Answer
$160$ ft-lb
Work Step by Step
We know that Hooke's Law states that $F=k x$
and work $=force \times distance$ or, $W=Fd$
Here, the water weight $=16-\dfrac{4}{5} x$
To calculate the work done, the limits for $x$ will be from $0$ to $20$ such that:
$W=\int_0^{20} (16-\dfrac{4}{5} x) dx=[(16x -\dfrac{4}{5}(x^2/2))]_0^{20}$
or, $W=[16(20-0)-\dfrac{2}{5}(400-0)]=320-160=160$ ft-lb
