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Three blocks of masses 1 kg, 2 kg, and 3 kg are connected by strings on a frictionless surface. A force of 12 N is applied to the 1 kg block, pulling the system. What is the tension in the string between the 1 kg and 2 kg blocks?

RankGuru · Accepted Answer

1. **System Acceleration**: Since all blocks move together, $a = F_{ext} / \sum m$.
   - $a = 12 / (1+2+3) = 12 / 6 = 2$ m/s$^2$.
2. **Free Body Diagram (2 & 3 together)**: The string between the 1kg and 2kg blocks must pull BOTH the 2kg and 3kg masses.
3. **Tension Formula**: $T_1 = (m_2 + m_3) 	imes a$.
4. **Calculation**:
   - $T_1 = (2 + 3) 	imes 2 = 5 	imes 2 = 10$ N.
5. **Alternative**: Look at the 1kg block: $F - T_1 = m_1 a \implies 12 - T_1 = 1 	imes 2 \implies T_1 = 10$ N.

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