To develop this problem it is necessary to apply the concepts related to Work and energy conservation.
By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,
[tex]W = F*d[/tex]
Where,
F= Force
d = Distance
On the other hand we know that the potential energy of a body is given based on height and weight, that is
[tex]PE = -mgh[/tex]
The total work done would be given by the conservation and sum of these energies, that is to say
[tex]W_{net} = W+PE[/tex]
PART A) Applying the work formula,
[tex]W = F*d\\W = 2750*1.25\\W = 3437.5J[/tex]
PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then
[tex]PE = -mgh*sin25[/tex]
[tex]PE = -16*9.81*1.15sin25[/tex]
[tex]PE = -82.9J[/tex]
The net work would then be given by
[tex]W_{net} = W+PE[/tex]
[tex]W_{net} = 3437.5J-82.9J[/tex]
[tex]W_{net} = 3354.6J[/tex]
Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J