Work, Energy and Power PDF

CHAPTER SIX WORK, ENERGY AND POWER 6.1 INTRODUCTION The terms ‘work’, ‘energy’ and ‘power’ are frequently used in everyday language. A farmer p...

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 6.1 Introduction construction worker carrying bricks, a student studying for a competitive examination, an artist painting a beautiful 6.2 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 6.3 Work day is said to have a large stamina or energy. We admire a 6.4 Kinetic energy long distance runner for her stamina or energy. Energy is 6.5 Work done by a variable thus our capacity to do work. In Physics too, the term ‘energy’ force is related to work in this sense, but as said above the term 6.6 The work-energy theorem for ‘work’ itself is defined much more precisely. The word ‘power’ a variable force is used in everyday life with different shades of meaning. In 6.7 The concept of potential karate or boxing we talk of ‘powerful’ punches. These are energy delivered at a great speed. This shade of meaning is close to 6.8 The conservation of the meaning of the word ‘power’ used in physics. We shall mechanical energy find that there is at best a loose correlation between the 6.9 The potential energy of a physical definitions and the physiological pictures these spring terms generate in our minds. The aim of this chapter is to 6.10 Various forms of energy : the develop an understanding of these three physical quantities. law of conservation of energy Before we proceed to this task, we need to develop a 6.11 Power mathematical prerequisite, namely the scalar product of two 6.12 Collisions vectors. Summary 6.1.1 The Scalar Product Points to ponder We have learnt about vectors and their use in Chapter 4. Exercises Physical quantities like displacement, velocity, acceleration, Additional exercises force etc. are vectors. We have also learnt how vectors are Appendix 6.1 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 WORK, ENERGY AND POWER 115 A dot B) is defined as ˜ A = Ax ˜i + A y ˜j + Az k A.B = A B cos θ (6.1a) ˜ B = B x ˜i + B y ˜j + B z k where θ is the angle between the two vectors as shown in Fig. 6.1(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 ˜i + Ay ˜j + Az k )( ˜. B ˜i + B ˜j + B k x y z ˜ ) but their scalar product does not have a = A x B x + Ay By + Az B z (6.1b) direction. From the definition of scalar product and From Eq. (6.1a), we have (Eq. 6.1b) we have : (i) A. A = A x A x + Ay 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 (6.1c) Geometrically, B cos θ is the projection of B onto since A.A = |A ||A| cos 0 = A2. A in Fig.6.1 (b) and A cos θ is the projection of A (ii) A.B = 0, if A and B are perpendicular. onto B in Fig. 6.1 (c). So, A.B is the product of the magnitude of A and the component of B along u Example 6.1 Find the angle between force A. Alternatively, it is the product of the ˆ unit and displacement F = (3 $i + 4 $j – 5 k) magnitude of B and the component of A along B. ˆ unit. Also find the d = (5 $i + 4 $j + 3 k) Equation (6.1a) 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, λ 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 $i, $j, k$ we have = 25 + 16 + 9 = 50 unit i$ ⋅ $i = $j ⋅ $j = k$ ⋅ k$ = 1 16 16 ∴ cos θ = = = 0.32 , i$ ⋅ $j = $j ⋅ k$ = k$ ⋅ $i = 0 50 50 50 Given two vectors θ = 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 θ. (b) B cos θ is the projection of B onto A. (c) A cos θ is the projection of A onto B. 116 PHYSICS 6.2 NOTIONS OF WORK AND KINETIC ENERGY: THE WORK-ENERGY THEOREM known to be proportional to the speed of the drop but is otherwise undetermined. The following relation for rectilinear motion under Consider a drop of mass 1.00 g falling from constant acceleration a has been encountered a height 1.00 km. It hits the ground with in Chapter 3, a speed of 50.0 m s-1. (a) What is the work v2 − u2 = 2 as done by the gravitational force ? What is where u and v are the initial and final speeds the work done by the unknown resistive and s the distance traversed. Multiplying both force? sides by m/2, we have Answer (a) The change in kinetic energy of the 1 1 drop is mv 2 − mu 2 = mas = Fs (6.2a) 2 2 1 ∆K = m v2 − 0 where the last step follows from Newton’s 2 Second Law. We can generalise Eq. (6.1) 1 to three dimensions by employing = × 10-3 × 50 × 50 2 vectors = 1.25 J v2 − u2 = 2 a.d where we have assumed that the drop is initially at rest. Once again multiplying both sides by m/2 , we Assuming that g is a constant with a value obtain 10 m/s2, the work done by the gravitational force 1 1 is, mv 2 − mu 2 = m a.d = F.d (6.2b) 2 2 Wg = mgh The above equation provides a motivation for = 10-3 ×10 ×103 the definitions of work and kinetic energy. The = 10.0 J left side of the equation is the difference in the (b) From the work-energy theorem quantity ‘half the mass times the square of the speed’ from its initial value to its final value. We ∆ K = W g + Wr call each of these quantities the ‘kinetic energy’, where Wr is the work done