In the figure assumes that there is negligible friction between the block and the table. Compute
the tension in the cord connecting M ₁ and the pulley and acceleration of M₂. if
M₁ = 300 g ,M₂=200 g and F = 0.40 N

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$\frac{12}{35} N, \frac{2}{7} {m/s}^{2}$

$\frac{2}{35} N, \frac{12}{7} {m/s}^{2}$

$\frac{7}{2} N, \frac{35}{12} {m/s}^{2}$

$\frac{7}{35} N, \frac{12}{25} {m/s}^{2}$

answer is A.

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Detailed Solution

$2 T \times a_{B} = T \times a_{A} a_{B} = \frac{1}{2} a_{A} 0.4 - 2 T = 0.2 a_{B} \to (1) T = 2 m_{1} a_{B} \to (2) (∵ a_{A} = 2 a_{B}) From 1 and 2, 0.4 - 4 \times \frac{300}{1000} a = 0.2 a \Rightarrow a = \frac{2}{7} {ms}^{- 2} T e n s i o n = 2 T = 4 m_{1} a = 4 \times \frac{300}{1000} \times \frac{2}{7} = \frac{12}{35}$