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While hovering, a typical flying insect applies an average force equal to twice its weight during each downward stroke. Take the mass of the insect to be 10 g, and assume the wings move an average downward distance of 1.0 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

Calculate the Weight of the Insect

First, we need to find the force due to gravity, acting on the insect. Given the mass of the insect is 10 g, convert this mass into kilograms to use the standard unit: \[ m = 10 \text{ g} = 0.01 \text{ kg} \]The gravitational acceleration \( g \) is approximately \( 9.8 \text{ m/s}^2 \).The weight \( W \) of the insect is given by:\[ W = m \cdot g = 0.01 \text{ kg} \times 9.8 \text{ m/s}^2 = 0.098 \text{ N} \]
02

Determine the Force Per Stroke

The average force applied by the insect during each downward stroke is twice its weight. Therefore, we calculate this force \( F_{stroke} \) as:\[ F_{stroke} = 2 \times W = 2 \times 0.098 \text{ N} = 0.196 \text{ N} \]
03

Compute the Work Done Per Stroke

Work done per stroke \( W_{stroke} \) is calculated by multiplying the force per stroke by the distance the wings move downward. Convert the distance from cm to meters for the calculation:\[ d = 1 \text{ cm} = 0.01 \text{ m} \]So,\[ W_{stroke} = F_{stroke} \times d = 0.196 \text{ N} \times 0.01 \text{ m} = 0.00196 \text{ J} \]
04

Calculate the Power Output

Power is the rate at which work is done. The power output \( P \) can be calculated by multiplying the work done per stroke by the number of strokes per second:\[ f = 100 \text{ strokes/second} \]\[ P = W_{stroke} \times f = 0.00196 \text{ J} \times 100 = 0.196 \text{ W} \]

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Key Concepts

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

Force Calculation
Insect flight involves interesting physics, especially when it comes to calculating the forces that enable hovering. Force calculation in this context is crucial for understanding how an insect maintains flight. Usually, insects exert force by flapping their wings. This force is often greater than their weight to achieve lift.
Let's consider our exercise: the insect applies a force twice its weight during each downward stroke. By knowing the insect's mass, we can calculate the gravitational force, commonly known as weight.To break it down:
  • First, we convert the mass from grams to kilograms: 10 g = 0.01 kg.
  • Then, by using the acceleration due to gravity (9.8 m/s²), we calculate the weight: \[ W = m \times g = 0.01 \text{ kg} \times 9.8 \text{ m/s}^2 = 0.098 \text{ N} \]
Here, weight acts as the downward force due to gravity. The insect needs to apply a force greater than this to raise itself upward, which, as given, is twice the weight. So, calculate:
  • The force per stroke: \[ F_{stroke} = 2 \times W = 0.196 \text{ N} \]
This explains how insects lift themselves by creating sufficient upward force through wing flapping.
Work and Energy
When an insect flaps its wings, it performs work. Work in physics relates to the amount of energy transferred by a force over a distance. In this exercise, the work done during each stroke involves the thrust created by the wings moving through the air.
Let's calculate the work done per wing stroke:
  • Distance the wings move downward is given as 1 cm. In meters, that's 0.01 m.
  • Use the formula for work done: \[ W_{stroke} = F_{stroke} \times d = 0.196 \text{ N} \times 0.01 \text{ m} = 0.00196 \text{ J} \]
The calculation tells us how much energy is transferred in each stroke to both overcome gravitational forces and sustain flight. Work is essentially the energy needed to move an object over a certain distance. Here, it measures the energy required for the wings' downward movement.
Power in Physics
Power in physics refers to how quickly work is done or energy is transferred. In our flying insect scenario, power tells us about the speed of energy usage during flight. Calculating power helps us understand efficiency and endurance of the insect in flight.
Power is calculated as work done per unit time, represented as:
  • Considering 100 strokes per second (frequency), we calculate the power output: \[ P = W_{stroke} \times f = 0.00196 \text{ J} \times 100 \text{ strokes/second} = 0.196 \text{ W} \]
This 0.196 W indicates how much energy the insect produces each second to maintain hovering. It highlights the remarkable energy efficiency of insects, despite their small size, and the rapidity with which their wings operate. Understanding these calculations offers insights into the power dynamics critical for sustaining insect flight.

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Most popular questions from this chapter

When its 75-kW (100-hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

A 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a) During one complete circle, starting anywhere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path.

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1200-kg car moving at 0.65 m/s is to compress the spring no more than 0.090 m before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 J of work when you compress the springs 0.200 m from their uncompressed length. (a) What magnitude of force must you apply to hold the platform in this position? (b) How much \(additional\) work must you do to move the platform 0.200 m \(farther\), and what maximum force must you apply?

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x\), a force along the \(x\)-axis with \(x\)-component \(F_x = kx - bx^2 + cx^3\) must be applied to the free end. Here \(k = 100 \, \mathrm {N/m}\), \(b = 700 \, \mathrm {N/m{^2}}\), and \(c = 12,000 \, \mathrm{N/m}^3\). Note that \(x > 0\) when the spring is stretched and \(x< 0\) when it is compressed. How much work must be done (a) to stretch this spring by 0.050 m from its unstretched length? (b) To \(compress\) this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_x\) on \(x\). (Many real springs behave qualitatively in the same way.)

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