by the resistive force denoted by K. The right side is a product of the on the raindrop. Thus displacement and the component of the force Wr = ∆K − Wg along the displacement. This quantity is called = 1.25 −10 ‘work’ and is denoted by W. Eq. (6.2b) is then = − 8.75 J Kf − Ki = W (6.3) is negative. ⊳ where Ki and Kf are respectively the initial and 6.3 WORK final kinetic energies of the object. Work refers As seen earlier, work is related to force and the to the force and the displacement over which it displacement over which it acts. Consider a acts. Work is done by a force on the body over constant force F acting on an object of mass m. a certain displacement. The object undergoes a displacement d in the Equation (6.2) is also a special case of the positive x-direction as shown in Fig. 6.2. 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 Fig. 6.2 An object undergoes a displacement d opposing resistive force. The latter is under the influence of the force F. WORK, ENERGY AND POWER 117 The work done by the force is defined to be Table 6.1 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 (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 ⊳ Example 6.3 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 6.3 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. (6.4) 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 6.4 KINETIC ENERGY by friction is negative (cos 180o = –1). From Eq. (6.4) 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 (6.5) 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 6.1. energy of an object is a measure of the work an 118 PHYSICS Table 6.2 Typical kinetic energies (K) object can do by the virtue of its motion. This This is illustrated in Fig. 6.3(a). Adding notion has been intuitively known for a long time. successive rectangular areas in Fig. 6.3(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 6.2 lists the kinetic energies for various W≅ ∑F (x )∆x xi (6.6) objects. where the summation is from the initial position xi to the final position xf. Example 6.4 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 6.2) 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. 6.3(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 (6.7) 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%). ⊳ 6.5 WORK DONE BY A VARIABLE FORCE A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 6.2 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. 6.3(a) WORK, ENERGY AND POWER 119 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. ⊳ 6.6 THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE Fig. 6.3 (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 6.5 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. 6.4 Plot of the force F applied by the woman and or K f − Ki = ∫ Fdx (6.8a) xi the opposing frictional force f versus From Eq. (6.7), it follows that displacement. Kf − Ki = W (6.8b) The plot of the applied force is shown in Fig. 6.4. 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 120 PHYSICS 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 6.6 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 ms–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 d x =− ∫ 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 =− (6.15) 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. 6.7(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 124 PHYSICS 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. 6.8. final displacement xf , the work done by the spring force Ws is xf k x i2 k x 2f Ws = − ∫ k x dx = − (6.17) 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 (6.18) Fig. 6.8 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. (6.17). Thus, the spring Example 6.8 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) = (6.19) constant 6.25 × 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. 6.7 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 2 1 2 m vm = k xm 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 xm, the potential energy V of the spring is equal to the kinetic k energy K of the moving car from the principle of or vm = xm m conservation of mechanical energy. Note that k/m has the dimensions of [T-2] and our equation is dimensionally correct. The 1 2 V = k xm kinetic energy gets converted to potential energy 2 WORK, ENERGY AND POWER 125 = 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. 6.9 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 6.9. 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 1/2 energy. For the constant gravitational force − µ m g +  µ2m 2 g 2 + m k v 2  mg, we took V = 0 on the earth’s surface. In xm = k 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 xm zero is best defined at an infinite distance is positive. Putting in numerical values we from the gravitational source. However, once obtain xm = 1.35 m the zero of the potential energy is fixed in a given discussion, it must be consistently which, as expected, is less than the result in adhered to throughout the discussion. You Example 6.8. cannot change horses in midstream ! If the two forces on the body consist of a conservative force Fc and a non-conservative ⊳ force Fnc , the conservation of mechanical energy Example 6.9 Consider Example 6.8 taking formula will have to be modified. By the WE the coefficient of friction, µ, to be 0.5 and theorem calculate the maximum compression of the 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. 