Kinetic friction on a moving object

[KingNothing]

Hey...this problem arose:

A force of 40N is required to start a 5 kg box moving across a horizontal concrete floor.
A) What is the coefficient of static friction between the box and the floor? (its .82, I got this no problem)

B) If the 40N force continues, the box accelerates at 0.7 m/s^2. What is the coefficient of kinetic friction?

I don't see how to do part B. I know it's .74, but I don't know how. Please explain in detail by showing equation(s).

 

[jamesrc]

Use Newton's second law. You know the net force is equal to the mass times the acceleration. Since the net force is equal to the applied force less the friction, you can calculate the coefficient of friction.

$$F_{net} = ma = 40 - f$$
$$f = \mu_k mg$$

 

[M98Ranger]

To calculate the coefficient of kinetic friction, we can use the equation F = μkN, where F is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force. In this case, the normal force is equal to the weight of the box, which is given by mg, where m is the mass of the box and g is the acceleration due to gravity (9.8 m/s^2). Therefore, we can rewrite the equation as F = μkmg.

We know that the applied force of 40N is equal to the force of kinetic friction (since the box is already in motion), so we can substitute this into our equation: 40N = μkmg. We also know that the acceleration of the box is 0.7 m/s^2, so we can substitute this into the equation as well: 40N = μk(5kg)(0.7 m/s^2)(9.8 m/s^2).

Solving for μk, we get μk = 40N / (5kg)(0.7 m/s^2)(9.8 m/s^2) = 0.74. Therefore, the coefficient of kinetic friction is 0.74. This means that it takes 74% of the normal force to keep the box in motion across the concrete floor.