Question

A laboratory cart (m1 500 g) rests on an inclined track (e 11°). It is connected to a lead weight (m2 150 g) suspended vertically off the end of a pulley as shown in the diagram below. (Assume the string and pulley contribute negligible mass to the system and that friction is kept low enough to be ignored.) If the lead weight falls 85 cm, determine the final speed of the system. string pulley cart (m,) track suspended mass (m) angle of inclination (0)

Answer 1

Given, m_1 = 500 g = 0.500 kg theta = 11 degree m_2 = 150 g = 0.150 kg From the free-body diagram: Let acceleration of the system is a Then for m_1 m_1 a = T - m_1 g sin theta ---(1) for m_2 m_2 a = m_2 g - T Therefore T = m_2 g - m_2 a ---(2) from (1) and (2) m_1 a = (m_2 g - m_2 a) - m_1 g sin theta Therefore m_1 a = m_2 g - m_2 a - m_1 g sin theta Therefore (m_1 + m_2)a = g(m_2 - m_1 sin theta) Therefore a = (m_2 - m_1 sin theta)/(m_1 + m_2) g Therefore a = [(0.150) - (0.500 times sin 11)]/(0.500 + 0.150) (9.8) Therefore a = 0.823 m/s^2 velocity after: s = 85 cm = 0.85 m Using equation: v^2 - u^2 = 2as Therefore v^2 = 2as = 2(0.823)(0.85 m) Therefore v^2 = 1.399 Therefore v = 1.183 m/s