1. **Problem 1: Speed of 12.0 kg bucket striking the floor** We have two buckets connected by a rope over a pulley. The 12.0 kg bucket is initially 2.00 m above the floor and is released from rest. We want to find the speed of this bucket when it hits the floor using conservation of energy. 2. **Relevant formula:** The principle of conservation of mechanical energy states: $$K_i + U_i = K_f + U_f$$ where $K$ is kinetic energy and $U$ is gravitational potential energy. Since the system starts from rest, initial kinetic energy $K_i = 0$. 3. **Set zero potential energy at the floor:** Initial potential energy of 12.0 kg bucket: $$U_i = mgh = 12.0 \times 9.8 \times 2.00 = 235.2\, \text{J}$$ Final potential energy at floor: $$U_f = 0$$ 4. **Energy conservation:** Initial total energy = Final total energy $$U_i = K_f$$ Kinetic energy when bucket hits floor: $$K_f = \frac{1}{2}mv^2$$ 5. **Solve for speed $v$:** $$12.0 \times 9.8 \times 2.00 = \frac{1}{2} \times 12.0 \times v^2$$ Simplify: $$235.2 = 6.0 v^2$$ $$v^2 = \frac{235.2}{6.0} = 39.2$$ $$v = \sqrt{39.2} \approx 6.26\, \text{m/s}$$ --- 6. **Problem 2: Two blocks connected over a pulley (8.00 kg and 6.00 kg)** (a) For a 0.200 m downward displacement of the 8.00 kg block, find the change in gravitational potential energy for each block. (b) Find the work done by the rope tension $T$ on each block. (c) Use conservation of energy to find the speed of the 8.00 kg block after descending 0.200 m. 7. **(a) Change in gravitational potential energy ($\Delta U$):** For 8.00 kg block moving down 0.200 m: $$\Delta U_8 = mgh = 8.00 \times 9.8 \times (-0.200) = -15.68\, \text{J}$$ For 6.00 kg block moving up 0.200 m: $$\Delta U_6 = 6.00 \times 9.8 \times 0.200 = 11.76\, \text{J}$$ 8. **(b) Work done by tension $T$ on each block:** Work done by tension is force times displacement in direction of force. For 8.00 kg block moving down, tension acts upward, so work done by tension: $$W_8 = -T \times 0.200$$ For 6.00 kg block moving up, tension acts upward, so work done by tension: $$W_6 = T \times 0.200$$ 9. **(c) Conservation of energy for the system:** Total work done by tension on system: $$W_{total} = W_8 + W_6 = -T \times 0.200 + T \times 0.200 = 0$$ Change in gravitational potential energy of system: $$\Delta U = \Delta U_8 + \Delta U_6 = -15.68 + 11.76 = -3.92\, \text{J}$$ This loss in potential energy converts to kinetic energy. 10. **Find speed $v$ of 8.00 kg block after descending 0.200 m:** Total kinetic energy gained by both blocks: $$K = \frac{1}{2} m_8 v^2 + \frac{1}{2} m_6 v^2 = \frac{1}{2} (8.00 + 6.00) v^2 = 7 v^2$$ Set kinetic energy equal to loss in potential energy: $$7 v^2 = 3.92$$ $$v^2 = \frac{3.92}{7} = 0.56$$ $$v = \sqrt{0.56} \approx 0.75\, \text{m/s}$$