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 = 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.

Answer 1

W= 4.92mJ

Step-by-step explanation:

the work is calculated by:

`W=-\int\limits^(x_f)_(x_0) {F} \, dx`

where
`x_f` is the stretch of the spring,
`x_0` the unstrech position of the sprinf and F the force applied.

so:

`W=-\int\limits^(x_f)_(x_0) {-ax+bx^3} \, dx`

`W=-\int\limits^(0.15)_(0) {-10x+850x^3} \, dx`

`W=5(0.15)^2-212.5(0.15)^4`

finally:

W= 4.92mJ