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Imagine a spring made of a material that is not very elastic, so that the spring force does not satisfy Hooke’s Law, but instead satisfies the equation F = −α x + β x3 , where α = 10 N/m and β = 850 N/m3. Calculate the work done by the spring when it is stretched from its equilibrium position to 0.15 m past its equilibrium. Answer in units of mJ.

Respuesta :

Answer:

W= 4.92mJ

Explanation:

the work is calculated by:

[tex]W=-\int\limits^{x_f}_{x_0} {F} \, dx[/tex]

where [tex]x_f[/tex] is the stretch of the spring, [tex]x_0[/tex] the unstrech position of the sprinf and F the force applied.

so:

[tex]W=-\int\limits^{x_f}_{x_0} {-ax+bx^3} \, dx[/tex]

[tex]W=-\int\limits^{0.15}_{0} {-10x+850x^3} \, dx[/tex]

[tex]W=5(0.15)^2-212.5(0.15)^4[/tex]

finally:

W= 4.92mJ

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