Question

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 = 5.2 N/m and β = 700 N/m3 . Calculate the work done by the spring when it is stretched from its equilibrium position to 0.12 m past its equilibrium. Answer in units of mJ.

Answer 1

Answer:

     W = -1.152 mJ

Explanation:

given,

F = −α x + β x³

α = 5.2 N/m and β = 700 N/m³

Work done = ?

length of stretch = ?

we know,

W = F . ds

$W = \intF . dx$

$W = \int_0^{0.12} (-\alpha x + \beta x^3).dx$

$W = \int_0^{0.12}(-5.2 x +700 x^3).dx$

$W = [\dfrac{-5.2x^2}{2} +700\dfrac{x^4}{4}]_0^{0.12}$

now,

$W = [\dfrac{-5.2(0.12)^2}{2} +700\dfrac{(0.12)^4}{4}]$

     W = -0.001152 J

    W = -1.152 mJ

work done by the spring is equal to W = -1.152 mJ