Three identical blocks of masses m = 2 kg are drawn by a force F = 10.2 N with an acceleration of 0.6 ms⁻² on a frictionless surface, then what is the tension (in N) in the string between the blocks B and C? (a) 9.2 (b) 3.4 (c) 4 (d) 7.8

Answer: (d). Force F acts on A; the string between B and C is behind A and B. For A and B together: F - TBC = (mA + mB)a. TBC = 10.2 - 4 × 0.6 = 7.8 N So the answer is (d) 7.8.

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Connected Bodies & Pulleys

Three identical blocks of masses $m = 2\text{ kg}$ are drawn by a force $F = 10.2\text{ N}$ with an acceleration of $0.6\text{ ms}^{-2}$ on a frictionless surface, then what is the tension (in $\text{N}$ ) in the string between the blocks $B$ and $C$ ?

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Answer: (d)

Force F acts on A; the string between B and C is behind A and B.
For A and B together: $F - T_{BC} = (m_A + m_B)a$ .
$T_{BC} = 10.2 - 4 \times 0.6 = 7.8\text{ N}$
So the answer is (d) 7.8.

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