The terms ‘work’, ‘energy’ and ‘power’ are frequently used in everyday language. A farmer ploughing the field, a construction worker carrying bricks, a student studying for a competitive examination, an artist painting a beautiful landscape, all are said to be working.
In physics, however, the word ‘Work’ covers a definite and precise meaning. Somebody who has the capacity to work for 14-16 hours a day is said to have a large stamina or energy. We admire a long distance runner for her stamina or energy.
Energy is thus our capacity to do work. In Physics too, the term ‘energy’ is related to work in this sense, but as said above the term ‘work’ itself is defined much more precisely. The word ‘power’ is used in everyday life with different shades of meaning. In karate or boxing we talk of ‘powerful’ punches. These are delivered at a great speed. This shade of meaning is close to the meaning of the word ‘power’ used in physics.
We shall find that there is at best a loose correlation between the physical definitions and the physiological pictures these terms generate in our minds. The aim of this chapter is to develop an understanding of these three physical quantities. Before we proceed to this task, we need to develop a mathematical prerequisite, namely the scalar product of two vectors.
NOTIONS OF WORK AND KINETIC ENERGY: THE WORK-ENERGY THEOREM
The following relation for rectilinear motion under constant acceleration a has been encountered in Chapter 3
v$^2$ − u$^2$ = 2as
where u and v are the initial and final speeds and s the distance traversed. Multiplying both sides by m/2, we have
$\frac{1}{2}mv^2-\frac{1}{2}mu^2$= mas = Fs
where the last step follows from Newton’s Second Law. We can generalise Eq. (6.2) to three dimensions by employing vectors
v$^2$ − u$^2$ = 2a.d
Here a and d are acceleration and displacement vectors of the object respectively. Once again multiplying both sides by m/2 , we obtain
$\frac{1}{2}mv^2-\frac{1}{2}mu^2$ = ma.d = F.d
The above equation provides a motivation for the definitions of work and kinetic energy. The left side of the equation is the difference in the quantity ‘half the mass times the square of the speed’ from its initial value to its final value. We call each of these quantities the ‘kinetic energy’, denoted by K. The right side is a product of the displacement and the component of the force along the displacement. This quantity is called ‘work’ and is denoted by W. Eq. (6.2b) is then
K$_f$ − K$_i$ = W
where K$_i$ and K$_f$ are respectively the initial and final kinetic energies of the object. Work refers to the force and the displacement over which it acts. Work is done by a force on the body over a certain displacement.
Equation (6.2) is also a special case of the work-energy (WE) theorem: The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to a varying force in a later section.
Example 1
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height 1.00 km. It hits the ground with a speed of 50.0 m.s$^{-1}$. (a) What is the work done by the gravitational force? What is the work done by the unknown resistive force?
Answer
(a) The change in kinetic energy of the drop is
$\Delta K=\frac{1}{2}mv^2-0$
$=\frac{1}{2}(10^{-3})(50)(50)$
= 1.25 J
where we have assumed that the drop is initially at rest.
Assuming that g is a constant with a value 10 m/s$^2$, the work done by the gravitational force is,
W$_g$ = mgh
= 10$^{-3}$ ×10 ×10$^3$
= 10.0 J
(b) From the work-energy theorem
∆K = W$_g$ + W$_r$
where Wr is the work done by the resistive force on the raindrop. Thus
W$_r$ = ∆K − W$_g$
= 1.25 −10
= − 8.75 J is negative.
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