Help understand how to solve question 4.---**Question:** 4. Block A of mass m rests on the plank B of mass 3m which is free to slide on a frictionless horizontal surface. The coefficient of friction between the block and plank is 0.2. If a horizontal force of magnitude 2mg is applied to the plank B, the acceleration of A relative to the plank and relative to the ground respectively, are: **Options:** (a) 0, g/2 (b) 0, 2g/3 (c) 3g/5, g/5 (d) 2g/5, g/5 **Diagram Description:** * Type: Schematic diagram illustrating a physical setup. * Elements: * Block A: A rectangular block labeled "A" is shown resting on top of Plank B. It is labeled with mass "m". * Plank B: A larger rectangular plank labeled "B" is shown below Block A. It is labeled with mass "3m". * Surface: Plank B is shown resting on a horizontal surface represented by horizontal lines with small diagonal lines indicating it is a surface. This surface is described as frictionless in the text. * Applied Force: A horizontal arrow points to the right, applied to Plank B. It is labeled with "2mg". * Relative Positions: Block A is centered on top of Plank B. Plank B is resting on the horizontal surface. * Direction: The applied force is horizontal and acts towards the right.

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Let's analyze this friction problem step by step. We have block A with mass m resting on plank B with mass 3m. A horizontal force of 2mg is applied to the plank. The coefficient of friction between the block and plank is 0.2. We need to find the acceleration of A relative to the plank and relative to the ground. First, let's check if the blocks move together. If they do, the total force 2mg acts on the combined mass of 4m, giving a common acceleration of g/2. For block A to have this acceleration, it needs a friction force of mg/2, which equals 0.5mg. However, the maximum static friction is only 0.2mg. Since 0.5mg is greater than 0.2mg, the required friction exceeds the maximum available friction. Therefore, slipping occurs between the blocks. Now let's calculate the individual accelerations. For block A, the only horizontal force is kinetic friction of 0.2mg to the right. Using Newton's second law, this gives block A an acceleration of g/5 to the right. For plank B, the net force is 2mg minus 0.2mg equals 1.8mg to the right. With mass 3m, this gives plank B an acceleration of 3g/5 to the right. Notice that plank B accelerates faster than block A, which confirms our slipping analysis. Finally, let's calculate the relative acceleration. The acceleration of A relative to the plank is the difference between their accelerations: g/5 minus 3g/5 equals negative 2g/5. The negative sign indicates that A accelerates backward relative to the plank. The magnitude is 2g/5. So the acceleration of A relative to the plank is 2g/5, and relative to the ground is g/5. The answer is option d: 2g/5, g/5. Let's summarize the key concepts for solving friction problems. First, always check if objects can move together by comparing required friction with maximum available friction. When slipping occurs, use kinetic friction and apply Newton's second law to each object separately. In this problem, block A accelerates at g/5 relative to ground, while its acceleration relative to the plank is 2g/5. This animation shows how plank B moves faster than block A, confirming our calculation that slipping occurs.