Introduction

The terms ‘work’, ‘energy’, and ‘power’ are frequently used in everyday language.  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 has large 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.  The aim of this chapter is to develop an understanding of these three physical quantities.

The Scalar Product

The scalar product or dot product of any two vectors \(\mathbf{A}\) and \(\mathbf{B}\), denoted as \(A \cdot B\) (read \(\mathbf{A}\) dot \(\mathbf{B}\)) is defined as
\(
\mathbf{A} \cdot \mathbf{B}=A B \cos \theta \dots(6.1a)
\)
where \(\theta\) is the angle between the two vectors as shown in Fig. 6.1(a). Since \(A, B\) and \(\cos \theta\) are scalars, the dot product of \(\mathbf{A}\) and \(\mathbf{B}\) is a scalar quantity. Each vector, \(\mathbf{A}\) and \(\mathbf{B}\), has a direction but their scalar product does nolt have a direction.
\(
\begin{aligned}
\mathbf{A} \cdot \mathbf{B} & =A(B \cos \theta) \\
& =B(A \cos \theta)
\end{aligned}
\)
Geometrically, \(B \cos \theta\) is the projection of \(\mathbf{B}\) onto A in Fig.6.1 (b) and \(A \cos \theta\) is the projection of \(\mathbf{A}\) onto B in Fig. 6.1 (c). So, A.B is the product of the magnitude of \(\mathbf{A}\) and the component of \(\mathbf{B}\) along A. Alternatively, it is the product of the magnitude of \(\mathbf{B}\) and the component of \(\mathbf{A}\) along \(\mathbf{B}\).
Equation (6.1a) shows that the scalar product follows the commutative law :
\(
\mathbf{A} \cdot \mathbf{B}=\mathbf{B} \cdot \mathbf{A}
\)
Scalar product obeys the distributive law:
\(
\mathbf{A} \cdot(\mathbf{B}+\mathbf{C})=\mathbf{A} \cdot \mathbf{B}+\mathbf{A} \cdot \mathbf{C}
\)
Further, \(\quad \mathbf{A} \cdot(\lambda \mathbf{B})=\lambda(\mathbf{A} \cdot \mathbf{B})\) where \(\lambda\) is a real number.
The proofs of the above equations are left to you as an exercise.
For unit vectors \(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\) we have
\(
\begin{aligned}
& \hat{\mathbf{i}} \cdot \hat{\mathbf{i}}=\hat{\mathbf{j}} \cdot \hat{\mathbf{j}}=\hat{\mathbf{k}} \cdot \hat{\mathbf{k}}=1 \\
& \hat{\mathbf{i}} \cdot \hat{\mathbf{j}}=\hat{\mathbf{j}} \cdot \hat{\mathbf{k}}=\hat{\mathbf{k}} \cdot \hat{\mathbf{i}}=0
\end{aligned}
\)
Given two vectors
\(
\begin{aligned}
& \mathbf{A}=A_x \hat{\mathbf{i}}+A_y \hat{\mathbf{j}}+A_z \hat{\mathbf{k}} \\
& \mathbf{B}=B_x \hat{\mathbf{i}}+B_y \hat{\mathbf{j}}+B_z \hat{\mathbf{k}}
\end{aligned}
\)
their scalar product is
\(
\begin{aligned}
& \mathbf{A} \cdot \mathbf{B}=\left(A_x \hat{\mathbf{i}}+A_y \hat{\mathbf{j}}+A_z \hat{\mathbf{k}}\right) \cdot\left(B_x \hat{\mathbf{i}}+B_y \hat{\mathbf{j}}+B_z \hat{\mathbf{k}}\right) \\
& =A_x B_x+A_y B_y+A_z B_z \dots(6.1b)
\end{aligned}
\)
From the definition of scalar product and (Eq. 6.1b) we have :
(1) A.A \(=A_x A_x+A_y A_y+A_z A_z\)
Or, \(\quad A^2=A_x^2+A_y^2+A_z^2 \dots(6.1c)\)
since \(\mathbf{A} \cdot \mathbf{A}=|\mathbf{A}||\mathbf{A}| \cos 0=A^2\).
(ii) \(\mathbf{A} \cdot \mathbf{B}=0\), if \(\mathbf{A}\) and \(\mathbf{B}\) are perpendicular.

Example 6.1:

\(
\begin{aligned}
& \text { Example } 6.1 \text { Find the angle between force } \\
& \mathbf{F}=(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}) \text { unit and displacement } \\
& \mathrm{d}=(5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \text { unit. Also, find the } \\
& \text { projection of } \mathbf{F} \text { on } \mathrm{d} \text {. }
\end{aligned}
\)

Answer:

\(
\begin{aligned}
\mathbf{F} \cdot \mathbf{d} & =F_x d_x+F_y d_y+F_z d_z \\
& =3(5)+4(4)+(-5)(3) \\
& =16 \text { unit }
\end{aligned}
\)
Hence \(\mathbf{F} \cdot \mathbf{d}=F d \cos \theta=16\) unit
Now \(
\mathbf{F} \cdot \mathbf{F}
\)
\(
\begin{aligned}
& =F^2=F_x^2+F_y^2+F_z^2 \\
& =9+16+25 \\
& =50 \text { unit }
\end{aligned}
\)
\(
\text { and } \mathbf{d} \cdot \mathbf{d} \quad=d^2=d_x^2+d_y^2+d_z^2
\)
\(
\begin{aligned}
& =25+16+9 \\
& =50 \text { unit }
\end{aligned}
\)
\(
\begin{aligned}
\therefore \cos \theta & =\frac{16}{\sqrt{50} \sqrt{50}}=\frac{16}{50}=0.32, \\
& \theta=\cos ^{-1} 0.32
\end{aligned}
\)