ANSWER
[tex]\begin{gathered} (b)\text{ }841J \\ \\ (c)\text{ }-454.72J \\ \\ (d)\text{ }386.28J \end{gathered}[/tex]EXPLANATION
Parameters given:
Depth of the well, hy = 14.5 m
Force of pull, F1 = 58 N
The total mass of the bucket of water, mb + mw = 3.2 kg
(b) The amount of work done by the farmer on the bucket, we have to apply the formula for work done i.e. the product of force applied and distance traveled:
[tex]W_f=F_1*h_y[/tex]Therefore, the work done by the farmer is:
[tex]\begin{gathered} W_f=58*14.5 \\ \\ W_f=841J \end{gathered}[/tex](c) To find the work done by gravity, apply the formula:
[tex]W_g=-(m_b+m_w)gh_y[/tex]where g = acceleration due to gravity
Therefore, the work done by gravity is:
[tex]\begin{gathered} W_g=-3.2*9.8*14.5 \\ \\ W_g=-454.72J \end{gathered}[/tex](d) The total work done by the two forces is the sum of the work done by the farmer and the work done by gravity:
[tex]W=W_f+W_g[/tex]Therefore, the total work done is:
[tex]\begin{gathered} W=841+(-454.72) \\ \\ W=841-454.72 \\ \\ W=386.28J \end{gathered}[/tex]