/*! 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 34 BIO An exercise program. A \(75 ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 34

BIO An exercise program. A \(75 \mathrm{~kg}\) person is put on an exercise program by a physical therapist, the goal being to burn up 500 food calories in each daily session. Recall that human muscles are about \(20 \%\) efficient in converting the energy they use up into mechanical energy. The exercise program consists of a set of consecutive high jumps, each one \(50 \mathrm{~cm}\) into the air (which is pretty good for a human) and lasting \(2.0 \mathrm{~s},\) on the average. How many jumps should the person do per session, and how much time should be set aside for each session? Do you think that this is a physically reasonable exercise session?

Short Answer

Expert verified
An impractical 28,467 jumps are needed, taking about 15.8 hours per session; not feasible.

Step by step solution

01

Calculate Total Energy Requirement

To burn 500 food calories, which equals 500 kilocalories (kcal), we must consider the inefficiency of human muscles. Since human muscles are only 20% efficient, the total energy expenditure is calculated as follows: \[ \text{Total Energy Required} = \frac{500}{0.2} = 2500 \text{ kcal} \]
02

Convert Kilocalories to Joules

To convert the energy from kilocalories into joules, use the conversion factor 1 kcal = 4184 J. Therefore, \[ 2500 \text{ kcal} = 2500 \times 4184 \text{ J} = 10,460,000 \text{ J} \]
03

Energy Per Jump

The energy required for each jump is equivalent to the gravitational potential energy change, calculated by:\[ E_{\text{jump}} = mgh \] where \( m = 75 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h = 0.5 \text{ m} \). Therefore, \[ E_{\text{jump}} = 75 \times 9.8 \times 0.5 = 367.5 \text{ J} \]
04

Calculate Number of Jumps

To find the number of jumps required, divide the total energy expenditure by the energy per jump:\[ \text{Number of Jumps} = \frac{10,460,000}{367.5} \approx 28,467 \text{ jumps} \]
05

Calculate Time for Session

Each jump lasts 2 seconds. Therefore, the time for the entire session is:\[ \text{Time Required} = 28,467 \times 2 \text{ s} \approx 56,934 \text{ s} \] Convert this into hours:\[ \text{Time in Hours} = \frac{56,934}{3600} \approx 15.8 \text{ hours} \]
06

Assess Reasonableness of the Session

Completing approximately 28,467 jumps in one session is extremely unreasonable for a daily exercise routine, as it requires about 15.8 hours. This amount of exercise is impractical.

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Key Concepts

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

Energy Conversion in Muscles
Muscles are essential for movement and exercise. They convert chemical energy stored in the body into mechanical energy, which powers our actions. However, human muscles have an efficiency rate of about 20% in these conversions. This means only a small part of the total energy used goes into performing useful work, like lifting weights or jumping.
  • The remaining 80% of the energy is unfortunately lost, mostly as heat in the body.
  • This inefficiency must be considered when determining the caloric burn and energy expenditure of an activity.
For instance, when someone aims to burn 500 calories (kcal) through exercise, the actual energy expenditure becomes much larger. With just 20% efficiency, the person would need to use up 2,500 kcal in total energy to achieve a net mechanical work equivalent to burning 500 kcal.
Gravitational Potential Energy
Gravitational potential energy is a type of energy that an object possesses due to its height above the ground. Every time you jump, your muscles do work against gravity, increasing your gravitational potential energy.
  • It can be calculated with the formula: \( E_{\text{GRAV}} = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
  • In the context of our exercise problem, a person weighing 75 kg jumping 0.5 meters high gains a gravitational potential energy of approximately 367.5 joules (J) per jump.
This formula is crucial in quantifying the energy change during each jump, assisting us in estimating the total energy expenditure of the exercise session.
Exercise Efficiency
Exercise efficiency is a term used to describe how effectively the body converts energy into work during physical activities. As discussed earlier, due to physiological constraints, only about 20% of the energy burned is converted into useful mechanical work.
  • When planning exercise regimens, understanding efficiency helps set realistic expectations about energy expenditure and potential weight loss.
  • This efficiency factor significantly influences how much energy needs to be consumed relative to the mechanical output, often requiring more calories to be burned than expected for weight loss purposes.
Improved understanding of exercise efficiency could lead to better tailored exercise programs, considering factors such as individual's metabolism and specific fitness goals.
Physical Feasibility of Exercise
When creating an exercise program, it is essential to consider the physical feasibility of the exercises involved. Simply put, can the exercise realistically be completed within the set time frame and with given physical ability levels?
  • In our example, performing 28,467 consecutive jumps, each 2 seconds long, is not feasible daily, as it requires over 15 hours of continuous exercise.
  • Feasibility includes considering the realistic amounts of time and energy a typical person can commit to workouts.
Therefore, exercise programs should always aim to balance intensity and duration with the participant's physical limits to make them sustainable.
Physics of High Jumps
The physics behind high jumps involves converting kinetic energy, generated as you push off the ground, into gravitational potential energy. This energy conversion dictates how high you can jump and impacts the overall effectiveness of the exercise session.
  • The jumping height directly ties into the energy required for each leap, calculated through the gravitational potential energy formula.
  • Thorough understanding of these principles helps in accurately estimating energy expenditures of different physical activities and forming effective exercise strategies.
For effective training, variations in jump technique and the integration of other activities could potentially elevate training effectiveness while also preventing monotony and promoting overall physical health.

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Most popular questions from this chapter

While a roofer is working on a roof that slants at \(36^{\circ}\) above the horizontal, he accidentally nudges his \(85.0 \mathrm{~N}\) toolbox, causing it to start sliding downward, starting from rest. If it starts \(4.25 \mathrm{~m}\) from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is \(22.0 \mathrm{~N} ?\)

DATA A physics student measures the energy stored in a spring as a function of the distance it is stretched beyond its undistorted length. Her data are given in the table. $$ \begin{array}{cc} x(\mathrm{~cm}) & \text { Energy (J) } \\ \hline 2.6 & 0.34 \\ 6.3 & 2.00 \\ 7.5 & 2.81 \\ 8.2 & 3.36 \end{array} $$ Draw a linearized graph of the data by plotting the spring's energy as a function of the square of the distance it is stretched. Using a linear "best fit" to the data, determine the force constant of the spring.

A tow truck pulls a car \(5.00 \mathrm{~km}\) along a horizontal roadway using a cable having a tension of \(850 \mathrm{~N}\). (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

A rope is tied to a box and used to pull the box \(1.5 \mathrm{~m}\) along a h zontal floor. The rope makes an angle of \(30^{\circ}\) with the horizontal has a tension of \(5 \mathrm{~N}\). The opposing friction force between the box the floor is \(1 \mathrm{~N}\). How much work does each of the following forces do on the box: (a) gravity, (b) the tension in the rope, (c) friction, and (d) the normal force? What is the total work done on the box?

Automotive power. A truck engine transmits \(28.0 \mathrm{~kW}(37.5 \mathrm{hp})\) to the driving wheels when the truck is traveling at a constant velocity of magnitude \(60.0 \mathrm{~km} / \mathrm{h}(37.7 \mathrm{mi} / \mathrm{h})\) on a level road. (a) What is the resisting force acting on the truck? (b) Assume that \(65 \%\) of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at \(30.0 \mathrm{~km} / \mathrm{h}\) ? At \(120.0 \mathrm{~km} / \mathrm{h} ?\) Give your answers in kilowatts and in horsepower.
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