c) **Water, bucket, rope**

The weight of the rope at height *x* is:

`F(x) = 0.08(-x + 20)`

The work done on the rope is:

`int_0^20 0.08(-x+20) dx`

`=0.08[-(x^2)/(2)+20x]_0^20`

`=-16+32`

`=16\ text[N.m]`

So the total work done on the water, bucket and rope is

`W` `=(20 + 100 + 16)\ "N.m"` `= 136\ "N.m"` (or `136\ "J"`).