ou are pushing a 20-kg box along a horizontal floor. Friction acts on the box. When you apply a horizontal force of magnitude 48 N, the box moves at a constant velocity. What is the magnitude of the frictional force on the box?

The magnitude of the frictional force is 48.02 N

Explanation:

Mass of box = 20 kg

Weight of the box (Normal reaction) = mass * acceleration due to gravity = 20 * 9.8 = 196 N

Horizontal force applied = 48 N

Coefficient of friction = horizontal force : normal reaction = 48 : 196 = 0.245

Frictional force = coefficient of friction * normal reaction = 0.245 * 196 N = 48.02 N

Frictional force is 48N.

Explanation:

The box is moving at constant velocity so the acceleration of the box is zero. When a body moves with a constant velocity it has no acceleration and the net force on it zero. This is from Newtown's first law of motion.

Let the force you apply in pushing the box be F and the frictional force be Ff.

From Newtown's first law,

F - Ff = 0

F = Ff or Ff = F

In the question the force applied is 48N

Therefore,

Ff = F = 48N

This tells us that only an amount of force that balances the frictional force is needed to keep the box moving at constant velocity.