Answer
(a) $4.9N$
(b) $15N$
Work Step by Step
(a) We can find the required tension in the string as follows:
$a=(\frac{m_3}{m_1+m_2+m_3})g$
We plug in the known values to obtain:
$a=(\frac{3.0Kg}{1.0Kg+2.0Kg+3.0Kg})(9.8m/s^2)$
$\implies a=4.9m/s^2$
Now $T_1=m_1a$
We plug in the known values to obtain:
$T_1=(1.0Kg)(4.9m/s^2)$
$T_1=4.9N$
(b) We know that
$T_2=m_3(g-a)$
We plug in the known values to obtain:
$T_2=(3.0)(9.81-4.91)$
$T_2=15N$