6.9. 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 126 PHYSICS unlike the conservative force, Wnc depends on Chemical energy arises from the fact that the the particular path i to f. ⊳ molecules participating in the chemical reaction have different binding energies. A stable chemical 6.10 VARIOUS FORMS OF ENERGY : THE LAW compound has less energy than the separated parts. OF CONSERVATION OF ENERGY A chemical reaction is basically a rearrangement In the previous section we have discussed of atoms. If the total energy of the reactants is more mechanical energy. We have seen that it can be than the products of the reaction, heat is released classified into two distinct categories : one based and the reaction is said to be an exothermic on motion, namely kinetic energy; the other on reaction. If the reverse is true, heat is absorbed and configuration (position), namely potential energy. the reaction is endothermic. Coal consists of Energy comes in many a forms which transform carbon and a kilogram of it when burnt releases into one another in ways which may not often about 3 × 107 J of energy. be clear to us. Chemical energy is associated with the forces 6.10.1 Heat that give rise to the stability of substances. These forces bind atoms into molecules, molecules into We have seen that the frictional force is not a polymeric chains, etc. The chemical energy conservative force. However, work is associated with the force of friction, Example 6.5. A block of arising from the combustion of coal, cooking gas, mass m sliding on a rough horizontal surface wood and petroleum is indispensable to our daily with speed v0 comes to a halt over a distance x0. existence. The work done by the force of kinetic friction f 6.10.3 Electrical Energy over x 0 is –f x0. By the work-energy theorem The flow of electrical current causes bulbs to m vo2/2 = f x 0. If we confine our scope to glow, fans to rotate and bells to ring. There are mechanics, we would say that the kinetic energy laws governing the attraction and repulsion of of the block is ‘lost’ due to the frictional force. charges and currents, which we shall learn On examination of the block and the table we later. Energy is associated with an electric would detect a slight increase in their current. An urban Indian household consumes temperatures. The work done by friction is not about 200 J of energy per second on an average. ‘lost’, but is transferred as heat energy. This raises the internal energy of the block and the 6.10.4 The Equivalence of Mass and Energy table. In winter, in order to feel warm, we Till the end of the nineteenth century, physicists generate heat by vigorously rubbing our palms believed that in every physical and chemical together. We shall see later that the internal process, the mass of an isolated system is energy is associated with the ceaseless, often conserved. Matter might change its phase, e.g. random, motion of molecules. A quantitative idea glacial ice could melt into a gushing stream, but of the transfer of heat energy is obtained by matter is neither created nor destroyed; Albert noting that 1 kg of water releases about 42000 J Einstein (1879-1955) however, showed that mass of energy when it cools by10 °C. and energy are equivalent and are related by 6.10.2 Chemical Energy the relation One of the greatest technical achievements of E = m c2 (6.20) humankind occurred when we discovered how where c, the speed of light in vacuum is to ignite and control fire. We learnt to rub two approximately 3 ×108 m s–1. Thus, a staggering flint stones together (mechanical energy), got amount of energy is associated with a mere them to heat up and to ignite a heap of dry leaves kilogram of matter (chemical energy), which then provided E = 1× (3 ×108)2 J = 9 ×1016 J. sustained warmth. A matchstick ignites into a This is equivalent to the annual electrical output bright flame when struck against a specially of a large (3000 MW) power generating station. prepared chemical surface. The lighted matchstick, when applied to a firecracker, 6.10.5 Nuclear Energy results in a spectacular display of sound and The most destructive weapons made by man, the light. fission and fusion bombs are manifestations of WORK, ENERGY AND POWER 127 Table 6.3 Approximate energy associated with various phenomena the above equivalence of mass and energy [Eq. ⊳ Example 6.10 Examine Tables 6.1-6.3 (6.20)]. On the other hand the explanation of the and express (a) The energy required to life-nourishing energy output of the sun is also break one bond in DNA in eV; (b) The based on the above equation. In this