Problem 89 - Physics for Scientists and Engineers with Modern Physics

Question

A \(65-\mathrm{kg}\) hiker climbs to the top of a 4200 -m-high mountain. The climb is made in \(5.0 \mathrm{~h}\) starting at an elevation of \(2800 \mathrm{~m} .\) Calculate \((a)\) the work done by the hiker against gravity, (b) the average power output in watts and in horsepower, and (c) assuming the body is \(15 \%\) efficient, what rate of energy input was required.

Step-by-Step Solution

Answer

(a) 892,400 J; (b) 49.58 W or 0.066 hp; (c) 330.53 W of energy input.

Step 1: Determine the Change in Elevation

To calculate the work done by the hiker against gravity, we first need to find the change in elevation. The hiker starts at an elevation of 2800 meters and climbs to 4200 meters. Thus, the change in elevation (height) is:\[ \Delta h = 4200 \, \text{m} - 2800 \, \text{m} = 1400 \, \text{m} \]

Step 2: Calculate Work Done Against Gravity

The work done against gravity is given by the formula:\[ W = m \cdot g \cdot \Delta h \]where:* \( m = 65 \, \text{kg} \) is the mass of the hiker,* \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity,* \( \Delta h = 1400 \, \text{m} \) is the change in elevation.Substitute in the values to find:\[ W = 65 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1400 \, \text{m} = 892,400 \, \text{J} \]

Step 3: Calculate Average Power Output in Watts

Power is the work done per unit time. Given the climb is made in 5 hours, convert the time into seconds:\[ t = 5 \, \text{hours} = 5 \times 3600 \, \text{s} = 18,000 \, \text{s} \]The average power output is then:\[ P = \frac{W}{t} = \frac{892,400 \, \text{J}}{18,000 \, \text{s}} = 49.58 \, \text{W} \]

Step 4: Convert Power from Watts to Horsepower

1 horsepower is approximately equal to 746 watts. Therefore, convert the power from watts to horsepower:\[ P_{hp} = \frac{49.58 \, \text{W}}{746 \, \text{W/hp}} \approx 0.066 \, \text{hp} \]

Step 5: Calculate the Energy Input Required

Given that the body is 15% efficient, only 15% of the energy input translates into useful work. Therefore, to find the rate of energy input, use the formula:\[ \text{Energy input rate} = \frac{\text{Useful power output}}{\text{Efficiency}} \]Convert the efficiency to a decimal (15% = 0.15):\[ \text{Energy input rate} = \frac{49.58 \, \text{W}}{0.15} \approx 330.53 \, \text{W} \]

Key Concepts

Work Done Against Gravity

When a hiker climbs a mountain, they are performing work against gravity. This is because they are moving their body against the force that pulls everything downwards – gravity. To calculate this work, we use the formula:
\[ W = m \cdot g \cdot \Delta h \]
where:

• \(W\) is the work done,
• \(m\) is the mass of the hiker,
• \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity, and
• \(\Delta h\) is the change in height (altitude) during the climb.

In essence, this formula helps us understand how much energy the hiker uses to overcome gravity to reach higher elevations. For a hiker with a mass of 65 kg climbing 1400 meters higher, the work done against gravity amounts to 892,400 joules. This is a measure of energy and highlights the physical effort required to make the climb.

Average Power Output

Average power output measures how fast work is being done over a period. It is calculated by dividing the total work done by the time taken. In our scenario, the hiker's climb took 5 hours, converted to seconds as 18,000 seconds, and the work done was 892,400 joules. Thus, the formula used is:
\[ P = \frac{W}{t} \]
where:

• \(P\) is the average power output,
• \(W\) is the work done,
• \(t\) is the time duration in seconds.

Substituting the given values gives us an average power output of approximately 49.58 watts. To provide more context, 1 horsepower is approximately 746 watts, making the power output about 0.066 horsepower. Expressing power in both watts and horsepower helps appreciate the energy expenditure from both a physical and mechanical perspective.

Energy Efficiency Calculations

Energy efficiency is a measure of how effectively energy is converted into work. Human bodies are only partially efficient, meaning not all energy consumed is used productively; the rest is lost as heat, etc. For our hiker exercise, with a body efficiency of 15%, the calculation involves determining the actual energy needed to sustain the climb. This is done by dividing the useful power output by the efficiency.
\[\text{Energy input rate} = \frac{\text{Useful power output}}{\text{Efficiency}}\]where:

• Efficiency is expressed as a decimal (15% = 0.15).
• Useful power output is the average power output calculated before.

The productive work output was calculated as 49.58 watts, therefore the energy input needed was about 330.53 watts. This calculation emphasizes the role of efficiency in human exertion, highlighting how much more energy is required than what is actually used for physical tasks, helping us understand metabolism better and the high demands of climbing in terms of energy.