JEE Mains Physics · Work, Energy and Power

Potential Energy and Conservation of Mechanical Energy

A conservative force stores the work it does as potential energy, with F = −dU/dx, and where only such forces do work, kinetic plus potential energy stays constant.

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Why this matters

Twenty-six PYQs, twenty-one of them multiple choice, and five from 2026; sixteen carry a figure. Eight are about conservative forces and potential energy, from F = −dU/dx and equilibrium to reading a U(x) graph; seven store energy in a spring, from a hanging mass to a block that slides or falls onto it; eleven conserve mechanical energy on tracks, pulleys, pendulums and vertical circles. The figure usually gives the heights, so read it before writing any equation.

Concept 1 of 3: Conservative forces and potential energy

The work done by a conservative force depends only on where the body starts and ends, not on the path between. That makes it possible to store the work as a potential energy U. The force then points downhill on the U curve, and its size is the slope.

Definition

$W_{\text{cons}} = -\Delta U = U_i - U_f$ , the same along every path, and zero round a closed path.
$F = -\dfrac{dU}{dx}$ ; in three dimensions $F_x = -\dfrac{\partial U}{\partial x}$ , and likewise for y and z.
Conservative: gravity, the spring force, the electrostatic force and any central force $F(r)\hat r$ . Not conservative: friction and air drag, which have no potential energy.
Equilibrium where $dU/dx = 0$ : stable at a minimum of U, unstable at a maximum.
On a U(x) graph, the size of the force is the size of the slope; a flat stretch has zero force.
$K = E - U$ : the body can be only where $U \le E$ , it stops where $U = E$ , and it is fastest where U is lowest.
Interatomic $U = \dfrac{A}{r^{n}} - \dfrac{B}{r^{m}}$ : set $dU/dr = 0$ for the equilibrium separation.

Force from potential energy

F = -\frac{dU}{dx} \qquad F_x = -\frac{\partial U}{\partial x} \qquad W_{\text{cons}} = -\Delta U

Worked example

The potential energy of two atoms at separation r is

U = \dfrac{a}{r^{12}} - \dfrac{b}{r^{6}}

, with a and b positive. Find the equilibrium separation.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q101Moderate

Example 1 · Work, Energy and Power · Potential Energy and Conservation of Mechanical Energy

Potential energy as a function of

r

is given by

U =\frac{A}{r^{10}}-\frac{B}{r^{5}}

, where

r

is the interatomic distance,

A

and

B

are positive constants. The equilibrium distance between the two atoms will be:

Work is the integral of F·dr; the minus sign belongs to ΔU

The work done by a force from r₁ to r₂ is W = ∫F·dr. It is the change in potential energy that carries the minus sign: ΔU = −∫F·dr. A statement that writes W = −∫F·dr is false.

The force is the slope, not the height

On a U(x) graph, a steep stretch means a strong force and a flat stretch means no force, however high it sits. Ranking forces by the height of U instead of the steepness gives the wrong order.

Friction has no potential energy

The work done by friction depends on the path, so no function U can store it. Coulomb, gravitational and spring forces all have a potential energy; friction does not.

Drill 7 more on conservative forces and potential energy

Concept 2 of 3: Potential energy of a spring

Stretching or compressing a spring by x stores ½kx² of energy. The force grows from zero to kx as the spring is pushed, so the energy is the area of a triangle, half of kx times x. A moving block that hits a spring turns its kinetic energy into this stored energy.

Definition

$U = \tfrac12 kx^{2}$ , with x measured from the natural length.
Stretching from $x_1$ to $x_2$ takes $\tfrac12 k(x_2^{2} - x_1^{2})$ .
A mass m hanging at rest stretches the spring by $mg/k$ and stores $\dfrac{m^{2}g^{2}}{2k}$ : for the same mass, $U \propto 1/k$ ; for the same spring, $U \propto m^{2}$ .
A block at speed u on a smooth floor: $\tfrac12 mu^{2} = \tfrac12 kx_{\max}^{2}$ . When its speed has fallen to v: $\tfrac12 m(u^{2} - v^{2}) = \tfrac12 kx^{2}$ .
A ball dropped from height h onto a platform on a spring falls $h + x$ : $mg(h + x) = \tfrac12 kx^{2}$ .
With friction on the way: $\tfrac12 kx^{2} =$ energy at the start $- \mu mg\,d$ .

Spring potential energy

U = \tfrac12 kx^{2} \qquad W_{x_1 \to x_2} = \tfrac12 k\left(x_2^{2} - x_1^{2}\right)

Worked example

A 2 kg block sliding at 4 m/s on a smooth floor hits a spring of force constant 800 N/m. Find (a) the maximum compression and (b) the compression when the block's speed has fallen to 2 m/s.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · JEE Mains 2022 — 25 June · Q22Moderate

Example 2 · Work, Energy and Power · Potential Energy and Conservation of Mechanical Energy

0.5\text{ }kg

block moving at a speed of

12{\text{ }ms}^{- 1}

compresses a spring through a distance

30\text{ }cm

when its speed is halved. The spring constant of the spring will be

\_\_\_\_

Nm^{- 1}

Spring energy is half of force times stretch

The spring force grows from 0 to kx as it is stretched, so the stored energy is ½kx², not kx·x. Writing kx² doubles every answer.

