Work, Energy and Power PDF

CHAPTER FIVE WORK, ENERGY AND POWER 5.1 INTRODUCTION The terms ‘work’, ‘energy’ and ‘power’ are frequently used in everyday language. A farmer ploughing the field, a...

CHAPTER FIVE WORK, ENERGY AND POWER 5.1 INTRODUCTION The terms ‘work’, ‘energy’ and ‘power’ are frequently used in everyday language. A farmer ploughing the field, a 5.1 Introduction construction worker carrying bricks, a student studying for a competitive examination, an artist painting a beautiful 5.2 Notions of work and kinetic landscape, all are said to be working. In physics, however, energy : The work-energy theorem the word ‘Work’ covers a definite and precise meaning. Somebody who has the capacity to work for 14-16 hours a 5.3 Work day is said to have a large stamina or energy. We admire a 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. Summary Before we proceed to this task, we need to develop a Points to ponder mathematical prerequisite, namely the scalar product of two 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 2024-25 72 PHYSICS A dot B) is defined as A = A x iɵ + Ay ɵj + Az kɵ A.B = A B cos θ (5.1a) B = B x iɵ + By ɵj + Bz kɵ where 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 quantity. Each vector, A and B, has a direction ( A.B = A x ˆi + Ay ˆj + Az k )( ˆ. B ˆi + B ˆj + B k x y z ˆ ) but their scalar product does not have a = A x B x + Ay By + Az 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 Ay + A z A z A.B = A (B cos θ ) = B (A cos θ ) Or, 2 2 A = A x + Ay + A z 2 2 (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 the magnitude of A and the component of B along u Example 5.1 Find the angle between force A. Alternatively, it is the product of the F = (3 ˆi + 4 ˆj - 5 k ˆ ) unit and displacement magnitude of B and the component of A along B. d = (5 ˆi + 4 ˆj + 3 k ˆ ) 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 = Fx d x + Fy d y + Fz d z Scalar product obeys the distributive = 3 (5) + 4 (4) + (– 5) (3) law: = 16 unit A. (B + C) = A.B + A.C Hence F.d = F d cos θ = 16 unit Further, A. (λ B) = λ (A.B) Now F.F = F 2 = Fx2 + Fy2 + Fz2 where λ is a real number. = 9 + 16 + 25 = 50 unit The proofs of the above equations are left to you as an exercise. and d.d = d 2 = d x2 + dy2 + dz2 For unit vectors ɵi, ɵj, kɵ we have = 25 + 16 + 9 = 50 unit iɵ ⋅ ɵi = ɵj ⋅ ɵj = kɵ ⋅ kɵ = 1 16 16 ∴ cos θ = = = 0.32 , iɵ ⋅ ɵj = ɵj ⋅ kɵ = kɵ ⋅ iɵ = 0 50 50 50 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 θ is the projection of B onto A. (c) A cos θ is the projection of A onto B. 2024-25 WORK, ENERGY AND POWER 73 5.2 NOTIONS OF WORK AND KINETIC ENERGY: THE WORK-ENERGY THEOREM to be proportional to the speed of the drop 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 1 1 mv 2 − mu 2 = mas = Fs (5.2a) ∆K = m v2 − 0 2 2 2 where the last step follows from Newton’s Second 1 Law. We can generalise Eq. (5.2) to three = × 10-3 × 50 × 50 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 Wg = mgh mv 2 − mu 2 = m a.d = F.d (5.2b) 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 + Wr 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 2024-25 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 (= 90o), cos (π/2) = 0. For the block moving on = – 2000 J a smooth horizontal table, the gravitational It is this negative work that brings the cycle force mg does no work since it acts at right to a halt in accordance with WE theorem. angles to the displacement. If we assume that (b) From Newton’s Third Law an equal and the moon’s orbits around the earth is opposite force acts on the road due to the perfectly circular then the earth’s cycle. Its magnitude is 200 N. However, the gravitational force does no work. The moon’s road undergoes no displacement. Thus, instantaneous displacement is tangential work done by cycle on the road is zero. ⊳ 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). From Eq. (5.4) it is clear that work and energy As noted earlier, if an object of mass m has have the same dimensions, [ML2T–2]. The SI unit velocity v, its kinetic energy K is of these is joule (J), named after the famous British 1 1 physicist James Prescott Joule (1811-1869). Since K = m v.v = mv 2 (5.5) work and energy are so widely used as physical 2 2 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 2024-25 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 xf ships employ the kinetic energy of the wind. Table 5.2 lists the kinetic energies for various W≅ ∑F (x )∆x xi (5.6) 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 ? lim xf Answer The initial kinetic energy of the bullet W = lim ∆x → 0 ∑F (x )∆x xi is mv2/2 = 1000 J. It has a final kinetic energy of 0.1×1000 = 100 J. If vf is the emergent speed xf of the bullet, = ∫ F ( x ) dx (5.7) 1 2 xi 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 vf = 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%). ⊳ 5.5 WORK DONE BY A VARIABLE FORCE A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 5.3 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 Fig. 5.3(a) 2024-25 76 PHYSICS The work done by the frictional force is Wf → area of the rectangle AGHI Wf = (−50) × 20 = − 1000 J The area on the negative side of the force axis has a negative sign. ⊳ 5.6 THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE Fig. 5.3 (a) The shaded rectangle represents the We are now familiar with the concepts of work work done by the varying force F(x), over and kinetic energy to prove the work-energy the small displacement ∆x, ∆W = F(x) ∆x. theorem for a variable force. We confine (b) adding the areas of all the rectangles we ourselves to one dimension. The time rate of find that for ∆x → 0, the area under the curve change of kinetic energy is is exactly equal to the work done by F(x). dK d 1 2 =  m v  ⊳ Example 5.5 A woman pushes a trunk on dt dt  2 a railway platform which has a rough dv surface. She applies a force of 100 N over a =m v distance of 10 m. Thereafter, she gets dt progressively tired and her applied force =F v (from Newton’s Second Law) reduces linearly with distance to 50 N. The dx total distance through which the trunk has =F dt been moved is 20 m. Plot the force applied Thus by the woman and the frictional force, which dK = Fdx is 50 N versus displacement. Calculate the Integrating from the initial position (x i ) to final work done by the two forces over 20 m. position ( x f ), we have Answer Kf xf ∫ dK = ∫ Fdx Ki xi where, Ki and K f are the initial and final kinetic energies corresponding to x i and x f. xf Fig. 5.4 Plot of the force F applied by the woman and or K f − Ki = ∫ Fdx (5.8a) xi the opposing frictional force f versus From Eq. (5.7), it follows that displacement. Kf − Ki = W (5.8b) The plot of the applied force is shown in Fig. 5.4. At x = 20 m, F = 50 N (≠ 0). We are given Thus, the WE theorem is proved for a variable that the frictional force f is |f|= 50 N. It opposes force. motion and acts in a direction opposite to F. It While the WE theorem is useful in a variety of is therefore, shown on the negative side of the problems, it does not, in general, incorporate the force axis. complete dynamical information of Newton’s The work done by the woman is second law. It is an integral form of Newton’s WF → area of the rectangle ABCD + area of second law. Newton’s second law is a relation the trapezium CEID between acceleration and force at any instant of 1 time. Work-energy theorem involves an integral WF = 100 × 10 + (100 + 50) × 10 over an interval of time. In this sense, the temporal 2 = 1000 + 750 (time) information contained in the statement of = 1750 J Newton’s second law is ‘integrated over’ and is 2024-25 WORK, ENERGY AND POWER 77 not available explicitly. Another observation is that are like ‘compressed springs’. They possess a Newton’s second law for two or three dimensions large amount of potential energy. An earthquake is in vector form whereas the work-energy results when these fault lines readjust. Thus, theorem is in scalar form. In the scalar form, potential energy is the ‘stored energy’ by virtue information with respect to directions contained of the position or configuration of a body. The in Newton’s second law is not present. body left to itself releases this stored energy in ⊳ the form of kinetic energy. Let us make our notion Example 5.6 A block of mass m = 1 kg, of potential energy more concrete. moving on a horizontal surface with speed The gravitational force on a ball of mass m is vi = 2 m s–1 enters a rough patch ranging mg. g may be treated as a constant near the earth from x = 0.10 m to x = 2.01 m. The retarding surface. By ‘near’ we imply that the height h of force Fr on the block in this range is inversely the ball above the earth’s surface is very small proportional to x over this range, compared to the earth’s radius RE (h 0 xm xm and Fs < 0 (c) For the compressed spring Ws = ∫ Fs dx = − ∫ kx dx x < 0 and Fs > 0.