Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Do devices with efficiencies of less than one violate the law of conservation of energy? Explain

Short Answer

Expert verified

No, devices with an efficiency of less than one do not violate the law of energy conservation.

Step by step solution

01

Step 1: Definition of Concepts

Efficiency: Efficiency is defined as the ratio of the total useful power to the total input power.

Mathematically,

η=totalusefuloutputpowertotalinputpower

Here, ηis the efficiency.

02

Explain the devices with efficiencies of less than one violate the law of conservation of energy

For a nonconservative force, all the work done by the nonconservative force adds to the mechanical energy.

Apart from kinetic energy, potential energy, and the work done by all nonconservative forces, there are other forms of output energy that cannot be used. But efficiency takes into account only the useful output power.

Hence, the devices with an efficiency of less than one do not violate the law of conservation of energy.

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 Vaia!

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 car advertisement claims that its \(900 - {\rm{kg}}\) car accelerated from rest to \(30.0{\rm{ m}}/{\rm{s}}\) and drove \(100{\rm{ km}}\), gaining \(3.00{\rm{ km}}\) in altitude, on \(1.0{\rm{ gal}}\) of gasoline. The average force of friction including air resistance was \(700{\rm{ N}}\). Assume all values are known to three significant figures.

(a) Calculate the car’s efficiency.

(b) What is unreasonable about the result?

(c) Which premise is unreasonable, or which premises are inconsistent?

How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure 7.36? Assume no friction acts on the wagon.

Figure 7.36 The boy does work on the system of the wagon and the child when he pulls them as shown.

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º with the horizontal. (See Figure 7.35.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.

Figure 7.35 A man pushes a crate up a ramp

Suppose a car travels\(108{\rm{ km}}\)at a speed of\(30.0\,{\rm{m}}/{\rm{s}}\), and uses\(2.0{\rm{ gal}}\)of gasoline. Only\(30\% \)30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.)

(a) What is the magnitude of the force exerted to keep the car moving at constant speed?

(b) If the required force is directly proportional to speed, how many gallons will be used to drive\(108{\rm{ km}}\) at a speed of \(28.0{\rm{ m}}/{\rm{s}}\)?

The 70.0-kg swimmer in Figure 7.44 starts a race with an initial velocity of 1.25 m/s and exerts an average force of 80.0 N backward with his arms during each 1.80 m long stroke.

(a) What is his initial acceleration if water resistance is 45.0 N?

(b) What is the subsequent average resistance force from the water during the 5.00 s it takes him to reach his top velocity of 2.50 m/s?

(c) Discuss whether water resistance seems to increase linearly with velocity.

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.

Sign-up for free