/*! 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 50 BIO Should You Walk or Run? It i... [FREE SOLUTION] | 91Ó°ÊÓ

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BIO Should You Walk or Run? It is 5.0 \(\mathrm{km}\) from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 \(\mathrm{km} / \mathrm{h}\) (which uses up energy at the rate of 700 \(\mathrm{W}\) ), or you could walk it leisurely at 3.0 \(\mathrm{km} / \mathrm{h}\) (which uses energy at 290 \(\mathrm{W}\) W). Which choice would burn up more energy, and how much energy (in joules) would it burn? Why is it that the more intense exercise actually burns up less energy than the less intense exercise?

Short Answer

Expert verified
Walking burns 1,740,000 J, more than running (1,260,000 J). Walking takes longer, burning more energy overall.

Step by step solution

01

Determine the time taken for running

First, calculate the time required to cover 5.0 km at a running speed of 10 km/h. Use the formula \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]So, the time for running is:\[ \frac{5.0 \text{ km}}{10 \text{ km/h}} = 0.5 \text{ hours} \]
02

Determine the energy used during running

Now, determine the energy used when running using the power consumption rate and the time calculated. Use the formula for energy \[ \text{Energy} = \text{Power} \times \text{Time} \]Convert the time into seconds for energy calculation. Since 1 hour = 3600 seconds, \[ 0.5 \text{ hours} = 1800 \text{ seconds} \]Thus, the energy used is:\[ 700 \text{ W} \times 1800 \text{ s} = 1,260,000 \text{ J} \]
03

Determine the time taken for walking

Next, calculate the time required to cover 5.0 km at a walking speed of 3.0 km/h. Again use\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]So, the time for walking is:\[ \frac{5.0 \text{ km}}{3.0 \text{ km/h}} = \frac{5}{3} \text{ hours} \approx 1.667 \text{ hours} \]
04

Determine the energy used during walking

Determine the energy used when walking using the power consumption rate and the time calculated. Again, convert time into seconds. \[ 1.667 \text{ hours} = 1.667 \times 3600 \text{ seconds} \approx 6000 \text{ seconds} \] Thus, the energy used is:\[ 290 \text{ W} \times 6000 \text{ s} = 1,740,000 \text{ J} \]
05

Compare the energy for both activities

Compare the energy expenditures calculated for running and walking. - Running energy: 1,260,000 J - Walking energy: 1,740,000 J We can see that walking consumes more energy. Despite its lower intensity, walking takes longer, thus burning more total energy.

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects. It describes how an object moves, but doesn't concern itself with the forces that cause this motion. In the given exercise, kinematics helps us to understand how long it takes to complete certain distances, whether walking or running.
Let's break it down. Kinematics involves essential parameters like distance, speed, and time. These are related through the equation:
  • \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
In the problem of whether to walk or run a 5 km distance, kinematics allows us to calculate the time taken for each activity.
Knowing the time is crucial for further calculations, like determining overall energy expenditure. This fundamental understanding of how motion's basic variables interact is a stepping stone in applications ranging from everyday activities to complex engineering problems.
Power and Energy
Power and energy are key concepts in physics that describe the rate at which work is done or energy is transferred. In simpler terms, power is the energy used per unit time.
The mathematical relationship between energy and power is:
  • \[ \text{Energy} = \text{Power} \times \text{Time} \]
The exercise provides us with the power consumption rates for walking (290 W) and running (700 W). W stands for watts, a unit of power. These rates represent how much energy is used every second during each activity.
For running, despite having a higher power rate, it is performed over less time, resulting in less total energy usage compared to walking. This example illustrates how power alone doesn't dictate energy consumption, as time also plays a critical role. Hence, understanding these relationships can help in efficiently planning physical activities or in designing machines and systems.
Energy Expenditure in Exercise
Energy expenditure refers to the amount of energy a person uses during physical activities. Measured in joules, it reflects our body's energy consumption over time.
In this exercise, energy expenditure differs between walking and running due to changes in both power and time. Walking takes more time to cover the same distance, thereby consuming more total energy despite having a lower power requirement.
To put it simply:
  • Walking for longer means more sustained energy use.
  • Running is quicker, thus less total energy use despite its higher power.
Analyzing energy expenditure helps in optimizing physical fitness plans. It elucidates why lower-intensity exercises over extended periods could consume more energy overall. This insight is essential for anyone aiming to regulate weight or enhance endurance through controlled energy expenditure.

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Most popular questions from this chapter

Use the work-energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0 -m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 \(\mathrm{m}\) into the air. How fast was the boulder moving just as it left the volcano? (c) A skier moving at 5.00 \(\mathrm{m} / \mathrm{s}\) encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. How far does she travel on this patch before stopping? (d) Suppose the rough patch in part (c) was only 2.90 m long? How fast would the skier be moving when she reached the end of the patch? (e) At the base of a frictionless icy hill that rises at \(25.0^{\circ}\) above the horizontal, a toboggan has a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) toward the hill. How high vertically above the base will it go before stopping?

A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be \(10 \mathrm{g},\) and assume the wings move an average downward distance of 1.0 \(\mathrm{cm}\) during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

Animal Energy. BIO Adult cheetahs, the fastest of the great cats, have a mass of about 70 \(\mathrm{kg}\) and have been clocked running at up to 72 \(\mathrm{mph}(32 \mathrm{m} / \mathrm{s})\) . (a) How many joules of kinetic energy does such a swift cheetah have? (b) By what factor would its kinetic energy change if its speed were doubled?

Some Typical Kinetic Energies. (a) In the Bohr model of the atom, the ground- state electron in hydrogen has an orbital speed of 2190 \(\mathrm{km} / \mathrm{s} .\) What is its kinetic energy? (Consult Appendix F.) (b) If you drop a 1.0-kg weight (about 2 lb) from a height of 1.0 \(\mathrm{m}\) , how many joules of kinetic energy will it have when it reaches the ground? (c) Is it reasonable that a \(30-\mathrm{kg}\) child could run fast enough to have 100 \(\mathrm{J}\) of kinetic energy?

The spring of a spring gun has force constant \(k=400 \mathrm{N} / \mathrm{m}\) and negligible mass. The spring is compressed \(6.00 \mathrm{cm},\) and a ball with mass 0.0300 \(\mathrm{kg}\) is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 \(\mathrm{cm}\) long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 \(\mathrm{N}\) acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel.)

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