Page 1 CHAPTER SIX WORK, ENERGY AND POWER 6.1 INTRODUCTION 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. 6.1.1 The Scalar Product We have learnt about vectors and their use in Chapter 4. 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 7. 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 6.1 Introduction 6.2 Notions of work and kinetic energy : The work-energy theorem 6.3 Work 6.4 Kinetic energy 6.5 Work done by a variable force 6.6 The work-energy theorem for a variable force 6.7 The concept of potential energy 6.8 The conservation of mechanical energy 6.9 The potential energy of a spring 6.10 Various forms of energy : the law of conservation of energy 6.11 Power 6.12 Collisions Summary Points to ponder Exercises Additional exercises Appendix 6.1 2020-21 Page 2 CHAPTER SIX WORK, ENERGY AND POWER 6.1 INTRODUCTION 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. 6.1.1 The Scalar Product We have learnt about vectors and their use in Chapter 4. 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 7. 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 6.1 Introduction 6.2 Notions of work and kinetic energy : The work-energy theorem 6.3 Work 6.4 Kinetic energy 6.5 Work done by a variable force 6.6 The work-energy theorem for a variable force 6.7 The concept of potential energy 6.8 The conservation of mechanical energy 6.9 The potential energy of a spring 6.10 Various forms of energy : the law of conservation of energy 6.11 Power 6.12 Collisions Summary Points to ponder Exercises Additional exercises Appendix 6.1 2020-21 A dot B) is defined as A . B = A B cos ? (6.1a) where ? is the angle between the two vectors as shown in Fig. 6.1(a). Since A, B and cos ? are scalars, the dot product of A and B is a scalar quantity. Each vector, A and B, has a direction but their scalar product does not have a direction. From Eq. (6.1a), we have A . B = A (B cos ? ) = B (A cos ? ) Geometrically, B cos ? is the projection of B onto A in Fig.6.1 (b) and A cos ? is the projection of A onto B in Fig. 6.1 (c). So, A . B is the product of the magnitude of A and the component of B along A. Alternatively, it is the product of the magnitude of B and the component of A along B. Equation (6.1a) shows that the scalar product follows the commutative law : A . B = B . A Scalar product obeys the distributive law: A . (B + C) = A . B + A . C Further, A . (? B) = ? (A . B) where ? is a real number. The proofs of the above equations are left to you as an exercise. For unit vectors i, j,k we have i i j j k k · = · = · = 1 i j j k k i · = · = · = 0 Given two vectors A i j k = + + A A A x y z B i j k = + + B B B x y z their scalar product is ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ . . x y z x y z A A A B B B = + + + + A B i j k i j k = + + A B A B A B x x y y z z (6.1b) From the definition of scalar product and (Eq. 6.1b) we have : ( i ) x x y y z z A A A A A A = + + A A . . . . Or, A A A A 2 x 2 y 2 z 2 = + + (6.1c) since A . A = |A ||A| cos 0 = A 2 . (ii) A . B = 0, if A and B are perpendicular. Example 6.1 Find the angle between force F = (3 +4 -5 ) ˆ ˆ ˆ i j k unit and displacement d = (5 + 4 +3 ) ˆ ˆ ˆ i j k unit. Also find the projection of F on d. Answer F . d = x x y y z z F d F d F d + + = 3 (5) + 4 (4) + (– 5) (3) = 16 unit Hence F . d = cos F d ? = 16 unit Now F . F = 2 2 2 2 x y z F F F F = + + = 9 + 16 + 25 = 50 unit and d . d = d 2 = 2 2 2 x y z d d d + + = 25 + 16 + 9 = 50 unit ? cos ? = 16 16 = = 0.32 50 50 50 , ? = cos –1 0.32 Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A . B = A B cos ?. (b) B cos ? is the projection of B onto A. (c) A cos ? is the projection of A onto B. WORK, ENERGY AND POWER 115 u 2020-21 Page 3 CHAPTER SIX WORK, ENERGY AND POWER 6.1 INTRODUCTION 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. 6.1.1 The Scalar Product We have learnt about vectors and their use in Chapter 4. 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 7. 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 6.1 Introduction 6.2 Notions of work and kinetic energy : The work-energy theorem 6.3 Work 6.4 Kinetic energy 6.5 Work done by a variable force 6.6 The work-energy theorem for a variable force 6.7 The concept of potential energy 6.8 The conservation of mechanical energy 6.9 The potential energy of a spring 6.10 Various forms of energy : the law of conservation of energy 6.11 Power 6.12 Collisions Summary Points to ponder Exercises Additional exercises Appendix 6.1 2020-21 A dot B) is defined as A . B = A B cos ? (6.1a) where ? is the angle between the two vectors as shown in Fig. 6.1(a). Since A, B and cos ? are scalars, the dot product of A and B is a scalar quantity. Each vector, A and B, has a direction but their scalar product does not have a