case kinetic energy of an air molecule (10—21 J) effectively four light hydrogen nuclei fuse to form in eV; (c) The daily intake of a human adult a helium nucleus whose mass is less than the in kilocalories. sum of the masses of the reactants. This mass difference, called the mass defect ∆m is the Answer (a) Energy required to break one bond source of energy (∆m)c2. In fission, a heavy of DNA is nucleus like uranium 235 , is split by a neutron 92 U 10 −20 J ~ 0.06 eV into lighter nuclei. Once again the final mass is 1.6 × 10 −19 J/eV less than the initial mass and the mass difference translates into energy, which can be tapped to provide electrical energy as in nuclear power Note 0.1 eV = 100 meV (100 millielectron volt). plants (controlled nuclear fission) or can be (b) The kinetic energy of an air molecule is employed in making nuclear weapons 10 −21 J (uncontrolled nuclear fission). Strictly, the energy ~ 0.0062 eV ∆E released in a chemical reaction can also be 1.6 × 10 −19 J/eV related to the mass defect ∆m = ∆E/c2. However, This is the same as 6.2 meV. for a chemical reaction, this mass defect is much (c) The average human consumption in a day is smaller than for a nuclear reaction. Table 6.3 lists the total energies for a variety of events and 107 J ~ 2400 kcal phenomena. 4.2 × 103 J/kcal 128 PHYSICS We point out a common misconception created W by newspapers and magazines. They mention Pav = food values in calories and urge us to restrict t diet intake to below 2400 calories. What they The instantaneous power is defined as the should be saying is kilocalories (kcal) and not limiting value of the average power as time calories. A person consuming 2400 calories a interval approaches zero, day will soon starve to death! 1 food calorie is dW 1 kcal. ⊳ P = (6.21) dt 6.10.6 The Principle of Conservation of The work dW done by a force F for a displacement Energy dr is dW = F.dr. The instantaneous power can also be expressed as We have seen that the total mechanical energy dr of the system is conserved if the forces doing work P = F. on it are conservative. If some of the forces dt involved are non-conservative, part of the = F.v (6.22) mechanical energy may get transformed into where v is the instantaneous velocity when the other forms such as heat, light and sound. force is F. However, the total energy of an isolated system Power, like work and energy, is a scalar does not change, as long as one accounts for all quantity. Its dimensions are [ML2T –3]. In the SI, forms of energy. Energy may be transformed from its unit is called a watt (W). The watt is 1 J s–1. one form to another but the total energy of an The unit of power is named after James Watt, isolated system remains constant. Energy can one of the innovators of the steam engine in the neither be created, nor destroyed. eighteenth century. Since the universe as a whole may be viewed There is another unit of power, namely the as an isolated system, the total energy of the horse-power (hp) universe is constant. If one part of the universe 1 hp = 746 W loses energy, another part must gain an equal This unit is still used to describe the output of amount of energy. automobiles, motorbikes, etc. The principle of conservation of energy cannot We encounter the unit watt when we buy be proved. However, no violation of this principle electrical goods such as bulbs, heaters and has been observed. The concept of conservation refrigerators. A 100 watt bulb which is on for 10 and transformation of energy into various forms hours uses 1 kilowatt hour (kWh) of energy. links together various branches of physics, 100 (watt) × 10 (hour) chemistry and life sciences. It provides a = 1000 watt hour unifying, enduring element in our scientific =1 kilowatt hour (kWh) pursuits. From engineering point of view all = 103 (W) × 3600 (s) electronic, communication and mechanical = 3.6 × 106 J devices rely on some forms of energy Our electricity bills carry the energy transformation. consumption in units of kWh. Note that kWh is 6.11 POWER a unit of energy and not of power. Often it is interesting to know not only the work u Example 6.11 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 WORK, ENERGY AND POWER 129 Answer The downward force on the elevator is by the second particle. F21 is likewise the force F = m g + Ff = (1800 × 10) + 4000 = 22000 N exerted on the second particle by the first particle. Now from Newton’s third law, F12 = − F21. This The motor must supply enough power to balance implies this force. Hence, ∆p1 + ∆p2 = 0 P = F. v = 22000 × 2 = 44000 W = 59 hp ⊳ The above conclusion is true even though the 6.12 COLLISIONS forces vary in a complex fashion during the In physics we study motion (change in position). collision time ∆t. Since the third law is true at At the same time, we try to discover physical every instant, the total impulse on the first object quantities, which do not change in a physical is equal and opposite to that on the second. process. The laws of momentum and energy