(d) The plot of Fs versus x. 0 0 The area of the shaded triangle represents the work done by the spring force. Due to the 2 k xm opposing signs of Fs and x, this work done is =− (5.15) 2 2 negative, Ws = −kx m / 2. This expression may also be obtained by considering the area of the triangle as in The same is true when the spring is Fig. 5.7(d). Note that the work done by the compressed with a displacement xc (< 0). The external pulling force F is positive since it overcomes the spring force. spring force does work Ws = − kx c2 / 2 while the 2024-25 WORK, ENERGY AND POWER 81 external force F does work + kx c2 / 2. If the block and vice versa, however, the total mechanical energy remains constant. This is graphically is moved from an initial displacement xi to a depicted in Fig. 5.8. final displacement xf , the work done by the spring force Ws is xf k x i2 k x 2f Ws = − ∫ k x dx = − (5.17) xi 2 2 Thus the work done by the spring force depends only on the end points. Specifically, if the block is pulled from xi and allowed to return to xi ; xi k x i2 k x i2 Ws = − ∫ k x dx = − xi 2 2 =0 (5.18) Fig. 5.8 Parabolic plots of the potential energy V and The work done by the spring force in a cyclic kinetic energy K of a block attached to a process is zero. We have explicitly demonstrated spring obeying Hooke’s law. The two plots are complementary, one decreasing as the that the spring force (i) is position dependent other increases. The total mechanical only as first stated by Hooke, (Fs = − kx); (ii) energy E = K + V remains constant. does work which only depends on the initial and final positions, e.g. Eq. (5.17). Thus, the spring Example 5.8 To simulate car accidents, auto ⊳ force is a conservative force. We define the potential energy V(x) of the spring manufacturers study the collisions of moving to be zero when block and spring system is in the cars with mounted springs of different spring equilibrium position. For an extension (or constants. Consider a typical simulation with compression) x the above analysis suggests that a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding kx 2 with a horizontally mounted spring of spring V(x) = (5.19) constant 5.25 × 103 N m–1. What is the 2 You may easily verify that − dV/dx = − k x, the maximum compression of the spring ? spring force. If the block of mass m in Fig. 5.7 is extended to xm and released from rest, then its Answer At maximum compression the kinetic total mechanical energy at any arbitrary point x, energy of the car is converted entirely into the where x lies between – xm and + xm, will be given by potential energy of the spring. The kinetic energy of the moving car is 1 2 1 1 k xm = k x 2 + m v2 1 2 2 2 K = mv 2 where we have invoked the conservation of 2 mechanical energy. This suggests that the speed 1 and the kinetic energy will be maximum at the = × 103 × 5 × 5 2 equilibrium position, x = 0, i.e., K = 1.25 × 104 J 1 1 2 m vm = k xm 2 where we have converted 18 km h–1 to 5 m s–1 [It is 2 2 useful to remember that 36 km h–1 = 10 m s–1]. where vm is the maximum speed. At maximum compression x m, the potential energy V of the spring is equal to the kinetic k or vm = xm energy K of the moving car from the principle of m conservation of mechanical energy. Note that k/m has the dimensions of [T-2] and our equation is dimensionally correct. The 1 V = k xm 2 kinetic energy gets converted to potential energy 2 2024-25 82 PHYSICS = 1.25 × 104 J We obtain xm = 2.00 m We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction. ⊳ We conclude this section by making a few Fig. 5.9 The forces acting on the car. remarks on conservative forces. (i) Information on time is absent from the above 1 discussions. In the example considered ∆K = Kf − Ki = 0 − m v 2 2 above, we can calculate the compression, but The work done by the net force is not the time over which the compression 1 occurs. A solution of Newton’s Second Law W =− kx m2 − µm g x m for this system is required for temporal 2 information. Equating we have (ii) Not all forces are conservative. Friction, for 1 1 example, is a non-conservative force. The m v 2 = k x m2 + µm g x m 2 2 principle of conservation of energy will have to be modified in this case. This is illustrated Now µmg = 0.5 × 103 × 10 = 5 × 103 N (taking in Example 5.9. g =10.0 m s -2). After rearranging the above (iii) The zero of the potential energy is arbitrary. equation we obtain the following quadratic equation in the unknown xm. It is set according to convenience. For the spring force we took V(x) = 0, at x = 0, i.e. the k x m2 + 2µm g x m − m v 2 = 0 unstretched spring had zero potential energy. For the constant gravitational force mg, we took V = 0 on the earth’s surface. In a later chapter we shall see that for the force due to the universal law of gravitation, the where we take the positive square root since zero is best defined at an infinite distance xm is positive. Putting in