Problem 89 - Physics for Scientists and Engineers with Modern Physics

Question

A 65 -kg hiker climbs to the top of a 4200 -m-high mountain. The climb is made in 5.0 h starting at an elevation of 2800 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

The hiker does 892,290 J of work, with an average power output of 49.57 W (0.0665 hp), and needs 330.48 W of energy input.

Step 1: Calculate the Change in Elevation

To find the work done by the hiker against gravity, we need to determine the change in elevation. Subtract the starting elevation from the final elevation: \[ \Delta h = 4200\, \text{m} - 2800\, \text{m} = 1400\, \text{m} \]

Step 2: Compute the Gravitational Work Done

The work done against gravity is calculated using the formula:\[ W = m \cdot g \cdot \Delta h \]where:\( m = 65\, \text{kg} \) is the mass of the hiker,\( g = 9.81\, \text{m/s}^2 \) is the acceleration due to gravity, and \( \Delta h = 1400\, \text{m} \) is the change in elevation.Substitute these values into the formula:\[ W = 65 \cdot 9.81 \cdot 1400 = 892290\, \text{Joules} \]

Step 3: Calculate Average Power Output in Watts

Power is the rate of work done over time, given by:\[ P = \frac{W}{t} \]where:\( W = 892290\, \text{J} \) is the work done,\( t = 5\, \text{hours} = 5 \times 3600\, \text{seconds} = 18000\, \text{s} \).Substitute these to find power:\[ P = \frac{892290}{18000} \approx 49.57\, \text{Watts} \]

Step 4: Convert Power to Horsepower

To convert watts to horsepower, use the conversion factor: 1 horsepower \( = 746\, \text{Watts} \).So, the power in horsepower is:\[ P_{hp} = \frac{49.57}{746} \approx 0.0665 \text{Horsepower} \]

Step 5: Calculate Rate of Energy Input

Assuming the body is only 15\% efficient, the total energy input is higher than the work done:\[ \text{Total Energy Input} = \frac{W}{\text{Efficiency}} = \frac{892290}{0.15} \approx 5948600\, \text{J} \]The rate of energy input is:\[ P_{\text{input}} = \frac{5948600}{18000} \approx 330.48\, \text{Watts} \]

Key Concepts

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. It depends on the height of the object above a reference point, usually the ground, and the object's mass. When a hiker climbs a mountain, they increase their gravitational potential energy because they are moving higher above the Earth's surface.

To calculate this change in energy, we use the formula:

• Work done (\( W \)) = mass (\( m \)) × gravitational acceleration (\( g \)) × change in height (\( \Delta h \))

In the example, the hiker with mass 65 kg climbs a mountain from 2800 m to 4200 m. The change in height is 1400 m. Therefore, we calculate:

• \( W = 65 \times 9.81 \times 1400 = 892,290 \text{ Joules} \)

This result shows the work done against gravity, indicating the amount of energy converted to gravitational potential energy.

Power Output Calculation

Power is defined as the rate of doing work or transferring energy. It's important to know how much energy is used per unit of time, especially in activities like hiking where stamina and resource management are key. The formula for power is:

• \( P = \frac{W}{t} \)

Where:

• \( W \) is the work done (892,290 Joules in our example)
• \( t \) is the time taken (5 hours or 18,000 seconds)

Using this formula, the average power output of the hiker is:

• \( P = \frac{892,290}{18,000} \approx 49.57 \text{ Watts} \)

To convert this to horsepower, we use the conversion factor (1 horsepower = 746 Watts):

• \( P_{hp} = \frac{49.57}{746} \approx 0.0665 \text{ Horsepower} \)

This demonstrates that the hiker's average power output during the climb is rather modest.

Energy Efficiency in Humans

Energy efficiency is a measure of how effectively energy is used to perform work. In humans, not all energy input from food is converted to mechanical work; some is lost as heat. Typically, the human body's efficiency is only about 15-25%.

In this exercise, we assume a 15% efficiency rate. This means that only 15% of the total energy we consume is used for mechanical work, such as hiking. To find the total energy input required, we divide the work done by the efficiency:

• Total Energy Input = \( \frac{W}{\text{Efficiency}} = \frac{892,290}{0.15} \approx 5,948,600 \text{ Joules} \)

The rate of energy input required is then calculated by dividing this total energy by time:

• \( P_{\text{input}} = \frac{5,948,600}{18,000} \approx 330.48 \text{ Watts} \)

This calculation shows that the hiker must metabolize food energy at a much higher rate than the work being done on the mountain.