# Pulley System Problem: Solution Given: - \( m_1 = 3.\, \text{kg} \) (left mass) - \( m_2 = 5.\, \text{kg} \) (right mass) - System starts from rest - Pulley and string are frictionless and massless - \( g = 9.8\, \text{m/s}^2 \) Let's solve for: 1. **The magnitude of the acceleration of the masses** 2. **The tension in the string** --- ## 1. Magnitude of the Acceleration of the Masses **Step 1: Write the Newton's Second Law equations for each mass.** - For \( m_1 \) (upward is positive): \[ T - m_1g = m_1a \] - For \( m_2 \) (downward is positive): \[ m_2g - T = m_2a \] **Step 2: Add the two equations to eliminate \( T \):** \[ (T - m_1g) + (m_2g - T) = m_1a + m_2a \] \[ m_2g - m_1g = (m_1 + m_2)a \] \[ a = \frac{m_2 - m_1}{m_1 + m_2} g \] **Step 3: Plug in the values:** \[ a = \frac{5. - 3.}{5. + 3.} \times 9.8 \] \[ a = \frac{2.}{8.} \times 9.8 \] \[ a = .25 \times 9.8 = 2.45\, \text{m/s}^2 \] **Final Answer (Acceleration):** \[ \boxed{2.45\, \text{m/s}^2} \] --- ## 2. Tension in the String **Step 1: Use one of the mass equations (let's use \( m_1 \)):** \[ T = m_1g + m_1a \] **Step 2: Plug in the numbers:** \[ T = 3. \times 9.8 + 3. \times 2.45 \] \[ T = 29.4 + 7.35 = 36.75\, \text{N} \] **Final Answer (Tension):** \[ \boxed{36.8\, \text{N}} \] (Rounded to 3 significant figures) --- ## **Summary Table** | Quantity | Value | |------------------|-----------------------| | Acceleration, \( a \) | \( 2.45\, \text{m/s}^2 \) | | Tension, \( T \) | \( 36.8\, \text{N} \) | --- If you need the step-by-step derivation or have more questions about the problem, let me know!