numerical values we from the gravitational source. However, once obtain the zero of the potential energy is fixed in a xm = 1.35 m given discussion, it must be consistently which, as expected, is less than the result in adhered to throughout the discussion. You Example 5.8. cannot change horses in midstream ! If the two forces on the body consist of a conservative force Fc and a non-conservative ⊳ Example 5.9 Consider Example 5.8 taking force Fnc , the conservation of mechanical energy the coefficient of friction, µ, to be 0.5 and formula will have to be modified. By the WE calculate the maximum compression of the theorem spring. (Fc+ Fnc ) ∆x = ∆K But Fc ∆x = − ∆V Answer In presence of friction, both the spring Hence, ∆(K + V) = Fnc ∆x force and the frictional force act so as to oppose ∆E = Fnc ∆x the compression of the spring as shown in where E is the total mechanical energy. Over Fig. 5.9. the path this assumes the form We invoke the work-energy theorem, rather Ef − Ei = Wnc than the conservation of mechanical energy. where W nc is the total work done by the The change in kinetic energy is non-conservative forces over the path. Note that 2024-25 WORK, ENERGY AND POWER 83 unlike the conservative force, Wnc depends on Our electricity bills carry the energy the particular path i to f. ⊳ consumption in units of kWh. Note that kWh is a unit of energy and not of power. 5.10 POWER Often it is interesting to know not only the work u Example 5.10 An elevator can carry a done on an object, but also the rate at which maximum load of 1800 kg (elevator + this work is done. We say a person is physically passengers) is moving up with a constant fit if he not only climbs four floors of a building speed of 2 m s–1. The frictional force opposing but climbs them fast. Power is defined as the the motion is 4000 N. Determine the time rate at which work is done or energy is minimum power delivered by the motor to transferred. the elevator in watts as well as in horse The average power of a force is defined as the power. ratio of the work, W, to the total time t taken Answer The downward force on the elevator is W Pav = F = m g + Ff = (1800 × 10) + 4000 = 22000 N t The instantaneous power is defined as the The motor must supply enough power to balance limiting value of the average power as time this force. Hence, interval approaches zero, P = F. v = 22000 × 2 = 44000 W = 59 hp ⊳ dW 5.11 COLLISIONS P = (5.20) dt In physics we study motion (change in position). The work dW done by a force F for a displacement At the same time, we try to discover physical dr is dW = F.dr. The instantaneous power can quantities, which do not change in a physical also be expressed as process. The laws of momentum and energy conservation are typical examples. In this dr P = F. section we shall apply these laws to a commonly dt encountered phenomena, namely collisions. = F.v (5.21) Several games such as billiards, marbles or carrom involve collisions.We shall study the where v is the instantaneous velocity when the collision of two masses in an idealised form. force is F. Consider two masses m1 and m2. The particle Power, like work and energy, is a scalar m1 is moving with speed v1i , the subscript ‘i’ quantity. Its dimensions are [ML2T – 3]. In the SI, implying initial. We can cosider m2 to be at rest. its unit is called a watt (W). The watt is 1 J s–1. No loss of generality is involved in making such The unit of power is named after James Watt, a selection. In this situation the mass m 1 one of the innovators of the steam engine in the collides with the stationary mass m2 and this eighteenth century. is depicted in Fig. 5.10. There is another unit of power, namely the horse-power (hp) 1 hp = 746 W This unit is still used to describe the output of automobiles, motorbikes, etc. We encounter the unit watt when we buy electrical goods such as bulbs, heaters and refrigerators. A 100 watt bulb which is on for 10 hours uses 1 kilowatt hour (kWh) of energy. 100 (watt) × 10 (hour) Fig. 5.10 Collision of mass m1, with a stationary mass m2. = 1000 watt hour The masses m 1 and m 2 fly-off in different =1 kilowatt hour (kWh) directions. We shall see that there are = 103 (W) × 3600 (s) relationships, which connect the masses, the = 3.6 × 106 J velocities and the angles. 