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.
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Here are the solutions for questions 12.a and 12.b: 12.a) Limiting friction is the maximum static friction that can be exerted between two surfaces before one surface begins to slide over the other. It is the maximum force that opposes the initiation of motion. 12.b) Free Body Diagrams (FBDs): For the 20 kg trolley (W): Weight (W_T): Acts vertically downwards. W_T = M g = 20 kg × 10 m/s^2 = 200 N. Normal Force (N): Acts vertically upwards, perpendicular to the surface. N = W_T = 200 N. Tension (T): Acts horizontally to the right, due to the string. Rolling Friction (f_r): Acts horizontally to the left, opposing the motion. f_r = _r N. For the 3 kg hanging block: Weight (W_B): Acts vertically downwards. W_B = m g = 3 kg × 10 m/s^2 = 30 N. Tension (T): Acts vertically upwards, due to the string. Acceleration of the system: Step 1: Calculate the rolling friction force. The normal force on the trolley is N = M g. N = 20 kg × 10 m/s^2 = 200 N The rolling friction force is f_r = _r N. f_r = 0.04 × 200 N = 8 N Step 2: Write down the equations of motion for each body using Newton's second law (F=ma). For the 20 kg trolley (horizontal motion): T - f_r = M a T - 8 N = (20 kg) a (1) For the 3 kg hanging block (vertical motion, downwards is positive): m g - T = m a 30 N - T = (3 kg) a (2) Step 3: Solve the system of equations for acceleration (a). Add equation (1) and equation (2): (T - 8 N) + (30 N - T) = (20 kg) a + (3 kg) a 22 N = (23 kg) a Step 4: Calculate the acceleration. a = 22 N23 kg a ≈ 0.9565 m/s^2 The acceleration of the block and the trolley system is 0.96 m/s^2. That's 2 down. 3 left today — send the next one.