2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

5.9 THE POTENTIAL ENERGY OF A SPRING The spring force is an example of a variable force which is conservative. Fig. 5.7 shows a block attached to a spring and resting on a smooth horizontal surface. The other end of the spring is attached to a rigid wall. The spring is light and may be treated as massless. In an ideal spring, the spring force Fs is proportional to x where x is the displacement of the block from the equilibrium position. The displacement could be either positive [Fig. 5.7(b)] or negative [Fig. 5.7(c)]. This force law for the spring is called Hooke’s law and is mathematically stated as Fs = − kx The constant k is called the spring constant. Its unit is N m-1. The spring is said to be stiff if k is large and soft if k is small. Fig. 5.7 Illustration of the spring force with a block Suppose that we pull the block outwards as in attached to the free end of the spring. Fig. 5.7(b). If the extension is xm, the work done by (a) The spring force Fs is zero when the the spring force is displacement x from the equilibrium position is zero. (b) For the stretched spring x > 0 xm xm and Fs < 0 (c) For the compressed spring d x x < 0 and Fs > 0.(d) The plot of Fs versus x. Fs d x = −∫kx W s = ∫ 0 0 The area of the shaded triangle represents the work done by the spring force. Due to the k x m2 opposing signs of Fs and x, this work done is = − (5.15) 2 2 negative, W s = −kx m / 2 . This expression may also be obtained by considering the area of the triangle as in The same is true when the spring is Fig. 5.7(d). Note that the work done by the compressed with a displacement xc (< 0). Theexternal pulling force F is positive since it overcomes the spring force. spring force does work Ws = − kx c2 / 2 while the Reprint 2025-26 WORK, ENERGY AND POWER 81 2 and vice versa, however, the total mechanical external force F does work + kxc / 2. If the block energy remains constant. This is graphically is moved from an initial displacement xi to a depicted in Fig. 5.8. final displacement xf , the work done by the spring force Ws is xf k x 2f k x i2 (5.17) k x d x = − Ws = − ∫ 2 2 x i 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 ; x i k x i2 k x i2 k x dx = − Ws = − ∫ 2 2 x i = 0 (5.18) Fig. 5.8 Parabolic plots of the potential energy V and The work done by the spring force in a cyclic kinetic energy K of a block attached to a spring obeying Hooke’s law. The two plotsprocess is zero. We have explicitly demonstrated are complementary, one decreasing as the that the spring force (i) is position dependent other increases. The total mechanical only as first stated by Hooke, (Fs = − kx); (ii) energy E = K + V remains constant. does work which only depends on the initial and final positions, e.g. Eq. (5.17). Thus, the spring ⊳ Example 5.8 To simulate car accidents, autoforce is a conservative force. manufacturers study the collisions of moving We define the potential energy V(x) of the spring cars with mounted springs of different springto be zero when block and spring system is in the constants. Consider a typical simulation withequilibrium position. For an extension (or a car of mass 1000 kg moving with a speedcompression) x the above analysis suggests that 18.0 km/h on a smooth road and colliding kx 2 with a horizontally mounted spring of spring V(x) = (5.19) constant 5.25 × 103 N m–1. What is the 2 maximum compression of the spring ?You may easily verify that − dV/dx = −k x, the spring force. If the block of mass m in Fig. 5.7 is extended to xm and released from rest, then its Answer At maximum compression the kinetic total mechanical energy at any arbitrary point x, energy of the car is converted entirely into the where x lies between – xm and + xm, will be given by potential energy of the spring. The kinetic energy of the moving car is 1 2 1 2 1 2 k x m = k x + m v 1 2 2 2 K = mv2 2where we have invoked the conservation of mechanical energy. This suggests that the speed 1 3 and the kinetic energy will be maximum at the = × 10 × 5 × 5 2 equilibrium position, x = 0, i.e., K = 1.25 × 104 J 1 2 1 2 m v m = k x m 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 or v m = x m energy K of the moving car from the principle of m conservation of mechanical energy. Note that k/m has the dimensions of [T-2] and our equation is dimensionally correct. The 1 2 V = k x m kinetic energy gets converted to potential energy 2 Reprint 2025-26 82 PHYSICS = 1.25 × 104 J We obtain xm = 2.00 m We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction. ⊳ We conclude this section by making a few Fig. 5.9 The forces acting on the car. remarks on conservative forces. (i) Information on time is absent from the above 1 2 ∆K = Kf − Ki = 0 − m v discussions. In the example considered 2 above, we can calculate the compression, but The work done by the net force is not the time over which the compression 1 2 occurs. A solution of Newton’s Second Law W = − kx m −µm g x m 2 for this system is required for temporal information. Equating we have (ii) Not all forces are conservative. Friction, for 1 2 1 2 example, is a non-conservative force. The m v = k x m + µm g x m 2 2 principle of conservation of energy will have Now µmg = 0.5 × 103 × 10 = 5 × 103 N (taking to be modified in this case. This is illustrated g =10.0 m s-2). After rearranging the above in Example 5.9. equation we obtain the following quadratic(iii) The zero of the potential energy is arbitrary. 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 2 2 k x m + 2µm g x m − m v = 0 unstretched spring had zero potential energy. For the constant gravitational force mg, we took V = 0 on the earth’s surface. In a later chapter we shall see that for the force where we take the positive square root since due to the universal law of gravitation, the zero is best defined at an infinite distance xm is positive. Putting in numerical values we obtain from the gravitational source. However, once the zero of the potential energy is fixed in a xm = 1.35 m given discussion, it must be consistently which, as expected, is less than the result in adhered to throughout the discussion. You Example 5.8. cannot change horses in midstream ! If the two forces on the body consist of a conservative force Fc and a non-conservative⊳ force Fnc , the conservation of mechanical energy Example 5.9 Consider Example 5.8 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. 5.9. the path this assumes the form We invoke the work-energy theorem, rather Ef − Ei = Wnc than the conservation of mechanical energy. where Wnc is the total work done by the The change in kinetic energy is non-conservative forces over the path. Note that Reprint 2025-26 WORK, ENERGY AND POWER 83 unlike the conservative force, Wnc depends on Our electricity bills carry the energy the particular path i to f. ⊳ consumption in units of kWh. Note that kWh is a unit of energy and not of power.