direction. From Eq. (6.1a), we have A . B = A (B cos ? ) = B (A cos ? ) Geometrically, B cos ? is the projection of B onto A in Fig.6.1 (b) and A cos ? is the projection of A onto B in Fig. 6.1 (c). So, A . B is the product of the magnitude of A and the component of B along A. Alternatively, it is the product of the magnitude of B and the component of A along B. Equation (6.1a) shows that the scalar product follows the commutative law : A . B = B . A Scalar product obeys the distributive law: A . (B + C) = A . B + A . C Further, A . (? B) = ? (A . B) where ? is a real number. The proofs of the above equations are left to you as an exercise. For unit vectors i, j,k we have i i j j k k · = · = · = 1 i j j k k i · = · = · = 0 Given two vectors A i j k = + + A A A x y z B i j k = + + B B B x y z their scalar product is ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ . . x y z x y z A A A B B B = + + + + A B i j k i j k = + + A B A B A B x x y y z z (6.1b) From the definition of scalar product and (Eq. 6.1b) we have : ( i ) x x y y z z A A A A A A = + + A A . . . . Or, A A A A 2 x 2 y 2 z 2 = + + (6.1c) since A . A = |A ||A| cos 0 = A 2 . (ii) A . B = 0, if A and B are perpendicular. Example 6.1 Find the angle between force F = (3 +4 -5 ) ˆ ˆ ˆ i j k unit and displacement d = (5 + 4 +3 ) ˆ ˆ ˆ i j k unit. Also find the projection of F on d. Answer F . d = x x y y z z F d F d F d + + = 3 (5) + 4 (4) + (– 5) (3) = 16 unit Hence F . d = cos F d ? = 16 unit Now F . F = 2 2 2 2 x y z F F F F = + + = 9 + 16 + 25 = 50 unit and d . d = d 2 = 2 2 2 x y z d d d + + = 25 + 16 + 9 = 50 unit ? cos ? = 16 16 = = 0.32 50 50 50 , ? = cos –1 0.32 Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A . B = A B cos ?. (b) B cos ? is the projection of B onto A. (c) A cos ? is the projection of A onto B. WORK, ENERGY AND POWER 115 u 2020-21 PHYSICS 116 t 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 2 1 0 2 K m v ? = - = 1 2 10 50 50 -3 × × × = 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 g r K W W ? = + where W r 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. t 6.3 WORK As seen earlier, work is related to force and the displacement over which it acts. Consider a constant force F acting on an object of mass m. The object undergoes a displacement d in the positive x-direction as shown in Fig. 6.2. Fig. 6.2 An object undergoes a displacement d under the influence of the force F. 6.2 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 = 2 as (6.2) where u and v are the initial and final speeds and s the distance traversed. Multiplying both sides by m/2, we have 2 2 1 1 2 2 mv mu mas Fs - = = (6.2a) 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 = 2 a.d Here a and d are acceleration and displacement vectors of the object respectively. Once again multiplying both sides by m/2 , we obtain 2 2 1 1 2 2 mv mu m - = = a.d F.d (6.2b) 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. (6.2b) is then K f - K i = W (6.3) 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 6.2 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 2020-21 Page 4 CHAPTER SIX WORK, ENERGY AND POWER 6.1 INTRODUCTION 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. 6.1.1 The Scalar Product We have learnt about vectors and their use in Chapter 4. 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 7. 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 6.1 Introduction 6.2 Notions of work and kinetic energy : The work-energy theorem 6.3 Work 6.4 Kinetic energy 6.5 Work done by a variable force 6.6 The work-energy theorem for a variable force 6.7 The concept of potential energy 6.8 The conservation of mechanical energy 6.9 The potential energy of a spring 6.10 Various forms of energy : the law of conservation of energy 6.11 Power 6.12 Collisions Summary Points to ponder Exercises Additional exercises Appendix 6.1 2020-21 A dot B) is defined as A . B = A B cos ? (6.1a) where ? is the angle between the two vectors as shown in Fig. 6.1(a). Since A, B and cos ? are scalars, the dot product of A and B is a scalar quantity. Each vector, A and B, has a direction but their scalar product does not have a direction. From Eq. (6.1a), we have A . B = A (B cos ? ) = B (A cos ? ) Geometrically, B cos ? is the projection of B onto A in Fig.6.1 (b) and A cos ? is the projection of A onto B in Fig. 6.1 (c). So, A . B is the product of the magnitude of A and the component of B along A. Alternatively, it is the product of the magnitude of B and the component of A along B. Equation (6.1a) shows that the scalar product follows the commutative law : A . B = B . A Scalar product obeys the distributive law: A . (B + C) = A . B + A . C Further, A . (? B) = ? (A . B) where ? is a real number. The proofs of the above