A 12.0 kg box is being pulled along level ground at constant velocity by a horizontal force of 38.0 N. What is

Question

the coefficient of kinetic friction between the box and the floor?

Answer 1

To find the coefficient of kinetic friction between the box and the floor, we start by using the relationship between the force of friction ($F_f$), the normal force ($F_N$), and the coefficient of kinetic friction (${\mu }_k$).

1. The equation for the force of friction is: $$F_f={\mu }_k{F}_N$$

2. On a horizontal surface, the normal force $F_N$ is equal to the weight of the box, which can be calculated using: $$F_N=mg$$ where $m$ is the mass of the box and $g$ is the acceleration due to gravity (approximately $9.81,{\text{m/s}}^2$).

3. Calculate the weight of the box: $$F_N=(12.0,\text{kg})(9.81,{\text{m/s}}^2)=117.72,\text{N}$$

4. Since the box is being pulled at a constant velocity, the force of friction is equal to the applied force, which is $38.0,\text{N}$. So: $$F_f=38.0,\text{N}$$

5. Now we can set up the equation: $$38.0,\text{N}={\mu }_k(117.72,\text{N})$$

6. Solve for ${\mu }_k$: $${\mu }_k=\frac{38.0,\text{N}}{117.72,\text{N}}\approx 0.322$$

Thus, the coefficient of kinetic friction between the box and the floor is approximately 0.322.