Answer:
The work done by the force is 820.745 joules.
Explanation:
Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

Where:
- Work done by the external force, measured in joules.
,
- Translational potential energy, measured in joules.
The work done by the external force is now cleared within:

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

Where:
- Mass of the object, measured in kilograms.
,
- Initial and final speeds of the object, measured in meters per second.
Now, each speed is the magnitude of respective velocity vector:
Initial velocity



Final velocity



Finally, if
,
and
, then the work done by the force is:
![W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]](https://tex.z-dn.net/?f=W_%7BF%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%283.5%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cleft%2833.121%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D-%5Cleft%2825.060%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D%5Cright%5D)

The work done by the force is 820.745 joules.