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

3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

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) Reprint 2025-26 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 We are now familiar with the concepts of workFig. 5.3 (a) The shaded rectangle represents the 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). d K d  1 2  =⊳ d t  2 m v  Example 5.5 A woman pushes a trunk on d t a railway platform which has a rough d v surface. She applies a force of 100 N over a = m v d t distance of 10 m. Thereafter, she gets progressively tired and her applied force = F v (from Newton’s Second Law) reduces linearly with distance to 50 N. The d x total distance through which the trunk has = F d t 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 K f x f F dx ∫ d K = ∫ K i x i where, Ki and K f are the initial and final kinetic energies corresponding to x i and x f. x f F d x or (5.8a) K f − K i = ∫ Fig. 5.4 Plot of the force F applied by the woman and x i 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 Reprint 2025-26 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 <<RE) so that −k we can ignore the variation of g near the earth’s Fr = for 0.1 < x < 2.01 m surface*. In what follows we have taken the x upward direction to be positive. Let us raise the = 0 for x < 0.1m and x > 2.01 m ball up to a height h. The work done by the external where k = 0.5 J. What is the final kinetic agency against the gravitational force is mgh. This energy and speed vf of the block as it work gets stored as potential energy. crosses this patch ? Gravitational potential energy of an object, as a function of the height h, is denoted by V(h) and it Answer From Eq. (5.8a) is the negative of work done by the gravitational 2.01 ( −k ) force in raising the object to that height. d x V (h) = mgh K f = K i + ∫ x 0.1 If h is taken as a variable, it is easily seen that the gravitational force F equals the negative of 1 2 2.01 = mv i − k ln ( x ) 0.1 the derivative of V(h) with respect to h. Thus, 2 d F = − V(h) = −m g 1 2 d h = mv i − k ln (2.01/0.1) 2 The negative sign indicates that the = 2 − 0.5 ln (20.1) gravitational force is downward. When released, the ball comes down with an increasing speed. = 2 − 1.5 = 0.5 J Just before it hits the ground, its speed is given v f = 2K f / m = 1 m s−1 by the kinematic relation, v2 = 2gh This equation can be written as Here, note that ln is a symbol for the natural 1logarithm to the base e and not the logarithm to the base 10 [ln X = loge X = 2.303 log10 X]. ⊳ 2 m v2 = m g h which shows that the gravitational potential5.7 THE CONCEPT OF POTENTIAL ENERGY energy of the object at height h, when the object The word potential suggests possibility or is released, manifests itself as kinetic energy of capacity for action. The term potential energy the object on reaching the ground. brings to one’s mind ‘stored’ energy. A stretched Physically, the notion of potential energy is bow-string possesses potential energy. When it applicable only to the class of forces where work is released, the arrow flies off at a great speed. done against the force gets ‘stored up’ as energy. The earth’s crust is not uniform, but has When external constraints are removed, it discontinuities and dislocations that are called manifests itself as kinetic energy. Mathematically, fault lines. These fault lines in the earth’s crust (for simplicity, in one dimension) the potential * The variation of g with height is discussed in Chapter 7 on Gravitation. Reprint 2025-26 78 PHYSICS energy V(x) is defined if the force F(x) can be which means that K + V, the sum of the kinetic written as and potential energies of the body is a constant. Over the whole path, xi to xf, this means that d V F ( x ) = − d x Ki + V(xi ) = Kf + V(xf) (5.11) The quantity K +V(x), is called the totalThis implies that mechanical energy of the system. Individually xf Vf the kinetic energy K and the potential energy ∫ F(x) d x = − ∫ d V = Vi − V f V(x) may vary from point to point, but the sum x i Vi is a constant. The aptness of the term The work done by a conservative force such as ‘conservative force’ is now clear. gravity depends on the initial and final positions Let us consider some of the definitions of a only. In the previous chapter we have worked conservative force. on examples dealing with inclined planes. If an l A force F(x) is conservative if it can be derived object of mass m is released from rest, from the from a scalar quantity V(x) by the relation top of a smooth (frictionless) inclined plane of given by Eq. (5.9). The three-dimensional height h, its speed at the bottom generalisation requires the use of a vector is 2 gh irrespective of the angle of inclination. derivative, which is outside the scope of this book.Thus, at the bottom of the inclined plane it l The work done by the conservative forceacquires a kinetic energy, mgh. If the work done depends only on the end points. This can be or the kinetic energy did depend on other factors seen from the relation, such as the velocity or the particular path taken W = Kf – Ki = V (xi) – V(xf)by the object, the force would be called non- which depends on the end points. conservative. l A third definition states that the work done The dimensions of potential energy are by this force in a closed path is zero. This is [ML2T –2] and the unit is joule (J), the same as once again apparent from Eq. (5.11) since kinetic energy or work. To reiterate, the change xi = xf .in potential energy, for a conservative force, ∆V is equal to the negative of the work done by Thus, the principle of conservation of total mechanical energy can be stated asthe force ∆V = − F(x) ∆x (5.9) The total mechanical energy of a system is In the example of the falling ball considered in conserved if the forces, doing work on it, are this section we saw how potential energy was conservative. The above discussion can be made moreconverted to kinetic energy. This hints at an concrete by considering the example of theimportant principle of conservation in mechanics, gravitational force once again and that of thewhich we now proceed to examine. spring force in the next section. Fig. 5.5 depicts