#### Answer

(a) T = 19.6 N
(b) The 4.0-kg block will move down the slope.
(c) T = 20.6 N

#### Work Step by Step

(a) If the blocks are at rest, we can consider the system of the 2.0-kg block to find the tension in the string. Let $m_1$ be the mass of the 2.0-kg block.
$\sum F = m_1~a$
$m_1~g -T = 0$
$T = m_1~g$
$T = (2.0~kg)(9.80~m/s^2)$
$T = 19.6 ~N$
(b) We can find the component of the 4.0-kg block's weight which is directed down the slope. Let $m_2$ be the mass of the 4.0-kg block.
$F_{slope} = m_2~g~sin(\theta)$
$F_{slope} = (4.0~kg)(9.80~m/s^2)~sin(35^{\circ})$
$F_{slope} = 22.48~N$
Since the component of the 4.0-kg block's weight directed down the slope is greater than the weight of the 2.0-kg block, the 4.0-kg block will move down the slope.
(c) We can set up a force equation for the 2.0-kg block.
$\sum F = m_1~a$
$T - m_1~g = m_1~a$
$a = \frac{T - m_1~g}{m_1}$
We can use this expression in the force equation for the 4.0-kg block. Note that the magnitude of acceleration is the same for both blocks.
$\sum F = m_2~a$
$m_2~g~sin(\theta) - T = m_2~a$
$m_2~g~sin(\theta) - T = m_2~(\frac{T - m_1~g}{m_1})$
$m_1~m_2~g~sin(\theta) - m_1~T = m_2~T - m_1~m_2~g$
$m_1~m_2~g~[1+sin(\theta)] = T~(m_1+m_2)$
$T = \frac{m_1~m_2~g~[1+sin(\theta)]}{m_1+m_2}$
$T = \frac{(2.0~kg)(4.0~kg)(9.80~m/s^2)~[1+sin(35^{\circ})]}{2.0~kg+4.0~kg}$
$T = 20.6~N$