A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the object for 2.0 s, after which the object's velocity is v 2 = (16.0 i^ + 29.0 j^) m/s. Required:
Find the work done by the force in joules.

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:

The work done by the force is 820.745 joules.

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.