[tex]Dane:\\k = 150\frac{N}{m}\\E_{ps} = 3 \ J\\Szukane:\\x = ?\\\\Rozwiazanie\\\\E_{ps} = \frac{kx^{2}}{2} \ \ |\cdot2\\\\kx^{2} = 2E_{ps} \ \ /:k\\\\x^{2} = \frac{2E_{ps}}{k}\\\\x = \sqrt{\frac{2E_{ps}}{k}}\\\\x = \sqrt{\frac{2\cdot 3 \ J}{150\frac{N}{m}} } = \sqrt{\frac{6 \ N\cdot m}{150\frac{N}{m}}} = \sqrt{0,04 \ m^{2}}\\\\\boxed{x = 0,2 \ m = 20 \ cm}[/tex]
Odp. Odkształcenie tej sprężyny wynosi 0,2 m (20 cm).
