1. **Problem statement:** A 3.0 kg block on a horizontal surface is connected by a rope over a frictionless pulley to a 5.0 kg ball hanging vertically. The coefficient of friction between the block and surface is 0.4. The system starts from rest. Find the speed of the 5.0 kg ball after it has fallen 1.5 m. 2. **Relevant formulas and concepts:** - Frictional force: $f = \mu mg$ where $\mu$ is coefficient of friction, $m$ is mass, $g$ is acceleration due to gravity. - Newton's second law for each mass. - Kinematic equation for velocity: $v^2 = u^2 + 2as$, with $u=0$ since starting from rest. 3. **Step 1: Calculate frictional force on the 3.0 kg block** $$f = \mu m_1 g = 0.4 \times 3.0 \times 9.8 = 11.76\,\text{N}$$ 4. **Step 2: Set up equations of motion** Let acceleration be $a$ and tension in the rope be $T$. For the 3.0 kg block (horizontal): $$T - f = m_1 a \implies T - 11.76 = 3.0 a$$ For the 5.0 kg ball (vertical downward): $$m_2 g - T = m_2 a \implies 5.0 \times 9.8 - T = 5.0 a$$ 5. **Step 3: Solve for $a$ and $T$** Add the two equations: $$5.0 \times 9.8 - 11.76 = 3.0 a + 5.0 a = 8.0 a$$ $$49.0 - 11.76 = 8.0 a$$ $$37.24 = 8.0 a$$ $$a = \frac{37.24}{8.0} = 4.655\,\text{m/s}^2$$ 6. **Step 4: Calculate final speed after falling 1.5 m** Using $v^2 = u^2 + 2as$ with $u=0$: $$v = \sqrt{2 \times 4.655 \times 1.5} = \sqrt{13.965} = 3.74\,\text{m/s}$$ **Final answer:** The speed of the 5.0 kg ball after falling 1.5 m is $3.74\,\text{m/s}$.