Potential energy of a spring

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NCERT Class 11 Physics Chapter 5 Section 5.9

Quick answer

The potential energy of an ideal Hookean spring stretched or compressed by x from its natural length is V = ½kx^², where k is its spring constant: zero at the natural length and positive for both stretching and compressing. It equals the work done against the spring force in deforming it slowly from its natural length; for an ideal spring that force obeys Hooke’s law, F = −kx, and is conservative. For a block attached to the spring on a smooth horizontal surface, ½mv^² + ½kx^² stays constant, and the speed is greatest at the natural length.

With your textbook

NCERT Physics Part I (Class 11), Chapter 5 “Work, Energy and Power”, Section 5.9 “The potential energy of a spring”, pages 80–83.

Open the official NCERT chapter (PDF, 2.0 MB, ncert.nic.in) (opens in a new tab)

Checked against the 2026-27 reprint on 19 Oct 2026.

What is the potential energy of a spring?

The spring force in NCERT’s words

The spring force is an example of a variable force which is conservative.

NCERT Physics Part I (Class 11), Section 5.9, page 80

In plain words: when you stretch or compress a spring, you do work against its force, and that work is stored in the spring. Let go, and the spring gives it back, for example as the kinetic energy of a block attached to it. The stored energy is the elastic potential energy of the spring. For an ideal Hookean spring with positive spring constant k, stretched or compressed by x from its natural length:

Potential energy of a spring

V(x) = \tfrac{1}{2}kx^{2}

It is zero when the spring is at its natural length, and positive for both stretching and compressing.

The spring force and Hooke’s law

Fix one end of a light spring to a wall, attach a block to the other end, and let the block rest on a smooth horizontal surface. Measure the block’s displacement x from the position where the spring has its natural length. For an ideal spring, the force it exerts is proportional to x and opposite to it:

Hooke’s law

F_s = -kx

This is Hooke’s law. Real springs obey it only for small stretches and compressions, up to the proportional limit (see Hooke’s law for solids). The minus sign says that the force always points back towards the natural length: pull the block out (x > 0) and the spring pulls it in; push it in (x < 0) and the spring pushes it out. That is why it is called a restoring force.

The constant k is the spring constant, in N m⁻¹: the force needed per metre of stretch.

The spring is said to be stiff if k is large and soft if k is small.

NCERT Physics Part I (Class 11), Section 5.9, page 80

Work done by the spring force

The spring force changes with x, so its work is an integral, the area under the force–displacement graph (see work done by a variable force). As the block is pulled out from x = 0 to x_m, the spring force does

\begin{aligned} W_s &= \int_0^{x_m} (-kx)\,dx \\ &= -\tfrac{1}{2}kx_m^{2} \end{aligned}

This is negative, because the spring force points against the stretch. On a graph of F_s against x, it is the area of a triangle below the axis. If no other force does work and the stretch is quasistatic, the person does $+\tfrac{1}{2}kx_m^{2}$ of work.

From any x_i to any x_f, the spring force does

Work done by the spring force

W_s = \tfrac{1}{2}kx_i^{2} - \tfrac{1}{2}kx_f^{2}

This depends only on the end points. If the block is pulled out and returns to where it started (x_f = x_i), the work is zero:

The work done by the spring force in a cyclic process is zero.

NCERT Physics Part I (Class 11), Section 5.9, page 81

So the spring force is conservative, and it has a potential energy. Taking V = 0 at the natural length, the potential energy is $\tfrac{1}{2}kx^{2}$ , and you can check that −dV/dx = −kx, the spring force.

Energy of a block on a spring

Pull the block out to x_m and release it from rest on the smooth surface. Only the spring force does work, so mechanical energy is conserved. At any position x between −x_m and x_m:

\tfrac{1}{2}mv^{2} + \tfrac{1}{2}kx^{2} = \tfrac{1}{2}kx_m^{2}

At the ends (x = ±x_m) the block is momentarily at rest and all the energy is potential. At x = 0 the spring has its natural length, the potential energy is zero, and the kinetic energy and speed are largest:

Greatest speed of the block

v_m = x_m\sqrt{\frac{k}{m}}

Energy against displacement x for a block on a spring: V = ½kx² is an upward parabola, zero at x = 0; K is a downward parabola, largest at x = 0 and zero at ±xm; their sum E is a constant line. — As the block moves, kinetic and potential energy trade off, but their sum E = ½k x_m² stays constant.

In $v_m = x_m\sqrt{k/m}$ , k/m has dimensions [T⁻²], so $\sqrt{k/m}$ has dimensions [T⁻¹] and v_m has the dimensions of speed, as it should.

