Question #bdced

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Answer 1 · 2017-03-23T19:57:35

$T = 2 g N, & a = \frac{g}{3} m s^{- 2}$ $T=2g N, " & " a=g/3 ms^(-2)$

we use Newton's 2nd law on both particles and solve the

eqns simultaneously

$* m_{1} *$ $**m_1**$

$T = 6 a$ $T=6a$

$* m_{2} *$ $**m_2**$

$3 g - T = 3 a$ $3g-T=3a$

substitute for $T$ $T$

$3 g - 6 a = 3 a$ $3g-6a=3a$

$9 a = 3 g$ $9a=3g$

$a = \frac{g}{3} m s^{- 2}$ $a=g/3 ms^(-2)$

$:. T=6xxg/3=2g N$