Answer
$3857$ $J$
Work Step by Step
At a height of $x$ meters $(0{\leq}x{\leq}12)$, the mass of the rope is $(0.8kg/m)(12-x)m$ = $(36-3x)$ kg.
The mass of the bucket is $10$ kg, so the total mass is
$(9.6-0.8x)+(36-3x)+10$ = $(55.6-3.8x)$ kg,
and hence, the total force is
$9.8(55.6-3.8x)$ N.
The work needed to lift the bucket $Δx$ m through the $i$th subinterval of $[0,12]$ is
$9.8(55.6-3.8x_i)Δx$
so the total work is
$$\begin{align*}
W& = \lim\limits_{n \to \infty}\Sigma_{i=1}^n{9.8(55.6-3.8x_i)Δx}\\
& = \int_0^{12}{9.8(55.6-3.8x)}dx\\
& = 9.8[55.6x-1.9x^{2}]_0^{12}\\
& = 9.8(393.6)\\
&\approx 3857 J
\end{align*}$$
