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.

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%). ⊳

5.8 THE CONSERVATION OF MECHANICAL a ball of mass m being dropped from a cliff of ENERGY height H. For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have, ∆K = F(x) ∆x If the force is conservative, the potential energy function V(x) can be defined such that − ∆V = F(x) ∆x The above equations imply that Fig. 5.5 The conversion of potential energy to kinetic ∆K + ∆V = 0 energy for a ball of mass m dropped from a ∆(K + V ) = 0 (5.10) height H. Reprint 2025-26 WORK, ENERGY AND POWER 79 The total mechanical energies E0, Eh, and EH Answer (i) There are two external forces on of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the level), h and H, are string. The latter does no work since the displacement of the bob is always normal to the EH = mgH (5.11 a) string. The potential energy of the bob is thus 1 2 E h = mgh + mv h (5.11 b) associated with the gravitational force only. The 2 2 total mechanical energy E of the system is E0 = (1/2) mvf (5.11 c) conserved. We take the potential energy of the The constant force is a special case of a spatially system to be zero at the lowest point A. Thus, dependent force F(x). Hence, the mechanical at A : energy is conserved. Thus EH = E0 1 2 1 2 E = mv0 (5.12) or, mgH = mv f 2 2 v f = 2 gH [Newton’s Second Law] a result that was obtained in section 5.7 for a where TA is the tension in the string at A. At thefreely falling body. highest point C, the string slackens, as the Further, tension in the string (TC) becomes zero. EH = Eh Thus, at Cwhich implies, 2 1 2 v h = 2 g(H − h) (5.11 d) E = mv c + 2mgL (5.13) 2 and is a familiar result from kinematics. At the height H, the energy is purely potential. mvc2It is partially converted to kinetic at height h and mg = [Newton’s Second Law] (5.14) L is fully kinetic at ground level. This illustrates the conservation of mechanical energy. where vC is the speed at C. From Eqs. (5.13) ⊳ and (5.14) Example 5.7 A bob of mass m is suspended 5 by a light string of length L . It is imparted a E = mgL horizontal velocity vo at the lowest point A 2 such that it completes a semi-circular Equating this to the energy at A trajectory in the vertical plane with the string 5 m 2 becoming slack only on reaching the topmost mgL = v 0 point, C. This is shown in Fig. 5.6. Obtain an 2 2 expression for (i) vo; (ii) the speeds at points or, v 0 = 5 gL B and C; (iii) the ratio of the kinetic energies (ii) It is clear from Eq. (5.14) (KB/KC) at B and C. Comment on the nature of the trajectory of the bob after it reaches vC = gL the point C. At B, the energy is 1 2 E = mv B + mgL 2 Equating this to the energy at A and employing the result from (i), namely v 02 = 5 gL , 1 2 1 2 mv B + mgL = mv 0 2 2 5 = m g L Fig. 5.6 2 Reprint 2025-26 80 PHYSICS ∴ vB = 3 gL k x m2 W = + (5.16) 2 (iii) The ratio of the kinetic energies at B and C is : 1 2 mv B K B 2 3 = = 2 1 K C 1 mvC 2 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. ⊳