Hooke's law and the restoring force
Consider a block resting on a smooth horizontal surface, attached to a light spring whose other end is fixed to a rigid wall. Take the block's equilibrium position as $x = 0$. NCERT §5.9 states that for an ideal spring the spring force $F_s$ is proportional to $x$, the displacement of the block from equilibrium. The displacement may be positive (the spring stretched) or negative (the spring compressed). This is Hooke's law:
$$F_s = -kx$$
The constant $k$ is the spring constant. The single most important feature of this law is the negative sign. It tells you the spring force is a restoring force — it always points back toward equilibrium, opposite to the displacement. Stretch the block out to $x > 0$ and $F_s$ is negative, pulling it back; compress to $x < 0$ and $F_s$ is positive, pushing it back out. Either way the spring fights the displacement and tries to return the block to $x = 0$.
The spring constant k and its units
The spring constant $k$ measures how much restoring force the spring produces per unit of displacement. Both NCERT and NIOS give its unit as the newton per metre, $\text{N m}^{-1}$. NCERT records the qualitative reading: the spring is said to be stiff if $k$ is large and soft if $k$ is small. A stiff spring resists deformation hard; a soft spring yields easily.
NIOS Activity 6.2 gives a clean way to measure $k$. Hang the spring vertically and attach a block of mass $m$. The spring extends until its restoring force balances the weight: $ks = mg$, so $k = mg/s$, where $s$ is the measured extension. The dimensional formula of $k$ follows from $\text{N m}^{-1}$, namely $[\text{MT}^{-2}]$.
| Quantity | Symbol | Relation | SI unit |
|---|---|---|---|
| Spring (restoring) force | $F_s$ | Fₛ = -kx | N |
| Spring constant | $k$ | k = |Fₛ| / x | N m⁻¹ |
| Displacement from equilibrium | $x$ | measured from $x=0$ | m |
| Elastic potential energy | $V(x)$ | V = ½kx² | J |
Deriving the spring PE = ½kx²
Because the spring force changes with position, it is a variable force, and the work it does must be found by integrating $F_s$ over the displacement — exactly the variable-force method of NCERT §5.5. Suppose the block is pulled outward to an extension $x_m$. The work done by the spring force is
$$W_s = \int_{0}^{x_m} F_s\,dx = -\int_{0}^{x_m} kx\,dx = -\frac{1}{2}kx_m^{2}$$
This work is negative because the spring force opposes the outward pull. The external agent that does the pulling does positive work of equal magnitude, $+\tfrac{1}{2}kx_m^2$, since it overcomes the spring force. That input is what gets stored in the spring.
The same answer drops out geometrically. Plot the magnitude $|F_s| = kx$ against $x$: it is a straight line from the origin. The work done up to $x_m$ is the area under that line — a right triangle of base $x_m$ and height $kx_m$, giving $\tfrac{1}{2}kx_m^2$. NIOS reaches the identical result by averaging the force: the force rises linearly from $0$ to $kx_m$, so the average force is $\tfrac{1}{2}kx_m$, and work $=$ average force $\times$ displacement $= \tfrac{1}{2}kx_m \cdot x_m = \tfrac{1}{2}kx_m^2$. Compression by a displacement $x_c$ gives the same magnitude, $\tfrac{1}{2}kx_c^2$.
NCERT then defines the elastic potential energy $V(x)$ to be zero when the block–spring system sits at equilibrium. For an extension or compression $x$,
$$\boxed{\,V(x) = \frac{1}{2}kx^{2}\,}$$
You can check the definition is self-consistent: for a conservative force $F = -\,dV/dx$, and indeed $-\dfrac{d}{dx}\!\left(\tfrac{1}{2}kx^2\right) = -kx = F_s$, recovering Hooke's law.
The potential-energy parabola V(x)
Because $V(x) = \tfrac{1}{2}kx^2$ depends on $x^2$, its graph is an upward-opening parabola, symmetric about $x = 0$. The vertex sits at the origin, where $V = 0$. Moving the block either way — stretch or compress — raises $V$ identically, because $(+x)^2 = (-x)^2$. The steepness of the parabola is governed by $k$: a stiff spring (large $k$) gives a narrow, steep bowl; a soft spring gives a shallow one.
NCERT presents this alongside a complementary curve for the kinetic energy. As the spring PE rises, the KE falls by the same amount, so the two parabolas are mirror images and their sum, the total mechanical energy $E = K + V$, stays flat. That flat line is the horizontal dashed line in Fig. 3, fixed at $\tfrac{1}{2}kx_m^2$ for a block released from rest at $x_m$.
Energy exchange in an oscillating block
Take the block of mass $m$, pull it out to $x_m$, and release it from rest. With a smooth surface and a massless spring, no non-conservative force acts, so mechanical energy is conserved. At any intermediate point $x$ (with $-x_m \le x \le +x_m$), NCERT writes
$$\frac{1}{2}kx_m^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$$
The left side is the energy you stored at release; the right side is its split into spring PE and block KE at position $x$. Two limits matter for NEET:
- At the extremes $x = \pm x_m$: the block is momentarily at rest, $v = 0$, and the energy is entirely potential, $\tfrac{1}{2}kx_m^2$. These are the turning points.
