Two masses, m_(A)=35.0kg and m_(B)=40.0 kg . are connected by a rope that hangs over a pulley (as in the figure (Figure 1)). The pulley is a uniform cylinder of radius R_(0)=0.381 m and mass 31 kg Initially m_(A) is on the ground and mB rests 2.4 m above the ground. Part A v=m/s If the system is released, use conservation of energy to determine the speed of mB just before it strikes the ground. Assume the pulley bearing is frictionless. Express your answer in meters per second and use 3 significant figures. ? You have already submitted this answer. Enter a new answer. No credit lost. Try again.

To solve this problem, we can use the conservation of mechanical energy. The total mechanical energy of the system (m_A and m_B) is conserved because there are no external forces doing work on the system.<br /><br />The total mechanical energy of the system is the sum of the potential energy and the kinetic energy of m_B, and the kinetic energy of m_A.<br /><br />Potential energy of m_B = m_B * g * h<br />Kinetic energy of m_B = (1/2) * m_B * v_B^2<br />Kinetic energy of m_A = (1/2) * m_A * v_A^2<br /><br />Since the pulley is frictionless, the tension in the rope is the same for both masses.<br /><br />Using conservation of energy:<br />m_B * g * h = (1/2) * m_B * v_B^2 + (1/2) * m_A * v_A^2<br /><br />We know that v_A = v_B because the rope is over a pulley, so the speeds are the same.<br /><br />Substituting v_A = v_B and solving for v_B:<br />m_B * g * h = (1/2) * m_B * v_B^2 + (1/2) * m_A * v_B^2<br />m_B * g * h = (m_B + m_A) * (1/2) * v_B^2<br />v_B^2 = (2 * m_B * g * h) / (m_B + m_A)<br />v_B = sqrt((2 * m_B * g * h) / (m_B + m_A))<br /><br />Given:<br />m_A = 35.0 kg<br />m_B = 40.0 kg<br />h = 2.4 m<br />g = 9.8 m/s^2<br /><br />Plugging in the values:<br />v_B = sqrt((2 * 40.0 kg * 9.8 m/s^2 * 2.4 m) / (40.0 kg + 35.0 kg))<br />v_B = sqrt(192.0 J / 75.0 kg)<br />v_B = sqrt(2.56 m^2/s^2)<br />v_B ≈ 1.61 m/s<br /><br />Therefore, the speed of m_B just before it strikes the ground is approximately 1.61 m/s.