5.2 NOTIONS OF WORK AND KINETIC to be proportional to the speed of the drop ENERGY: THE WORK-ENERGY THEOREM but is otherwise undetermined. Consider The following relation for rectilinear motion under a drop of mass 1.00 g falling from a height constant acceleration a has been encountered 1.00 km. It hits the ground with a speed of in Chapter 3, 50.0 m s-1. (a) What is the work done by the v2 − u2 = 2 as (5.2) gravitational force ? What is the work done where u and v are the initial and final speeds by the unknown resistive force? and s the distance traversed. Multiplying both Answer (a) The change in kinetic energy of the sides by m/2, we have drop is 1 2 1 2 1 2 mv − mu = mas = Fs (5.2a) ∆ K = m v − 0 2 2 2 where the last step follows from Newton’s Second 1 -3 = × 10 × 50 × 50 Law. We can generalise Eq. (5.2) to three 2 dimensions by employing vectors = 1.25 J v2 − u2 = 2 a.d where we have assumed that the drop is initially at rest. Here a and d are acceleration and displacement Assuming that g is a constant with a value vectors of the object respectively. 10 m/s2, the work done by the gravitational force Once again multiplying both sides by m/2 , we obtain is, 1 1 mv 2 − mu 2 = m a.d = F.d (5.2b) Wg = mgh 2 2 = 10-3 ×10 ×103 The above equation provides a motivation for = 10.0 J the definitions of work and kinetic energy. The (b) From the work-energy theorem left side of the equation is the difference in the quantity ‘half the mass times the square of the ∆ K = W g + W r speed’ from its initial value to its final value. We where Wr is the work done by the resistive force call each of these quantities the ‘kinetic energy’, on the raindrop. Thus denoted by K. The right side is a product of the Wr = ∆K − Wg displacement and the component of the force = 1.25 −10 along the displacement. This quantity is called = − 8.75 J ‘work’ and is denoted by W. Eq. (5.2b) is then is negative. ⊳ Kf − Ki = W (5.3) 5.3 WORK where Ki and Kf are respectively the initial and As seen earlier, work is related to force and the final kinetic energies of the object. Work refers displacement over which it acts. Consider a to the force and the displacement over which it constant force F acting on an object of mass m. acts. Work is done by a force on the body over The object undergoes a displacement d in the a certain displacement. positive x-direction as shown in Fig. 5.2. Equation (5.2) 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 5.2 It is well known that a raindrop falls under the influence of the Fig. 5.2 An object undergoes a displacement d downward gravitational force and the under the influence of the force F. opposing resistive force. The latter is known Reprint 2025-26 74 PHYSICS The work done by the force is defined to be Table 5.1 Alternative Units of Work/Energy in J the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = (F cos θ)d = F.d (5.4) We see that if there is no displacement, there is no work done even if the force is large. Thus, when you push hard against a rigid brick wall, ⊳ the force you exert on the wall does no work. Yet Example 5.3 A cyclist comes to a skidding your muscles are alternatively contracting and stop in 10 m. During this process, the force relaxing and internal energy is being used up on the cycle due to the road is 200 N and and you do get tired. Thus, the meaning of work is directly opposed to the motion. (a) How in physics is different from its usage in everyday much work does the road do on the cycle ? language. (b) How much work does the cycle do on the road ? No work is done if : (i) the displacement is zero as seen in the example above. A weightlifter holding a 150 Answer Work done on the cycle by the road is kg mass steadily on his shoulder for 30 s the work done by the stopping (frictional) force does no work on the load during this time. on the cycle due to the road. (ii) the force is zero. A block moving on a smooth (a) The stopping force and the displacement make horizontal table is not acted upon by a an angle of 180o (π rad) with each other. horizontal force (since there is no friction), but Thus, work done by the road, may undergo a large displacement. Wr = Fd cosθ (iii) the force and displacement are mutually = 200 × 10 × cos π perpendicular. This is so since, for θ= π/2 rad = – 2000 J (= 90o), cos (π/2) = 0. For the block moving on It is this negative work that brings the cycle a smooth horizontal table, the gravitational to a halt in accordance with WE theorem. force mg does no work since it acts at right (b) From Newton’s Third Law an equal and angles to the displacement. If we assume that opposite force acts on the road due to the the moon’s orbits around the earth is cycle. Its magnitude is 200 N. However, the perfectly circular then the earth’s road undergoes no displacement. Thus, gravitational force does no work. The moon’s work done by cycle on the road is zero. ⊳ instantaneous displacement is tangential while the earth’s force is radially inwards and The lesson of Example 5.3 is that though the θ = π/2. force on a body A exerted by