[tex]Dane:\\v_{o} = 20\frac{km}{h}\\t = 10 \ s\\v = 30\frac{km}{h}\\m = 2 \ kg\\Szukane:\\F = ?[/tex]
Rozwiązanie
Z II zasady dynamiki Newtona:
[tex]\underline{F = m\cdot a}\\\\a = \frac{\Delta v}{t}\\\\\Delta v = v-v_{o} = (30-20)\frac{km}{h} = 10\frac{km}{h} = 10\cdot\frac{1000 \ m}{3600 \ s} = 2,(7)\frac{m}{s}\approx2,8\frac{m}{s}\\\\a = \frac{2,8\frac{m}{s}}{10 \ s} =0,28\frac{m}{s^{2}}\\\\\\F = m\cdot a\\\\F = 2 \ kg\cdot0,28\frac{m}{s^{2}}\\\\\boxed{F = 0,56 \ N}[/tex]
Odp. Szukana siła ma wartość 0,56 N (niutona).
