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Chapter 5: Problem 57

Calculate What is the power output of a 1.4-g fly as it walks straight up a windowpane at \(2.3 \mathrm{~cm} / \mathrm{s}\) ?

Short Answer

Expert verified
The power output of the fly is approximately 0.32 mW.

Step by step solution

01

Understand the Problem

We are tasked with finding the power output of a fly moving upwards. The fly's mass is given as 1.4 grams, and its velocity is 2.3 cm/s. Power is the rate at which work is done, typically expressed in watts.
02

Convert Units

Convert the fly's mass from grams to kilograms to keep units consistent. The mass of the fly is 1.4 grams, so in kilograms, it is \(1.4 \text{ g} = 0.0014 \text{ kg}\). The velocity should be maintained in meters per second, thus \(2.3 \text{ cm/s} = 0.023 \text{ m/s}\).
03

Calculate the Force

The force exerted by the fly can be found using the formula for the force due to gravity: \( F = m \cdot g \), where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Substituting the values: \( F = 0.0014 \text{ kg} \times 9.8 \text{ m/s}^2 = 0.01372 \text{ N}\).
04

Calculate Work Done Per Second (Power)

Power is defined as the rate of doing work. Since work is force times distance, and the fly moves with constant velocity, the power can be calculated as \( P = F \times v \), where \( F \) is the force and \( v \) is the velocity. Thus, \( P = 0.01372 \text{ N} \times 0.023 \text{ m/s} = 0.00031556 \text{ W}\). This value can be rounded to \( 0.00032 \text{ W} \), or \( 0.32 \text{ mW} \).

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Key Concepts

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

Force Calculation
To determine the power output of a fly walking up a surface, you first need to calculate the force it exerts. Force is a vector quantity, meaning it has both magnitude and direction. In this context, force is influenced by gravity.To calculate the force, use the equation:
  • Force (\( F \)) = Mass (\( m \)) 脳 Gravitational Acceleration (\( g \))
The gravitational acceleration (\( g \)) is typically given as \( 9.8 \, \text{m/s}^2 \). This value is standard on Earth's surface, where gravity's effect is constant.By multiplying the fly's mass by the gravitational constant, you can discover how much force gravity applies to the fly, giving you a fundamental part of calculating power.
Unit Conversion
Dealing with physics requires meticulous unit conversions. It's essential to maintain consistency in units, ensuring correct calculations and interpretations.In our example, the mass of the fly is initially given in grams but needs to be converted to kilograms since the standard unit of mass in the International System of Units (SI) is kilograms.
  • \( 1.4 \, \text{g} = 0.0014 \, \text{kg} \)
Similarly, for velocity, converting centimeters per second to meters per second ensures harmony in equations because force, work, and power calculations are usually expressed in meters.
  • \( 2.3 \, \text{cm/s} = 0.023 \, \text{m/s} \)
These conversions are vital to achieving accurate calculations and conclusions.
Gravitational Force
Gravitational force is a critical component in our calculations as it defines the force acting on the fly due to Earth's gravity. It is represented by the force formula:
  • \( F = m \cdot g \)
In this formula:
  • \( F \) is the force in newtons (N)
  • \( m \) is the mass in kilograms (kg)
  • \( g \) is the gravitational acceleration \( 9.8 \, \text{m/s}^2 \)
Using the mass of the fly, you'll calculate the gravitational force it experiences. It's significant because this force is a primary contributor to determining the work done as the fly climbs upwards against gravitational pull.
Work and Energy
In physics, work and energy are intimately linked, as work is defined as the force exerted over a distance. As the fly walks up a windowpane, it does work against gravitational force.Work (\( W \)) is calculated using the formula:
  • \( W = F \times d \)
Where:
  • \( F \) is force
  • \( d \) is distance
However, since power is what we are interested in, which is the rate of work done over time, we use a different expression:
  • \( P = F \times v \)
Here, \( v \) represents velocity. In our case, multiplying the gravitational force by the velocity gives us the power output as the fly climbs.
Velocity Measurement
Velocity is a measure of how fast something is moving in a specified direction, key to calculating power output. It鈥檚 vectorial, possessing both magnitude and direction, which are essential in work and power calculations.For objects moving at constant velocity, such as the fly in this case, calculations are simplified. Here鈥檚 how:
  • Convert velocity into appropriate units (meters/second) for consistent calculations.
  • \( 2.3 \, \text{cm/s} = 0.023 \, \text{m/s} \)
With constant velocity, we determine how much work is done per second, directly reflecting the power generated by the fly climbing against gravity.

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Most popular questions from this chapter

At \(t=1.0 \mathrm{~s}\), a \(0.40-\mathrm{kg}\) object is falling with a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). At \(t=2.0 \mathrm{~s}\), it has a kinetic energy of \(25 \mathrm{~J}\). (a) What is the kinetic energy of the object at \(t=1.0 \mathrm{~s}\) ? (b) What is the speed of the object at \(t=2.0 \mathrm{~s}\) ? (c) How much work was done on the object between \(t=1.0 \mathrm{~s}\) and \(t=2.0 \mathrm{~s}\) ?

Analyze Discuss the various energy conversions that occur when a person performs a pole vault. Include as many conversions as you can. Be sure to consider times before, during, and after the vault itself.

Predict \& Explain When a ball of mass \(m\) is dropped from rest from a height \(h\), its kinetic energy just before landing is KE. Now, suppose a second ball of mass \(4 m\) is dropped from rest from a height \(h / 4\). (a) Just before ball 2 lands, is its kinetic energy \(4 K E, 2 K E, K E, K E / 2\), or \(K E / 4\) ? (b) Choose the best explanation from among the following: A. The two balls have the same initial energy. B. The more massive ball will have the greater kinetic energy. C. The lower drop height results in a reduced kinetic energy.

Think \& Calculate To clean a floor, a janitor pushes on a mop handle with a force of \(43 \mathrm{~N}\). (a) If the mop handle is at an angle of \(55^{\circ}\) above the horizontal, how much work is required to push the mop \(0.50 \mathrm{~m}\) ? (b) If the angle the mop handle makes with the horizontal is increased to \(65^{\circ}\), does the work done by the janitor increase, decrease, or stay the same? Explain.

After hitting a long fly ball that goes over the right fielder's head and lands in the outfield, a batter decides to keep going past second base and try for third base. The \(62-\mathrm{kg}\) player begins sliding \(3.4 \mathrm{~m}\) from the base with a speed of \(4.5 \mathrm{~m} / \mathrm{s}\). (a) If the player comes to rest at third base, how much work was done on the player by friction with the ground? (b) What was the coefficient of kinetic friction between the player and the ground?
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