Answer
$12\ kJ$
Work Step by Step
a) We call $x_0$ the initial stretch length. (The book calls this $L_0$, but this is the same idea.) We know that work is equal to the integral of the force with respect to x. Thus:
$W = \int_{x_0}^{2x_0}(-kx-bx^2-cx^3-dx^4)dx$
$W=.5kx_0^2+.33bx_0^3+.25cx_0^4+.2dx_0^5 $
b) Plugging in the given values, we find that the work done is $12\ kJ$.
