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Chapter 6: Problem 89

BIO Power of the Human Heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about \(7500 \mathrm{~L}\) of blood. Assume that the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American woman ( \(1.63 \mathrm{~m}\) ). The density (mass per unit volume) of blood is \(1.05 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?

Short Answer

Expert verified
The heart does about 125353.275 joules of work in a day and has a power output of 1.45 watts.

Step by step solution

01

Calculate the mass of blood

Using the formula \(mass = density \times volume\), multiply the given volume of blood with its density to get the mass of blood the heart pumps in a day. That is, \(mass = (1.05 \times 10^{3} \mathrm{~kg/m^{3}}) \times (7500 \mathrm{~L}\) = 7875000 \mathrm{~kg}\). We convert the volume from litres to cubic meters by remembering that 1L = \(0.001 \mathrm{~m^{3}}\), so our final calculated mass is 7875 \mathrm{~kg}.
02

Determine the work done

The work done 'W' is given by the formula \(W = m \times g \times h\), where 'm' is the mass, 'g' is the acceleration due to gravity (approximated as \(9.81 \, \mathrm{m/s^{2}}\)), and 'h' is the height. Substituting our calculated mass value and the given height, the work done by the heart in a day is \(W = 7875 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^{2}} \times 1.63 \, \mathrm{m} = 125353.275 \, \mathrm{J}\) (joules).
03

Calculate the heart's power output

Power is defined as work done per unit time. The heart's power output is determined by dividing the work done by the number of seconds in a day (\(24 \times 60 \times 60 = 86400\) seconds). \(Power = \frac{125353.275 \, \mathrm{J}}{86400 \, \mathrm{s}} = 1.45 \, \mathrm{W}\).

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Key Concepts

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

Human Physiology
The human heart is a fascinating organ in the field of human physiology. As a relentless and efficient pump, it sustains life by circulating around 7500 liters of blood daily, which is roughly equivalent to the volume of a medium-sized swimming pool. This task involves the efficient coordination of muscle contractions known as heartbeats.

Each heartbeat is controlled by the sinoatrial node, often referred to as the heart's natural pacemaker. This electrical impulse initiates the contractions that propel blood through the circulatory system. Given that the heart beats approximately 100,000 times in a day, it performs strenuous and tireless work constantly, demonstrating its critical role in human physiology.
Density of Blood
Understanding the density of blood is crucial when delving into cardiovascular functions or related physics problems. Blood density is a measure of mass per unit volume, typically denoted in units like kg/m鲁. The average density of blood is around 1.05 脳 10鲁 kg/m鲁, which can slightly vary depending on various factors such as the concentration of red blood cells and plasma.

This density is an essential factor in calculating mass, which is used to determine how much work the heart does in moving blood. Blood's density enables the conversion of volume into mass, which is why we multiply the given volume by its density to find out the mass the heart must move.
Physics of the Heart
The heart operates on basic physics principles, notably Newton鈥檚 laws of motion, when moving blood throughout the body. As the heart contracts, it generates a force that propels blood against resistance throughout the vascular system.

The force of heart contractions, translated into pressure within the cardiovascular system, is managed by the heart鈥檚 structure. The heart鈥檚 left ventricle is especially muscular, more so than the right, since it powers blood through the entire systemic circuit. This efficient utilization of energy echoes concepts in basic physics, as the heart maximizes output economic with each contraction.
Work-Energy Principle
The work-energy principle is a core concept used in calculating the work done by the human heart. This principle states that work is equal to the force applied times the distance over which it鈥檚 applied. In terms of the heart, the distance corresponds to the height of an average woman, or 1.63 meters, as per the example.

To find the work done, multiply the mass of blood (7875 kg), gravitational acceleration (9.81 m/s虏), and the given height. This gives the daily work output at 125353.275 Joules. To find the heart鈥檚 power in watts, this total is divided by the number of seconds in a day (86400), yielding 1.45 watts. This showcases how power quantifies energy per unit time, allowing us to appreciate the heart's sustained effort in maintaining circulation.

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Most popular questions from this chapter

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of \(1.80 \times 10^{6} \mathrm{~N}\), one \(14^{\circ}\) west of north and the other \(14^{\circ}\) east of north, as they pull the tanker \(0.75 \mathrm{~km}\) toward the north. What is the total work they do on the supertanker?

Proton Bombardment. A proton with mass \(1.67 \times 10^{-27} \mathrm{~kg}\) is propelled at an initial speed of \(3.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\) directly toward a uranium nucleus \(5.00 \mathrm{~m}\) away. The proton is repelled by the uranium nucleus with a force of magnitude \(F=\alpha / x^{2},\) where \(x\) is the separation between the two objects and \(\alpha=2.12 \times 10^{-26} \mathrm{~N} \cdot \mathrm{m}^{2}\). Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10} \mathrm{~m}\) from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get? (c) What is the speed of the proton when it is again \(5.00 \mathrm{~m}\) away from the uranium nucleus?

A \(5.00 \mathrm{~kg}\) package slides \(2.80 \mathrm{~m}\) down a long ramp that is inclined at \(24.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_{\mathrm{k}}=0.310 .\) Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of \(2.20 \mathrm{~m} / \mathrm{s}\) at the top of the ramp, what is its speed after it has slid \(2.80 \mathrm{~m}\) down the ramp?

A block of ice with mass \(2.00 \mathrm{~kg}\) slides \(1.35 \mathrm{~m}\) down an inclined plane that slopes downward at an angle of \(36.9^{\circ}\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

BIO Animal Energy. Adult cheetahs, the fastest of the great cats, have a mass of about \(70 \mathrm{~kg}\) and have been clocked to run at up to \(72 \mathrm{mi} / \mathrm{h}(32 \mathrm{~m} / \mathrm{s}) .\) (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?
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