On the other hand, the total kinetic energy of conservation are typical examples. In this the system is not necessarily conserved. The section we shall apply these laws to a commonly impact and deformation during collision may encountered phenomena, namely collisions. generate heat and sound. Part of the initial kinetic Several games such as billiards, marbles or energy is transformed into other forms of energy. carrom involve collisions.We shall study the A useful way to visualise the deformation during collision of two masses in an idealised form. collision is in terms of a ‘compressed spring’. If Consider two masses m1 and m2. The particle the ‘spring’ connecting the two masses regains m1 is moving with speed v1i , the subscript ‘i’ its original shape without loss in energy, then implying initial. We can cosider m2 to be at rest. the initial kinetic energy is equal to the final No loss of generality is involved in making such kinetic energy but the kinetic energy during the a selection. In this situation the mass m 1 collision time ∆t is not constant. Such a collision collides with the stationary mass m2 and this is called an elastic collision. On the other hand is depicted in Fig. 6.10. the deformation may not be relieved and the two bodies could move together after the collision. A collision in which the two particles move together after the collision is called a completely inelastic collision. The intermediate case where the deformation is partly relieved and some of the initial kinetic energy is lost is more common and is appropriately called an inelastic collision. 6.12.2 Collisions in One Dimension Consider first a completely inelastic collision Fig. 6.10 Collision of mass m1, with a stationary mass m2. in one dimension. Then, in Fig. 6.10, The masses m 1 and m 2 fly-off in different θ1 =θ2 = 0 directions. We shall see that there are relationships, which connect the masses, the m1v1i = (m1+m2)vf (momentum conservation) velocities and the angles. m1 vf = v1i (6.23) 6.12.1 Elastic and Inelastic Collisions m1 + m 2 In all collisions the total linear momentum is The loss in kinetic energy on collision is conserved; the initial momentum of the system is equal to the final momentum of the system. 1 1 One can argue this as follows. When two objects ∆K = m1v1i2 − (m1 + m 2 )v 2f 2 2 collide, the mutual impulsive forces acting over the collision time ∆t cause a change in their 1 1 m12 respective momenta : = m1v12i − v12i [using Eq. (6.23)] 2 2 m1 + m 2 ∆p1 = F12 ∆t ∆p2 = F21 ∆t 1  m1  = m1v12i 1 −  where F12 is the force exerted on the first particle 2  m1 + m 2  130 PHYSICS An experiment on head-on collision In performing an experiment on collision on a horizontal surface, we face three difficulties. One, there will be friction and bodies will not travel with uniform velocities. Two, if two bodies of different sizes collide on a table, it would be difficult to arrange them for a head-on collision unless their centres of mass are at the same height above the surface. Three, it will be fairly difficult to measure velocities of the two bodies just before and just after collision. By performing this experiment in a vertical direction, all the three difficulties vanish. Take two balls, one of which is heavier (basketball/football/volleyball) and the other lighter (tennis ball/rubber ball/table tennis ball). First take only the heavier ball and drop it vertically from some height, say 1 m. Note to which it rises. This gives the velocities near the floor or ground, just before and just after the bounce (by using v 2 = 2 gh ). Hence you will get the coefficient of restitution. Now take the big ball and a small ball and hold them in your hands one over the other, with the heavier ball below the lighter one, as shown here. Drop them together, taking care that they remain together while falling, and see what happens. You will find that the heavier ball rises less than when it was dropped alone, while the lighter one shoots up to about 3 m. With practice, you will be able to hold the ball properly so that the lighter ball rises vertically up and does not fly sideways. This is head-on collision. You can try to find the best combination of balls which gives you the best effect. You can measure the masses on a standard balance. We leave it to you to think how you can determine the initial and final velocities of the balls. (m 1 − m 2 ) 1 m1m 2 2 v1 f = v1i (6.27) = v1i m1 + m 2 2 m1 + m 2 2m1v1i which is a positive quantity as expected. and v2 f = (6.28) m1 + m 2 Consider next an elastic collision. Using the Thus, the ‘unknowns’ {v1f, v2f} are obtained in above nomenclature with θ 1 = θ 2 = 0, the terms of the ‘knowns’ {m1, m2, v1i}. Special cases momentum and kinetic energy conservation of our analysis are interesting. equations are Case I : If the two masses are equal m1v1i = m1v1f + m2v2f (6.24) v1f = 0 v2f = v1i m v =m v 2 1 1i 2 1 1f +m v 2 2 2f (6.25) The first mass comes to rest and pushes off the From Eqs. (6.24) and (6.25) it follows