A boy pushes a 500kg box along a floor with a force of 2000N. If the velocity of the box is uniform, the…

Analyze the question: A box of mass 500 kg is pushed with a force of 2000 N at uniform (constant) velocity. We need the coefficient of friction between the box and the floor.

Concept: Uniform velocity means acceleration = 0, so the net force is zero. The applied force must therefore equal the frictional force.

Friction formula: F = μR, where R is the normal reaction. On a horizontal floor, R = mg.

Solution:

1. Applied force = Frictional force = 2000 N
2. Normal reaction: R = mg = 500 × 10 = 5000 N (using g = 10 m/s²)
3. Coefficient of friction:

μ = F / R = 2000 / 5000 = 0.4

Answer: The coefficient of friction is 0.4, which is option D.

Tip: Whenever you see "uniform velocity" or "constant speed", think "forces are balanced". For horizontal motion, push = friction, so μ = F/(mg). Also, JAMB usually lets you take g = 10 m/s², which makes the numbers clean. If you got 0.41 using g = 9.8, that is a sign to round to the nearest option.