/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 mmh One ounce of a well-known br... [FREE SOLUTION] | 91影视

91影视

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

Short Answer

Expert verified
The heaviest barbell is approximately 449.5 kg.

Step by step solution

01

Calculate Energy in Joules

First, we need to convert the Calories in the cereal into Joules. We know that 1 Calorie equals 4186 Joules. Thus, \[ E = 110 \times 4186 = 460460 \text{ J} \].
02

Determine Usable Energy for Work

We are given that only 2.0% of this energy is converted into work. Therefore, the usable energy is \( 0.02 \times 460460 = 9209.2 \text{ J} \).
03

Calculate the Work Done

The work done lifting the barbell is given by the formula \( W = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity \( (9.8 \text{ m/s}^2) \), and \( h \) is the height lifted (2.1 m).
04

Set Usable Energy to Work Done Formula

Using the work done formula, set \( W = 9209.2 \text{ J} \). Therefore, \[ 9209.2 = m \times 9.8 \times 2.1 \].
05

Solve for Mass

Rearrange to find \( m \): \[ m = \frac{9209.2}{9.8 \times 2.1} \approx 449.5 \text{ kg} \].

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 91影视!

Key Concepts

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

Calories to Joules
To understand the energy involved in lifting a barbell, we first need to convert Calories, a common food energy unit, into Joules, the standard metric unit for measuring energy.
One food Calorie is equivalent to 4186 Joules. When converting Calories to Joules, simply multiply the number of Calories by 4186.
For instance, if a breakfast cereal contains 110 Calories, the energy content is calculated as follows:
  • Energy in Joules: \( 110 \times 4186 = 460460 \text{ J} \)
Understanding this conversion is essential for calculating the potential energy transformations that occur within the body or during physical activities.
Work Done Formula
The concept of work in physics is tied to energy transfer. Specifically, it is the energy transferred to an object when a force moves it over a distance. The formula for work done is given by:
  • \( W = mgh \)
where:
  • \( W \) is the work done
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity \( (9.8 \text{ m/s}^2) \)
  • \( h \) is the height the object is lifted
Essentially, this formula helps you calculate the amount of energy used (or work done) while lifting an object vertically. By understanding how work is calculated, you can determine energy expenditure in any activity involving vertical movement.
Acceleration Due to Gravity
A key component in the physics of lifting objects is gravity, which exerts a force on objects pulling them towards the earth鈥檚 center. This force is characterized by the acceleration due to gravity, typically denoted as \( g \), and approximated as \( 9.8 \text{ m/s}^2 \).
Gravity鈥檚 constancy means that for each kilogram of mass, an object experiences a force of approximately 9.8 Newtons downward.
This predictable force is crucial in calculating work done during lifting, allowing us to compute the exact force needed to overcome gravitational pull at a given height. Using \( g \) ensures our calculations for work or energy are accurate and consistent across different scenarios.
Energy Efficiency in Biomechanics
When discussing energy efficiency in physical activities, it is vital to understand how efficiently the human body converts consumed energy into work.
In biomechanics, energy efficiency refers to the percentage of energy intake that translates into actual physical work. Due to various physiological processes, the human body is not 100% efficient.
In the provided exercise, only 2% of the consumed energy is used effectively to lift a barbell, which is a typical reflection of energy conversion efficiency in human muscles during physical exertion.
Improving energy efficiency can enhance performance in sports and daily activities, making such calculations valuable for athletes and fitness enthusiasts alike.

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

A spherical brass shell has an interior volume of \(1.60 \times 10^{-3} {m}^{3}\) Within this interior volume is a solid steel ball that has a volume of \(0.70 \times 10^{-3} {m}^{3}\) . The space between the steel ball and the inner surface of the brass shell is filled completely with mercury. A small hole is drilled through the brass, and the temperature of the arrangement is increased by 12 \({C}^{\circ}\). What is the volume of the mercury that spills out of the hole?

An insulated container is partly filled with oil. The lid of the container is removed, 0.125 kg of water heated to \(90.0^{\circ} {C}\) is poured in, and the lid is replaced. As the water and the oil reach equilibrium, the volume of the oil increases by \(1.20 \times 10^{-5} {m}^{3}\) . The density of the oil is 924 \({kg} / {m}^{3}\) , its specific heat capacity is \(1970{J} /({kg} \cdot {C}^{\circ}),\) and its coefficient of volume expansion is \(721 \times 10^{-6}({C}^{0})^{-1}.\) What is the temperature when the oil and the water reach equilibrium?

When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature \((37^{\circ} {C})\) by warming up the water in your stomach. Could drinking ice water, then, substitute for exercise as a way to 鈥渂urn calories?鈥 Suppose you expend 430 kilocalories during a brisk hour-long walk. How many liters of ice water \((0^{\circ} {C})\)) would you have to drink in order to use up 430 kilocalories of metabolic energy? For comparison, the stomach can hold about 1 liter.

ssm On the Rankine temperature scale, which is sometimes used in engineering applications, the ice point is at \(491.67^{\circ}{R}\) and the steam point is at \(671.67^{\circ}{R}\). Determine a relationship (analogous to Equation 12.1) between the Rankine and Fahrenheit temperature scales.

You are sick, and your temperature is 312.0 kelvins. Convert this temperature to the Fahrenheit scale.

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.