2 \ kg[/tex]
Odp. Masa ciała m = 2 kg.
9.
[tex]Dane:\\F_{c} = 10 \ N\\m = 3 \ kg\\a = 2,5\frac{m}{s^{2}}\\Szukane:\\F_{T} = ?\\\\Rozwiazanie\\\\Z \ II \ zasady \ dynamiki\\\\a = \frac{F_{w}}{m}\\\\F_{w} = F_{c}-F_{T}\\\\a = \frac{F_{c}-F_{T}}{m} \ \ /\cdot m\\\\m\cdot a = F_{c}-F_{T}\\\\F_{T} = F_{c}-m\cdot a\\\\F_{T} = 10 \ N - 3 \ kg\cdot2,5\frac{m}{s^{2}}\\\\F_{T} = 10 \ N - 7,5 \ N\\\\F_{T} = 2,5 \ N[/tex]
Odp. Wartość siły tarcia w tym ruchu wynosi 2,5 N.
