/*! 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 7 BIO Human Energy vs. Insect Ener... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

BIO Human Energy vs. Insect Energy. For its size, the common flea is one of the most accomplished jumpers in the animal world. A 2.0 -mm-long, \(0.50 \mathrm{mg}\) flea can reach a height of \(20 \mathrm{~cm}\) in a single leap. (a) Ignoring air drag, what is the takeoff speed of such a flea? (b) Calculate the kinetic energy of this flea at takeoff and its kinetic energy per kilogram of mass. (c) If a \(65 \mathrm{~kg}, 2.0-\mathrm{m}\) -tall human could jump to the same height compared with his length as the flea jumps compared with its length, how high could the human jump, and what takeoff speed would the man need? (d) Most humans can jump no more than \(60 \mathrm{~cm}\) from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a \(65 \mathrm{~kg}\) person? (e) Where does the flea store the energy that allows it to make sudden leaps?

Short Answer

Expert verified
1. The takeoff speed of the flea is about \(0.89 \, m/s\). 2: The flea's kinetic energy at takeoff is about \(2 \times 10^{-7} \, J\) and its kinetic energy per kilogram of mass is approximately \(400 \, J/kg\). 3. If a human could jump equivalent to the flea's jumping height compared with its length, he could jump 65 meters high and would need a takeoff speed of approximately \(36 \, m/s\). 4: The kinetic energy at takeoff per kilogram of mass for a human, who can jump no more than 60 cm, is about \(294 \, J/kg\). 5: The flea stores its energy in the protein resillin contained in its legs, which allows it to make sudden leaps.

Step by step solution

01

Calculate the takeoff speed of the flea

We'll start by using the principle of energy conservation. When the flea jumps to its maximum height, all its initial kinetic energy will be transformed into potential energy. Using the formula for potential energy \(PE = m \cdot g \cdot h\) and kinetic energy \(KE = \frac{1}{2} m v^2\), we get \(m \cdot g \cdot h = \frac{1}{2} m \cdot v^2 \). Solve for \(v\), the speed of the flea.
02

Calculate the kinetic energy of the flea at takeoff

Using the speed calculated in the previous step, calculate the kinetic energy of the flea using the formula \(KE = \frac{1}{2} m v^2\).
03

Calculate kinetic energy per kilogram

To calculate the kinetic energy per kilogram, the kinetic energy calculated in the previous step must be divided by the mass of the flea in kilograms.
04

Calculate height and takeoff speed for human

This step requires knowledge of ratios. The human should be able to jump to the same height compared with his length as the flea jumps compared with its length. Using this ratio, calculate the height to which the human can jump. Using the method from step 1, we can then calculate the takeoff speed.
05

Calculate kinetic energy per kilogram for human

Assuming that the human can jump no more than 60 cm from a crouched start, use the formula for kinetic energy and the method from step 3 to calculate kinetic energy per kilogram for the human.
06

Explain the energy storage in flea

This question asks for a knowledge of biology rather than physics or mathematics. However, we can provide an explanation based on the basic physics principle of energy conservation.

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 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy associated with an object in motion. It is given by the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
In the example exercise about the flea, we start by calculating its takeoff speed. This is found by converting all kinetic energy into potential energy at the highest point of the jump.
The kinetic energy of an object, such as the flea at the moment of takeoff, is crucial to understanding how much energy it can direct into its movement.
Understanding how kinetic energy works helps in determining just how much energy peak speed can convert into height for any given mass. Once you know the takeoff speed, you can compute kinetic energy that helps predict how it will move and leap, employing the formula consistently across different scenarios.
Potential Energy
Potential energy is the stored energy of position possessed by an object. For a flea jump, the potential energy can be understood as the energy stored due to its height. The formula used is \( PE = m \cdot g \cdot h \), where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is height.
When talking about the flea's leap, at its peak (20 cm height), potential energy is at its maximum because its kinetic energy has been entirely converted into potential energy.
This concept is not just limited to vertical movements; it is foundational to understanding energy transformations in physics. In daily phenomena, potential energy allows us to explain how energy is stored, waiting for conversion back into kinetic energy for motion or work.
Conservation of Energy
The law of conservation of energy is a critical principle of physics, stating that energy cannot be created or destroyed, only transformed from one form to another.
In our flea exercise, the conversation between kinetic and potential energy exemplifies this law. As the flea leaps upward, its initial kinetic energy converts into potential energy at the jump's apex. The total energy remains constant throughout the jump.
This principle widely applies in physics, from large systems like planets to small organisms like fleas, helping us understand mechanical systems and biological processes alike.
  • It highlights efficiency in energy use across systems.
  • Ensures energy balance in calculations and theoretical predictions.

By following this law, we accurately predict motion dynamics in various contexts.
Energy Storage in Biology
Biological systems, such as fleas, have evolved to efficiently use and store energy for various activities, including remarkable feats like jumping.
The flea, for example, stores energy in resilin, a protein found in its legs. This protein acts like an elastic spring, storing muscular energy until it is released in a powerful leap.
Understanding biological energy storage provides insight into:
  • How organisms convert and store chemical energy from nutrients.
  • How energy storage methods, like spring mechanisms, enhance organism survival.

This biological paradigm showcases nature's ingenuity in using available resources for maximizing physical performance and adaptation.

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 small rock with mass 0.20 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R = 0.50 m (Fig. E7.9). Assume that the size of the rock is small com-pared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl has magnitude 0.22 J. (a) Between points A and B, how much work is done on the rock by (i) the normal force and (ii) gravity? (b) What is the speed of the rock as it reaches point B? (c) Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain. (d) Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl?

A block with mass \(m=\) \(\begin{array}{lll}0.200 \mathrm{~kg} & \text { is placed against a com- }\end{array}\) pressed spring at the bottom of a ramp that is at an angle of \(53.0^{\circ}\) above the horizontal. The spring has \(8.00 \mathrm{~J}\) of elastic potential energy stored in it. The spring is released, and the block moves up the incline. After the block has traveled a distance of \(3.00 \mathrm{~m},\) its speed is \(4.00 \mathrm{~m} / \mathrm{s} .\) What is the magnitude of the friction force that the ramp exerts on the block while the block is moving?

CALC The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=\left(a / r^{12}\right)-\left(b / r^{6}\right),\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r .\) Draw two graphs: one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\). (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the cquilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{~m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{~J}\) per molecule. Find the values of the constants \(a\) and \(b\).

A baseball is thrown from the roof of a 22.0-m-tall building with an initial velocity of magnitude 12.0 m>s and directed at an angle of 53.1° above the horizontal. (a) What is the speed of the ball just be- fore it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of 53.1° below the horizontal? (c) If the effects of air resistance are included, will part (a) or (b) give the higher speed?

BIO Tendons. Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke's law. In laboratory tests on a particular tendon, it was found that, when a \(250 \mathrm{~g}\) object was hung from it, the tendon stretched \(1.23 \mathrm{~cm}\). (a) Find the force constant of this tendon in \(\mathrm{N} / \mathrm{m}\). (b) Because of its thickness, the maximum tension this tendon can support without rupturing is 138 N. By how much can the tendon stretch without rupturing. and how much energy is stored in it at that point?

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.