Answer
(a) The acceleration of each block is $a = 2.21~m/s^2$.
(b) T = 2.27 N
(c) If the positions of the blocks are reversed, then the 4.00-kg block would accelerate faster than the 8.00-kg block. In this situation, the tension in the string would be zero.
The acceleration of the 4.00-kg block would be $2.78~m/s^2$.
The acceleration of the 8.00-kg block would be $1.93~m/s^2$.
Work Step by Step
(a) Since $\mu_k$ is less for the 4.00-kg block $m_1$ than the 8.00-kg block $m_2$, the 4.00-kg block would accelerate faster down the incline if they were not tied together. However, since the two blocks are tied together, the tension in the string keeps the acceleration for the two blocks equal.
$\sum F = (m_1+m_2)~a$
$(m_1)(g)~sin(\theta) - (m_1)(g)~cos(\theta)(0.25) + (m_2)(g)~sin(\theta) - (m_2)(g)~cos(\theta)(0.35)= (m_1+m_2)~a$
$a = \frac{(m_1)(g)~sin(\theta) - (m_1)(g)~cos(\theta)(0.25) + (m_2)(g)~sin(\theta) - (m_2)(g)~cos(\theta)(0.35)}{(m_1+m_2)}$
$a = \frac{(4.00~kg)(9.80~m/s^2)~sin(30.0^{\circ}) - (4.00~kg)(9.80~m/s^2)~cos(30.0^{\circ})(0.25) + (8.00~kg)(9.80~m/s^2)~sin(30.0^{\circ}) - (8.00~kg)(9.80~m/s^2)~cos(30.0^{\circ})(0.35)}{(4.00~kg+8.00~kg)}$
$a = 2.21~m/s^2$
The acceleration of each block is $a = 2.21~m/s^2$.
(b) We can use the 4.00-kg block to find the tension $T$.
$\sum F = m_1~a$
$(m_1)(g)~sin(\theta) - (m_1)(g)~cos(\theta)(0.25) - T = m_1~a$
$T = (m_1)(g)~sin(\theta) - (m_1)(g)~cos(\theta)(0.25) - m_1~a$
$T = (4.00~kg)(9.80~m/s^2)~sin(30.0^{\circ}) - (4.00~kg)(9.80~m/s^2)~cos(30.0^{\circ})(0.25) - (4.00~kg)(2.21~m/s^2)$
$T = 2.27~N$
(c) If the positions of the blocks are reversed, then the 4.00-kg block would accelerate faster than the 8.00-kg block because $\mu_k$ is less for the 4.00-kg block. In this situation, the tension in the string would be zero.
We can find the acceleration $a_1$ of the 4.00-kg block.
$\sum F = m_1~a_1$
$(m_1)(g)~sin(\theta) - (m_1)(g)~cos(\theta)(\mu_k) = m_1~a_1$
$a_1 = g~sin(\theta) - g~cos(\theta)(\mu_k)$
$a_1 = (9.80~m/s^2)~sin(30.0^{\circ}) - (9.80~m/s^2)~cos(30.0^{\circ})(0.25)$
$a_1 = 2.78~m/s^2$
The acceleration of the 4.00-kg block would be $2.78~m/s^2$.
We can find the acceleration $a_2$ of the 8.00-kg block.
$\sum F = m_2~a_2$
$(m_2)(g)~sin(\theta) - (m_2)(g)~cos(\theta)(\mu_k) = m_2~a_2$
$a_2 = g~sin(\theta) - g~cos(\theta)(\mu_k)$
$a_2 = (9.80~m/s^2)~sin(30.0^{\circ}) - (9.80~m/s^2)~cos(30.0^{\circ})(0.35)$
$a_2 = 1.93~m/s^2$
The acceleration of the 8.00-kg block would be $1.93~m/s^2$.
