\( \begin{array}{ll}5.1 .1 & \text { Calcuate the accelaration of the } 20 \mathrm{~kg} \text { mass. } \\ 5.12 & \text { Calculate the speed of the } 20 \mathrm{~kg} \text { mass as it strices the ground } \\ 5.1 .3 & \text { At what minimum distance from the pulley should the } 5 \mathrm{~kg} \text { mass be place } \\ \text { inilialy, so that the } 20 \mathrm{~kg} \text { mass just strices the ground? }\end{array} \)

Sure, let's tackle **Problem 5.1.1** step by step. ### **Problem 5.1.1** **Calculate the acceleration of the 20 kg mass.** #### **Assumptions:** 1. **System Setup:** We have two masses connected by a rope over a frictionless and massless pulley. - **Mass \( m_1 = 20 \, \text{kg} \)** - **Mass \( m_2 = 5 \, \text{kg} \)** 2. **Direction of Motion:** The heavier mass (\( m_1 \)) will accelerate downward, while the lighter mass (\( m_2 \)) will accelerate upward. 3. **Negligible Pulley and Rope Mass:** The pulley is frictionless and massless, and the rope is massless and inextensible. 4. **Gravitational Acceleration:** \( g = 9.8 \, \text{m/s}² \) #### **Applying Newton’s Second Law:** For the **20 kg mass (\( m_1 \))**: \[ m_1 g - T = m_1 a \quad \text{(1)} \] Where: - \( T \) is the tension in the rope. - \( a \) is the acceleration. For the **5 kg mass (\( m_2 \))**: \[ T - m_2 g = m_2 a \quad \text{(2)} \] #### **Solving the Equations:** 1. **Add Equations (1) and (2)** to eliminate \( T \): \[ m_1 g - T + T - m_2 g = m_1 a + m_2 a \] \[ (m_1 - m_2)g = (m_1 + m_2)a \] 2. **Solve for \( a \):** \[ a = \frac{(m_1 - m_2)g}{m_1 + m_2} \] 3. **Plug in the Given Values:** \[ a = \frac{(20 \, \text{kg} - 5 \, \text{kg}) \times 9.8 \, \text{m/s}²}{20 \, \text{kg} + 5 \, \text{kg}} = \frac{15 \times 9.8}{25} = \frac{147}{25} = 5.88 \, \text{m/s}² \] #### **Final Answer:** The acceleration of the 20 kg mass is **5.88 m/s² downward**. ### **Summary:** \[ a = 5.88 \, \text{m/s}² \quad \text{(downward)} \]