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Chapter 12: Problem 53

One ounce of a well-known breakfast cereal contains 110 Calories (1 food Calorie \(=4186 \mathrm{J}\) ). If \(2.0 \%\) of this energy could be converted by a weight lifter's body into work done in lifting a barbell, what is the heaviest barbell that could be lifted a distance of \(2.1 \mathrm{m} ?\)

Short Answer

Expert verified
Approximately 447.5 kilograms.

Step by step solution

01

Convert Calories to Joules

First, convert the energy content of the cereal from Calories to Joules. Since 1 food Calorie equals 4186 Joules, the energy in one ounce of cereal is calculated as follows: \[ 110 \text{ Calories} \times 4186 \frac{\text{J}}{\text{Calorie}} = 460460 \text{ J} \]
02

Calculate Useful Energy

Only 2% of the total energy is used for work. To find the useful energy available for lifting the barbell, multiply the total energy by 0.02:\[ 460460 \text{ J} \times 0.02 = 9209.2 \text{ J} \]
03

Use Work Formula

The work done in lifting the barbell is given by the formula: \[ \text{Work} = \text{Force} \times \text{Distance} \]where the force equals the weight of the barbell (mass \(m\) times gravity \(g\)), and the distance lifted is 2.1 meters. The work done is the useful energy calculated in Step 2, so:\[ 9209.2 = m \times 9.8 \times 2.1 \]
04

Solve for Mass

Rearrange the equation to solve for the mass of the barbell:\[ m = \frac{9209.2}{9.8 \times 2.1} \]\[ m = \frac{9209.2}{20.58} \approx 447.5 \text{ kg} \]
05

Present the Final Answer

The heaviest barbell that the weight lifter could lift using the energy from 2% of the cereal is approximately 447.5 kilograms.

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Key Concepts

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

Calories to Joules Conversion
In the world of physics, understanding how to convert energy measures into usable units is essential. When dealing with nutritional energy, such as food calories, we often need to convert them into joules for practical calculations. The standard conversion is that one food Calorie equals 4186 joules. This means that the energy we derive from food can be converted into mechanical energy, such as in the process of lifting weights or doing other physical work. To express this conversion in a mathematical equation, we use:

\[ ext{Energy in Joules} = ext{Energy in Calories} imes 4186 \]

This conversion helps us to quantify the energy intake from food in terms of the energy that can be used to perform mechanical work.
Work and Energy
Work and energy are closely linked in physics. Work is done when a force causes an object to move, and it is calculated as the product of force and distance. In mathematical terms, the work is expressed as:

\[ ext{Work} = ext{Force} imes ext{Distance} \]

The force here is usually the weight of the object, calculated by multiplying its mass by the acceleration due to gravity (approximately 9.8 m/s虏 on Earth).

In the exercise, only a small part of the energy consumed (2%) is actually translated into the kind of work that helps lift a barbell. This concept clarifies how not all energy intake translates directly into physical work because the body uses a significant portion to sustain vital processes and functions. The remaining portion is all that can be used for tasks requiring exertion, like weight lifting.
Physics Problems
Physics problems often require applying simple formulas and concepts to arrive at a solution. In the given problem, the objective was to determine how heavy a barbell a person could lift using a limited amount of energy. This type of problem requires a step-by-step approach, ensuring quantities are in the correct units and understanding which physics principles apply.

The main steps involve:
  • Converting energy intake from Calories to Joules, as our calculations need energy in joules.
  • Finding out the portion of useful energy for work
  • Applying the work formula (\[ ext{Work} = ext{Force} imes ext{Distance} \])
  • Solving the equation for the unknown, such as the mass of the barbell.
Understanding these parts and how they fit together helps in solving complex physics problems with a methodical and logical approach.

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Most popular questions from this chapter

An unknown material has a normal melting/freezing point of \(-25.0^{\circ} \mathrm{C},\) and the liquid phase has a specific heat capacity of \(160 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) One-tenth of a kilogram of the solid at \(-25.0{ }^{\circ} \mathrm{C}\) is put into a \(0.150-\mathrm{kg}\) aluminum calorimeter cup that contains \(0.100 \mathrm{kg}\) of glycerin. The temperature of the cup and the glycerin is initially \(27.0^{\circ} \mathrm{C} .\) All the unknown material melts, and the final temperature at equilibrium is \(20.0^{\circ} \mathrm{C} .\) The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, \(0.6 \mathrm{kg}\) of blood flows to the body's surface and releases \(2000 \mathrm{J}\) of energy. The blood arriving at the surface has the temperature of the body's interior, \(37.0^{\circ} \mathrm{C} .\) Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.

A thermos contains \(150 \mathrm{cm}^{3}\) of coffee at \(85^{\circ} \mathrm{C}\). To cool the coffee, you drop two \(11-\mathrm{g}\) ice cubes into the thermos. The ice cubes are initially at \(0^{\circ} \mathrm{C}\) and melt completely. What is the final temperature of the coffee? Treat the coffee as if it were water.

On the moon the surface temperature ranges from \(375 \mathrm{K}\) during the day to \(1.00 \times 10^{2} \mathrm{K}\) at night. What are these temperatures on the (a) Celsius and (b) Fahrenheit scales?

A thick, vertical iron pipe has an inner diameter of \(0.065 \mathrm{m}\). A thin aluminum disk, heated to a temperature of \(85^{\circ} \mathrm{C},\) has a diameter that is \(3.9 \times 10^{-5} \mathrm{m}\) greater than the pipe's inner diameter. The disk is laid on top of the open upper end of the pipe, perfectly centered on it, and allowed to cool. What is the temperature of the aluminum disk when the disk falls into the pipe? Ignore the temperature change of the pipe.
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