Work, Energy and Power

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.

Summary Points to ponder Exercises

We have learnt about vectors and their use in Chapter 3. Physical quantities like displacement, velocity, acceleration, force etc. are vectors. We have also learnt how vectors are added or subtracted. We now need to know how vectors are multiplied. There are two ways of multiplying vectors which we shall come across : one way known as the scalar product gives a scalar from two vectors and the other known as the vector product produces a new vector from two vectors. We shall look at the vector product in Chapter 6. Here we take up the scalar product of two vectors. The scalar product or dot product of any two vectors A and B, denoted as A · B (read

A dot B) is defined as

$$\vec{A}\cdot \vec{B}=AB\cos \theta (5.1a)$$

where $\theta$ is the angle between the two vectors as shown in Fig. 5.1(a). Since $A$, $B$ and $\cos \theta$ are scalars, the dot product of $\vec{A}$ and $\vec{B}$ is a scalar quantity. Each vector, $\vec{A}$ and $\vec{B}$, has a direction but their scalar product does not have a direction.

From Eq. (5.1a), we have

$$\vec{A}\cdot \vec{B}=A(B\cos \theta )$$

$=B(A\cos \theta )$ Geometrically, $B\cos \theta$ is the projection of $\vec{B}$ onto $\vec{A}$ in Fig. 5.1(b) and $A\cos \theta$ is the projection of $\vec{A}$ onto $\vec{B}$ in Fig. 5.1(c). So, $\vec{A}\cdot \vec{B}$ is the product of the magnitude of $\vec{A}$ and the component of $\vec{B}$ along $\vec{A}$. Alternatively, it is the product of the magnitude of $\vec{B}$ and the component of $\vec{A}$ along $\vec{B}$.

Equation (5.1a) shows that the scalar product follows the commutative law :

$$\vec{A}\cdot \vec{B}=\vec{B}\cdot \vec{A}$$

Scalar product obeys the distributive law: $$\vec{A}\cdot (\vec{B}+\vec{C})=\vec{A}\cdot \vec{B}+\vec{A}\cdot \vec{C}$$

$$\vec{A}\cdot (\lambda \vec{B})=\lambda (\vec{A}\cdot \vec{B})$$ Further,

where $\lambda$ is a real number.

The proofs of the above equations are left to you as an exercise.

For unit vectors $\hat{i},\hat{j},\hat{k}$ we have

$$\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$$ $$\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$$ Given two vectors

$$\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$$

$$\vec{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$$

their scalar product is

$$\vec{A}\cdot \vec{B}=(A_x\hat{i}+A_y\hat{j}+A_z\hat{k})\cdot (B_x\hat{i}+B_y\hat{j}+B_z\hat{k})$$

$$=A_x{B}_x+A_y{B}_y+A_z{B}_z(5.1b)$$ From the definition of scalar product and (Eq. 5.1b) we have :

$$\vec{A}\cdot \vec{A}=A_x{A}_x+A_y{A}_y+A_z{A}_z(\mathrm{i})$$

Or, $$A^2=A_x^2+A_y^2+A_z^2(5.1c)$$

since $\vec{A}\cdot \vec{A}=|\vec{A}||\vec{A}|\cos 0=A^2$. $\vec{A}\cdot \vec{B}=0$, if $\vec{A}$ and $\vec{B}$ are perpendicular. (ii)

Example 5.1 Find the angle between force

$\vec{F}=(3\hat{i}+4\hat{j}-5\hat{k})$ unit and displacement

$\vec{d}=(5\hat{i}+4\hat{j}+3\hat{k})$ unit. Also find the

projection of $\vec{F}$ on $\vec{d}$.

Answer $$\vec{F}\cdot \vec{d}=F_x{d}_x+F_y{d}_y+F_z{d}_z$$

$$=3(5)+4(4)+(-5)(3)=16\text{unit}$$

Hence $$\cos \theta =\frac{\vec{F}\cdot \vec{d}}{Fd}=\frac{16}{Fd}$$

Now $$\vec{F}\cdot \vec{F}=F_x^2+F_y^2+F_z^2=F^2$$

$$=9+16+25=50\text{unit}$$

and $$\vec{d}\cdot \vec{d}=d^2=d_x^2+d_y^2+d_z^2$$

$$=25+16+9=50\text{unit}$$

$$\frac{16}{\sqrt{50}\sqrt{50}}=\frac{16}{50}=0.32$$

$\therefore \cos \theta =$

$$\theta ={\cos }^{-1}0.32$$

The following relation for rectilinear motion under constant acceleration $a$ has been encountered in Chapter 3, $$v^2-u^2=2as(5.2)$$ where $u$ and $v$ are the initial and final speeds and $s$ the distance traversed. Multiplying both sides by $m/2$, we have

$$\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=mas=Fs(5.2a)$$

where the last step follows from Newton’s Second Law. We can generalise Eq. (5.2) to three dimensions by employing vectors

$$v^2-u^2=2\vec{a}\cdot \vec{d}$$

Here $\vec{a}$ and $\vec{d}$ are acceleration and displacement vectors of the object respectively. Once again multiplying both sides by $m/2$, we obtain

$$\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=m\vec{a}\cdot \vec{d}=\vec{F}\cdot \vec{d}(5.2b)$$

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. (5.2b) is then

$$K_f-K_i=W(5.3)$$

Example 5.2 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