A 500N box rests on a horizontal floor. A constant horizontal force is exerted on the box so that it moves…

Analyze the question

A box weighing 500 N is pushed along a horizontal floor through 8 m. The coefficient of kinetic friction is 0.22. We need the work done on the box by the applied force.

Concept

• On a horizontal floor, the normal reaction equals the weight: N = 500 N.
• Kinetic friction: F_f = μN
• Work done = Force × distance moved in the direction of the force.
• If the box moves at constant velocity (the usual assumption in this kind of question), the applied force equals the friction force.

Solution

1. Friction force: F_f = μN = 0.22 × 500 = 110 N

2. Applied force (constant speed): F = 110 N

3. Work done: W = F × d = 110 × 8 = 880 J

Answer

The work done is 880 J, which matches option A.

The 110 in option D is only the friction force (not the work), and 440 J would come from using the wrong distance (4 m). Always multiply the force by the full distance.

Tip

For friction-work questions, follow this chain: N → μN → F × d. Check that your answer is in joules (N × m) and not just newtons.