1. **Problem 1: Wrecking Ball Energies and Work** Given: - State A: Potential Energy (PE) = 6000 J - State B: Kinetic Energy (KE) = ? - State C: Work done from B to C = ? We assume the lowest point (State B) has zero potential energy, so all PE at A converts to KE at B. **Step 1:** Use conservation of mechanical energy: $$PE_A + KE_A = PE_B + KE_B$$ Since at A, KE_A = 0 (at highest point, momentarily at rest), and at B, PE_B = 0 (lowest point), $$6000 + 0 = 0 + KE_B$$ **Step 2:** Solve for kinetic energy at B: $$KE_B = 6000 \text{ J}$$ **Step 3:** Work done from B to C is the change in mechanical energy from B to C. At C, the ball comes to rest, so KE_C = 0 and PE_C = 0 (assuming resting at lowest point or ground). Work done on the ball from B to C equals the change in kinetic energy: $$W = KE_C - KE_B = 0 - 6000 = -6000 \text{ J}$$ Negative work means energy is removed from the ball (e.g., by the building). 2. **Problem 2: Work, Kinetic Energy, and Velocity of Mug** Given: - Force $F = 4.0$ N - Mass $m = 1.0$ kg - Displacement $d = 1.0$ m - Initial velocity $v_0 = 0$ m/s **Step 1:** Calculate work done by Pete: $$W = F \times d = 4.0 \times 1.0 = 4.0 \text{ J}$$ **Step 2:** Work-energy theorem states work done equals change in kinetic energy: $$W = \Delta KE = KE_{final} - KE_{initial}$$ Since initial KE is zero (at rest), $$KE_{final} = 4.0 \text{ J}$$ **Step 3:** Calculate final velocity using kinetic energy formula: $$KE = \frac{1}{2} m v^2 \Rightarrow v = \sqrt{\frac{2 KE}{m}} = \sqrt{\frac{2 \times 4.0}{1.0}} = \sqrt{8} = 2.83 \text{ m/s}$$ **Final answers:** - Kinetic energy at State B (wrecking ball): $6000$ J - Work done from B to C (wrecking ball): $-6000$ J - Work done on mug: $4.0$ J - Final kinetic energy of mug: $4.0$ J - Final velocity of mug: $2.83$ m/s