that, second mass with its initial speed on collision. m 1v1i (v 2 f − v1i ) = m1v1 f (v 2 f − v1 f ) Case II : If one mass dominates, e.g. m2 > > m1 v1f ~ − v1i v2f ~ 0 or, v 2 f (v1i − v1 f ) = v12i − v12f The heavier mass is undisturbed while the lighter mass reverses its velocity. = (v1i − v1 f )(v1i + v1 f ) Example 6.12 Slowing down of neutrons: ⊳ Hence, ∴ v 2 f = v1i + v1 f (6.26) In a nuclear reactor a neutron of high speed (typically 107 m s–1) must be slowed Substituting this in Eq. (6.24), we obtain WORK, ENERGY AND POWER 131 to 103 m s–1 so that it can have a high 6.12.3 Collisions in Two Dimensions probability of interacting with isotope 235 Fig. 6.10 also depicts the collision of a moving 92 U 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 in an elastic collision with a light nuclei Since momentum is a vector this implies three like deuterium or carbon which has a mass equations for the three directions {x, y, z}. of only a few times the neutron mass. The Consider the plane determined by the final material making up the light nuclei, usually velocity directions of m1 and m2 and choose it to heavy water (D2O) or graphite, is called a be the x-y plane. The conservation of the moderator. z-component of the linear momentum implies that the entire collision is in the x-y plane. The Answer The initial kinetic energy of the neutron x- and y-component equations are is m1v1i = m1v1f cos θ 1 + m2v2f cos θ 2 (6.29) 1 0 = m1v1f sin θ1 − m2v2f sin θ2 K1i = m1v12i (6.30) 2 One knows {m1, m2, v1i} in most situations. There while its final kinetic energy from Eq. (6.27) are thus four unknowns {v1f , v2f , θ1 and θ2}, and  m − m2  2 2 only two equations. If θ 1 = θ 2 = 0, we regain 1 1 K1 f = m 1v12f = m1  1 v1i Eq. (6.24) for one dimensional collision. 2 2  m1 + m 2  If, further the collision is elastic, 1 1 1 The fractional kinetic energy lost is m1v1i 2 = m1v1 f 2 + m 2v2 f 2 (6.31) 2 2 2 2 K1 f m − m2  We obtain an additional equation. That still f1 = = 1 K1i  m 1 + m 2  leaves us one equation short. At least one of while the fractional kinetic energy gained by the the four unknowns, say θ 1, must be made known moderating nuclei K2f /K1i is for the problem to be solvable. For example, θ1 can be determined by moving a detector in an f2 = 1 − f1 (elastic collision) angular fashion from the x to the y axis. Given 4m1m 2 {m1, m2, v1i , θ1} we can determine {v1f , v2f , θ2} = from Eqs. (6.29)-(6.31). (m1 + m 2 )2 ⊳ Example 6.13 Consider the collision One can also verify this result by substituting depicted in Fig. 6.10 to be between two from Eq. (6.28). billiard balls with equal masses m1 = m2. For deuterium m 2 = 2m 1 and we obtain The first ball is called the cue while the f1 = 1/9 while f2 = 8/9. Almost 90% of the second ball is called the target. The neutron’s energy is transferred to deuterium. For billiard player wants to ‘sink’ the target carbon f1 = 71.6% and f2 = 28.4%. In practice, ball in a corner pocket, which is at an however, this number is smaller since head-on angle θ2 = 37°. Assume that the collision collisions are rare. ⊳ is elastic and that friction and rotational If the initial velocities and final velocities of motion are not important. Obtain θ 1. both the bodies are along the same straight line, Answer From momentum conservation, since then it is called a one-dimensional collision, or the masses are equal head-on collision. In the case of small spherical bodies, this is possible if the direction of travel v1i = v1f + v 2f of body 1 passes through the centre of body 2 which is at rest. In general, the collision is two- or ( )( v 1i 2 = v1 f + v 2 f ⋅ v1 f + v 2 f ) dimensional, where the initial velocities and the final velocities lie in a plane. = v1 f 2 + v 2 f 2 + 2v1 f.v2 f 132 PHYSICS The matter simplifies greatly if we consider = {v 1f 2 + v 2 f 2 + 2v1 f v 2 f cos (θ1 + 37° ) } (6.32) 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 (6.33) In our everyday world, collisions take place only Comparing Eqs. (6.32) and (6.33), 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. 6. 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. WORK, ENERGY AND POWER 133 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 6.2 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. 6. 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. 134 PHYSICS EXERCISES 6.1 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. 6.2 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. 6.3 Given in Fig. 6.11 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. 6.11 WORK, ENERGY AND POWER 135 6.4 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. 6.12. Show that a particle of total energy 1 J moving under this Fig. 6.12 potential must ‘turn back’ when it reaches x = ± 2 m. 6.5 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. 6.13 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. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13(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 ? 