8.2 STRESS AND STRAIN forces are applied parallel to the cross-sectional When forces are applied on a body in such a area of the cylinder, as shown in Fig. 8.1(b), manner that the body is still in static equilibrium, there is relative displacement between the it is deformed to a small or large extent depending opposite faces of the cylinder. The restoring force upon the nature of the material of the body and per unit area developed due to the applied the magnitude of the deforming force. The tangential force is known as tangential or deformation may not be noticeable visually in shearing stress. many materials but it is there. When a body is As a result of applied tangential force, there subjected to a deforming force, a restoring force is a relative displacement ∆x between opposite is developed in the body. This restoring force is faces of the cylinder as shown in the Fig. 8.1(b). equal in magnitude but opposite in direction to The strain so produced is known as shearing the applied force. The restoring force per unit area strain and it is defined as the ratio of relative is known as stress. If F is the force applied normal displacement of the faces ∆x to the length of the to the cross–section and A is the area of cross cylinder L. section of the body, ∆x Magnitude of the stress = F/A (8.1) Shearing strain = = tan θ (8.3) L The SI unit of stress is N m–2 or pascal (Pa) and its dimensional formula is [ ML–1T–2 ]. where θ is the angular displacement of the There are three ways in which a solid may cylinder from the vertical (original position of the change its dimensions when an external force cylinder). Usually θ is very small, tan θ acts on it. These are shown in Fig. 8.1. In is nearly equal to angle θ, (if θ = 10°, for Fig.8.1(a), a cylinder is stretched by two equal example, there is only 1% difference between θ forces applied normal to its cross-sectional area. and tan θ). The restoring force per unit area in this case is It can also be visualised, when a book is called tensile stress. If the cylinder is pressed with the hand and pushed horizontally, compressed under the action of applied forces, as shown in Fig. 8.2 (c). the restoring force per unit area is known as Thus, shearing strain = tan θ ≈ θ (8.4) compressive stress. Tensile or compressive In Fig. 8.1 (d), a solid sphere placed in the fluid stress can also be termed as longitudinal stress. under high pressure is compressed uniformly on In both the cases, there is a change in the all sides. The force applied by the fluid acts in length of the cylinder. The change in the length perpendicular direction at each point of the ∆L to the original length L of the body (cylinder surface and the body is said to be under in this case) is known as longitudinal strain. hydraulic compression. This leads to decrease (a) (b) (c) (d) Fig. 8.1 (a) A cylindrical body under tensile stress elongates by ∆L (b) Shearing stress on a cylinder deforming it by an angle θ(c) A body subjected to shearing stress (d) A solid body under a stress normal to the surface at every point (hydraulic stress). The volumetric strain is ∆V/V, but there is no change in shape. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 169 in its volume without any change of its compression and shear stress may also be geometrical shape. obtained. The stress-strain curves vary from The body develops internal restoring forces material to material. These curves help us to that are equal and opposite to the forces applied understand how a given material deforms with by the fluid (the body restores its original shape increasing loads. From the graph, we can see and size when taken out from the fluid). The that in the region between O to A, the curve is internal restoring force per unit area in this case linear. In this region, Hooke’s law is obeyed. is known as hydraulic stress and in magnitude The body regains its original dimensions when is equal to the hydraulic pressure (applied force the applied force is removed. In this region, the per unit area). solid behaves as an elastic body. The strain produced by a hydraulic pressure is called volume strain and is defined as the ratio of change in volume (∆V) to the original volume (V). ∆V Volume strain = (8.5) V Since the strain is a ratio of change in dimension to the original dimension, it has no units or dimensional formula.