Lesson 10 of 13
In Progress

Potential Energy of a Spring

The spring force is an example of a variable force which is conservative. Fig. 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. 7(b)] or negative [Fig. 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 Nm-1. The spring is said to be stiff if k is small.

Suppose that we pull the block outwards as in Fig. 7(b). If the extension is xm, the work done by the spring force is

Ws = ∫0xmF5dx =-∫0xmkxdx

= –k(xm)2/2………………………………………………………………………………………………(15)

This expression may also be obtained by considering the area of the triangle as in Fig. 7(d). Note that the work done by the external pulling force F is positive since it overcomes the spring force.

W = +k(xm)2/2………………………………………………………………………………………………(16)

Fig. 7 Illustration of the spring force with a block attached to the free end of the spring.

(a) The spring force Fs is zero when the displacement x from the equilibrium position is zero.
(b) For the stretched spring x > 0 and Fs< 0
(c) For the compressed spring x < 0 and Fs> 0
(d) The plot of Fs versus x.
The area of the shaded triangle represents work done by the spring force. Due to the opposing signs of Fs and x this work done is negative, Ws = -k(xm)2/2.

The same is true when the spring is compressed with a displacement xc(<0). The spring force does work Ws = –k(xc)2/2 while the external force F does the work +k(xc)2/2. If the block is moved from an initial displacement xi to a final displacement xf , work done by the spring force Ws = -∫xixfkxdx = k (xi)2/2 – k (xf)2/2 ………………………………..(17)

Thus the work done by the spring force depends only on the endpoints. Specifically, if the block is pulled from xi and allowed to return to xi :

Ws = -∫xixikxdx = k (xi)2/2 – k (xi)2/2 = 0 ………………………………..(18)

The work done by the spring force in a cyclic process is zero. We have explicitly demonstrated that the spring force (i) is position dependent only as first stated by Hooke. (Fs = -kx): (ii) does work which only depends on the initial and final positions, e.g. Eq. (17). Thus, the spring force is a conservative force.

We define the potential energy V(x) of the spring to be zero when block and spring system is in the equilibrium position. For an extension (or compression) x the above analysis suggests that

V(x) = kx2/2 ……………………………………………………………………….(19)

You may easily verify that -dV/dx = –kx, the spring force. If the block of mass m in Fig. 7 is extended to xi and released from rest, then its total mechanical energy at any arbitrary point x, where x lies between –xm and +xm will be given by

1/2k(xm)2= 1/2kx2 + 1/2mv2

where we have invoked the conservation of mechanical energy. This suggests that the speed and kinetic energy will be maximum at the equilibrium position, x = 0. i.e.,

1/2m(vm)2= 1/2k(xm)2

where vm is the maximum speed.

or vm =[ √(k/m)]xm

Note that k/m has the dimensions of [ T]2 and our equation is dimensionally correct. The kinetic energy converted to potential energy and vice versa, however, the total mechanical energy remains constant. This is graphically depicted in Fig. 8.

Example 8
To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 × 10³ C. What is the maximum compression of the spring.


Fig. 8 Parabolic plots of the potential energy V and kinetic energy K of a block attached to a spring obeying Hooke’s law.

The two plots are complementary, one decreasing as the other increases. The total mechanical energy E = K +V remains constant.

Answer

At maximum compression the kinetic energy of the car is converted entirely in to the potential energy of the spring.

The kinetic energy of the moving car is

K = 1/2mv2

= 1/2 × 10³ × 5 × 5

K = 1.25 × 104 J

where we have converted 18 km h-1 to 5 m s-1 [It is useful to remember that 36 km h-1 = 10 m s-1 ]. At maximum compression xm, the potential energy V of the spring is equal to the kinetic energy K of the moving car from the principle of conservation of mechanical energy,

V = 1/2k(xm)2

= 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 remarks on conservative forces.

  1. Information on time is absent from the above discussions. In the example considered above, we can calculate the compression, but not the time over which the compression occurs. A solution of Newton’s Second Law for this system is required for temporal information.
  2. Not all forces are conservative. Friction, for example, is a non-conservative force. The principle of conservation of energy will have to be modified in this case. This is illustrated in Example 9.
  3. The zero of the potential energy is arbitrary, It is set according to convenience. For the spring force we took V(x) = 0, at x = 0. i.e., the unstretched spring had zero potential energy. For the constant gravitational force mg, we took V = 0 on the earth’s surface. In a later lesson we shall see that the force due to the universal law of gravitation, the zero is the best defined at an infinite distance from the gravitational source. However, once the zero of the potential energy is fixed in a given discussion, it must be consistently adhered to throughout the discussion. You cannot change the horses in the midstream!

Example 9
Consider Example 7 taking the coefficient of friction, μ, to be 0.5 and calculate the maximum compression of the spring.

Answer

In presence of friction, both the spring force and the frictional force act so as to oppose the compression of the spring as shown in Fig. 9. We invoke the work-energy theorem, rather than the conservation of mechanical energy.

The change in kinetic energy is

ΔK = Kf – Ki = 0 – 1/2 mv2

The work done by the net force is

W = -1/2 k(xm)2 μmgxm

Equating we have,

1/2 mv2 = 1/2 k(xm)2 + μmgxm

Now μmg = 0.5 × 10³ × 10 = 5 × 10³ N (taking g = 10.0 m s-2 ). After rearranging the above equation, we obtain the following quadratic equation in the unknown xm.

k(xm)2 +2μmgxm – mv2 = 0

xm = {-μmg + [μ2m2g2 + mkv2]1/2}/k where we take the positive square root since xm is positive. Putting in numerical values we obtain

xm = 1.35 m

which, as expected, is less than the result in Example 8.

If the two forces on the body consist of a conservative force Fc and a non-conservative force Fnc , the conservation of mechanical energy formula will have to be modified. By the WE theorem,

(Fc + Fnc x = ΔK

But Fc Δx = -ΔV

Hence, Δ(K+V) = Fnc Δx

ΔE = Fnc Δx

where E is the total mechanical energy. Over the path this assumes the form

Ef – Ei = Wnc

where Wnc is the total work done by the non-conservative forces over the path. Note that unlike the conservative force, Wnc depends on the partiular path i to f.