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

11.4 HEAT, INTERNAL ENERGY AND WORK associated with various random motions of its The Zeroth Law of Thermodynamics led us to molecules. We will see in the next chapter that the concept of temperature that agrees with our in a gas this motion is not only translational commonsense notion. Temperature is a marker (i.e. motion from one point to another in the of the ‘hotness’ of a body. It determines the volume of the container); it also includes rotational and vibrational motion of thedirection of flow of heat when two bodies are molecules (Fig. 11.3).placed in thermal contact. Heat flows from the body at a higher temperature to the one at lower temperature. The flow stops when the temperatures equalise; the two bodies are then in thermal equilibrium. We saw in some detail how to construct temperature scales to assign temperatures to different bodies. We now describe the concepts of heat and other relevant quantities like internal energy and work. Fig. 11.3 (a) Internal energy U of a gas is the sum of the kinetic and potential energies of its The concept of internal energy of a system is molecules when the box is at rest. Kinetic not difficult to understand. We know that every energy due to various types of motion bulk system consists of a large number of (translational, rotational, vibrational) is to molecules. Internal energy is simply the sum of be included in U. (b) If the same box is the kinetic energies and potential energies of moving as a whole with some velocity, these molecules. We remarked earlier that in the kinetic energy of the box is not to be thermodynamics, the kinetic energy of the included in U. system, as a whole, is not relevant. Internal energy is thus, the sum of molecular kinetic and potential energies in the frame of reference relative to which the centre of mass of the system is at rest. Thus, it includes only the (disordered) energy associated with the random motion of molecules of the system. We denote the internal energy of a system by U. Though we have invoked the molecular picture to understand the meaning of internal energy, as far as thermodynamics is concerned, U is simply a macroscopic variable of the system. The important thing about internal energy is that it depends only on the state of the system, not on how that state was achieved. Internal energy U of a system is an example of a thermodynamic ‘state variable’ – its value depends only on the given state of the system, not on history i.e. not on the ‘path’ taken to arrive at that state. Thus, the internal energy of a given mass of gas depends on its state described by Fig. 11.4 Heat and work are two distinct modes of specific values of pressure, volume and energy transfer to a system that results in temperature. It does not depend on how this change in its internal energy. (a) Heat is energy transfer due to temperaturestate of the gas came about. Pressure, volume, difference between the system and the temperature, and internal energy are surroundings. (b) Work is energy transfer thermodynamic state variables of the system brought about by means (e.g. moving the (gas) (see section 11.7). If we neglect the small piston by raising or lowering some weight intermolecular forces in a gas, the internal connected to it) that do not involve such a energy of a gas is just the sum of kinetic energies temperature difference. Reprint 2025-26 230 PHYSICS What are the ways of changing internal 11.5 FIRST LAW OF THERMODYNAMICS energy of a system ? Consider again, for We have seen that the internal energy U of a simplicity, the system to be a certain mass of system can change through two modes of energy gas contained in a cylinder with a movable transfer : heat and work. Let piston as shown in Fig. 11.4. Experience shows ∆Q = Heat supplied to the system by thethere are two ways of changing the state of the surroundingsgas (and hence its internal energy). One way is to put the cylinder in contact with a body at a ∆W = Work done by the system on the higher temperature than that of the gas. The surroundings temperature difference will cause a flow of ∆U = Change in internal energy of the system energy (heat) from the hotter body to the gas, The general principle of conservation of thus increasing the internal energy of the gas. energy then implies that The other way is to push the piston down i.e. to ∆Q = ∆U + ∆W (11.1) do work on the system, which again results in i.e. the energy (∆Q) supplied to the system goesincreasing the internal energy of the gas. Of in partly to increase the internal energy of thecourse, both these things could happen in the system (∆U) and the rest in work on thereverse direction. With surroundings at a lower environment (∆W). Equation (11.1) is known astemperature, heat would flow from the gas to the First Law of Thermodynamics. It is simplythe surroundings. Likewise, the gas could push the general law of conservation of energy applied the piston up and do work on the surroundings. to any system in which the energy transfer from In short, heat and work are two different modes or to the surroundings is taken into account. of altering the state of a thermodynamic system Let us put Eq. (11.1) in the alternative form and changing its internal energy. The notion of heat should be carefully ∆Q – ∆W = ∆U (11.2) distinguished from the notion of internal energy. Heat is certainly energy, but it is the energy in Now, the system may go from an initial state transit. This is not