## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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