1. **Stating the problem:** We have four blocks with masses 10 kg, 3 kg, 5 kg, and 2 kg connected by cords and pulled to the right by a force \( \vec{F} \) on the 2 kg block. We want to find: (a) The total mass accelerated by the force. (b) The total mass accelerated by cord 3. (c) The total mass accelerated by cord 1. (d) Order the blocks by their acceleration, greatest first. (e) Order the cords by their tension, greatest first. 2. **Key concepts and formulas:** - The system is pulled without friction, so all blocks accelerate together with the same acceleration \( a \). - Newton's second law: \( F = ma \), where \( m \) is the total mass being accelerated by the force or tension. - The tension in each cord equals the force needed to accelerate the blocks to the right of that cord. 3. **Calculations:** - Total mass of all blocks: \( 10 + 3 + 5 + 2 = 20 \) kg. (a) Force \( \vec{F} \) pulls all four blocks, so total mass accelerated by \( \vec{F} \) is \( 20 \) kg. (b) Cord 3 is between the 5 kg and 2 kg blocks, so it pulls the 2 kg block only. Mass accelerated by cord 3 is \( 2 \) kg. (c) Cord 1 is between the 10 kg and 3 kg blocks, so it pulls the 3 kg, 5 kg, and 2 kg blocks. Mass accelerated by cord 1 is \( 3 + 5 + 2 = 10 \) kg. (d) Since all blocks are connected and move together, their acceleration is the same. So the order by acceleration is: \[ 10\,\text{kg} = 3\,\text{kg} = 5\,\text{kg} = 2\,\text{kg} \] (e) Tension in cords depends on the mass they pull: - Cord 1 pulls 10 kg total (3+5+2), so tension \( T_1 \) is proportional to 10 kg. - Cord 2 pulls 7 kg total (5+2), so tension \( T_2 \) is proportional to 7 kg. - Cord 3 pulls 2 kg total, so tension \( T_3 \) is proportional to 2 kg. Order of tensions from greatest to smallest: \[ T_1 > T_2 > T_3 \] **Final answers:** (a) 20 kg (b) 2 kg (c) 10 kg (d) All blocks have the same acceleration. (e) Tensions ordered as \( T_1 > T_2 > T_3 \).