A dropped ball falls the extra compression too

A ball dropped from h onto a spring platform loses mg(h + x) of potential energy by the lowest point, not mgh. Using mgh gives a spring constant that is too small.

Energy between two stretches uses x₂² − x₁²

Stretching from 2 cm to 4 cm takes ½k(4² − 2²) in cm², not ½k(4 − 2)². Every extension is measured from the natural length.

Drill 6 more on potential energy of a spring

Concept 3 of 3: Conservation of mechanical energy

Normal reactions and the tension in an inextensible string are always at right angles to the motion, so they do no work. When only gravity and springs do work, kinetic plus potential energy stays the same. The speed at a point then depends only on how far the body has dropped, not on the shape of the track.

Definition

$K_1 + U_1 = K_2 + U_2$ ; on a smooth track $v^{2} = u^{2} + 2g(h_1 - h_2)$ .
Dropped from height S: at height h, $v^{2} = 2g(S - h)$ . K equals n times U at $h = \dfrac{S}{n + 1}$ .
Pendulum at angle θ from the vertical: it is $l(1 - \cos\theta)$ above the lowest point.
Projectile: at the top $K = \tfrac12 m(u\cos\theta)^{2}$ , so the kinetic energy lost on the way up is $\tfrac12 mu^{2}\sin^{2}\theta$ .
Two masses over a light pulley: $(m_1 - m_2)gh = \tfrac12 (m_1 + m_2)v^{2}$ ; both masses move.
Vertical circle on a string (from Laws of Motion: at the top $T + mg = mv^{2}/L$ ): just completing it needs $v_{\text{top}}^{2} = gL$ and $v_{\text{bottom}}^{2} = 5gL$ , so $K_{\text{bottom}} : K_{\text{top}} = 5 : 1$ .
On a rigid rod the top speed can be zero, so $v_{\text{bottom}}^{2} = 4gL$ is enough.
A point at angle θ from the bottom of a circle of radius R is $R(1 - \cos\theta)$ above the bottom.
A fraction f of the energy kept (the rest lost to air): $f\,mgh = \tfrac12 mv^{2}$ .

Mechanical energy conserved

\tfrac12 mv_1^{2} + mgh_1 = \tfrac12 mv_2^{2} + mgh_2 \qquad v_{\text{bottom}}^{2} = 5gL

Worked example

A 2 kg bob hangs on a string 1.6 m long. It is pulled aside until the string makes 60° with the vertical and released. Find its speed (a) at the lowest point and (b) when the string makes 37° with the vertical (cos 37° = 0.8). (

g = 10

m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q16Moderate

Example 3 · Work, Energy and Power · Potential Energy and Conservation of Mechanical Energy

A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

A rod is not a string

A bob on a string needs a speed of √(gL) at the top, so √(5gL) at the bottom. A bob on a rigid rod can reach the top with zero speed, so √(4gL) at the bottom is enough.

Measure every height from one level

Pick the lowest point as zero and keep it. On a circle of radius R, a point at angle θ from the bottom is R(1 − cos θ) up; a point at angle θ from the top is R(1 + cos θ) up.

Both masses on a pulley carry kinetic energy

When one mass falls and the other rises, the energy released, (m₁ − m₂)gh, is shared by both: ½(m₁ + m₂)v². Giving it all to the falling mass makes the speed too large.

Drill 10 more on conservation of mechanical energy

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Conservative forces and potential energy
Force from potential energy
$F = -\frac{dU}{dx} \qquad F_x = -\frac{\partial U}{\partial x} \qquad W_{\text{cons}} = -\Delta U$
Potential energy of a spring
Spring potential energy
$U = \tfrac12 kx^{2} \qquad W_{x_1 \to x_2} = \tfrac12 k\left(x_2^{2} - x_1^{2}\right)$
Conservation of mechanical energy
Mechanical energy conserved
$\tfrac12 mv_1^{2} + mgh_1 = \tfrac12 mv_2^{2} + mgh_2 \qquad v_{\text{bottom}}^{2} = 5gL$

Watch out for (9)

Work is the integral of F·dr; the minus sign belongs to ΔU
→ Conservative forces and potential energy
The force is the slope, not the height
→ Conservative forces and potential energy
Friction has no potential energy
→ Conservative forces and potential energy
Spring energy is half of force times stretch
→ Potential energy of a spring
A dropped ball falls the extra compression too
→ Potential energy of a spring
Energy between two stretches uses x₂² − x₁²
→ Potential energy of a spring
A rod is not a string
→ Conservation of mechanical energy
Measure every height from one level
→ Conservation of mechanical energy
Both masses on a pulley carry kinetic energy
→ Conservation of mechanical energy

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