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A certain nonlinear spring has a force function given by F=−ax2−b, where x is the displacement of the spring from equilibrium, a=3.0Nm2, and b=4.0N. The change in elastic potential energy of the spring as it is stretched from x=0m to x=2.0m is

Respuesta :

Answer:

16 J

Explanation:

[tex]\int{x^n} \, dx = \frac{x^n^+^1}{n+1} \\\int\limits^a_b {f(x)} \, dx \\\int\limits^2_0 ({-ax^2-b}) \, dx \\\int\limits^2_0 ({-3(\frac{x^3}{3} )-4x})\\(-3\frac{2^3}{3} -4(2)) - (-3\frac{0^3}{3} - 4(0)) = (-3\frac{8}{3} -8) - 0\\= (-8 - 8)\\= 16 J[/tex]

I messed up the sign somewhere, but the answer should be 16 J

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