Let's break down the problem step by step: **Given:** - Mass of box, \( m = 46 \) kg - Ramp angle, \( \theta = 10^\circ \) - Applied force, \( F_{\text{push}} = 300 \) N (horizontal) - Box moves at constant velocity (i.e., net force along the ramp is zero) - We need to find the coefficient of kinetic friction, \( \mu_k \). --- ## **Step 1: Draw a Free-Body Diagram** Let's identify the forces acting on the box: - Weight, \( mg \), vertically downward - Normal force, \( N \), perpendicular to the ramp - Friction force, \( f_k = \mu_k N \), opposite to motion, parallel to the ramp - Applied force, \( F_{\text{push}} \), horizontal (not parallel to the ramp) --- ## **Step 2: Resolve Forces** Let's align the axes with the ramp: - **x-axis**: along the ramp (up the ramp) - **y-axis**: perpendicular to the ramp (into the ramp) ### **Breakdown of Forces:** **Weight (\(mg\)):** - Along ramp (down): \( mg \sin \theta \) - Perpendicular to ramp: \( mg \cos \theta \) **Applied Force (\(F_{\text{push}}\)):** - Along ramp: \( F_{\text{push}} \cos \theta \) - Perpendicular to ramp: \( F_{\text{push}} \sin \theta \) --- ## **Step 3: Write Equations** ### **Along the Ramp (x-axis):** Since the box moves at constant velocity, net force is zero: \[ F_{\text{push}} \cos \theta - mg \sin \theta - f_k = 0 \] \[ F_{\text{push}} \cos \theta - mg \sin \theta = f_k \] \[ F_{\text{push}} \cos \theta - mg \sin \theta = \mu_k N \] ### **Perpendicular to the Ramp (y-axis):** Net force perpendicular to the ramp is zero: \[ N + F_{\text{push}} \sin \theta = mg \cos \theta \] \[ N = mg \cos \theta - F_{\text{push}} \sin \theta \] --- ## **Step 4: Plug in Values** Given: - \( m = 46 \) kg - \( g = 9.8 \) m/s² - \( \theta = 10^\circ \) - \( F_{\text{push}} = 300 \) N Calculate the trigonometric values: - \( \sin 10^\circ \approx 0.1736 \) - \( \cos 10^\circ \approx 0.9848 \) ### **Calculate the Normal Force:** \[ N = mg \cos \theta - F_{\text{push}} \sin \theta \] \[ N = (46)(9.8)(0.9848) - 300(0.1736) \] \[ N = 446.3376 - 52.08 = 394.2576 \text{ N} \] ### **Calculate Forces Along the Ramp:** \[ F_{\text{push}} \cos \theta = 300 \times 0.9848 = 295.44 \text{ N} \] \[ mg \sin \theta = 46 \times 9.8 \times 0.1736 = 78.217 \text{ N} \] \[ F_{\text{push}} \cos \theta - mg \sin \theta = 295.44 - 78.217 = 217.223 \text{ N} \] --- ## **Step 5: Solve for \( \mu_k \)** Recall: \[ F_{\text{push}} \cos \theta - mg \sin \theta = \mu_k N \] \[ \mu_k = \frac{F_{\text{push}} \cos \theta - mg \sin \theta}{N} \] \[ \mu_k = \frac{217.223}{394.2576} \approx 0.551 \] --- ## **Final Answer:** \[ \boxed{\mu_k \approx 0.55} \] **The coefficient of kinetic friction between the box and the ramp is approximately 0.55.**