Let's break the problem into **2 steps** and explain each step clearly. --- ## **Step 1: Find the total force needed to accelerate the block** The block of mass m = .8\,text{kg} \) must accelerate at \( a = 2.00\,\text{m/s}^2 \). **Force needed to accelerate the block:** \[ F = m a = 2.8 \times 2. = 5.6\,\text{N} \] **Explanation:** This is the force the string must apply to the block to achieve the desired acceleration. --- ## **Step 2: Relate the force to the torque applied by the motor** ### **a. Torque due to the block:** The tension in the string creates a torque at the pulley's edge: \[ \tau_{\text{block}} = F \times R = 5.6 \times .12 = .672\,\text{N}\cdot\text{m} \] ### **b. Torque to accelerate the pulley:** The pulley has a moment of inertia \( I = .040\,\text{kg}\cdot\text{m}^2 \). First, relate the block's acceleration to the pulley's angular acceleration: \[ a = \alpha R \implies \alpha = \frac{a}{R} = \frac{2.}{.12} = 16.67\,\text{rad/s}^2 \] Torque to accelerate the pulley: \[ \tau_{\text{pulley}} = I \alpha = .040 \times 16.67 = .667\,\text{N}\cdot\text{m} \] ### **c. Include frictional torque:** The frictional torque is given as \( \tau_f = .05\,\text{N}\cdot\text{m} \). ### **d. Total torque required:** \[ \tau_{\text{total}} = \tau_{\text{block}} + \tau_{\text{pulley}} + \tau_f = .672 + .667 + .05 = 1.389\,\text{N}\cdot\text{m} \] **Explanation:** We add the torques needed to (1) pull the block, (2) accelerate the pulley, and (3) overcome friction. --- ## **Final Answer** \[ \boxed{1.39 \;\text{N}\cdot\text{m}} \] The motor must apply a torque of **1.39 N·m** to accelerate the block at \(2.00 \;\text{m/s}^2\).