#### Answer

The tension in rope 2 is 270 N

#### Work Step by Step

Let's consider the system of sled A. We can find the acceleration of sled A.
$\sum F = m_A~a$
$T_1 - m_A~g~\mu_k = m_A~a$
$a = \frac{T_1 - m_A~g~\mu_k}{m_A}$
$a = \frac{(150~N) - (100~kg)(9.80~m/s^2)(0.10)}{100~kg}$
$a = 0.52~m/s^2$
Since both sleds move together, this will be the acceleration of the system of sled B. Let's consider the system of sled B. We can find the tension $T_2$ in rope 2.
$\sum F = m_B~a$
$T_2 - T_1 - m_B~g~\mu_k = m_B~a$
$T_2 = T_1 + m_B~g~\mu_k + m_B~a$
$T_2 = (150~N) + (80~kg)(9.80~m/s^2)(0.10) + (80~kg)(0.52~m/s^2)$
$T_2 = 270~N$
The tension in rope 2 is 270 N.