2024-25 84 PHYSICS 5.11.1 Elastic and Inelastic Collisions The loss in kinetic energy on collision is In all collisions the total linear momentum is 1 1 conserved; the initial momentum of the system ∆K = m1v1i2 − (m1 + m 2 )v 2f 2 2 is equal to the final momentum of the system. One can argue this as follows. When two objects 1 1 m12 collide, the mutual impulsive forces acting over = m1v12i − v12i [using Eq. (5.22)] the collision time ∆t cause a change in their 2 2 m1 + m 2 respective momenta : 1  m1  ∆p1 = F12 ∆t = m1v12i 1 −  2  m 1 + m2  ∆p2 = F21 ∆t where F12 is the force exerted on the first particle 1 m1m 2 2 by the second particle. F21 is likewise the force = v1i exerted on the second particle by the first particle. 2 m1 + m 2 Now from Newton’s third law, F12 = − F21. This implies which is a positive quantity as expected. Consider next an elastic collision. Using the ∆p1 + ∆p2 = 0 above nomenclature with θ 1 = θ 2 = 0, the The above conclusion is true even though the momentum and kinetic energy conservation forces vary in a complex fashion during the equations are collision time ∆t. Since the third law is true at every instant, the total impulse on the first object m1v1i = m1v1f + m2v2f (5.23) is equal and opposite to that on the second. m 1v12i = m 1v12f + m 2v 22 f (5.24) On the other hand, the total kinetic energy of the system is not necessarily conserved. The From Eqs. (5.23) and (5.24) it follows that, impact and deformation during collision may generate heat and sound. Part of the initial kinetic m 1v1i (v 2 f − v1i ) = m1v1 f (v 2 f − v1 f ) energy is transformed into other forms of energy. A useful way to visualise the deformation during or, v 2 f (v1i − v1 f ) = v12i − v12f collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains = (v1i − v1 f )(v1i + v1 f ) its original shape without loss in energy, then the initial kinetic energy is equal to the final Hence, ∴ v 2 f = v1i + v1 f (5.25) kinetic energy but the kinetic energy during the collision time ∆t is not constant. Such a collision Substituting this in Eq. (5.23), we obtain is called an elastic collision. On the other hand (m 1 − m 2 ) v1 f = v1i (5.26) the deformation may not be relieved and the two m1 + m 2 bodies could move together after the collision. A 2m1v1i collision in which the two particles move together and v2 f = (5.27) after the collision is called a completely inelastic m1 + m 2 collision. The intermediate case where the Thus, the ‘unknowns’ {v1f, v2f} are obtained in deformation is partly relieved and some of the terms of the ‘knowns’ {m1, m2, v1i}. Special cases initial kinetic energy is lost is more common and of our analysis are interesting. is appropriately called an inelastic collision. Case I : If the two masses are equal 5.11.2 Collisions in One Dimension v1f = 0 Consider first a completely inelastic collision v2f = v1i in one dimension. Then, in Fig. 5.10, The first mass comes to rest and pushes off the θ1 =θ2 = 0 second mass with its initial speed on collision. m1v1i = (m1+m2)vf (momentum conservation) Case II : If one mass dominates, e.g. m2 > > m1 v1f ~ − v1i v2f ~ 0 m1 The heavier mass is undisturbed while the vf = v1i (5.22) m1 + m 2 lighter mass reverses its velocity. 2024-25 WORK, ENERGY AND POWER 85 ⊳ dimensional, where the initial velocities and the Example 5.11 Slowing down of neutrons: final velocities lie in a plane. In a nuclear reactor a neutron of high speed (typically 107 m s–1) must be slowed 5.11.3 Collisions in Two Dimensions to 103 m s–1 so that it can have a high Fig. 5.10 also depicts the collision of a moving probability of interacting with isotope U 235 92 mass m1 with the stationary mass m2. Linear and causing it to fission. Show that a momentum is conserved in such a collision. neutron can lose most of its kinetic energy Since momentum is a vector this implies three in an elastic collision with a light nuclei equations for the three directions {x, y, z}. like deuterium or carbon which has a mass Consider the plane determined by the final of only a few times the neutron mass. The velocity directions of m1 and m2 and choose it to material making up the light nuclei, usually be the x-y plane. The conservation of the heavy water (D2O) or graphite, is called a z-component of the linear momentum implies moderator. that the entire collision is in the x-y plane. The x- and y-component equations are Answer The initial kinetic energy of the neutron m1v1i = m1v1f cos θ 1 + m2v2f cos θ 2 (5.28) is 0 = m1v1f sin θ1 − m2v2f sin θ2 (5.29) 1 K1i = m1v12i 2 One knows {m1, m2, v1i} in most situations. There are thus four unknowns {v1f , v2f , θ1 and θ2}, and while its final kinetic energy from Eq. (5.26) only two equations. If θ 1 = θ 2 = 0, we regain 2 1 1  m − m2  2 Eq. (5.23) for one dimensional collision. K1 f = m 1v12f = m1  1 v1i 2 2  m1 + m 2  If, further the collision is elastic, 1 1 1 m1v1i 2 = m1v1 f 2 + m2v2 f 2 (5.30) The fractional kinetic energy