Answer
$400 ft-lb$
Work Step by Step
Hooke's Law states that $F=k x$
and $W=Fd$
The water weight is found as: $=2(20-x)$
Use formula: $\int x^n=\dfrac{x^{n+1}}{n+1}+C$
This implies that
$W=\int_0^{(20)} (2(20-x)) dx$
Use formula: $\int x^n=\dfrac{x^{n+1}}{n+1}+C$
Then $W=\int_0^{20} (40-2x) dx=[40x-\dfrac{2x^{1+1}}{2}]_0^{20}$
Hence, $W=[40x-\dfrac{2x^{2}}{2}]_0^{(20)}=[40(20-0)-(400-0)]=400 ft-lb$
