a) Water only.

Work done by variable force - integration

The force required to lift the water is the water's weight.

When the bucket is x metres off the ground, the expression for the water's weight can be found by first graphing it:

Work done by variable force - integration

So, using y = mx + c, we see that the slope is

`m = -2/20 = -1/10` and

the y-intercept is `c = 2`

So we can write the function for the weight at height x as:

`F(x)=-x/10 + 2`

Then

`"Work"=int_a^bF(x)\ dx`

`=int_0^20(-x/10+2) dx`

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

`=-20+40`

`=20 text[ N.m (20 joules)]`

Alternatively, we could have found the function as follows:

`F(x) = 2(20-x)/20 "N"`

where `2` is the original weight of the water, and `(20-x)/20` is the proportion of water left at height `t`.

This simplifies to `F(x)=(2-x/10)\ "N"`.

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