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The \(food \, calorie\), equal to 4186 J, is a measure of how much energy is released when the body metabolizes food. A certain fruitandcereal bar contains 140 food calories. (a) If a 65kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical, only 20% of the food calories go into mechanical energy, what would be the answer to part (a)? (\(Note\): In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.)

Short Answer

Expert verified
(a) 919.6 m, (b) 183.9 m.

Step by step solution

01

Convert Food Calories to Joules

First, we need to convert the energy content of the fruit and cereal bar from food calories to joules. The bar contains 140 food calories. Given that 1 food calorie is equivalent to 4186 joules, the total energy in joules is calculated as follows:\[ 140 \, \text{food calories} \times 4186 \, \text{J/food calorie} = 586040 \, \text{J} \]
02

Calculate Height for 100% Mechanical Efficiency

Assuming that all the consumed energy is converted into gravitational potential energy, we use the formula for gravitational potential energy \(E_p = mgh\), where \(m\) is the mass of the hiker (65 kg), \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)), and \(h\) is the height:\[ E_p = 65 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times h = 586040 \, \text{J} \]Solving for \(h\), we get:\[ h = \frac{586040 \, \text{J}}{65 \, \text{kg} \times 9.81 \, \text{m/s}^2} = 919.6 \, \text{m} \]
03

Calculate Height for 20% Mechanical Efficiency

If only 20% of the energy consumed is used for mechanical work, we need to calculate the effective energy for potential work:\[ \text{Effective Energy} = 0.20 \times 586040 \, \text{J} = 117208 \, \text{J} \]Using this energy in the potential energy formula with 20% efficiency:\[ 65 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times h = 117208 \, \text{J} \]Solving for \(h\), we get:\[ h = \frac{117208 \, \text{J}}{65 \, \text{kg} \times 9.81 \, \text{m/s}^2} = 183.9 \, \text{m} \]

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy an object gains because of its position relative to the Earth, typically its height above the ground. It's a key player in many physical phenomena and is particularly relevant when considering how energy is converted from one form to another.To calculate gravitational potential energy (E_p), you can use the formula:\[ E_p = mgh \]where:
  • \( m \) is the mass of the object in kilograms.
  • \( g \) is the acceleration due to gravity, usually 9.81 \( \, \text{m/s}^2 \).
  • \( h \) is the height in meters.
This formula tells us how much energy is stored due to the object's elevation. In the context of the exercise, when a hiker climbs, their body's energy is transformed into gravitational potential energy, raising them higher against Earth's pull. This concept demonstrates energy conversion, as chemical energy from food is converted to mechanical work, moving the hiker upwards.
Mechanical Efficiency
Mechanical efficiency refers to how effectively a mechanism or body can convert energy from one form to another, specifically from chemical to mechanical energy in this context.Not all the energy consumed by our bodies is used for physical work. In many cases, like with our hiker, only a fraction of the energy from food is transformed into mechanical work. The concept of mechanical efficiency highlights this by considering energy losses such as those due to metabolism or other bodily processes.To calculate the useful energy from total energy, mechanical efficiency is applied:\[ \text{Efficiency} = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \]In the exercise, for a typical 20% mechanical efficiency, only 20% of the energy from the food is used to actually move the hiker up the mountain. This means if the hiker consumes energy equivalent to 586040 Joules from a cereal bar, only 117208 Joules are available to overcome Earth's gravitational force. This highlights the importance of recognizing that energy transformations are not always 100% efficient.
Calorimetry
Calorimetry is the science of measuring the heat of chemical reactions or physical changes. In nutrition, it's used to quantify the energy content of food, typically expressed in calories. A food calorie (or kilocalorie) is equivalent to 4186 Joules, which is crucial for understanding how much work the body can perform by metabolizing it. When we consume food, the body converts chemical energy into various types of energy needed for bodily functions, like maintaining body temperature and allowing muscle activity. In the given problem, calorimetry is involved when determining how many calories from a cereal bar can be used as energy for an activity. The specific task asks how high one could climb a mountain if all the calorimeter-measured energy went into increasing gravitational potential energy, emphasizing the connection between energy consumed and physical work done. Understanding calorimetry helps us appreciate energy expenditures in physical activities, leading to better insights into dietary needs and energy budgeting in everyday life.

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Most popular questions from this chapter

A 0.150-kg block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 m above the floor. The spring has force constant 1900 N/m and is initially compressed 0.045 m. The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

You are designing an amusement park ride. A cart with two riders moves horizontally with speed \(v = 6.00\) m/s. You assume that the total mass of cart plus riders is 300 kg. The cart hits a light spring that is attached to a wall, momentarily comes to rest as the spring is compressed, and then regains speed as it moves back in the opposite direction. For the ride to be thrilling but safe, the maximum acceleration of the cart during this motion should be 3.00\(g\). Ignore friction. What is (a) the required force constant of the spring, (b) the maximum distance the spring will be compressed?

A conservative force \(\overrightarrow{F}\) is in the \(+x\)-direction and has magnitude \(F(x) = a/(x + x_0)^2\), where \(\alpha = 0.800\) N \(\cdot\) m\(^2\) and \(x_0 = 0.200\) m. (a) What is the potential-energy function \(U(x)\) for this force? Let \(U(x) \rightarrow 0\) as \(x \rightarrow \infty\). (b) An object with mass \(m = 0.500\) kg is released from rest at \(x = 0\) and moves in the \(+x\)-direction. If \(\overrightarrow{F}\) is the only force acting on the object, what is the object's speed when it reaches \(x = 0.400\) m?

Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke's law. In laboratory tests on a particular tendon, it was found that, when a 250-g object was hung from it, the tendon stretched 1.23 cm. (a) Find the force constant of this tendon in N/m. (b) Because of its thickness, the maximum tension this tendon can support without rupturing is 138 N. By how much can the tendon stretch without rupturing, and how much energy is stored in it at that point?

A basket of negligible weight hangs from a vertical spring scale of force constant 1500 N/m. (a) If you suddenly put a 3.0-kg adobe brick in the basket, find the maximum distance that the spring will stretch. (b) If, instead, you release the brick from 1.0 m above the basket, by how much will the spring stretch at its maximum elongation?

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