[tex]dane:\\m = 900 \ kg\\v_1 = 36\frac{km}{h} = 36\cdot\frac{1000 \ m}{3600 \ s} = 10\frac{m}{s}\\v_2 = 54\frac{km}{h} = 54\cdot\frac{1000 \ m}{3600 \ s} = 15\frac{m}{s}\\szukane:\\\Delta E_{k} = ?\\\\Rozwiazanie\\\\E_{k} = \frac{mv^{2}}{2}\\\\\Delta E_{k} = E_{k_2}-E_{k_1}=\frac{mv_2^{2}}{2}-\frac{mv_1^{2}}2}=\frac{1}{2}m(v_2^{2}-v_1^{2})\\\\\Delta E_{k} = \frac{1}{2}\cdot900 \ kg\cdot[(15\frac{m}{s})^{2}-(10\frac{m}{s})^{2}]\\\\\Delta E_{k} =450 \ kg\cdot(225\frac{m^{2}}{s^{2}}-100\frac{m^{2}}{s^{2}})[/tex]
[tex]\Delta E_{k} = 56 \ 250 \ J = 56,25 \ kJ[/tex]