Springs around us

Crash testing. NCERT Example 5.8 describes how car makers study collisions of cars with mounted springs: the car’s kinetic energy becomes the spring’s potential energy at maximum compression.
Shock absorbers and mattresses. Springs store energy briefly and spread the impact over a longer time, or turn the energy into heat with dampers.
Spring balances. The stretch is proportional to the force, so the scale can be marked evenly in newtons.
Toy guns, staplers and retractable pens. A compressed spring stores energy that is released as the kinetic energy of the dart, the staple or the pen’s button.

Solved examples

Example 1: Force and stored energy

A spring of spring constant 400 N m^⁻¹ is stretched by 5 cm. Find (a) the force it exerts, (b) the energy stored in it, and (c) the extra work needed to stretch it slowly from 5 cm to 10 cm, with no other work or energy losses.

(a) The size of the force is kx = 400 × 0.05, that is 20 N, directed back towards the natural length.

(b)

\begin{aligned} V &= \tfrac{1}{2}(400)(0.05)^{2} \\ &= 0.5\ \text{J} \end{aligned}

(c) The extra work is the increase in stored energy:

\begin{aligned} W &= \tfrac{1}{2}(400)(0.10^{2} - 0.05^{2}) \\ &= 200 \times 0.0075 = 1.5\ \text{J} \end{aligned}

Answer: 20 N, 0.5 J, and 1.5 J more. Doubling the stretch makes the stored energy four times as large (2 J), so the second 5 cm takes three times the work of the first.

Example 2: Speed of a block on a spring

A 0.5 kg block on a smooth floor is attached to a spring with k = 200 N m⁻¹. It is pulled out 10 cm and released. Find (a) its greatest speed and (b) its speed when it is 5 cm from the natural length.

(a)

\begin{aligned} v_m &= x_m\sqrt{\frac{k}{m}} = 0.1\sqrt{400} \\ &= 2\ \text{m s}^{-1} \end{aligned}

(b) From energy conservation:

v^{2} = \frac{k}{m}\,(x_m^{2} - x^{2})

\begin{aligned} v &= \sqrt{400 \times (0.01 - 0.0025)} \\ &= \sqrt{3} \approx 1.73\ \text{m s}^{-1} \end{aligned}

Answer: 2 m s^⁻¹ at the natural length, and about 1.73 m s^⁻¹ at 5 cm.

Example 3: A car hits a spring buffer

A car of mass 800 kg moving at 10 m s^⁻¹ (36 km/h) on a level road runs into a spring buffer of spring constant 2.0 × 10⁴ N m^⁻¹. Find the greatest compression (a) if the road is smooth, and (b) if the coefficient of kinetic friction between the tyres and the road is 0.5, with the wheels locked. Take g = 9.8 m s⁻².

(a) All the kinetic energy becomes spring energy:

\begin{aligned} \tfrac{1}{2}kx_m^{2} &= \tfrac{1}{2}mv^{2} \\ x_m &= v\sqrt{\frac{m}{k}} = 10\sqrt{0.04} \\ &= 2.0\ \text{m} \end{aligned}

(b) The friction force has size μmg, that is 0.5 × 800 × 9.8, or 3920 N, and it does work −3920 x_m. By the work-energy theorem, the change in kinetic energy, $0 - \tfrac{1}{2}mv^{2}$ , equals the work done by the spring and by friction, so

\tfrac{1}{2}kx_m^{2} + \mu mg\,x_m = \tfrac{1}{2}mv^{2}

\begin{aligned}&10\,000\,x_m^{2} + 3920\,x_m \\ &= 40\,000\end{aligned}

Taking the positive root of this quadratic:

\begin{aligned} x_m &= \frac{-3920 + \sqrt{D}}{20\,000} \\ &= \frac{-3920 + 40\,192}{20\,000} \\ &\approx 1.81\ \text{m} \end{aligned}

where

D = 3920^{2} + 4 \times 10^{4} \times 4 \times 10^{4}

Answer: 2.0 m on a smooth road and about 1.81 m with friction.

NCERT Examples 5.8 and 5.9 (pages 81–82) do the same for a 1000 kg car at 18 km/h, finding 2.00 m without friction and 1.35 m with μ = 0.5 (NCERT takes g = 10 m s⁻²). NCERT prints the spring constant as 5.25 × 10³ N m⁻¹, but its answers match 6.25 × 10³ N m⁻¹; with 5.25 × 10³, the compressions would be about 2.18 m and 1.43 m.

Common mistakes to avoid

Common mistake

Using ½k(x_₂ − x_₁)^² for the work between two stretches

The energy stored goes as x², so the work from x₁ to x₂ is $\tfrac{1}{2}k(x_2^{2} - x_1^{2})$ ⁠. In Example 1(c), (x₂ − x₁)² would give 0.5 J instead of 1.5 J.

Common mistake

Getting the sign of the spring’s work wrong

The spring force does negative work when the spring is stretched or compressed further, and positive work when it returns towards its natural length. For a quasistatic stretch with no other work, the person does the opposite work.

