Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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:\[ h = \frac{117208 \, \text{J}}{65 \, \text{kg} \times 9.81 \, \text{m/s}^2} = 183.9 \, \text{m} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In an experiment, one of the forces exerted on a proton is \(\overrightarrow{F}\) \(= -a x^2 \hat{\imath}\), where \(\alpha = 12 \mathrm{N/m}^2\). (a) How much work does \(\overrightarrow{F}\) do when the proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.40 m)? (b) Along the straight-line path from the point (0.10 m, 0) to the point (0.30 m, 0)? (c) Along the straight-line path from the point (0.30 m, 0) to the point (0.10 m, 0)? (d) Is the force \(\overrightarrow{F}\) conservative? Explain. If \(\overrightarrow{F}\) is conservative, what is the potential-energy function for it? Let \(U =\) 0 when \(x =\) 0.

A 3.00-kg block is connected to two ideal horizontal springs having force constants \(k_1 = 25.0\) N/cm and \(k_2 = 20.0\) N/cm (\(\textbf{Fig. P7.62}\)). The system is initially in equilibrium on a horizontal, frictionless surface. The block is now pushed 15.0 cm to the right and released from rest. (a) What is the maximum speed of the block? Where in the motion does the maximum speed occur? (b) What is the maximum compression of spring 1?

A crate of mass \(M\) starts from rest at the top of a frictionless ramp inclined at an angle \(\alpha\) above the horizontal. Find its speed at the bottom of the ramp, a distance \(d\) from where it started. Do this in two ways: Take the level at which the potential energy is zero to be (a) at the bottom of the ramp with \(y\) positive upward, and (b) at the top of the ramp with y positive upward. (c) Why didn't the normal force enter into your solution?

A 28-kg rock approaches the foot of a hill with a speed of 15 m/s. This hill slopes upward at a constant angle of 40.0\(^\circ\) above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20, respectively. (a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) Will the rock remain at rest at its highest point, or will it slide back down the hill? (c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

A 10.0-kg microwave oven is pushed 6.00 m up the sloping surface of a loading ramp inclined at an angle of 36.9\(^\circ\) above the horizontal, by a constant force \(\overrightarrow{F}\) with a magnitude 110 N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250. (a) What is the work done on the oven by the force \(\overrightarrow{F}\)? (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy. (e) Use \(\sum \overrightarrow{F} = m\overrightarrow{a}\) to calculate the oven's acceleration. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after the oven has traveled 6.00 m. From this, compute the increase in the oven's kinetic energy, and compare it to your answerfor part (d).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free