Answer
$0.08 j$ or 0.08 N-m
Work Step by Step
Hooke's Law states that $F=k x$
or, $k=\dfrac{F}{x}=\dfrac{2}{(0.02)}= 100$ N-m
When we stretch the rubber band, the limits for $x$ become from $0$ to $x=\dfrac{4}{100}=0.04 m$
The work done can be found as:
$W=\int_0^{(0.04)} (100 x) dx$
Use formula: $\int x^n=\dfrac{x^{n+1}}{n+1}+C$
This implies that $W= 100[ \dfrac{x^2}{2}]_0^{(0.04)}=50 (0.04)^2 =0.08 J$ or 0.08 N-m
