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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.

Short Answer

Expert verified
The average power output of the insect is 0.196 W.

Step by step solution

01

Understand the Problem

We are asked to determine the average power output of an insect during hovering, where the insect applies twice its weight in force. The insect moves its wings downward 1 cm with each stroke and completes 100 strokes per second. The mass of the insect is 10 g, or 0.01 kg.
02

Calculate the Weight of the Insect

The weight of the insect is calculated using the equation for gravitational force, \( F_w = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (\( 9.8 \ \text{m/s}^2 \)). Therefore, \( F_w = 0.01 \ ext{kg} \times 9.8 \ ext{m/s}^2 = 0.098 \ ext{N} \).
03

Determine the Force Applied

The problem states the insect applies a force equal to twice its weight during each stroke. Thus, the force applied, \( F_a = 2 \times 0.098\ ext{N} = 0.196 \ ext{N} \).
04

Calculate Work Done per Stroke

Work is calculated using \( W = F \times d \), where \( F \) is the force and \( d \) is the distance. Here, \( d = 1.0 \ ext{cm} = 0.01 \ ext{m} \). Thus, \( W = 0.196 \ ext{N} \times 0.01 \ ext{m} = 0.00196 \ ext{J} \) per stroke.
05

Calculate Power Output

Power is calculated using \( P = \frac{W}{t} \), where \( t \) is time. Here, \( t = \frac{1}{100} \ ext{seconds} \) per stroke. Using the work done from Step 4, \( P = 0.00196 \ ext{J} \times 100 \ ext{strokes/sec} = 0.196 \ ext{W} \).

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Key Concepts

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

Work Done
To understand how work is done by the insect's wings while hovering, we first consider the concept of work in physics. Work is defined as the energy transferred to or from an object via the application of force along a displacement. When the insect moves its wings downward, it applies a force to move through a certain distance.

The mathematical formula for work is \( W = F \times d \), where \( W \) denotes work, \( F \) represents the applied force, and \( d \) is the distance over which the force acts. Here, the insect applies a force twice its weight over a downward stroke distance of 1 cm, leading to a calculated work done per stroke.
  • For each stroke, the work done is \( 0.00196 \text{ J} \).
  • This repetition over multiple strokes results in the insect's power output.
The work done by the wings in this context affects the insect's ability to maintain hovering flight.
Force and Motion
Force and motion are deeply interconnected in the study of physics. They help us describe how objects move in our universe. In the scenario of a hovering insect, understanding how force contributes to motion is vital.

The force applied by the insect can be determined to be twice its weight. This extra force is essential to overcome gravitational pull and allow motion of its wings. It ensures that with each stroke, the wings can propel the insect upwards sufficiently to counteract the forces of gravity. This cycle is repeated with a frequency of 100 strokes per second.
  • The balance between applied force and gravitational pull allows for consistent hovering.
  • Motion is maintained through the repeated force application during each stroke.
This repeated action transfers energy through motion to keep the insect aloft.
Gravitational Force
Gravitational force is the fundamental force acting on the insect due to the presence of Earth. It draws the insect downwards with an acceleration of approximately \( 9.8 \, \text{m/s}^2 \). This force, also known as weight, is calculated by the formula \( F_w = mg \), where \( m \) is mass, and \( g \) is the gravitational acceleration.

For an insect weighing 0.01 kg, this translates into a gravitational force of 0.098 N. To maintain a hovering position, the insect must exert a net upward force that counteracts this gravitational pull. That means applying twice this gravitational force during each downward stroke.
  • Achieving balance between gravitational force and applied force is key to hovering.
  • Twice the gravitational force amount is necessary to manage this balance effectively.
The interplay between gravitational force and motion determines the energy required for hovering.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in a system, related to the position and motion of objects. For the hovering insect scenario, mechanical energy is an essential part of maintaining flight.

Every wing stroke involves converting potential energy (position of the wings) and kinetic energy (motion of the wings). This conversion allows the insect to maintain its altitude and speed during the hovering process.
  • The insect’s energy arises from rapid wing movements, maintaining mechanical energy balance.
  • Every stroke completes an energy cycle, where energy is spent and regained through motion.
Understanding how mechanical energy is utilized by the insect aids in understanding the efficiency and power needed to sustain its flight via intricate balances of forces and energy.

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Most popular questions from this chapter

A \(4.00-\mathrm{kg}\) block of ice is placed against a horizontal spring that has force constant \(k=200 \mathrm{N} / \mathrm{m}\) and is compressed 0.025 \(\mathrm{m}\) . The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its nncompressed length. (b) What is the speed of the block after it leaves the spring?

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Pushing a Cat. Your cat "Ms" (mass 7.00 \(\mathrm{kg}\) ) is trying to make it to the top of a frictionless ramp 2.00 \(\mathrm{m}\) long and inclined upward at \(30.0^{\circ}\) above the horizontal. Since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by excring a constant \(100-\mathrm{N}\) force parallel to the ramp. If Ms. takes a running start so that she is moving at 2.40 \(\mathrm{m} / \mathrm{s}\) at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.

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