## Step 1: Write Forces for Block For each block, resolve forces parallel to the ramp: - **Gravity down the ramp:** \( m_i g \sin\theta \) (where \( i = A \) or \( B \)) - **Friction up the ramp:** \( f_i = \mu_i m_i g \cos\theta \) - **Tension in the cord:** \( T \), but since both blocks are moving together, treat them as a system. **Explanation:** Each block experiences gravity pulling it down the incline, and friction (opposing motion) up the incline. The tension will be internal when summing both blocks as a single system. --- ## Step 2: Write Net Force Equation for the System Sum forces for both blocks (mass \( m_A + m_B \)), since they accelerate together: \[ \text{Net force} = (m_A g \sin\theta - \mu_A m_A g \cos\theta) + (m_B g \sin\theta - \mu_B m_B g \cos\theta) \] \[ = (m_A + m_B) g \sin\theta - (\mu_A m_A + \mu_B m_B) g \cos\theta \] Set net force equal to total mass times acceleration (\( a \)): \[ (m_A + m_B)a = (m_A + m_B) g \sin\theta - (\mu_A m_A + \mu_B m_B) g \cos\theta \] \[ a = g \sin\theta - \frac{\mu_A m_A + \mu_B m_B}{m_A + m_B} g \cos\theta \] **Explanation:** We sum the forces for both blocks and set equal to the total mass times acceleration to solve for \( a \). --- ## Final Answer \[ \boxed{ a = g \sin\theta - \frac{\mu_A m_A + \mu_B m_B}{m_A + m_B} g \cos\theta } \] This is the acceleration of the blocks sliding down the ramp.