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Give step-by-step solution with explanation and final answer:A 4 kg block and a 9 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless pulley that has a radius of 25 cm and a mass of 2 kg. The pulley's moment of inertia is 7 = SMR? . Find: (a) the magnitude of the tension (in N) of the rope on the end with the lighter block (1 point); (b) the magnitude of the tension (in N) of the rope on the end with the heavier block (1 point); (c) the magnitude of the acceleration (in m/s?) of the 9 kg block (1 point).

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Give step-by-step solution with explanation and final answer:Uploaded ImageA 4 kg block and a 9 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless pulley that has a radius of 25 cm and a mass of 2 kg. The pulley's moment of inertia is 7 = SMR? . Find: (a) the magnitude of the tension (in N) of the rope on the end with the lighter block (1 point); (b) the magnitude of the tension (in N) of the rope on the end with the heavier block (1 point); (c) the magnitude of the acceleration (in m/s?) of the 9 kg block (1 point).

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Asked by: gvr
Created at: 2025-06-18 05:56:00

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Let's solve the problem step by step. **Given:** - Mass of lighter block (\( m_1 \)) = 4 kg - Mass of heavier block (\( m_2 \)) = 9 kg - Radius of pulley (\( R \)) = 25 cm = 0.25 m - Mass of pulley (\( M \)) = 2 kg - Pulley's moment of inertia (\( I \)) = \(\frac{1}{2}MR^2\) Let \( T_1 \) = tension on lighter block side. Let \( T_2 \) = tension on heavier block side. Let \( a \) = acceleration of blocks (same for both, as rope is light and inextensible). Let \( g \) = acceleration due to gravity = 9.8 m/s². --- ### 1. **Write equations of motion for each block:** #### For the lighter block (4 kg, moving up): \[ T_1 - m_1g = m_1 a \implies T_1 = m_1 a + m_1g \] #### For the heavier block (9 kg, moving down): \[ m_2g - T_2 = m_2 a \implies T_2 = m_2g - m_2a \] --- ### 2. **Write equation for the Pulley (rotational motion):** Net torque on the pulley = \( (T_2 - T_1)R \) Moment of inertia \( I = \frac{1}{2}MR^2 \) Angular acceleration \( \alpha = \frac{a}{R} \) \[ (T_2 - T_1)R = I \alpha = \frac{1}{2}MR^2 \cdot \frac{a}{R} \] \[ T_2 - T_1 = \frac{1}{2}M a \] --- ### 3. **Combine equations:** From above: - \( T_1 = m_1 a + m_1g \) - \( T_2 = m_2g - m_2a \) - \( T_2 - T_1 = \frac{1}{2}M a \) Plug in the values of \( T_1 \) and \( T_2 \) into the third equation: \[ (m_2g - m_2a) - (m_1a + m_1g) = \frac{1}{2} M a \] \[ m_2g - m_2a - m_1a - m_1g = \frac{1}{2} M a \] \[ (m_2 - m_1)g - (m_1 + m_2)a = \frac{1}{2} M a \] \[ (m_2 - m_1)g = (m_1 + m_2 + \frac{1}{2} M)a \] \[ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2} M} \] --- ### 4. **Plug in the numbers:** - \( m_1 = 4 \) kg - \( m_2 = 9 \) kg - \( M = 2 \) kg - \( g = 9.8 \) m/s² \[ a = \frac{(9 - 4) \times 9.8}{4 + 9 + \frac{1}{2} \times 2} = \frac{5 \times 9.8}{13 + 1} = \frac{49}{14} = 3.5~\text{m/s}^2 \] --- ### 5. **Find Tensions:** **For lighter block:** \[ T_1 = m_1 a + m_1 g = 4 \times 3.5 + 4 \times 9.8 = 14 + 39.2 = 53.2~\text{N} \] **For heavier block:** \[ T_2 = m_2 g - m_2 a = 9 \times 9.8 - 9 \times 3.5 = 88.2 - 31.5 = 56.7~\text{N} \] --- ### **Final Answers** **(a) Tension in rope on end with lighter block:** \[ \boxed{53.2~\text{N}} \] **(b) Tension in rope on end with heavier block:** \[ \boxed{56.7~\text{N}} \] **(c) Magnitude of acceleration of the 9 kg block:** \[ \boxed{3.5~\text{m/s}^2} \]

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