Question

The average power generated by a 90 kg mountain climber who climbs a summit of height 600 m in 90 minutes is (Acceleration due to gravity \( g = 10 \) m/s\(^2\)):

• A. \(100 \text{ W} \)
• B. \(25 \text{ W} \)
• C. \(200 \text{ W} \)
• D. \(50 \text{ W} \)

Hint:

Power is the rate of doing work, and for climbing problems, it is calculated using \( P = \frac{mgh}{t} \), where \( t \) must be in seconds.

Answer 1

The Correct Option is A

Step 1: Compute Work Done Against Gravity The work done in climbing the height is equivalent to the gain in potential energy: \[ W = mgh. \] Substituting the values: \[ W = 90 \times 10 \times 600. \] \[ W = 540000 \text{ J}. \]
Step 2: Compute Power Output Power is defined as work done per unit time: \[ P = \frac{W}{t}. \] The climber takes 90 minutes, which is: \[ t = 90 \times 60 = 5400 \text{ s}. \] \[ P = \frac{540000}{5400} = 100 \text{ W}. \] % Final Answer Thus, the correct answer is option (1): \( 100 \) W.

Answer 2

Step 1: Calculate the Work Done Against Gravity The work done by the climber to ascend a height \( h \) is equal to the increase in gravitational potential energy: \[ W = mgh. \] Given mass \( m = 90 \, \text{kg} \), gravitational acceleration \( g = 10 \, \text{m/s}^2 \), and height \( h = 600 \, \text{m} \), substitute these values: \[ W = 90 \times 10 \times 600 = 540000 \text{ J}. \]
Step 2: Determine the Power Output Power is the rate at which work is done, calculated by: \[ P = \frac{W}{t}. \] The time taken is 90 minutes, which converts to seconds as: \[ t = 90 \times 60 = 5400 \text{ s}. \] Therefore, the power output is: \[ P = \frac{540000}{5400} = 100 \text{ W}. \]
Final Answer: Thus, the correct answer is option (1): \[ \boxed{100 \text{ W}}. \]