[tex]Dane:\\m = 2 \ 000 \ kg\\v_1 = 10\frac{m}{s}\\v_2 = 15\frac{m}{s}\\Szukane:\\W= ?\\\\Rozwiazanie\\\\W = \Delta E_{k}\\\\E_{k} = \frac{1}{2}mv^2[/tex]
[tex]E_{k_1} = \frac{1}{2}mv_1^{2} =\frac{1}{2}\cdot2000 \ kg\cdot(10\frac{m}{s})^{2} = 100 \ 000 \ J = 100 \ kJ\\\\E_{k_2} = \frac{1}{2}mv_2^{2} = \frac{1}{2}\cdot2000 \ kg\cdot(15\frac{m}{s})^{2} = 225 \ 000 \ J = 225 \ kJ\\\\W = E_{k_2}-E_{k_1}\\\\W = 225 \ kJ -100 \ kJ\\\\\boxed{W = 125 \ kJ}[/tex]
Odp. Siła wypadkowa musi wykonać pracę W = 125 kJ.
