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:
[tex]K_{1} + W_{F} = K_{2}[/tex]
Where:
[tex]W_{F}[/tex] - Work done by the external force, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Translational potential energy, measured in joules.
The work done by the external force is now cleared within:
[tex]W_{F} = K_{2} - K_{1}[/tex]
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:
[tex]W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})[/tex]
Where:
[tex]m[/tex] - Mass of the object, measured in kilograms.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the object, measured in meters per second.
Now, each speed is the magnitude of respective velocity vector:
Initial velocity
[tex]v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}[/tex]
[tex]v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v_{1} \approx 25.060\,\frac{m}{s}[/tex]
Final velocity
[tex]v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}[/tex]
[tex]v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v_{2} \approx 33.121\,\frac{m}{s}[/tex]
Finally, if [tex]m = 3.5\,kg[/tex], [tex]v_{1} \approx 25.060\,\frac{m}{s}[/tex] and [tex]v_{2} \approx 33.121\,\frac{m}{s}[/tex], then the work done by the force is:
[tex]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][/tex]
[tex]W_{F} = 820.745\,J[/tex]
The work done by the force is 820.745 joules.
