THE IDEA OF ENERGY

The Industrial Revolution started in the late $18th$ century in Britain. Today, many other countries have become or are becoming industrialised (Figure 5.1). Industrialisation is the development of new machines capable of doing the work of hundreds of craftsmen and labourers. At first, people used water and wind to power machines. Water stored behind a dam was used to turn a wheel, which turned many machines. Steam engines were developed, initially for pumping water out of mines. Steam engines use a fuel such as coal; there is much more energy stored in $1 kg$ of coal than in $1 kg$ of water held behind a dam.
Nowadays, most factories rely on electrical power, generated by burning coal or gas at a power station.
High-pressure steam is generated, and this turns a turbine that turns a generator. Even in the most efficient coal-fired power station, only about $40\% $ of the energy from the fuel is transferred to the electrical energy that the station supplies to the electricity grid.

Engineers worked hard to develop machines that made the most efficient use of the energy supplied to them. At the same time, scientists were working out the basic ideas of energy transfer and energy transformations. The idea of energy itself had to be developed; it was not obvious at first that heat, light, electrical energy were all forms of the same thing: energy. What is the history of your country in developing the use of machines, generating electrical power and increasing efficiency?
The earliest steam engines had very low efficiencies–many converted less than 1% of the energy supplied to them into useful work. The understanding of the relationship between work and energy led to many clever ways of making the most of the energy supplied by fuel.

The weight-lifter shown in Figure 5.3 has powerful muscles. They can provide the force needed to lift a large weight above her head – about $2 m$ above the ground. The force exerted by the weight-lifter transfers energy from her to the weights. We know that the weights have gained energy because, when the athlete releases them, they come crashing down to the ground.
As the athlete lifts the weights and transfers energy to them, we say that her lifting force is doing work.
‘Doing work’ is a way of transferring energy from one object to another. In fact, if you want to know the scientific meaning of the word ‘energy’, we have to say it is ‘that which is transferred when a force moves through a distance’. So, work and energy are two closely linked concepts.

Table 5.1 describes some situations that illustrate the meaning of doing work in physics.
It is important to understand that our bodies sometimes mislead us. If you hold a heavy weight above your head for some time, your muscles will get tired. However, you are not doing any work on the weights, because you are not transferring energy to the weights once they are above your head. Your muscles get tired because they are constantly relaxing and contracting, and this uses energy, but none of the energy is being transferred to the weights.

Calculating work done

Because doing work defines what we mean by energy, we start this chapter by considering how to calculate work done.
There is no doubt that you do work if you push a car along the road. A force transfers energy from you to the car. But how much work do you do? Figure 5.4 shows the two factors involved:
the size of the force F – the bigger the force, the greater the amount of work you do the distance s you push the car – the further you push it, the greater the amount of work done.
So, the bigger the force, and the further it moves, the greater the amount of work done.

The work done by a force is defined as the product of the force and the distance moved in the direction of the force:
$W = F \times s$
where s is the distance moved in the direction of the force.
In the example shown in Figure 5.4, $F = 300 N$ and $s = 5.0 m$, so:
work done $W = F \times s = 300 \times 5.0 = 1500J$

Energy transferred

Doing work is a way of transferring energy. For both energy and work the correct SI unit is the joule (J).
The amount of work done, calculated using $W = F \times s$, shows the amount of energy transferred:

work done = energy transferred

Newtons, metres and joules

From the equation $W = Fs$ we can see how the unit of force (the newton), the unit of distance (the metre) and the unit of work or energy (the joule) are related.
The joule is defined as the amount of work done when a force of 1 newton moves a distance of 1 metre in the direction of the force. Since work done = energy transferred, it follows that a joule is also the amount of energy transferred when a force of 1 newton moves a distance of 1 metre in the direction of the force.

Force, distance and direction

It is important to understand that, for a force to do work, there must be movement in the direction of the force. Both the force F and the distance s moved in the direction of the force are vector quantities, so you should know that their directions are likely to be important. To illustrate this, we will consider three examples involving gravity (Figure 5.5). In the equation for work done, $W = F \times s$, the distance moved s is the displacement in the direction of the force.

Suppose that the force F moves through a distance s that is at an angle $\theta $ to F, as shown in Figure 5.6.
To determine the work done by the force, it is simplest to determine the component of F in the direction of s. This component is $F\,\cos \,\theta $, and so we have:

or simply:

$work\,down\, = \,Fs\,\cos \theta $

Worked example 1 shows how to use this.

WORKED EXAMPLE

1) A man pulls a box along horizontal ground using a rope (Figure 5.7). The force provided by the rope is $200 N$, at an angle of ${30^ \circ }$ to the horizontal.

Calculate the work done if the box moves $5.0 m$ along the ground.
Step 1: Calculate the component of the force in the direction in which the box moves. This is the horizontal component of the force:
$horizontal\,component\,of\,the\,force\, = \,200\,\cos {30^ \circ }\, \approx \,173N$
Hint: F cos θ is the component of the force at an angle θto the direction of motion.
Step 2: Now calculate the work done:
$\begin{array}{l}
work\,done\, = \,force\, \times \,dis\tan ce\,moved\\
 = 173\, \times 5.0 = 865J
\end{array}$
Hint: Note that we could have used the equation work done $ = Fs\,\cos \theta \,$ to combine the two
steps into one.

Questions

 

4) The crane shown in Figure 5.8 lifts its $500 N$ load to the top of the building from A to B. Distances are as shown on the diagram. Calculate how much work is done by the crane.

5) Figure 5.9 shows the forces acting on a box that is being pushed up a slope. Calculate the work done by each force if the box moves $0.50 m$ up the slope.