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Which of the following is a correct unit of energy? a) \(\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}\) c) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\) e) \(\mathrm{kg}^{2} \mathrm{~m}^{2} / \mathrm{s}^{2}\) b) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}\) d) \(\mathrm{kg}^{2} \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
a) kg m / s² b) kg m² / s c) kg m² / s² d) kg² m / s² e) kg² m² / s² Answer: c) kg m² / s²

Step by step solution

01

Write the equation for work (energy) in terms of force and displacement

The work done \(W\) is given by the force \(F\) applied to the object multiplied by the displacement \(d\): \(W = F \cdot d\)
02

Replace force with its unit

The force \(F\) has the unit of newtons (\(\mathrm{N}\)), which is defined as: \(\mathrm{N} = \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{2}\) The work equation now becomes: \(W = \left(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{2}\right) \cdot d\)
03

Replace displacement with its unit

The displacement \(d\) has the unit of meters (\(\mathrm{m}\)), so the work equation becomes: \(W = \left(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{2}\right) \cdot \mathrm{m}\)
04

Multiply the units

Multiply the units together: \(W = \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2}\)
05

Match the result with the given options

Now that we have the correct unit for energy, let's compare it with the given options: a) \(\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}\) c) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\) e) \(\mathrm{kg}^{2} \mathrm{~m}^{2} / \mathrm{s}^{2}\) b) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}\) d) \(\mathrm{kg}^{2} \mathrm{~m} / \mathrm{s}^{2}\) The correct unit of energy matches option c: \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\). So, the correct answer is (c).

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Key Concepts

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

Work-Energy Principle
The work-energy principle is fundamental in physics and ties together two crucial concepts: work and energy. When work is done on an object, it results in a change in the object's energy. The work performed by a force acting on an object is calculated as the product of the force applied and the displacement of the object in the direction of the force.

For instance, when you lift a book off the ground, you're doing work against gravity. The energy that you transfer to the book is equal to the work done. This energy is called potential energy because it can be converted back into work - in this case, when the book falls back to the ground. Understanding the work-energy principle helps us to analyze how objects store and transfer energy in physical processes.
Unit Conversion
Unit conversion is a critical skill in physics and engineering, which involves transforming a quantity expressed in one unit to another unit while maintaining the same quantity. Energy can be expressed in various units such as joules, calories, or kilowatt-hours, depending on the context.

Converting between these units requires an understanding of the relationship between them. For example, 1 joule is equivalent to 0.239 calories. Therefore, knowing how to convert between units is essential for interpreting and applying energy values correctly in different scientific and engineering contexts.
Dimensional Analysis
Dimensional analysis is the process of checking the correctness of an equation or conversion by ensuring that the dimensional units on both sides of an equation balance each other out. In physics, every measurement has dimensions such as length, mass, time, etc. Dimensional analysis helps to derive relationships between physical quantities and to convert from one system of units to another.

By using dimensional analysis, it is possible to confirm that the unit of energy, generally given in joules (or \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\)) in SI units, is correctly derived from the fundamental units for mass (kg), length (m), and time (s). This technique serves as a powerful tool to validate equations and conversions in physics and engineering.
Mechanics in Physics
Mechanics is a branch of physics concerned with the motion of objects and the forces that cause this motion. It's divided into two main branches: statics, which studies objects at rest, and dynamics, which studies objects in motion. The analysis of energy and work is a critical component of dynamics.

Mechanics employs physical quantities and principles like the work-energy principle to understand how and why objects move. A thorough comprehension of mechanics is essential for solving problems in physics and engineering, providing a foundation for further study in other areas of science and technology. Recognizing units of energy and how they relate to the forces and motions of objects is a vital element of practicing mechanics.

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Most popular questions from this chapter

A \(95.0-\mathrm{kg}\) refrigerator rests on the floor. How much work is required to move it at constant speed for \(4.00 \mathrm{~m}\) along the floor against a friction force of \(180 .\) N?

A variable force is given by \(F(x)=A x^{6},\) where \(A=13.75 \mathrm{~N} / \mathrm{m}^{6} .\) This force acts on an object of mass \(3.433 \mathrm{~kg}\) that moves on a frictionless surface. Starting from rest, the object moves from \(x=1.105 \mathrm{~m}\) to a new position, \(x\). The object gains \(5.662 \cdot 10^{3} \mathrm{~J}\) of kinetic energy. What is the new position \(x ?\)

Santa's reindeer pull his sleigh through the snow at a speed of \(2.791 \mathrm{~m} / \mathrm{s}\). Assuming that the reindeer can provide a total power of 3.182 hp and the coefficient of friction between the runners of the sleigh and the snow is \(0.1595,\) what is the mass of the sleigh, including Santa and the presents?

The damage done by a projectile on impact is correlated with its kinetic energy. Calculate and compare the kinetic energies of these three projectiles: a) a \(10.0-\mathrm{kg}\) stone at \(30.0 \mathrm{~m} / \mathrm{s}\) b) a 100.0 -g baseball at \(60.0 \mathrm{~m} / \mathrm{s}\) c) a \(20.0-\mathrm{g}\) bullet at \(300 . \mathrm{m} / \mathrm{s}\)

At sea level, a nitrogen molecule in the air has an average kinetic energy of \(6.2 \cdot 10^{-21} \mathrm{~J}\). Its mass is \(4.7 \cdot 10^{-26} \mathrm{~kg}\). If the molecule could shoot straight up without colliding with other molecules, how high would it rise? What percentage of the Earth's radius is this height? What is the molecule's initial speed? (Assume that you can use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\); although we'll see in Chapter 12 that this assumption may not be justified for this situation.)

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