equations are left to you as an exercise. For unit vectors i, j,k we have i i j j k k · = · = · = 1 i j j k k i · = · = · = 0 Given two vectors A i j k = + + A A A x y z B i j k = + + B B B x y z their scalar product is ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ . . x y z x y z A A A B B B = + + + + A B i j k i j k = + + A B A B A B x x y y z z (6.1b) From the definition of scalar product and (Eq. 6.1b) we have : ( i ) x x y y z z A A A A A A = + + A A . . . . Or, A A A A 2 x 2 y 2 z 2 = + + (6.1c) since A . A = |A ||A| cos 0 = A 2 . (ii) A . B = 0, if A and B are perpendicular. Example 6.1 Find the angle between force F = (3 +4 -5 ) ˆ ˆ ˆ i j k unit and displacement d = (5 + 4 +3 ) ˆ ˆ ˆ i j k unit. Also find the projection of F on d. Answer F . d = x x y y z z F d F d F d + + = 3 (5) + 4 (4) + (– 5) (3) = 16 unit Hence F . d = cos F d ? = 16 unit Now F . F = 2 2 2 2 x y z F F F F = + + = 9 + 16 + 25 = 50 unit and d . d = d 2 = 2 2 2 x y z d d d + + = 25 + 16 + 9 = 50 unit ? cos ? = 16 16 = = 0.32 50 50 50 , ? = cos –1 0.32 Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A . B = A B cos ?. (b) B cos ? is the projection of B onto A. (c) A cos ? is the projection of A onto B. WORK, ENERGY AND POWER 115 u 2020-21 PHYSICS 116 t 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 2 1 0 2 K m v ? = - = 1 2 10 50 50 -3 × × × = 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 g r K W W ? = + where W r 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. t 6.3 WORK As seen earlier, work is related to force and the displacement over which it acts. Consider a constant force F acting on an object of mass m. The object undergoes a displacement d in the positive x-direction as shown in Fig. 6.2. Fig. 6.2 An object undergoes a displacement d under the influence of the force F. 6.2 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 = 2 as (6.2) where u and v are the initial and final speeds and s the distance traversed. Multiplying both sides by m/2, we have 2 2 1 1 2 2 mv mu mas Fs - = = (6.2a) 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 = 2 a.d Here a and d are acceleration and displacement vectors of the object respectively. Once again multiplying both sides by m/2 , we obtain 2 2 1 1 2 2 mv mu m - = = a.d F.d (6.2b) 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. (6.2b) is then K f - K i = W (6.3) 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 6.2 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 2020-21 WORK, ENERGY AND POWER 117 t Table 6.1 Alternative Units of Work/Energy in J Example 6.3 A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle ? (b) How much work does the cycle do on the road ? Answer Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road. (a) The stopping force and the displacement make an angle of 180 o (p rad) with each other. Thus, work done by the road, W r = Fd cos? = 200 × 10 × cos p = – 2000 J It is this negative work that brings the cycle to a halt in accordance with WE theorem. (b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is 200 N. However, the road undergoes no displacement. Thus, work done by cycle on the road is zero. t The lesson of Example 6.3 is that though the force on a body A exerted by the body B is always equal and opposite to that on B by A (Newton’s Third Law); the work done on A by B is not necessarily equal and opposite to the work done on B by A. 6.4 KINETIC ENERGY As noted earlier, if an object of mass m has velocity v, its kinetic energy K is 2 K m mv 1 1 = = 2 2 v v . (6.5) Kinetic energy is a scalar quantity. The kinetic energy of an object is a measure of the work an The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = (F cos ? )d = F.d (6.4) 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 your muscles are alternatively contracting and relaxing and internal energy is being used up and you do get tired. Thus, the meaning of work in physics is different from its usage in everyday language. No work is done if : (i) the displacement is zero as seen in the example above. A weightlifter holding a 150 kg mass steadily on his shoulder for 30 s does no work on the load during this time. (ii) the force is zero. A block moving on a smooth horizontal table is not acted upon by a horizontal force (since there is no friction), but may undergo a large displacement. (iii) the force and displacement are mutually perpendicular. This is so since, for ? = p/2 rad (= 90 o ), cos (p/2) = 0. For the block moving on a smooth horizontal table, the gravitational force mg does no work since it acts at right angles to the displacement. If we assume that the moon’s orbits around the earth is perfectly circular then the earth’s gravitational force does no work. The moon’s instantaneous displacement is tangential while the earth’s force is radially inwards and ? = p/2. Work can be both positive and negative. If ? is between 0 o and 90 o , cos ? in Eq. (6.4) is positive. If ? is between 90 o and 180 o , cos ? is negative. In many examples the frictional force opposes displacement and ? = 180 o . Then the work done by friction is negative (cos 180 o = –1). From Eq. (6.4) it is clear that work and energy have the same dimensions, [ML 2 T –2 ]. The SI unit of these is joule (J), named after the famous British physicist James Prescott Joule (1811-1869). Since work and energy are so widely used as physical concepts, alternative units abound and some of these are listed in Table 6.1. 