just a play of words. The to the final state in a number of ways. For distinction is of basic significance. The state of example, to change the state of a gas from a thermodynamic system is characterised by its (P1, V1) to (P2, V2), we can first change the volume of the gas from V1 to V2, keeping itsinternal energy, not heat. A statement like ‘a pressure constant i.e. we can first go the stategas in a given state has a certain amount of heat’ is as meaningless as the statement that (P1, V2) and then change the pressure of the gas from P1 to P2, keeping volume constant, to‘a gas in a given state has a certain amount take the gas to (P2, V2). Alternatively, we canof work’. In contrast, ‘a gas in a given state first keep the volume constant and then keep has a certain amount of internal energy’ is a the pressure constant. Since U is a state perfectly meaningful statement. Similarly, the variable, ∆U depends only on the initial and statements ‘a certain amount of heat is final states and not on the path taken by the supplied to the system’ or ‘a certain amount gas to go from one to the other. However, ∆Q of work was done by the system’ are perfectly and ∆W will, in general, depend on the path meaningful. taken to go from the initial to final states. From To summarise, heat and work in the First Law of Thermodynamics, Eq. (11.2), thermodynamics are not state variables. They it is clear that the combination ∆Q – ∆W, is are modes of energy transfer to a system however, path independent. This shows that resulting in change in its internal energy, if a system is taken through a process in which ∆U = 0 (for example, isothermal expansion ofwhich, as already mentioned, is a state variable. an ideal gas, see section 11.8), In ordinary language, we often confuse heat with internal energy. The distinction between ∆Q = ∆W them is sometimes ignored in elementary physics books. For proper understanding of i.e., heat supplied to the system is used up thermodynamics, however, the distinction is entirely by the system in doing work on the crucial. environment. Reprint 2025-26 THERMODYNAMICS 231 If the system is a gas in a cylinder with a If the amount of substance is specified in movable piston, the gas in moving the piston does terms of moles µ (instead of mass m in kg ), we work. Since force is pressure times area, and can define heat capacity per mole of the area times displacement is volume, work done substance by by the system against a constant pressure P is S 1 ∆Q C = = (11.6) ∆W = P ∆V µ µ ∆T C is known as molar specific heat capacity of where ∆V is the change in volume of the gas. the substance. Like s, C is independent of the Thus, for this case, Eq. (11.1) gives amount of substance. C depends on the nature ∆Q = ∆U + P ∆V (11.3) of the substance, its temperature and the conditions under which heat is supplied. The As an application of Eq. (11.3), consider the unit of C is J mo1–1 K–1. As we shall see later (in change in internal energy for 1 g of water when connection with specific heat capacity of gases), we go from its liquid to vapour phase. The additional conditions may be needed to define measured latent heat of water is 2256 J/g. i.e., C or s. The idea in defining C is that simple for 1 g of water ∆Q = 2256 J. At atmospheric predictions can be made in regard to molar pressure, 1 g of water has a volume 1 cm3 in specific heat capacities. liquid phase and 1671 cm3 in vapour phase. Table 11.1 lists measured specific and molar heat capacities of solids at atmospheric pressure Therefore, and ordinary room temperature. ∆W =P (Vg –Vl ) = 1.013 ×105 × (1671 × 10–6) =169.2 J We will see in Chapter 12 that predictions of specific heats of gases generally agree with Equation (11.3) then gives experiment. We can use the same law of equipartition of energy that we use there to ∆U = 2256 – 169.2 = 2086.8 J predict molar specific heat capacities of solids We see that most of the heat goes to increase (See Section 12.5 and 12.6). Consider a solid of the internal energy of water in transition from N atoms, each vibrating about its mean the liquid to the vapour phase. position. An oscillator in one dimension has

11.3 ZEROTH LAW OF THERMODYNAMICS (a) Imagine two systems A and B, separated by an adiabatic wall, while each is in contact with a third system C, via a conducting wall [Fig. 11.2(a)]. The states of the systems (i.e., their macroscopic variables) will change until both A and B come to thermal equilibrium with C. After this is achieved, suppose that the adiabatic wall between A and B is replaced by a conducting wall and C is insulated from A and B by an adiabatic wall [Fig.11.2(b)]. It is found that the states of A and B change no (b) further i.e. they are found to be in thermal Fig. 11.2 (a) Systems A and B are separated by an equilibrium with each other. This observation adiabatic wall, while each is in contact forms the basis of the Zeroth Law of with a third system C via a conducting Thermodynamics, which states that ‘two wall. (b) The adiabatic wall between A systems in thermal equilibrium with a third and B is replaced by a conducting wall, system separately are in thermal equilibrium while C is insulated from A and B by an adiabatic wall.with each other’. R.H. Fowler formulated this * Both the variables need not change. It depends on the constraints. For instance, if the gases are in containers of fixed volume, only the pressures of the gases would change to achieve thermal equilibrium. Reprint 2025-26 THERMODYNAMICS 229