/*! 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 For its size, the common flea is... [FREE SOLUTION] | 91Ó°ÊÓ

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For its size, the common flea is one of the most accomplished jumpers in the animal world. A 2.0-mm-long, 0.50-mg flea can reach a height of 20 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-kg, 2.0-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 cm from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a 65-kg person? (e) Where does the flea store the energy that allows it to make sudden leaps?

Short Answer

Expert verified
The flea's takeoff speed is about 1.98 m/s. Humans would need 62.6 m/s to jump proportionally. Fleas store energy in resilin.

Step by step solution

01

Understand the Problem

We have a flea that can jump to a certain height, and we need to calculate its takeoff speed, kinetic energy, and compare it with a human's jumping ability. Additionally, we should explore where the flea stores energy for its jumps. We will address these sub-questions in sequence.
02

Solve for Takeoff Speed of the Flea

Using the kinematic equation for vertical motion, \( v^2 = u^2 + 2as \), where the final velocity \( v \) at the peak is zero, the acceleration \( a \) due to gravity is \(-9.81\, m/s^2\), and \( s \) is 0.20 m, we solve for \( u \), the takeoff speed:\[ 0 = u^2 - 2(9.81)(0.20) \]\[ u^2 = 3.924 \]\[ u = \sqrt{3.924} \approx 1.98 \, m/s \]Thus, the takeoff speed of the flea is approximately 1.98 m/s.
03

Calculate Kinetic Energy of the Flea

Kinetic Energy (KE) is given by \( KE = \frac{1}{2}mv^2 \). Given flea's mass \( m = 0.50 \times 10^{-3} \) kg and takeoff speed \( v = 1.98 \) m/s, we find:\[ KE = \frac{1}{2}(0.50 \times 10^{-3})(1.98)^2 \]\[ KE \approx 9.801 \times 10^{-4} \text{ J} \]The KE per kilogram is:\[ \text{KE per kg} = \frac{9.801 \times 10^{-4}}{0.50 \times 10^{-3}} \approx 1.96 \, \text{J/kg} \]
04

Compare Human Jumping Ability

First, calculate how high a human could jump if they could jump as high as the flea relative to their body size. The flea's jump height is 100 times its length (0.20 m / 0.002 m). For a 2.0 m tall human, this is:\[ \text{Human jump height} = 100 \times 2.0 = 200 \text{ m} \]Next, use the kinematic equation to find the required takeoff speed to jump 200 m.\[ v^2 = 2(9.81)(200) \]\[ v = \sqrt{3924} \approx 62.6 \text{ m/s} \]
05

Kinetic Energy per Kilogram for a Human's Jump

If a human can jump 60 cm, calculate the kinetic energy per kilogram:\( s = 0.60 \text{ m}, m = 65 \text{ kg} \)First, find takeoff speed for 60 cm:\[ v^2 = 2gh = 2(9.81)(0.60) \]\[ v = \sqrt{11.772} \approx 3.43 \text{ m/s} \]KE at takeoff:\[ KE = \frac{1}{2}(65)(3.43)^2 \approx 382.4 \text{ J} \]KE per kilogram:\[ \text{KE/kg} = \frac{382.4}{65} = 5.88 \text{ J/kg} \]
06

Energy Storage in Fleas

Fleas store energy in a spring-like protein called resilin found in their legs. This protein can quickly release stored energy, allowing fleas to execute powerful, rapid jumps.

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Key Concepts

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

Kinematic Equations
Kinematic equations are fundamental tools in physics for analyzing motion. They relate several important quantities like initial and final velocities, acceleration, time, and displacement. In the problem of assessing the flea's jump, we use the equation \( v^2 = u^2 + 2as \), where:
  • \( u \) is the initial velocity or takeoff speed,
  • \( v \) is the final velocity,
  • \( a \) is acceleration (in this case, acceleration due to gravity \(-9.81\, \text{m/s}^2\)),
  • \( s \) is the displacement, which is the height the flea reaches.
Using these equations, we can determine that when reaching the peak of its jump, the flea's final velocity \( v \) becomes zero. By rearranging the formula, we solve for \( u \), the takeoff speed, which is crucial to understanding the flea's motion and energy required to make the leap.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is calculated using the formula \( KE = \frac{1}{2}mv^2 \), where:
  • \( m \) is the object's mass,
  • \( v \) is the object's velocity.
For the flea, with a mass of 0.50 mg and a takeoff speed of approximately 1.98 m/s, the kinetic energy can be calculated easily. Upon calculation, it's evident that despite its tiny size, the flea's kinetic energy is quite significant relative to its mass. This shows how efficient fleas are at converting stored energy into motion. Comparing the flea's kinetic energy per kilogram of its mass to that of a human illustrates how fleas are specialists in utilizing kinetic energy for jumps.
Energy Storage in Animals
The mystery of how fleas can achieve such impressive jumps lies in their unique bio-mechanical anatomy. Fleas have a specialized form of energy storage located in their legs, specifically in a protein called resilin. Resilin acts much like a biological spring. It stores potential energy when compressed and then releases this energy rapidly:
  • Provides elasticity and stores energy during muscle contraction,
  • Releases energy explosively to facilitate powerful jumps.
This system ensures maximal efficiency, conserving muscle energy while enabling swift and high jumps. This natural adaptation is not only fascinating but also serves as inspiration in designing advanced mechanical systems.
Gravity and Motion
Gravity is one of the key forces influencing motion on Earth. It continuously acts on objects, pulling them towards the Earth's center. When an object like a flea jumps, it works against gravity to reach a certain height. The acceleration due to gravity is denoted as \(-9.81\, \text{m/s}^2\), which is a constant value around the Earth's surface.
Understanding gravity helps in calculating the motion parameters like takeoff speed and height. For example, when solving the flea's leap height, we essentially calculate how much initial speed is required to overcome the gravitational pull and achieve the desired height. This principle of motion under gravity is universal, applying to everything from flea jumps to spacecraft launches.

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Most popular questions from this chapter

A spring of negligible mass has force constant \(k =\) 800 N/m. (a) How far must the spring be compressed for 1.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then lay a 1.60-kg book on top of the spring and release the book from rest. Find the maximum distance the spring will be compressed.

A wooden block with mass 1.50 kg is placed against a compressed spring at the bottom of an incline of slope 30.0\(^\circ\) (point \(A\)). When the spring is released, it projects the block up the incline. At point \(B\), a distance of 6.00 m up the incline from A, the block is moving up the incline at 7.00 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_k =\) 0.50. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

A small block with mass 0.0400 kg slides in a vertical circle of radius \(R =\) 0.500 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point \(A\), the normal force exerted on the block by the track has magnitude 3.95 N. In this same revolution, when the block reaches the top of its path, point \(B\), the normal force exerted on the block has magnitude 0.680 N. How much work is done on the block by friction during the motion of the block from point \(A\) to point \(B\)?

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You are an industrial engineer with a shipping company. As part of the package-handling system, a small box with mass 1.60 kg is placed against a light spring that is compressed 0.280 m. The spring has force constant \(k = 45.0\) N/m. The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is \(\mu_k = 0.300\). When the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. (a) What is the speed of the box at the instant when it leaves the spring? (b) What is the maximum speed of the box during its motion?

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