A box is pulled a distance s along a smooth horizontal floor by a force of magnitude F, inclined at an angle θ to the horizontal. The work done is

Analyze the question A force F acts at an angle θ to the horizontal, and the box moves a distance s horizontally. We need the work done by F.

Concept Work done is the product of the force component in the direction of motion and the distance moved:

W = F s cos φ

where φ is the angle between the force and the direction of displacement.

Solution

1. The displacement is horizontal.
2. The force is inclined at θ to the horizontal, so the angle between F and the displacement is θ.
3. The horizontal component of F is F cos θ. This is the only part that helps move the box along the floor.
4. The vertical component, F sin θ, acts perpendicular to the motion, so it does no work.
5. Therefore:

W = (F cos θ) × s = Fs cos θ

The floor is smooth, so there is no friction. We don't have to subtract any frictional work, and the work done by F is just Fs cos θ.

Answer: A. Fs cos θ

Tip: Work uses the component of force along the displacement. If the angle is measured from the direction of motion, use cos. Also remember that a force perpendicular to the motion (like the vertical component here, or the normal reaction) does zero work, since cos 90° = 0.