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In Fig. assume that there is negligible friction between the blocks and table. Compute the tension in the cord connecting $m_{2}$ and the pulley and acceleration of $m_{2}$ if \ $(m_{1} = 300 g, m_{2} = 200 g)$ and $F = 0.40 N$ .

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

$\frac{2}{7} m / s^{2}, \frac{42}{35} N$

$\frac{5}{7} m / s^{2}, \frac{12}{35} N$

$\frac{2}{8} m / s^{2}, \frac{72}{35} N$

answer is C.

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Detailed Solution

$T = m_{1} (2 a) = (0.3) (2 a) = 0.6 a F - 2 T = m_{2} a 2 T = 0.4 - 0.2 a$

On solving these two equations, we get

$a = \frac{2}{7} m / s^{2}$

$and 2 T = \frac{12}{35} N$

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