The spring force is an example of a variable force which is conservative. Fig. 1 shows a block attached to a spring and resting on a smooth horizontal surface.
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| Fig.1 |
Fig.1: Illustration of the spring force with a block attached to the free end of the spring. (a) The spring force $F_s$ is zero when the displacement x from the equilibrium position is zero. (b) For the stretched spring x > 0 and $F_s$ < 0 (c) For the compressed spring x < 0 and $F_s$ > 0.
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 F$_s$ is proportional to x where x is the displacement of the block from the equilibrium position. The displacement could be either positive [Fig. 1(b)] or negative [Fig. 1(c)]. This force law for the spring is called Hooke’s law and is mathematically stated as
F$_s$ = − 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.
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| Fig.1 |
Suppose that we pull the block outwards as in Fig. 1(b). If the extension is x$_m$ , the work done by the spring force is
$W_f=\int_{0}^{x_m}F_sdx=-\int_{0}^{x_m}kxdx$
$W_f=-\frac{1}{2}kx_m^2$
This expression may also be obtained by considering the area of the triangle as in Fig. 1(d). Note that the work done by the external pulling force F is positive since it overcomes the spring force.
$W_f=+\frac{1}{2}kx_m^2$
The same is true when the spring is compressed with a displacement x$_c$ (< 0). The spring force does work $W_c=-\frac{1}{2}kx_c^2$ while the external force F does work $=+\frac{1}{2}kx_c^2$. If the block is moved from an initial displacement x$_i$ to a final displacement x$_f$ , the work done by the spring force W$_s$ is
$W=-\int_{x_i}^{x_f}kxdx=\frac{1}{2}kx_i^2-\frac{1}{2}kx_f^2$
Thus the work done by the spring force depends only on the end points. Specifically, if the block is pulled from x$_i$ and allowed to return to x$_i$;
$W=-\int_{x_i}^{x_f}kxdx=\frac{1}{2}kx_i^2-\frac{1}{2}kx_i^2=0$
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, (F$_s$ = − kx); (ii) does work which only depends on the initial and final positions, e.g. Eq. (6.17). 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
$V(x)=\frac{1}{2}kx^2$
You may easily verify that − dV/dx = − kx, the spring force. If the block of mass m in Fig. 1 is extended to x$_m$ and released from rest, then its total mechanical energy at any arbitrary point x, where x lies between – x$_m$ and +x$_m$, will be given by
$\frac{1}{2}kx_m^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2$
where we have invoked the conservation of mechanical energy. This suggests that the speed and the kinetic energy will be maximum at the equilibrium position, x = 0, i.e.,
$\frac{1}{2}kx_m^2=\frac{1}{2}mv^2$
where v$_m$ is the maximum speed.
$v_m=\sqrt{\frac{k}{m}}x_m$
Note that k/m has the dimensions of [$T^{-2}$] and our equation is dimensionally correct. The kinetic energy gets converted to potential energy and vice versa, however, the total mechanical energy remains constant. This is graphically depicted in Fig. 2.
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| Fig.2 |
Example 1
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 of 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 × 10$^3$ N.$m^{–1}$. What is the maximum compression of the spring?
Answer
At maximum compression the kinetic energy of the car is converted entirely into the potential energy of the spring.
The kinetic energy of the moving car is
$K=\frac{1}{2}mv^2=\frac{1}{2}(10^3)(5)(5)$
K = $1.25 \times 10^4$ J
where we have converted 18 km/h to 5 m/s [It is useful to remember that 36 km/h = 10 m/s]. At maximum compression x$_m$, 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.
$V=\frac{1}{2}kx_m^2$
V = $1.25 \times 10^4$ J
We obtain
x$_m$ = 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.
(i) 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.
(ii) 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 6.9.
(iii) 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 chapter we shall see that for the force due to the universal law of gravitation, the zero is 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 horses in midstream!
Example 2
Consider Example 1 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. 3.
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| Fig.3 The forces acting on the car. |
We invoke the work-energy theorem, rather than the conservation of mechanical energy.
The change in kinetic energy is
∆K = K$_f$ − K$_i$ = 0 − $\frac{1}{2}mv^2$
The work done by the net force is
$W=-\frac{1}{2}kx_m^2-\mu mgx_m$
Equating we have
$\frac{1}{2}mv^2=\frac{1}{2}kx_m^2+\mu mgx_m$
Now µmg = 0.5 × 10$^3$ × 10 = 5 × 10$^3$ N (taking g =10.0 m.s$^{-2}$). After rearranging the above equation we obtain the following quadratic equation in the unknown x$_m$.
$kx_m^2+2 \mu mgx_m-mv^2=0$
$x_m=\frac{-\mu mg+\sqrt{\mu^2m^2g^2+mkv^2}}{k}$
where we take the positive square root since x$_m$ is positive. Putting in numerical values we obtain
$x_m$ = 1.35 m
which, as expected, is less than the result in Example 1.
If the two forces on the body consist of a conservative force $F_c$ and a non-conservative force $F_{nc}$, the conservation of mechanical energy formula will have to be modified. By the WE theorem
(F$_c$ + F$_{nc}$)∆x = ∆K
But F$_c$∆x = −∆V
Hence, ∆(K + V) = F$_{nc}$∆x
∆E = F$_{nc}$∆x
where E is the total mechanical energy. Over the path this assumes the form
E$_f$ − E$_i$ = W$_{nc}$
where W$_{nc}$ is the total work done by the non-conservative forces over the path. Note that unlike the conservative force, Wnc depends on the particular path i to f.
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