A 20Kg box is pushed across the floor at a constant velocity with a force of 200N. What is the kinetic friction coefficient between the box and the floor?

The kinetic coefficient of friction between the box and the floor is 1.020.

Explanation:

Let suppose that box is on horizontal ground. According to the Newton's Laws, an object has a net acceleration of zero when either is at rest or moves at constant velocity. Friction is a reaction to the external force that moves the box. Hence, the equation of equilibrium for the 20-Kg box is:

(Eq. 1)

Where:

- External force, measured in newtons.

- Friction force, measured in newtons.

If we know that , then the magnitude of the kinetic friction force is:

In addition, friction force is represented by the following formula:

(Eq. 2)

Where:

- Kinetic coefficient of friction, dimensionless.

- Mass, measured in kilograms.

- Gravitational acceleration, measured in meters per square second.

Now we clear the kinetic coefficient of friction:

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

The kinetic coefficient of friction between the box and the floor is 1.020.

multiple

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s