- At the equilibrium position $x = 0$: the spring PE is zero, so the entire energy is kinetic, and the speed is maximum. Setting $\tfrac{1}{2}mv_m^2 = \tfrac{1}{2}kx_m^2$ gives
$$v_m = x_m\sqrt{\frac{k}{m}}$$
NCERT notes that $k/m$ has dimensions of $[\text{T}^{-2}]$, so the expression is dimensionally consistent. The kinetic energy converts to potential energy and back, over and over, while the total stays constant — the block oscillates. This continuous handing-back-and-forth is precisely the spring–mass simple harmonic motion studied in Chapter 13.
The quantity $\sqrt{k/m}$ that sets the maximum speed is the angular frequency $\omega$ of the oscillation — see spring–block SHM in Oscillations for the full time-dependence.
Why spring PE is always positive
The elastic potential energy $V(x) = \tfrac{1}{2}kx^2$ can never be negative. The spring constant $k$ is a positive material property, and $x^2 \ge 0$ for any real displacement, so the product is non-negative whether the spring is stretched ($x>0$) or compressed ($x<0$). Its only minimum value, zero, is reached at $x = 0$ — the equilibrium position, which NCERT fixes as the reference where $V = 0$.
NCERT is careful to add that the zero of potential energy is arbitrary and set by convenience; for the spring, the natural choice is $V = 0$ at the unstretched length. Once that zero is fixed, it must be held throughout a problem. With this convention spring PE is positive everywhere except at equilibrium, which is why the parabola never dips below the $x$-axis.
Worked examples
A mass of 2 kg is attached to a light spring of force constant $k = 100~\text{N m}^{-1}$. Calculate the work done by an external force in stretching the spring by 10 cm. What is the work done by the restoring force in the spring?
Work by the external force is the energy stored, $V = \tfrac{1}{2}kx^2$. With $x = 0.10~\text{m}$: $V = \tfrac{1}{2}\times 100 \times (0.10)^2 = 50 \times 0.01 = 0.5~\text{J}$.
Work by the restoring (spring) force has the same magnitude and the opposite sign, $W_s = -0.5~\text{J}$, because it opposed the stretching. The mass (2 kg) is a decoy here — work to stretch depends only on $k$ and $x$.
To simulate car accidents, a car of mass 1000 kg moving at 18.0 km/h on a smooth road collides head-on with a horizontally mounted spring of spring constant $5.25\times 10^{3}~\text{N m}^{-1}$. What is the maximum compression of the spring?
Energy conservation. At maximum compression the car is momentarily at rest, so all of its kinetic energy has become spring PE. First convert the speed: $18~\text{km h}^{-1} = 5~\text{m s}^{-1}$ (recall $36~\text{km h}^{-1} = 10~\text{m s}^{-1}$).
KE of the car: $K = \tfrac{1}{2}mv^2 = \tfrac{1}{2}\times 1000 \times 5^2 = 1.25\times 10^{4}~\text{J}$.
Set $\tfrac{1}{2}kx_m^2 = K$: $\;x_m = \sqrt{\dfrac{2K}{k}} = \sqrt{\dfrac{2\times 1.25\times 10^{4}}{5.25\times 10^{3}}}$, giving $x_m = 2.00~\text{m}$. (NCERT idealises the spring as massless and the surface as frictionless; with friction the compression would be smaller.)
A block of mass 3 kg moving with a velocity $20~\text{m s}^{-1}$ collides with a spring of force constant $1200~\text{N m}^{-1}$. Calculate the maximum compression of the spring.
Same conservation idea. All kinetic energy converts to spring PE at maximum compression: $\tfrac{1}{2}mv^2 = \tfrac{1}{2}kx_m^2$, so $x_m = v\sqrt{m/k}$.
$x_m = 20\sqrt{\dfrac{3}{1200}} = 20\sqrt{\dfrac{1}{400}} = 20 \times \dfrac{1}{20} = 1~\text{m}$. The block momentarily halts at $1~\text{m}$ of compression, then the spring drives it back — kinetic and potential energy trading places.
Link to Oscillations
The spring potential energy is the structural backbone of simple harmonic motion. A force of the form $F = -kx$ — exactly Hooke's law — is the defining condition of SHM, and the parabolic $V(x) = \tfrac{1}{2}kx^2$ is the SHM potential well. NCERT Exercise 5.4 makes this explicit: it gives the SHM potential as $V(x) = kx^2/2$ and asks you to show that a particle of total energy 1 J in a spring potential with $k = 0.5~\text{N m}^{-1}$ must turn back at $x = \pm 2~\text{m}$ — its turning points, where $E = V$ and $K = 0$.
The maximum-speed result $v_m = x_m\sqrt{k/m}$ carries straight into Chapter 13, where $\sqrt{k/m}$ is named the angular frequency $\omega$, the amplitude is $x_m$, and the energy oscillates between $K$ and $V$ at twice the frequency of the displacement. Mastering the spring PE here is the cleanest preparation for the energy methods of Oscillations.