lost is 2 2 2 2 We obtain an additional equation. That still K1 f m − m2  f1 = = 1 leaves us one equation short. At least one of K1i  m 1 + m 2  the four unknowns, say θ 1, must be made known while the fractional kinetic energy gained by the for the problem to be solvable. For example, θ1 moderating nuclei K2f /K1i is can be determined by moving a detector in an angular fashion from the x to the y axis. Given f2 = 1 − f1 (elastic collision) {m1, m2, v1i , θ1} we can determine {v1f , v2f , θ2} 4m1m 2 from Eqs. (5.28)-(5.30). = (m1 + m 2 )2 ⊳ Example 5.12 Consider the collision depicted in Fig. 5.10 to be between two One can also verify this result by substituting billiard balls with equal masses m1 = m2. from Eq. (5.27). The first ball is called the cue while the For deuterium m 2 = 2m 1 and we obtain second ball is called the target. The f1 = 1/9 while f2 = 8/9. Almost 90% of the billiard player wants to ‘sink’ the target neutron’s energy is transferred to deuterium. For ball in a corner pocket, which is at an carbon f1 = 71.6% and f2 = 28.4%. In practice, angle θ2 = 37°. Assume that the collision is however, this number is smaller since head-on elastic and that friction and rotational collisions are rare. ⊳ motion are not important. Obtain θ 1. If the initial velocities and final velocities of both the bodies are along the same straight line, Answer From momentum conservation, since the masses are equal then it is called a one-dimensional collision, or head-on collision. In the case of small spherical v1i = v1f + v 2f bodies, this is possible if the direction of travel of body 1 passes through the centre of body 2 or ( )( v 1i 2 = v1 f + v 2 f ⋅ v1 f + v 2 f ) which is at rest. In general, the collision is two- = v1 f 2 + v 2 f 2 + 2 v1 f.v2 f 2024-25 86 PHYSICS = {v 1f 2 + v 2 f 2 + 2v1 f v 2 f cos (θ1 + 37° ) } (5.31) The matter simplifies greatly if we consider spherical masses with smooth surfaces, and Since the collision is elastic and m1 = m2 it follows assume that collision takes place only when the from conservation of kinetic energy that bodies touch each other. This is what happens in the games of marbles, carrom and billiards. v1i 2 = v1 f 2 + v 2 f 2 (5.32) In our everyday world, collisions take place only Comparing Eqs. (5.31) and (5.32), we get when two bodies touch each other. But consider a comet coming from far distances to the sun, or cos (θ1 + 37°) = 0 alpha particle coming towards a nucleus and or θ1 + 37° = 90° going away in some direction. Here we have to deal with forces involving action at a distance. Thus, θ1 = 53° Such an event is called scattering. The velocities This proves the following result : when two equal and directions in which the two particles go away masses undergo a glancing elastic collision with depend on their initial velocities as well as the one of them at rest, after the collision, they will type of interaction between them, their masses, move at right angles to each other. ⊳ shapes and sizes. SUMMARY 1. The work-energy theorem states that the change in kinetic energy of a body is the work done by the net force on the body. Kf - Ki = Wnet 2. A force is conservative if (i) work done by it on an object is path independent and depends only on the end points {xi, xj}, or (ii) the work done by the force is zero for an arbitrary closed path taken by the object such that it returns to its initial position. 3. For a conservative force in one dimension, we may define a potential energy function V(x) such that dV ( x ) F (x ) = − dx xf or Vi − V f = ∫ F ( x ) dx xi 4. The principle of conservation of mechanical energy states that the total mechanical energy of a body remains constant if the only forces that act on the body are conservative. 5. The gravitational potential energy of a particle of mass m at a height x about the earth’s surface is V(x) = m g x where the variation of g with height is ignored. 5. The elastic potential energy of a spring of force constant k and extension x is 1 V (x ) = k x2 2 7. The scalar or dot product of two vectors A and B is written as A.B and is a scalar quantity given by :A.B = AB cos θ, where θ is the angle between A and B. It can be positive, negative or zero depending upon the value of θ. The scalar product of two vectors can be interpreted as the product of magnitude of one vector and component of the other vector along the first vector. For unit vectors : ˆi ⋅ ˆi = ˆj ⋅ ˆj = k ˆ ⋅k ˆ = 1 and ˆi ⋅ ˆj = ˆj ⋅ k ˆ =k ˆ ⋅ ˆi = 0 Scalar products obey the commutative and the distributive laws. 2024-25 WORK, ENERGY AND POWER 87 POINTS TO PONDER 1. The phrase ‘calculate the work done’ is incomplete. We should refer (or imply clearly by context) to the work done by a specific force or a group of forces on a given body over a certain displacement. 