1. **State the problem:** A 62 kg person is in an elevator moving upward at a constant speed of 4.0 m/s for 5.0 s. We need to find the work done by the normal force on the person. 2. **Relevant formula:** Work done by a force is given by $$W = F \cdot d \cdot \cos(\theta)$$ where $F$ is the magnitude of the force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement directions. 3. **Analyze the forces:** The normal force acts upward, supporting the person. The displacement is also upward since the elevator moves up. 4. **Calculate displacement:** The elevator moves at constant speed $v = 4.0$ m/s for time $t = 5.0$ s, so displacement $$d = v \times t = 4.0 \times 5.0 = 20.0 \text{ m}$$ 5. **Calculate normal force:** Since the elevator moves at constant speed, acceleration is zero, so normal force equals gravitational force: $$F_N = mg = 62 \times 9.8 = 607.6 \text{ N}$$ 6. **Calculate work done:** The angle $\theta = 0^\circ$ because force and displacement are in the same direction. $$W = F_N \times d \times \cos 0^\circ = 607.6 \times 20.0 \times 1 = 12152 \text{ J}$$ **Final answer:** The work done by the normal force on the person is $$\boxed{12152 \text{ J}}$$.