2020-21 Page 5 CHAPTER SIX WORK, ENERGY AND POWER 6.1 INTRODUCTION 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. 6.1.1 The Scalar Product We have learnt about vectors and their use in Chapter 4. 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 7. 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 6.1 Introduction 6.2 Notions of work and kinetic energy : The work-energy theorem 6.3 Work 6.4 Kinetic energy 6.5 Work done by a variable force 6.6 The work-energy theorem for a variable force 6.7 The concept of potential energy 6.8 The conservation of mechanical energy 6.9 The potential energy of a spring 6.10 Various forms of energy : the law of conservation of energy 6.11 Power 6.12 Collisions Summary Points to ponder Exercises Additional exercises Appendix 6.1 2020-21 A dot B) is defined as A . B = A B cos ? (6.1a) where ? is the angle between the two vectors as shown in Fig. 6.1(a). Since A, B and cos ? are scalars, the dot product of A and B is a scalar quantity. Each vector, A and B, has a direction but their scalar product does not have a direction. From Eq. (6.1a), we have A . B = A (B cos ? ) = B (A cos ? ) Geometrically, B cos ? is the projection of B onto A in Fig.6.1 (b) and A cos ? is the projection of A onto B in Fig. 6.1 (c). So, A . B is the product of the magnitude of A and the component of B along A. Alternatively, it is the product of the magnitude of B and the component of A along B. Equation (6.1a) shows that the scalar product follows the commutative law : A . B = B . A Scalar product obeys the distributive law: A . (B + C) = A . B + A . C Further, A . (? B) = ? (A . B) where ? is a real number. The proofs of the above equations are left to you as an exercise. For unit vectors i, j,k we have i i j j k k · = · = · = 1 i j j k k i · = · = · = 0 Given two vectors A i j k = + + A A A x y z B i j k = + + B B B x y z their scalar product is ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ . . x y z x y z A A A B B B = + + + + A B i j k i j k = + + A B A B A B x x y y z z (6.1b) From the definition of scalar product and (Eq. 6.1b) we have : ( i ) x x y y z z A A A A A A = + + A A . . . . Or, A A A A 2 x 2 y 2 z 2 = + + (6.1c) since A . A = |A ||A| cos 0 = A 2 . (ii) A . B = 0, if A and B are perpendicular. Example 6.1 Find the angle between force F = (3 +4 -5 ) ˆ ˆ ˆ i j k unit and displacement d = (5 + 4 +3 ) ˆ ˆ ˆ i j k unit. Also find the projection of F on d. Answer F . d = x x y y z z F d F d F d + + = 3 (5) + 4 (4) + (– 5) (3) = 16 unit Hence F . d = cos F d ? = 16 unit Now F . F = 2 2 2 2 x y z F F F F = + + = 9 + 16 + 25 = 50 unit and d . d = d 2 = 2 2 2 x y z d d d + + = 25 + 16 + 9 = 50 unit ? cos ? = 16 16 = = 0.32 50 50 50 , ? = cos –1 0.32 Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A . B = A B cos ?. (b) B cos ? is the projection of B onto A. (c) A cos ? is the projection of A onto B. WORK, ENERGY AND POWER 115 u 2020-21 PHYSICS 116 t 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 2 1 0 2 K m v ? = - = 1 2 10 50 50 -3 × × × = 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 g r K W W ? = + where W r 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. t 6.3 WORK As seen earlier, work is related to force and the displacement over which it acts. Consider a constant force F acting on an object of mass m. The object undergoes a displacement d in the positive x-direction as shown in Fig. 6.2. Fig. 6.2 An object undergoes a displacement d under the influence of the force F. 6.2 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 = 2 as (6.2) where u and v are the initial and final speeds and s the distance traversed. Multiplying both sides by m/2, we have 2 2 1 1 2 2 mv mu mas Fs - = = (6.2a) 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 = 2 a.d Here a and d are acceleration and displacement vectors of the object respectively. Once again multiplying both sides by m/2 , we obtain 2 2 1 1 2 2 mv mu m - = = a.d F.d (6.2b) 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. (6.2b) is then K f - K i = W (6.3) 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 