Average power in vertical climbing (work–energy over time): A 50 kg student climbs a vertical rope of 8 m in 10 s (neglect air resistance). Estimate the average power developed (use g ≈ 9.81 m/s^2).

Introduction / Context:
Average power in mechanics equals total work done divided by time. When lifting a body vertically at roughly constant speed, work equals the increase in gravitational potential energy. This calculation is common in biomechanics, sports science, and machine sizing problems for hoists and winches.

Given Data / Assumptions:

• Mass m = 50 kg; height h = 8 m; time t = 10 s.
• Acceleration due to gravity g ≈ 9.81 m/s^2.
• Assume steady climbing (kinetic energy change negligible).

Concept / Approach:
Total mechanical work against gravity is W = m * g * h. Average power P_avg = W / t. Units: joules for work (N*m) and watts (J/s) for power. Minor numerical rounding is acceptable for practical answers (e.g., g = 9.8 m/s^2).

Step-by-Step Solution:

Verification / Alternative check:
If g is rounded to 10 m/s^2, W = 50 * 10 * 8 = 4000 J and P = 400 W—same rounded result.

Why Other Options Are Wrong:

• ≈ 500 W: requires higher speed or mass; overestimates work done.
• ≈ 4000 W: off by a factor of 10 (confuses work in joules with power).
• “None”: incorrect because 400 W is a reasonable estimate.
• ≈ 250 W: underestimates required power.

Common Pitfalls:
Mixing joules and watts; ignoring time; using weight in kgf without consistent conversion.

Final Answer:
≈ 400 W