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Un ressort de constante de rappel k = 10 N\/m est \u00e9tir\u00e9 de x = 2,0 cm. Voici la r\u00e9ponse d\u2019un \u00e9l\u00e8ve qui veut calculer l\u2019\u00e9nergie potentielle \u00e9lastique emmagasin\u00e9e dans ce ressort :
\n\\(E_{pe}=\\frac{1}{2}kx^2 = \\frac{1}{2}\\times(10\\;\\text{N\/m)} \\times (2{,}0\\; \\text{cm})^2=20 \\text{ J}\\)<\/p>\n\n

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