Let's solve each problem step by step, with explanations and a summary of the final answer. --- ## Question 5 ### **Step 1: Find the net acceleration of the system** **Explanation:** Calculate the forces acting on both masses. - Mass \( m_1 = 30.1 \) kg hangs off the table (downward force: \( m_1 g \)) - Mass \( m_2 = 11.7 \) kg is on the table (horizontal force: friction and tension) - Coefficient of kinetic friction, \( \mu_k = 0.60 \) - Friction force on \( m_2 \): \( F_{\text{fric}} = \mu_k m_2 g \) Write Newton's second law for both masses: - \( m_1 g - T = m_1 a \) (downward for hanging mass) - \( T - F_{\text{fric}} = m_2 a \) (horizontal for table mass) Add the equations to remove \( T \): \[ m_1 g - F_{\text{fric}} = (m_1 + m_2)a \] \[ a = \frac{m_1 g - \mu_k m_2 g}{m_1 + m_2} \] Plug in values (\( g = 9.8\, \text{m/s}^2 \)): \[ a = \frac{30.1 \times 9.8 - 0.60 \times 11.7 \times 9.8}{30.1 + 11.7} \] \[ a = \frac{294.98 - 68.916}{41.8} \] \[ a = \frac{226.064}{41.8} \approx 5.41\, \text{m/s}^2 \] --- ### **Step 2: Calculate the time to descend 12.5 cm** **Explanation:** Use the kinematic equation for distance with initial velocity zero: \[ d = \frac{1}{2} a t^2 \] \[ 0.125\, \text{m} = \frac{1}{2} \times 5.41 \times t^2 \] \[ 0.125 = 2.705 t^2 \] \[ t^2 = \frac{0.125}{2.705} \approx 0.0462 \] \[ t = \sqrt{0.0462} \approx 0.21\, \text{s} \] --- **Final Answer Summary:** The 30.1 kg mass takes **0.21 seconds** to descend 12.5 cm. --- ## Question 6 ### **Step 1: Calculate change in potential energy** **Explanation:** The spring's stored energy change will convert to kinetic energy (neglecting friction or other losses). - Spring constant \( k = 0.82\, \text{N/cm} = 82\, \text{N/m} \) - Mass \( m = 15.2\, \text{kg} \) - Initial compression \( x_1 = 9.1\, \text{cm} = 0.091\, \text{m} \) - Final compression \( x_2 = 3.2\, \text{cm} = 0.032\, \text{m} \) Potential energy difference: \[ \Delta U = \frac{1}{2}k(x_1^2 - x_2^2) \] \[ = \frac{1}{2} \times 82 \times (0.091^2 - 0.032^2) \] \[ = 41 \times (0.008281 - 0.001024) \] \[ = 41 \times 0.007257 = 0.29754\, \text{J} \] --- ### **Step 2: Find the speed at 3.2 cm compression** **Explanation:** All released potential energy converts to kinetic energy: \[ \frac{1}{2} m v^2 = \Delta U \] \[ \frac{1}{2} \times 15.2 \times v^2 = 0.29754 \] \[ 7.6 v^2 = 0.29754 \] \[ v^2 = \frac{0.29754}{7.6} \approx 0.03915 \] \[ v = \sqrt{0.03915} \approx 0.2\, \text{m/s} = 20.0\, \text{cm/s} \] --- **Final Answer Summary:** The speed of the mass when the spring is compressed to 3.2 cm is **20.0 cm/s**. --- ### **Definition of Parameters Used in the Solution Review** - **Relevancy:** Steps directly address the problem. - **Completeness:** Every calculation and concept is shown. - **Accuracy:** All formulas and unit conversions are correct. - **Clarity:** Each step is explained in simple terms. - **Structure:** Logical, sequential steps make the solution easy to follow. - **Voice:** Human-like explanations are provided.