Let’s solve the problem step by step! **Given:** - Block A: \( m_A = 12 \) kg - Block B: \( m_B = 18 \) kg - Tension between B and C: \( T = 60 \) N - Acceleration of the system: \( a = 2.5 \) m/s² - **Pulley and rope are massless and frictionless.** - **Force \( P \) acts downward on A.** Let's consider the forces acting on each block: --- ### **Step 1: Block C (hanging vertically from Block B)** Let \( m_C \) be the mass of block C (not given, but let's see if it's needed). But we know: - The tension in the rope between B and C, \( T_{BC} = 60 \) N. - The acceleration is upward for C (since the whole system accelerates downward for A, so C moves up). **Upward is positive (for C):** \[ T_{BC} - m_C g = m_C a \] \[ 60 - m_C (9.8) = m_C (2.5) \] \[ 60 = m_C (9.8 + 2.5) = m_C (12.3) \] \[ m_C = \frac{60}{12.3} \approx 4.88 \text{ kg} \] --- ### **Step 2: Block B (horizontally, mass = 18 kg)** Block B is **pulled to the right** by the tension \( T_{BC} = 60 \) N (from C), and to the left by the tension \( T_{AB} \) (from A). The net force on B causes it to accelerate at \( a \): \[ T_{AB} - 60 = m_B a \] \[ T_{AB} = 60 + m_B a = 60 + (18)(2.5) = 60 + 45 = 105\, \text{N} \] --- ### **Step 3: Block A (vertical, mass = 12 kg)** For Block A, the **external force \( P \) acts downward**. The tension \( T_{AB} \) acts upward. \[ P - m_A g - T_{AB} = m_A a \] \[ P = m_A g + T_{AB} + m_A a \] \[ P = (12)(9.8) + 105 + (12)(2.5) \] \[ P = 117.6 + 105 + 30 = 252.6\, \text{N} \] --- **Closest answer:** \[ \boxed{250\, \text{N}} \] --- **So, the correct choice is:** **250 N**

Certainly! Here's a detailed, step-by-step explanation suitable for a Chegg-style solution, with clear reasoning at each stage: --- **Given Data:** - Block A mass, \( m_A = 12\, \text{kg} \) - Block B mass, \( m_B = 18\, \text{kg} \) - Tension between Block B and C, \( T_{BC} = 60\, \text{N} \) - Acceleration of the system, \( a = 2.5\, \text{m/s}^2 \) - The external force \( P \) acts downward on Block A. - Assumption: The pulley and ropes are massless and frictionless. --- ### **Step 1: Determine the mass of Block C** Since the tension \( T_{BC} \) acts on C and the system accelerates at \( a \), we analyze Block C's vertical motion. - **For Block C (hanging vertically):** The forces on C are: - Upward: tension \( T_{BC} = 60\, \text{N} \) - Downward: weight \( m_C g \) Applying Newton's second law (\( \text{upward positive} \)): \[ T_{BC} - m_C g = m_C a \] Rearranged: \[ T_{BC} = m_C g + m_C a = m_C (g + a) \] Substitute known values: \[ 60 = m_C (9.8 + 2.5) = m_C (12.3) \] Solve for \( m_C \): \[ m_C = \frac{60}{12.3} \approx 4.88\, \text{kg} \] **Interpretation:** Block C has an approximate mass of **4.88 kg**. --- ### **Step 2: Find the tension \( T_{AB} \) between Blocks A and B** Next, analyze Block B, which is connected to Block C via \( T_{BC} \). - **For Block B (horizontal):** The tensions acting are: - To the right: \( T_{BC} = 60\, \text{N} \) (pulling B towards C) - To the left: \( T_{AB} \) (pulling B towards A) The net force causes B to accelerate at \( a = 2.5\, \text{m/s}^2 \): \[ T_{AB} - T_{BC} = m_B a \] Solve for \( T_{AB} \): \[ T_{AB} = T_{BC} + m_B a = 60 + (18)(2.5) = 60 + 45 = 105\, \text{N} \] **Interpretation:** The tension pulling B towards A is **105 N**. --- ### **Step 3: Calculate the external force \( P \) acting downward on Block A** Now, analyze Block A, which experiences: - Downward external force \( P \), - Upward tension \( T_{AB} \), - Its weight \( m_A g \). Applying Newton's second law in the vertical direction (taking downward as positive): \[ \text{Downward forces} = P + m_A g \] \[ \text{Upward force} = T_{AB} \] \[ \Rightarrow P + m_A g - T_{AB} = m_A a \] Rearranged to solve for \( P \): \[ P = m_A g + T_{AB} + m_A a \] Substitute the known values: \[ P = (12)(9.8) + 105 + (12)(2.5) \] \[ P = 117.6 + 105 + 30 = 252.6\, \text{N} \] **Approximate \( P \) as:** **253 N** --- ### **Final step: Choose the closest answer** Looking at the multiple-choice options: - 190 N - 250 N - 210 N - 170 N - 230 N **The closest value to our calculation (~253 N) is 250 N.** --- ### **Conclusion:** **The external force \( P \) applied downward on Block A is approximately \(\boxed{250\, \text{N}}\).** This detailed analysis considers all forces acting on each block, applies Newton's laws, and correctly accounts for the system's acceleration and tension forces to find \( P \).