the body B is always Work can be both positive and negative. If θ is equal and opposite to that on B by A (Newton’s between 0o and 90o, cos θ in Eq. (5.4) is positive. Third Law); the work done on A by B is not If θ is between 90o and 180o, cos θ is negative. necessarily equal and opposite to the work done In many examples the frictional force opposes on B by A. displacement and θ = 180o. Then the work done 5.4 KINETIC ENERGY by friction is negative (cos 180o = –1). As noted earlier, if an object of mass m has From Eq. (5.4) it is clear that work and energy velocity v, its kinetic energy K ishave the same dimensions, [ML2T–2]. The SI unit of these is joule (J), named after the famous British 1 1 2physicist James Prescott Joule (1811-1869). Since K = m v. v = mv (5.5) 2 2work and energy are so widely used as physical concepts, alternative units abound and some of Kinetic energy is a scalar quantity. The kinetic these are listed in Table 5.1. energy of an object is a measure of the work an Reprint 2025-26 WORK, ENERGY AND POWER 75 Table 5.2 Typical kinetic energies (K) object can do by the virtue of its motion. This This is illustrated in Fig. 5.3(a). Adding notion has been intuitively known for a long time. successive rectangular areas in Fig. 5.3(a) we The kinetic energy of a fast flowing stream get the total work done as has been used to grind corn. Sailing x f ships employ the kinetic energy of the wind. Table W ≅ F (x )∆x (5.6) ∑

5.4 Kinetic energy long distance runner for her stamina or energy. Energy is 5.5 Work done by a variable force thus our capacity to do work. In Physics too, the term ‘energy’ 5.6 The work-energy theorem for is related to work in this sense, but as said above the term a variable force ‘work’ itself is defined much more precisely. The word ‘power’ 5.7 The concept of potential is used in everyday life with different shades of meaning. In energy karate or boxing we talk of ‘powerful’ punches. These are 5.8 The conservation of delivered at a great speed. This shade of meaning is close to mechanical energy the meaning of the word ‘power’ used in physics. We shall 5.9 The potential energy of a find that there is at best a loose correlation between the spring physical definitions and the physiological pictures these 5.10 Power terms generate in our minds. The aim of this chapter is to 5.11 Collisions develop an understanding of these three physical quantities. Before we proceed to this task, we need to develop a Summary mathematical prerequisite, namely the scalar product of two Points to ponder vectors. Exercises 5.1.1 The Scalar Product 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 Reprint 2025-26 72 PHYSICS A dot B) is defined as ɵ ɵ ɵ A = A x i + Ay j + A z k A.B = A B cos θ (5.1a) ɵ ɵ ɵ B = B x i + B y j + B z kwhere θ is the angle between the two vectors as shown in Fig. 5.1(a). Since A, B and cos θ are their scalar product is scalars, the dot product of A and B is a scalar A .B = A x ˆi + Ay ˆj + A z kˆ . B x ˆi + B y ˆj + B z kˆquantity. Each vector, A and B, has a direction ( ) ( ) but their scalar product does not have a = A x B x + Ay B y + A z B z (5.1b) direction. From the definition of scalar product and From Eq. (5.1a), we have (Eq. 5.1b) we have : ( i ) A . A = A x A x + A y A y + A z A z A.B = A (B cos θ) = B (A cos θ ) Or, A 2 = A 2x + Ay2 + A z2 (5.1c) Geometrically, B cos θ is the projection of B onto since A.A = |A ||A| cos 0 = A2. A in Fig.5.1 (b) and A cos θ is the projection of A (ii) A.B = 0, if A and B are perpendicular. onto B in Fig. 5.1 (c). So, A.B is the product of u Example 5.1 Find the angle between forcethe magnitude of A and the component of B along A. Alternatively, it is the product of the F = (3 ˆi + 4 ˆj -5 kˆ ) unit and displacement magnitude of B and the component of A along B. d = (5 ˆi + 4 ˆj +3 kˆ ) unit. Also find the Equation (5.1a) shows that the scalar product follows the commutative law : projection of F on d. A.B = B.A Answer F.d = F x d x + Fy d y + F z d z = 3 (5) + 4 (4) + (– 5) (3)Scalar product obeys the distributive = 16 unitlaw: Hence F.d = F d cosθ = 16 unit A. (B + C) = A.B + A.C F.F = F 2 = F x2 + Fy2 + F z2Further, A. (λ B) = λ (A.B) Now = 9 + 16 + 25where λ is a real number. = 50 unit The proofs of the above equations are left to you as an exercise. and d.d = d2 = d x 2 + dy2 + dz2 ɵ ɵ ɵ = 25 + 16 + 9 For unit vectors i, j,k we have = 50 unit ɵ ɵ ɵ ɵ ɵ ɵ i ⋅ i = j ⋅ j = k ⋅ k = 1 16 16 ɵ ɵ ɵ ɵ ɵ ɵ ∴ cos θ = 50 50 = 50 = 0.32 , i ⋅ j = j ⋅ k = k ⋅ i = 0 Given two vectors θ= cos–1 0.32 Fig. 5.1 (a) The scalar product of two vectors A and B is a scalar : A.B = A B cos θ. (b) B cos θis the projection of B onto A. (c) A cos θ is the projection of A onto B. Reprint 2025-26 WORK, ENERGY AND POWER 73

5.2 Notions of work and kinetic landscape, all are said to be working. In physics, however, energy : The work-energy the word ‘Work’ covers a definite and precise meaning. theorem Somebody who has the capacity to work for 14-16 hours a5.3 Work day is said to have a large stamina or energy. We admire a