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# Mechanical Energy Study Guide

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## Introduction

In this lesson, we will pull all the knowledge about mass, motion, and forces together in a concept that, while very abstract, is at the essence of science and nature – energy, In this lesson, we will concentrate on a form of energy that characterizes mechanical properties, We will define mechanical energy and its two forms: kinetic and potential, We will wrap up our lesson with a study of the law of conservation of energy in a mechanical system.

## Work

Work is a quantity that measures the effect of moving an object when a force is applied. When a constant force (no variation in time or space) is exerted but the object does not move, as in the case when you would push against a wall, we say that no work has been done.

To summarize in an expression the definition of work, we have:

W = (F · cos α) · d

The unit for work is Newtons times meters (N · m) and is known as a Joule (in honor of James Joule 1818–1889).Work is a scalar quantity although it is the result of two vectors—force and displacement. In other words: The most general expression of the work done is the dot product of force and displacement; and the result of a dot product, although starting with two vectors, is a scalar.

### Work

Work is proportional to the product of the distance over which an object is moved, the displacement, and the constant force along that direction.

#### Example 1

Your car needs to be moved for a short distance, but it has no gas in the tank, so it will need to be pushed. You apply a force of 1,000 N and move it for about 180 em. What is the work done?

#### Solution 1

First, convert the units into SI (remember, 1 m = 100cm). Next, construct a diagram of the problem, an then proceed to find the answer.

d = 180 cm = 180 cm · 1 m/l00 cm = 1.8 m

F = 1,000 N

W = ?

As Figure 7.1 shows, the force and the displacement are in the same direction, so then the angle between F and d is zero. Since cos 0° = 1, the expression of the work becomes:

W = (F · cos α) · d = (F · cos 0°) · d = F · d

W = 1,000 N · 1.8 m = 1,800 J = 1.8 kJ

#### Example 2

Now imagine we are using the same force to push a big pile of boxes around, but the force is exerted so that the angle with the horizontal is 60°. Find out the work performed on the boxes by the force of 1,000 N in a distance of 180 cm.

#### Solution 2

First, we convert the units to SI (1 m = 100 cm). Next, construct a diagram of the problem, and then proceed to find the answer.

As Figure 7.2 shows, the force and the displacement are not in the same direction, but they make an angle of 60°. Because cos 60° = 0.5, the expression of the work becomes:

W = (F· cos α) · d = (F · cos 60°) · d = F · d/2

W = 1,000 N · 1.8 m/2 = 900 J = 0.9 kJ

### Note

If you are to consider a vector force, the equation for work shows that only the component of the force along displacement is important in measuring the work done on an object.

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