3.7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of vr and vw is R as shown in the figure. The magnitude of R is 2 2 2 2 −1 −1 R = v r + v w = 35 + 12 m s = 37 m s The direction θ that R makes with the vertical is given by v w 12 tan θ = = = 0.343 Fig. 3.8 (a) Two non-colinear vectors a and b. v r 35 (b) Resolving a vector A in terms of vectors Or, θ = tan-1 ( 0.343 ) = 19° a and b. Therefore, the boy should hold his umbrella We say that A has been resolved into two in the vertical plane at an angle of about 19o component vectors λ a and µ b along a and b with the vertical towards the east. ⊳ Reprint 2025-26 32 PHYSICS respectively. Using this method one can resolve and A2 is parallel to ɵj, we have :a given vector into two component vectors along a set of two vectors – all the three lie in the same A1= Ax ɵi , A2 = Ay ɵj (3.11) plane. It is convenient to resolve a general vector where Ax and Ay are real numbers.along the axes of a rectangular coordinate system using vectors of unit magnitude. These Thus, A = Ax ɵi + Ay ɵj (3.12) are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and This is represented in Fig. 3.9(c). The quantities points in a particular direction. It has no Ax and Ay are called x-, and y- components of the dimension and unit. It is used to specify a vector A. Note that Ax is itself not a vector, but direction only. Unit vectors along the x-, y- and ɵi is a vector, and so is Ay ɵj. Using simplez-axes of a rectangular coordinate system are Ax trigonometry, we can express Ax and Ay in terms denoted by ɵi , ɵj and ˆk , respectively, as shown of the magnitude of A and the angle θ it makes in Fig. 3.9(a). with the x-axis : Since these are unit vectors, we have Ax = A cos θ Ay = A sin θ (3.13) ˆi  = ˆj  = ˆk =1 (3.9) As is clear from Eq. (3.13), a component of a These unit vectors are perpendicular to each vector can be positive, negative or zero other. In this text, they are printed in bold face depending on the value of θ. with a cap (^) to distinguish them from other Now, we have two ways to specify a vector A vectors. Since we are dealing with motion in two in a plane. It can be specified by : dimensions in this chapter, we require use of (i) its magnitude A and the direction θ it makes only two unit vectors. If we multiply a unit vector, with the x-axis; or say ˆn by a scalar, the result is a vector (ii) its components Ax and Ay λ = λ ˆn. In general, a vector A can be written as If A and θ are given, Ax and Ay can be obtained using Eq. (3.13). If Ax and Ay are given, A and θ A = |A| ˆn (3.10) can be obtained as follows : where ˆn is a unit vector along A. 2 2 2 2 2 2 A x + A y = A cos θ + A sin θ We can now resolve a vector A in terms = A2 of component vectors that lie along unit vectors iˆ and ɵj. Consider a vector A that lies in x-y Or, A = A 2x + Ay2 (3.14) plane as shown in Fig. 3.9(b). We draw lines from the head of A perpendicular to the coordinate Ay A y tan θ = , θ = tan− 1 axes as in Fig. 3.9(b), and get vectors A1 and A2 And A x A x (3.15) such that A1 + A2 = A. Since A1 is parallel to ɵi Fig. 3.9 (a) Unit vectors ɵi , ɵj and ɵk lie along the x-, y-, and z-axes. (b) A vector A is resolved into its components Ax and Ay along x-, and y- axes. (c) A1 and A2 expressed in terms of ɵi and ɵj . Reprint 2025-26 MOTION IN A PLANE 33 So far we have considered a vector lying in ɵ ɵ B = B x i + B y jan x-y plane. The same procedure can be used Let R be their sum. We haveto resolve a general vector A into three components along x-, y-, and z-axes in three R = A + B dimensions. If α, β, and γ are the angles* ɵ ɵ ɵ ɵ = + (3.19a) ( A x i + A y j ) ( B x i + B y j )between A and the x-, y-, and z-axes, respectively [Fig. 3.9(d)], we have Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (3.19a) as convenient to us : ɵ ɵ j (3.19b) R = ( A x + B x ) i + ( A y + B y ) ɵ ɵ SinceR = R x i + R y j (3.20) we have, R x = A x + B x , R y = A y + B y (3.21) Thus, each component of the resultant vector R is the sum of the corresponding components of A and B. In three dimensions, we have ɵ ɵ ɵ A = A x i + Ay j + A z k ɵ ɵ ɵ B = B x i + B y j + B z k (d) ɵ ɵ ɵ R = A + B = R x i + R y j + R z kFig. 3.9 (d) A vector A resolved into components along x-, y-, and z-axes with R x = A x + B x y = A y + B yA x = A cos α, A y = A cos β, A z = A cos γ (3.16a) R In general, we have R z = A z + B z (3.22) A = Ax ˆi + Ay ˆj + Az kˆ (3.16b) This method can be extended to addition and The magnitude of vector A is subtraction of any number of vectors. For A = A x2 + Ay2 + A z2 (3.16c) example, if vectors a, b and c are given as ɵ ɵ ɵ A position vector r can be expressed as a = a x i + a y j + a z k ɵ ɵ ɵ r = x i + y j + z k (3.17) ɵ ɵ ɵ b = b x i + b y j + b z k where x, y, and z are the components of r along ɵ ɵ ɵ x-, y-, z-axes, respectively. c = c x i + c y j + c z k (3.23a)

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) ∑

5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position ⊳ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )∆xAnswer The initial kinetic energy of the bullet ∆ x → 0 ∑ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1×1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = ∫F ( i 1 2 x mv f = 100 J where ‘lim’ stands for the limit of the sum when 2 ∆x tends to zero. Thus, for a varying force 2 × 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m s–1 The speed is reduced by approximately 68% (not 90%). ⊳