Question

A 5.00 kg box is being pulled at constant speed up a 30° incline by a force of 45.0 N parallel to the incline.
Determine the coefficient of kinetic friction between the box and the plane.

Answer 1

To determine the coefficient of kinetic friction between the box and the inclined plane, we first need to analyze the forces acting on the box.

1. **Identify the forces:**
- Weight (\(W\)) of the box: \(W = mg\), where \(m = 5.00 \, \text{kg}\) and \(g = 9.81 \, \text{m/s}^2\). Therefore, \(W = 5.00 \times 9.81 = 49.05 \, \text{N}\).
- The gravitational force acting parallel to the incline: \(W_{\parallel} = mg \sin(\theta)\), where \(\theta = 30^\circ\).
- The normal force (\(N\)).
- The applied force (\(F_{\text{applied}} = 45.0 \, \text{N}\)).
- The frictional force (\(F_{\text{friction}} = \mu_k N\)), where \(\mu_k\) is the coefficient of kinetic friction.

2. **Calculate the components of the weight:**
\[
W_{\parallel} = mg \sin(\theta) = 5.00 \times 9.81 \times \sin(30^\circ) = 5.00 \times 9.81 \times 0.5 = 24.525 \, \text{N}
\]
\[
W_{\perpendicular} = mg \cos(\theta) = 5.00 \times 9.81 \times \cos(30^\circ) = 5.00 \times 9.81 \times \frac{\sqrt{3}}{2} \approx 42.435 \, \text{N}
\]

3. **Apply equilibrium conditions:**
Since the box is moving at a constant speed, the sum of the forces parallel to the incline must equal zero:
\[
F_{\text{applied}} - W_{\parallel} - F_{\text{friction}} = 0
\]
Substituting \(F_{\text{friction}} = \mu_k N\):
\[
45.0 - 24.525 - \mu_k N = 0
\]
\[
\mu_k N = 20.475 \, \text{N} \tag{1}
\]

4. **Find the normal force (\(N\)):**
The normal force is equal to the perpendicular component of the weight:
\[
N = W_{\perpendicular} = 42.435 \, \text{N}
\]

5. **Substitute \(N\) in equation (1):**
\[
\mu_k (42.435) = 20.475
\]
\[
\mu_k = \frac{20.475}{42.435} \approx 0.482
\]

Thus, the coefficient of kinetic friction between the box and the incline is approximately:
\[
\boxed{0.482}
\]