6.2 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 2020-21 WORK, ENERGY AND POWER 117 t Table 6.1 Alternative Units of Work/Energy in J Example 6.3 A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle ? (b) How much work does the cycle do on the road ? Answer Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road. (a) The stopping force and the displacement make an angle of 180 o (p rad) with each other. Thus, work done by the road, W r = Fd cos? = 200 × 10 × cos p = – 2000 J It is this negative work that brings the cycle to a halt in accordance with WE theorem. (b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is 200 N. However, the road undergoes no displacement. Thus, work done by cycle on the road is zero. t The lesson of Example 6.3 is that though the force on a body A exerted by the body B is always equal and opposite to that on B by A (Newton’s Third Law); the work done on A by B is not necessarily equal and opposite to the work done on B by A. 6.4 KINETIC ENERGY As noted earlier, if an object of mass m has velocity v, its kinetic energy K is 2 K m mv 1 1 = = 2 2 v v . (6.5) Kinetic energy is a scalar quantity. The kinetic energy of an object is a measure of the work an The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = (F cos ? )d = F.d (6.4) 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 your muscles are alternatively contracting and relaxing and internal energy is being used up and you do get tired. Thus, the meaning of work in physics is different from its usage in everyday language. No work is done if : (i) the displacement is zero as seen in the example above. A weightlifter holding a 150 kg mass steadily on his shoulder for 30 s does no work on the load during this time. (ii) the force is zero. A block moving on a smooth horizontal table is not acted upon by a horizontal force (since there is no friction), but may undergo a large displacement. (iii) the force and displacement are mutually perpendicular. This is so since, for ? = p/2 rad (= 90 o ), cos (p/2) = 0. For the block moving on a smooth horizontal table, the gravitational force mg does no work since it acts at right angles to the displacement. If we assume that the moon’s orbits around the earth is perfectly circular then the earth’s gravitational force does no work. The moon’s instantaneous displacement is tangential while the earth’s force is radially inwards and ? = p/2. Work can be both positive and negative. If ? is between 0 o and 90 o , cos ? in Eq. (6.4) is positive. If ? is between 90 o and 180 o , cos ? is negative. In many examples the frictional force opposes displacement and ? = 180 o . Then the work done by friction is negative (cos 180 o = –1). From Eq. (6.4) it is clear that work and energy have the same dimensions, [ML 2 T –2 ]. The SI unit of these is joule (J), named after the famous British physicist James Prescott Joule (1811-1869). Since work and energy are so widely used as physical concepts, alternative units abound and some of these are listed in Table 6.1. 2020-21 PHYSICS 118 t object can do by the virtue of its motion. This notion has been intuitively known for a long time. The kinetic energy of a fast flowing stream has been used to grind corn. Sailing ships employ the kinetic energy of the wind. Table 6.2 lists the kinetic energies for various objects. Example 6.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 m s -1 (see Table 6.2) on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet ? Answer The initial kinetic energy of the bullet is mv 2 /2 = 1000 J. It has a final kinetic energy of 0.1×1000 = 100 J. If v f is the emergent speed of the bullet, 1 2 = mv f 2 100 J kg 05 . 0 J 100 2 × = f v = 63.2 m s –1 The speed is reduced by approximately 68% (not 90%). t 6.5 WORK DONE BY A VARIABLE FORCE A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 6.3 is a plot of a varying force in one dimension. If the displacement ?x is small, we can take the force F (x) as approximately constant and the work done is then ?W =F (x) ?x Table 6.2 Typical kinetic energies (K) This is illustrated in Fig. 6.3(a). Adding successive rectangular areas in Fig. 6.3(a) we get the total work done as ( ) ? ? ? f i x x x x F W (6.6) where the summation is from the initial position x i to the final position x f . If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 6.3(b). Then the work done is = ( ) ? F x x x x i f d (6.7) where ‘lim’ stands for the limit of the sum when ?x tends to zero. Thus, for a varying force the work done can be expressed as a definite integral of force over displacement (see also Appendix 3.1). lim W = lim x ? ? ( ) ? ? f i x x x x F 0 Fig. 6.3(a) 2020-21Read More

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