400 N pushes on a 25-kg box. Starting from rest, the box achieves a velocity of 2.0 m/s in a time of 4.0 s. Find the coefficient of kinetic friction between box and
floor.

The kinetic coefficient of friction between box and floor is 0.443.

Explanation:

Let suppose that the box is accelerated uniformly, the acceleration (), measured in meters per square second, is determined by the following kinematic formula:

(1)

Where:

, - Initial and final speeds of the box, measured in meters per second.

- Time, measured in seconds.

If we know that , and , then the acceleration experimented by the box is:

Based on the body diagram, we proceed to form the equations of equilibrium:

(1)

(2)

Where:

- Magnitude of the force exerted on the box, measured in newtons.

- Direction of the force exerted on the box, measured in sexagesimal degrees.

- Kinetic coefficient of friction between box and floor, dimensionless.

- Normal force from the floor to the box, measured in newtons.

- Mass of the block, measured in kilograms.

- Gravitational acceleration, measured in meters per square second.

Now we proceed to solve the system of equation for the kinetic coefficient of friction between box and floor:

By (1)

By (2)

And then, we get the resulting expression:

(3)

If we know that , , , and , then the kinetic coefficient of friction between box and floor is:

The kinetic coefficient of friction between box and floor is 0.443.

b would be the answer