Answer to Question #139943 in Physics for Jacob Sarthin

Question

Consider the two blocks shown above, connected together by a massless rope around a massless, frictionless pulley. The mass of the top block is m1 = 5.70 kg and the mass of the bottom block is m2 = 11.0 kg, and all surfaces have a coefficient of kinetic friction mukmuk = 0.450. If the bottom block is pulled to the left with a force of T = 136 N, what is the acceleration of the bottom block?

Answer 1

$$m_2a=T-\mu N_2-T'\\m_2a=T-\mu (m_1+m_2)g-(m_2a+\mu m_1g)\\ 11a=136-0.45(5.1+11)(9.8)\\-(5.1a+0.45 (5.1))(9.8)\\a=0.697\frac{m}{s^2}$$