Question
A 12.0 kg box is being pulled along level ground at constant velocity by a horizontal force of 38.0 N. What is
the coefficient of kinetic friction between the box and the floor?
the coefficient of kinetic friction between the box and the floor?
Answers
GPT-4o mini
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**.
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**.