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

W = -1.152 mJ

Step-by-step 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 = [(-5.2x^2)/(2) +700(x^4)/(4)]_0^(0.12)`

now,

`W = [(-5.2(0.12)^2)/(2) +700((0.12)^4)/(4)]`

W = -0.001152 J

W = -1.152 mJ

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