6.6 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. 6.7 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. 6.8 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 ? 136 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). 6.9 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 6.10 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 6.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = −˜i + 2 ˜j + 3 k ˜N where ˜i, ˜j, k ˜ 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 ? 6.12 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). 6.13 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 ? 6.14 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 ? 6.15 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 ? 6.16 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. 6.14) is a possible result after collision ? Fig. 6.14 WORK, ENERGY AND POWER 137 6.17 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. 6.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic. 6.18 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 ? 6.19 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. 6.15 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 ? 6.20 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? 6.21 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 ? 6.22 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? 6.23 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. Additional Exercises 6.24 A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block. 6.25 Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ1 = 300, θ2 = 600, and h = 10 m, what are the speeds and times taken by the two stones ? Fig. 6.16 138 PHYSICS 6.26 A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless. Fig. 6.17 6.27 A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ? 6.28 A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s–1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley ? How much has the trolley moved from the time the child begins to run ? 6.29 Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls ? Here r is the distance between centres of the balls. Fig. 6.18 6.30 Consider the decay of a free neutron at rest : n g p + e– WORK, ENERGY AND POWER 139 Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus (Fig. 6.19). Fig. 6.19 [Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin ½ (like e—, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is : n g p + e – + ν ] 140 PHYSICS APPENDIX 6.1 : POWER CONSUMPTION IN WALKING The table below lists the approximate power expended by an adult human of mass 60 kg. Table 6.4 Approximate power consumption Mechanical work must not be confused with the everyday usage of the term work. A woman standing with a very heavy load on her head may get very tired. But no mechanical work is involved. That is not to say that mechanical work cannot be estimated in ordinary human activity. Consider a person walking with constant speed v0. The mechanical work he does may be estimated simply with the help of the work-energy theorem. Assume : (a) The major work done in walking is due to the acceleration and deceleration of the legs with each stride (See Fig. 6.20). (b) Neglect air resistance. (c) Neglect the small work done in lifting the legs against gravity. (d) Neglect the swinging of hands etc. as is common in walking. As we can see in Fig. 6.20, in each stride the leg is brought from rest to a speed, approximately equal to the speed of walking, and then brought to rest again. Fig. 6.20 An illustration of a single stride in walking. While the first leg is maximally off the round, the second leg is on the ground and vice-versa The work done by one leg in each stride is m l v 02 by the work-energy theorem. Here ml is the mass of the leg. Note m l v 02 /2 energy is expended by one set of leg muscles to bring the foot from rest to speed v0 while an additional m l v 02 /2 is expended by a complementary set of leg muscles to bring the foot to rest from speed v0. Hence work done by both legs in one stride is (study Fig. 6.20 carefully) Ws =2m l v02 (6.34) Assuming ml = 10 kg and slow running of a nine-minute mile which translates to 3 m s-1 in SI units, we obtain Ws = 180 J / stride If we take a stride to be 2 m long, the person covers 1.5 strides per second at his speed of 3 m s-1. Thus the power expended J stride P = 180 ×1.5 stride second = 270 W We must bear in mind that this is a lower estimate since several avenues of power loss (e.g. swinging of hands, air resistance etc.) have been ignored. The interesting point is that we did not worry about the forces involved. The forces, mainly friction and those exerted on the leg by the muscles of the rest of the body, are hard to estimate. Static friction does no work and we bypassed the impossible task of estimating the work done by the muscles by taking recourse to the work-energy theorem. We can also see the advantage of a wheel. The wheel permits smooth locomotion without the continual starting and stopping in mammalian locomotion.