We are given that 221J of work is required to lift a mass of 4.75 meters. The amount of work is determined by the following equation:
![W=Fd](https://img.qammunity.org/qa-images/2023/formulas/physics/college/e84ag4tarqp04c5dkq79.png)
Where:
![\begin{gathered} F=\text{ force} \\ d=\text{ distance} \end{gathered}](https://img.qammunity.org/qa-images/2023/formulas/physics/college/90kraxgyhxm5tranp10w.png)
The force is equivalent to the combined weight if the dog and the basket multiplied by the acceleration of gravity:
![F=mg](https://img.qammunity.org/qa-images/2023/formulas/physics/college/ikjwv8bafvw4bolytzsc.png)
The distance is the height that the basket is lifted:
![d=h](https://img.qammunity.org/qa-images/2023/formulas/physics/college/ooqewl0lad3x69ut1pgz.png)
Now, we substitute in the equation:
![W=mgh](https://img.qammunity.org/qa-images/2023/formulas/physics/high-school/7wwbtv56t78bz10ipza2.png)
Now, we divide by "gh":
![(W)/(gh)=m](https://img.qammunity.org/qa-images/2023/formulas/physics/college/5b96drmggf0fxd1ll6ty.png)
Now, we substitute the values:
![(221J)/((9.8(m)/(s^2))(4.75m))=m](https://img.qammunity.org/qa-images/2023/formulas/physics/college/sl27itsyp8x3rsh5m0o1.png)
Solving the operations:
![4.75kg=m](https://img.qammunity.org/qa-images/2023/formulas/physics/college/nhu3naz5ullcso6tgtel.png)
Therefore, the combined mass is 4.75 kg.