Common mistake

Thinking a compressed spring has negative potential energy

$\tfrac{1}{2}kx^{2}$ is positive whether x is positive (stretched) or negative (compressed). A compressed spring stores energy just like a stretched one.

Common mistake

Thinking the speed is always greatest at the natural length

That is true for a horizontal spring. For an ideal undamped vertical spring with fixed support, gravity also acts, and the speed is greatest where the spring force balances the weight, at a stretch of mg/k. A mass attached to the unstretched spring and let go suddenly falls until the stretch is 2mg/k, not mg/k.

Common mistake

Changing the zero of potential energy halfway

NCERT puts it memorably:

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 !

NCERT Physics Part I (Class 11), Section 5.9, page 82

Quick revision summary

An ideal spring, or a real spring within its proportional limit, obeys Hooke’s law: its force is proportional to the stretch or compression and points back towards the natural length.
The spring constant k is the force per unit stretch, in N m^⁻¹; a stiff spring has a large k.
The spring force is conservative: its work depends only on the end points, and is zero over a cycle.
A spring stretched or compressed by x stores potential energy ½kx^², zero at its natural length.
For a block on a spring on a smooth horizontal surface, K + V is constant; the speed is greatest at the natural length.

Formula sheet

F = −kx (Hooke’s law)
V = ½kx^²
W_s = ½kx_i² − ½kx_f²
½mv² + ½kx² = ½kx_m²
v_m = x_m√(k/m)

Questions students ask

What is the SI unit and dimensional formula of the spring constant?

The SI unit of the spring constant k is the newton per metre (N m^⁻¹), the same as kg s^⁻², because k is the force per unit stretch, |F|/|x|. Its dimensional formula is [M T^⁻²]. A stiff spring has a large k and a soft one a small k; joining springs in series or in parallel, or cutting a spring, changes the spring constant (see combinations of springs).

How do you derive V = ½kx^² for a spring?

The potential energy of a spring, ½kx^², is the work done against the spring force in stretching or compressing it slowly from its natural length by x. The force needed grows steadily from 0 to kx, so the work is the area of the triangle under the force–extension line: ½ × x × kx = ½kx^². Because the spring force is conservative, this work is stored, and the spring gives it back as it returns to its natural length.

Why is the spring force conservative?

The spring force is conservative because its work depends only on the initial and final stretch, W_s = ½kx_i² − ½kx_f², not on the path in between. Over any cycle that brings the spring back to its starting stretch its work is zero, so the work done against it is stored as potential energy and can be fully recovered. Friction is different: its work around a closed path is negative, and the energy turns into heat.

Practice questions

Try each question first, then open the answer to check your working.

Q1. A force of 50 N stretches a spring by 2 cm. Find its spring constant and the energy stored in it.

Show answerHide answer

k = 50/0.02, that is 2500 N m^⁻¹, and the energy is
$\begin{aligned} V &= \tfrac{1}{2}(2500)(0.02)^{2} \\ &= 0.5\ \text{J} \end{aligned}$
Q2. The extension of a spring is doubled. By what factor does its stored energy change?

Show answerHide answer

It goes as x², so it becomes four times as large.
Q3. A 0.2 kg ball is pushed against a spring with k = 500 N m⁻¹, compressing it by 4 cm on a smooth horizontal surface, and released. With what speed does it leave the spring?

Show answerHide answer

All the stored energy becomes kinetic energy:
$\begin{aligned} v &= x\sqrt{\frac{k}{m}} = 0.04\sqrt{2500} \\ &= 2\ \text{m s}^{-1} \end{aligned}$

Potential energy of a spring

What is the potential energy of a spring?

The spring force and Hooke’s law

Work done by the spring force

Energy of a block on a spring

Springs around us

Solved examples

Example 1: Force and stored energy

Example 2: Speed of a block on a spring

Example 3: A car hits a spring buffer

Common mistakes to avoid

Quick revision summary

Questions students ask

What is the SI unit and dimensional formula of the spring constant?

How do you derive V = ½kx^² for a spring?

Why is the spring force conservative?

Practice questions

Sources and review

Topics in Work, Energy and Power

What is the potential energy of a spring?

The spring force and Hooke’s law

Work done by the spring force

Energy of a block on a spring

Springs around us

Solved examples

Example 1: Force and stored energy

Example 2: Speed of a block on a spring

Example 3: A car hits a spring buffer

Common mistakes to avoid

Quick revision summary

Questions students ask

What is the SI unit and dimensional formula of the spring constant?

How do you derive V = ½kx² for a spring?

Why is the spring force conservative?

Practice questions

Sources and review

Topics in Work, Energy and Power

How do you derive V = ½kx^² for a spring?