5) In the scenario where the object Kon a frictional surface is released from rest along with 2 kg keobjects, and their motion is analyzed, given that the gravitational acceleration the objects is g=10m/s^2 friction mu =0.1 a) Draw the free-body diagrams of the 2 kg, 3 kg. and 5 kg objects (5 points) b) Calculate the magnitude of the acceleration for each of the 2 kg, 3 kg, and 5 kg objects. (5 points) c) Determine the tension forces in the ropes connected to the 2 kg and 3 kg objects. (10 points)

Let's analyze the given problem step by step.<br /><br />### Part (a): Free-Body Diagrams<br /><br />1. **2 kg Object:**<br /> - Weight (\(W_2\)): \(2 \text{ kg} \times 10 \text{ m/s}^2 = 20 \text{ N}\) downward.<br /> - Normal force (\(N_2\)): 20 N upward.<br /> - Frictional force (\(f_2\)): \(\mu \times N_2 = 0.1 \times 20 \text{ N} = 2 \text{ N}\) opposite to the direction of motion.<br /> - Tension force (\(T_2\)) in the rope connected to the 3 kg object.<br /><br />2. **3 kg Object:**<br /> - Weight (\(W_3\)): \(3 \text{ kg} \times 10 \text{ m/s}^2 = 30 \text{ N}\) downward.<br /> - Normal force (\(N_3\)): 30 N upward.<br /> - Frictional force (\(f_3\)): \(\mu \times N_3 = 0.1 \times 30 \text{ N} = 3 \text{ N}\) opposite to the direction of motion.<br /> - Tension force (\(T_3\)) in the rope connected to the 5 kg object.<br /> - Tension force (\(T_2\)) from the rope connected to the 2 kg object.<br /><br />3. **5 kg Object:**<br /> - Weight (\(W_5\)): \(5 \text{ kg} \times 10 \text{ m/s}^2 = 50 \text{ N}\) downward.<br /> - Normal force (\(N_5\)): 50 N upward.<br /> - Frictional force (\(f_5\)): \(\mu \times N_5 = 0.1 \times 50 \text{ N} = 5 \text{ N}\) opposite to the direction of motion.<br /> - Tension force (\(T_3\)) from the rope connected to the 3 kg object.<br /><br />### Part (b): Calculate the Magnitude of the Acceleration<br /><br />Assuming all objects are connected and move together with the same acceleration \(a\):<br /><br />The net force acting on the system is the sum of the forces minus the frictional forces:<br />\[ F_{\text{net}} = (m_2 + m_3 + m_5) \cdot a \]<br /><br />The total mass of the system:<br />\[ m_{\text{total}} = 2 \text{ kg} + 3 \text{ kg} + 5 \text{ kg} = 10 \text{ kg} \]<br /><br />The total frictional force:<br />\[ f_{\text{total}} = f_2 + f_3 + f_5 = 2 \text{ N} + 3 \text{ N} + 5 \text{ N} = 10 \text{ N} \]<br /><br />Since there is no external force pulling the system, the net force is due to the frictional forces only:<br />\[ F_{\text{net}} = -f_{\text{total}} = -10 \text{ N} \]<br /><br />Using Newton's second law:<br />\[ F_{\text{net}} = m_{\text{total}} \cdot a \]<br />\[ -10 \text{ N} = 10 \text{ kg} \cdot a \]<br />\[ a = -1 \text{ m/s}^2 \]<br /><br />The negative sign indicates that the acceleration is opposite to the assumed direction of motion.<br /><br />### Part (c): Determine the Tension Forces in the Ropes<br /><br />1. **Tension in the rope connected to the 2 kg object (\(T_2\)):**<br /><br />For the 2 kg object:<br />\[ T_2 - f_2 = m_2 \cdot a \]<br />\[ T_2 - 2 \text{ N} = 2 \text{ kg} \cdot (-1 \text{ m/s}^2) \]<br />\[ T_2 - 2 \text{ N} = -2 \text{ N} \]<br />\[ T_2 = 0 \text{ N} \]<br /><br />2. **Tension in the rope connected to the 3 kg object (\(T_3\)):**<br /><br />For the 3 kg object:<br />\[ T_3 - T_2 - f_3 = m_3 \cdot a \]<br />\[ T_3 - 0 \text{ N} - 3 \text{ N} = 3 \text{ kg} \cdot (-1 \text{ m/s}^2) \]<br />\[ T_3 - 3 \text{ N} = -3 \text{ N} \]<br />\[ T_3 = 0 \text{ N} \]<br /><br />Therefore, the tensions in the ropes are:<br />- \( T_2 = 0 \text{ N} \)<br />- \( T_3 = 0 \text{ N} \)<br /><br />This result suggests that the objects are not accelerating relative to each other, and the entire system is moving as a single unit under the influence of friction.