D.
Since the net force is pointing east, the resulting acceleration is also pointing east.
E.
The work can be calculated with the formula below:
![W=F\cdot d](https://img.qammunity.org/2023/formulas/physics/college/kgx983ena5n58pjn3ga2ml2ln51sad2jev.png)
Since Gabrielle's force is to the east (same direction of the acceleration), so let's use a positive force:
![W=40\cdot4=160\text{ J}](https://img.qammunity.org/2023/formulas/physics/college/wsk72lh2akxboffozwfnb4ccoiluluxkh2.png)
This work is positive.
F.
Pierre pushes the box in the opposite direction the box is moving, so his force is negative:
![W=-32\cdot4=-128\text{ J}](https://img.qammunity.org/2023/formulas/physics/college/tqb0tsppgunj2d8ztz5wdm01kkypyf1yxr.png)
G.
The friction force is acting on the opposite direction of the movement, so it is negative:
![W=-3\cdot4=-12\text{ J}](https://img.qammunity.org/2023/formulas/physics/college/z9pyjx1ffm3sdy0xp5leknvo0rtvmrbjw9.png)
H.
adding all works, we have:
![160-128-12=20\text{ J}](https://img.qammunity.org/2023/formulas/physics/college/vxig23b1n0md1jq7eb6peokvz81wig5uu1.png)
I.
To find the net work in horizontal, let's use the net force in horizontal:
![\begin{gathered} F_(net)=40-32-3=5\text{ N}\\ \\ W_(net)=5\cdot4=20\text{ J} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/d7hdik5gq1azbevsk7mopb8yiyzv9ntc4q.png)
J.
To find the net work in vertical, let's use the net force in vertical:
![\begin{gathered} F_(net)=F_(weight)-F_(normal)=0\text{ N}\\ \\ W_(net)=0\cdot4=0\text{ J} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/rf4i6f6udgy4v4nthegti45sad67utrgvz.png)