This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Here are the solutions for questions 3.1.1, 3.1.2, and 3.1.3. Given: Mass m = 10 kg Angle of inclination = 30^ Applied force F = 120 N (up the incline) Coefficient of kinetic friction _k = 0.2 Acceleration due to gravity g = 9.8 m/s^2 3.1.1. The magnitude of the normal force Step 1: Identify the forces perpendicular to the inclined plane. The normal force N acts perpendicular to the surface, balancing the component of the gravitational force perpendicular to the plane. N = mg Step 2: Substitute the given values and calculate N. N = (10 kg)(9.8 m/s^2) 30^ N = 98 N × sqrt(3)2 N = 49sqrt(3) N N ≈ 49 × 1.73205 N N ≈ 84.87 N The magnitude of the normal force is 49sqrt(3) N or approximately 84.87 N. 3.1.2. The magnitude of the kinetic friction force Step 1: Use the formula for kinetic friction force. The kinetic friction force f_k is given by the product of the coefficient of kinetic friction and the normal force. f_k = _k N Step 2: Substitute the values for _k and N (from 3.1.1) and calculate f_k. f_k = 0.2 × (49sqrt(3) N) f_k = 9.8sqrt(3) N f_k ≈ 0.2 × 84.87 N f_k ≈ 16.97 N The magnitude of the kinetic friction force is 9.8sqrt(3) N or approximately 16.97 N. 3.1.3. The magnitude of the net force Step 1: Identify all forces acting parallel to the inclined plane. • Applied force F = 120 N (up the incline) • Component of gravitational force parallel to the incline mg (down the incline) • Kinetic friction force f_k (down the incline, as the applied force F is greater than mg , indicating upward motion or tendency to move upward). Step 2: Calculate the component of gravitational force parallel to the incline. mg = (10 kg)(9.8 m/s^2) 30^ mg = 98 N × (1)/(2) mg = 49 N Step 3: Calculate the net force by summing the forces parallel to the incline. Taking the upward direction along the incline as positive: F_net = F - mg - f_k F_net = 120 N - 49 N - 9.8sqrt(3) N F_net = 71 N - 9.8sqrt(3) N F_net ≈ 71 N - 16.97 N F_net ≈ 54.03 N The magnitude of the net force is (71 - 9.8sqrt(3)) N or approximately 54.03 N. That's 2 down. 3 left today — send the next one.