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Chapter 3: Problem 14

A person's metabolic processes can usually operate at a power of \(6 \mathrm{W} / \mathrm{kg}\) of body mass for several hours at a time. If a \(60-\mathrm{kg}\) woman carrying a \(12-\mathrm{kg}\) pack is walking uphill with an energy- conversion efficiency of 20 percent, at what rate, in meters/hour, does her altitude increase?

Short Answer

Expert verified
Her altitude increases at a rate of 441 m/hr.

Step by step solution

01

- Identify Total Mass

Calculate the total mass the woman is carrying. The woman's mass is 60 kg, and she is carrying a 12 kg pack. Therefore, total mass = 60 kg + 12 kg = 72 kg.
02

- Calculate Metabolic Power

Metabolic power per kg of body mass = 6 W/kg. Therefore, total metabolic power = 6 W/kg * 72 kg = 432 W.
03

- Account for Energy-Conversion Efficiency

The energy-conversion efficiency is 20 percent, which means only 20 percent (0.2) of the metabolic power is used for climbing. Power used for climbing = 0.2 * 432 W = 86.4 W.
04

- Relate Power to Gravitational Potential Energy

Power used to climb at rate of altitude increase (denoted as rate) can be expressed as Power = rate * total mass * gravitational acceleration (g), where g = 9.8 m/s^2. Therefore, 86.4 W = rate * 72 kg * 9.8 m/s^2.
05

- Solve for Rate

Solve for the rate: rate = \( \frac{86.4 \text{ W}}{72 \text{ kg} * 9.8 \text{ m/s}^2} \) = 0.1225 m/s.
06

- Convert to Meters per Hour

Convert the rate from meters per second (m/s) to meters per hour (m/hr): rate = 0.1225 m/s * 3600 s/hr = 441 m/hr.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Conversion Efficiency
When someone talks about energy conversion efficiency, they are referring to how well energy from one form is turned into another. In our problem, the human body converts the energy from food into physical activity. Here, only 20% of the woman's metabolic energy is used for climbing. The rest is lost as heat and other bodily functions. Think of it like a car: if a car has an efficiency of 20%, only 20% of the fuel's energy propels the car forward, while the rest is wasted as heat and friction. So, for every 5 units of energy, only 1 unit is effectively used to lift the woman's body uphill.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy something has because of its height above the ground. It depends on three factors: mass, height, and gravity. The formula for GPE is GPE = mass * gravitational acceleration * height. In our case, as the woman climbs uphill, she gains gravitational potential energy because she is increasing her height. Gravity is always 9.8 m/s虏 on the surface of the earth. This means heavier objects or greater heights result in more GPE. So, as she climbs higher, her energy spent (if efficiency were perfect) would be converted into this type of energy.
Rate of Altitude Increase
The rate of altitude increase tells us how quickly height changes over time. In physics problems like this, it's often given in units like meters per second (m/s) or meters per hour (m/hr). To find this rate, we need to know how much power is dedicated to climbing (after accounting for efficiency) and then use this power to calculate the rate at which altitude changes. By relating power, mass, gravitational acceleration, and this rate, we can solve for how fast the woman is climbing. After calculation, we find that the woman climbs at a rate of 441 meters per hour when we account for all factors.
Metabolic Processes in Humans
Human metabolic processes convert food into energy, which the body uses to perform various activities. The amount of power our bodies can sustain over time varies, but in this problem, we've assumed a rate of 6 W/kg. This means each kilogram of body mass can generate 6 watts of power. For someone weighing 72 kg (including their pack), that totals up to 432 watts. However, not all this power is used for movement; only part of it (20% in our case) is used for climbing. During physical activities like climbing, the conversion of energy is crucial to understand how effectively we use our stored energy from food.
Physics Problem-Solving
Solving physics problems often involves breaking down a complex scenario into simpler parts. Here, we start by identifying the total mass and then calculating the total metabolic power. Next, we account for energy-conversion efficiency to find out how much power is actually used for climbing. By relating this power to gravitational potential energy, we can solve for the rate of altitude increase. It involves understanding formulas and how different concepts like energy, mass, and height interact. This structured approach helps to systematically tackle the problem and find the solution step-by-step.

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Most popular questions from this chapter

In an effort to lose weight, a person runs \(5 \mathrm{km}\) per day at a speed of \(4 \mathrm{m} / \mathrm{s}\). While running, the person's body processes consume energy at a rate of \(1.4 \mathrm{kW}\). Fat has an energy content of about \(40 \mathrm{kJ} / \mathrm{g}\) How many grams of fat are metabolized during each run?

Two identical balls move down a tilted board. Ball A slides down without friction and ball B rolls down. Which ball reaches the bottom first? Why?

A total of \(490 \mathrm{J}\) of work is needed to lift a body of unknown mass through a height of \(10 \mathrm{m}\). What is its mass?

One kilogram of water at \(0^{\circ} \mathrm{C}\) contains 335 kJ of energy more than 1 kg of ice at \(0^{\circ} \mathrm{C}\). What is the mass equivalent of this amount of energy?

An \(800-\mathrm{kg}\) car coasts down a hill \(40 \mathrm{m}\) high with its engine off and the driver's foot pressing on the brake pedal. At the top of the hill the car's speed is \(6 \mathrm{m} / \mathrm{s}\) and at the bottom it is \(20 \mathrm{m} / \mathrm{s}\). How much energy was converted into heat on the way down?
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