inal kinetic energy of the box in terms of the given variables.
Firstly, let us declare the symbols. Let's represent the mass of the box as M, the applied force as F, the distance moved by the box as L, and the coefficient of kinetic friction as u.
Let's start with finding the frictional force. The frictional force equals the coefficient of kinetic friction (u) times the normal force. The normal force in this case (since we are on a level floor) equals the weight of the box, which is mass (M) times gravity. Considering the gravity as 9.8 m/s², the frictional force is calculated as u * M * 9.8 or (μMg).
Next, we calculate the work done by the force F. The work done by the force equals the force times the distance that the box moves. Therefore, the work done by F is F * L.
The work done by the frictional force equals the frictional force calculated above times the distance the box moves in the opposite direction. Hence, the work done by friction is frictional force * L which gives us u * M * 9.8 * L.
Subtracting the work done by friction from the work done by the force F gives us the net work done. Therefore, the net work done is equal to (F * L) - (u * M * 9.8 * L).
The net work done on the box becomes its kinetic energy, since the box was initially at rest (the work-energy theorem). Thus, final kinetic energy of the box is given by (F * L) - (u * M * 9.8 * L), which is the final result.