Answer
(a) Less
(b) $4.36\frac{m}{s^2}$
(c) $15.3N$
Work Step by Step
(a) The hanging block can move freely (that is, it has acceleration). Thus, the tension in the string is less than the weight of the hanging block.
(b) For the first block: $\Sigma F_x=T=m_1a$.......eq(1)
For the second block: $\Sigma F_x=-T+m_2g=m_2a$.........eq(2)
adding eq(1) and eq(2), we obtain:
$m_2g=(m_1+m_2)a$
This simplifies to:
$a=(\frac{m_2}{m_1+m_2})g$
We plug in the known values to obtain:
$a=(\frac{2.80}{6.30})(9.81)$
$a=4.36\frac{m}{s^2}$
(c) We can find the tension in the string as
$T=m_1a$
We plug in the known values to obtain:
$T=(3.50)(4.36)$
$T=15.3N$