2. Work done is a scalar quantity. It can be positive or negative unlike mass and kinetic energy which are positive scalar quantities. The work done by the friction or viscous force on a moving body is negative. 3. For two bodies, the sum of the mutual forces exerted between them is zero from Newton’s Third Law, F12 + F21 = 0 But the sum of the work done by the two forces need not always cancel, i.e. W12 + W21 ≠ 0 However, it may sometimes be true. 4. The work done by a force can be calculated sometimes even if the exact nature of the force is not known. This is clear from Example 5.2 where the WE theorem is used in such a situation. 5. The WE theorem is not independent of Newton’s Second Law. The WE theorem may be viewed as a scalar form of the Second Law. The principle of conservation of mechanical energy may be viewed as a consequence of the WE theorem for conservative forces. 5. The WE theorem holds in all inertial frames. It can also be extended to non- inertial frames provided we include the pseudoforces in the calculation of the net force acting on the body under consideration. 7. The potential energy of a body subjected to a conservative force is always undetermined upto a constant. For example, the point where the potential energy is zero is a matter of choice. For the gravitational potential energy mgh, the zero of the potential energy is chosen to be the ground. For the spring potential energy kx2/2 , the zero of the potential energy is the equilibrium position of the oscillating mass. 8. Every force encountered in mechanics does not have an associated potential energy. For example, work done by friction over a closed path is not zero and no potential energy can be associated with friction. 9. During a collision : (a) the total linear momentum is conserved at each instant of the collision ; (b) the kinetic energy conservation (even if the collision is elastic) applies after the collision is over and does not hold at every instant of the collision. In fact the two colliding objects are deformed and may be momentarily at rest with respect to each other. 2024-25 88 PHYSICS EXERCISES 5.1 The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest. 5.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results. 5.3 Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant. Fig. 5.11 2024-25 WORK, ENERGY AND POWER 89 5.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this Fig. 5.12 potential must ‘turn back’ when it reaches x = ± 2 m. 5.5 Answer the following : (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not Fig. 5.13 normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ? (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ? 5.6 Underline the correct alternative : (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. 5.7 State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. 5.8 Answer carefully, with reasons : (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ? 2024-25 90 PHYSICS (c) What are the answers to (a) and (b) for an inelastic collision ? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy). 5.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = −ˆi + 2 ˆj + 3 k ˆ N where ˆi, ˆj, k ˆ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ? 5.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J). 5.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ? 5.14 A molecule in a gas container hits a horizontal wall with speed 200 m s–1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ? 5.15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ? 5.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision ? Fig. 5.14 2024-25 WORK, ENERGY AND POWER 91 5.17 The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic. 5.18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ? 5.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a Fig. 5.15 frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty ? 5.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x3/2 where a = 5 m–1/2 s–1. What is the work done by the net force during its displacement from x = 0 to x=2m? 5.21 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ? (b) What is the kinetic energy of the air ? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ? 5.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force ? (b) Fat supplies 3.8 × 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up? 5.23 A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house. 2024-25