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Chapter 5: Problem 52

BIO While running, a person dissipates about \(0.60 \mathrm{~J}\) of mechanical energy per step per kilogram of body mass. If a 60 -kg person develops a power of \(70 \mathrm{~W}\) during a race, how fast is the person running? (Assume a running step is \(1.5 \mathrm{~m}\) long.)

Short Answer

Expert verified
The person is running at a speed of approximately \(2.92 \mathrm{~m/s}\).

Step by step solution

01

Calculate the Energy per Step

First we need to find out the total mechanical energy dissipated by the runner per step. It's given that \(0.60 \mathrm{~J}\) of mechanical energy is dissipated per step per kilogram of body mass. So, the energy per step (\(E_{step}\)) can be calculated as follows: \(E_{step} = 0.60 \mathrm{~J/kg} \times 60 \mathrm{~kg} = 36 \mathrm{~J}\).
02

Calculate Work Done

The power (\(P_{race}\)) developed during the race by the person is given as 70 W. Power is defined as the work done per unit time. We know that, energy dissipated in each step is equivalent to the work done in each step. Hence we can find the time taken for each step (\(t_{step}\)) by rearranging the formula for power: \(t_{step} = \frac{Work}{P_{race}}\), substituting the values, we get \(t_{step} = \frac{36 \mathrm{~J}}{70 \mathrm{~W}} = 0.514 \mathrm{~s}\).
03

Calculate Speed

The speed of the runner can be calculated as the distance covered per unit time. In this case, the distance covered is per one running step, which is given as 1.5 meters. Hence, speed (\(v\)) is given by the equation: \(v = \frac{Distance}{Time}\). Substitute the given distance and calculated time to find out the speed: \(v = \frac{1.5 \mathrm{~m}}{0.514 \mathrm{~s}}\), which simplifies to a speed of approximately \(2.92 \mathrm{~m/s}\).

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics that relates the work done on an object to the change in its kinetic energy. When a runner is in motion, their legs do work against the ground to help propel them forward. According to this principle, the work done by the runner's legs as they push off the ground is equal to the change in their kinetic energy.

In the context of our exercise, each running step by a 60-kg person dissipates mechanical energy, specifically, 0.60 Joules (J) per kilogram of body mass per step. Therefore, the total energy (\(E_{step}\)) used in a step is calculated by the amount of work done by the runner for each step, which in this case is 36 J. This translates to the energy being transferred from the runner to the environment (usually in the form of heat or sound), and it's equivalent to the work done to overcome various forces like friction and air resistance while running.
Power and Energy Relationship
Understanding the relationship between power and energy is crucial for solving problems related to the rate of work done, as seen in our running example. Power, measured in Watts (W), is defined as the amount of work done per unit of time. For a runner who develops 70 W of power during a race, this indicates the rate at which the runner is using energy to maintain speed.

To find out how fast energy is used (the power output), we look at the runner's energy dissipation per step and the duration of each step. In this example, the time taken per step (\(t_{step}\)) is derived by rearranging the power formula \(P = \frac{Work}{Time}\) and finding the time. This relation can then link how fast the runner is expending energy to their pace during the race.
Calculating Speed
In physical terms, speed is the distance traveled divided by the time it takes to travel that distance. To calculate the runner's speed, as we did in the exercise, we use the distance of one running step (\(1.5 \text{ meters (m)}\)) and the time spent per step, found from the power and energy relationship.

The resulting calculation gives us the speed of the runner in meters per second (\(m/s\)). This speed helps us understand how quickly the runner is moving over the ground with each step. It's the final piece of the puzzle linking the runner's mechanical energy output to their movement efficiency, indicating how effective the runner's energy expenditure is in terms of distance covered.

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Most popular questions from this chapter

An older-model car accelerates from 0 to speed \(v\) in 10 s. A newer, more powerful sports car of the same mass accelerates from 0 to \(2 v\) in the same time period. Assuming the energy coming from the engine appears only as kinetic energy of the cars, compare the power of the two cars.

A \(3.50-\mathrm{kN}\) piano is lifted by three workers at constant speed to an apartment \(25.0 \mathrm{~m}\) above the street using a pulley system fastened to the roof of the building. Each worker is able to deliver 165 W of power, and the pulley system is \(75.0 \%\) efficient (so that \(25.0 \%\) of the mechanical energy is lost due to friction in the pulley). Neglecting the mass of the pulley, find the time required to lift the piano from the street to the apartment.

Hooke's law describes a certain light spring of unstretched length \(35.0 \mathrm{~cm}\). When one end is attached to the top of a door frame and a \(7.50-\mathrm{kg}\) object is hung from the other end, the length of the spring is \(41.5 \mathrm{~cm}\). (a) Find its spring constant. (b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of \(190 \mathrm{~N}\). Find the length of the spring in this situation.

Tarzan swings on a \(30.0\)-m-long vine initially inclined at an angle of \(37.0^{\circ}\) with the vertical. What is his speed at the bottom of the swing (a) if he starts from rest? (b) If he pushes off with a speed of \(4.00 \mathrm{~m} / \mathrm{s}\) ?

BIO In bicycling for aerobic exercise, a woman wants her heart rate to be between 136 and 166 beats per minute. Assume that her heart rate is directly proportional to her mechanical power output. Ignore all forces on the woman- plus-bicycle system, except for static friction forward on the drive wheel of the bicycle and an air resistance force proportional to the square of the bicycler's speed. When her speed is \(22.0 \mathrm{~km} / \mathrm{h}\), her heart rate is \(90.0\) beats per minute. In what range should her speed be so that her heart rate will be in the range she wants?
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