Work, Energy and Power Formula Sheet — JEE Main Physics

Q: What are the most important Work, Energy and Power formulas for JEE Main?

This Work, Energy and Power formula sheet covers all the high-yield Physics formulas, definitions and theorems you need for JEE Main, across Work done by a constant force and a variable force, Kinetic energy, Work-energy theorem, Power, Notion of potential energy — each shown with the key result and, where useful, a worked example.

Q: Is this Work, Energy and Power formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

Q: How should I revise Work, Energy and Power formulas?

Blurt the Work, Energy and Power formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Every key Work, Energy and Power formula, definition and theorem for JEE Main Physics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 11 sub-topics

Work done by a constant force and a variable force
Kinetic energy
Work-energy theorem
Power
Notion of potential energy
Potential energy of a spring
Conservative forces
Conservation of mechanical energy
Non-conservative forces
Motion in a vertical circle
Elastic and inelastic collisions in one and two dimensions

Work

Work:

W = F \cdot d = F d cos θ

— a

scalar

; the component of force along the displacement times the displacement.

Sign of work

▸ $W > 0$ for $0^{\circ} \leq θ < 9 0^{\circ}$ ; $W < 0$ for $9 0^{\circ} < θ \leq 18 0^{\circ}$ .
▸ $W = 0$ if $θ = 9 0^{\circ}$ , or $F = 0$ , or $d = 0$ (e.g. holding a weight, or circular-orbit gravity).

Variable force

W = \int_{x_{i}}^{x_{f}} F \cdot d x

F =

force (may depend on x),

x_{i}, x_{f} =

start/end positions;

W =

area under the F–x graph

Force vs displacement curve; shaded area = work done — Work $=$ area under the $F$ – $x$ curve.

Units

▸SI: joule (J) $=$ N m. $1 erg = 1 0^{- 7}$ J.
▸ $1 eV = 1.6 \times 1 0^{- 19}$ J, $1 cal = 4.186$ J, $1 kWh = 3.6 \times 1 0^{6}$ J.

🚫 Examiner Trap · Work

(1) Use the angle

between

F

and

d

, not between

F

and the horizontal. (2) Work can be negative (friction, braking) — KE then

decreases

. (3) For variable force you MUST integrate

\int F d x

; you cannot use

W = F d

with the average force unless F is linear in x. (4) Centripetal force does

zero

work (

F ⊥ v

Work by special forces

Gravity

W_{g} = - mgh

(rising),

+ mgh

(falling)

m =

mass,

h =

vertical rise/fall; path-independent (conservative)

Spring

W_{s p r in g} = - \frac{1}{2} k (x_{f}^{2} - x_{i}^{2})

k =

spring constant,

x =

deformation from natural length; work to stretch/compress by x (external)

= + \frac{1}{2} k x^{2}

Friction

W_{f} = - f d = - μ N d

f = μ N =

kinetic friction,

N =

normal reaction,

d =

path length; always negative, path-dependent (non-conservative)

Comparative: work by the three forces

Force	Work	Sign	Path?
Gravity	$- m g h$ / $+ m g h$	$\pm$	independent
Spring	$- \frac{1}{2} k (x_{f}^{2} - x_{i}^{2})$	$\pm$	independent
Friction	$- μ N d$	always $-$	dependent

⚡ Shortcut · Spring sign in 1 step

Spring (and gravity) work depend only on

end points

W = U_{i} - U_{f} = - Δ U

. Moving the block back to start always undoes the work — round-trip

= 0

. For friction, round-trip work is

- 2 μ N (half-path) \neq = 0

★ Remember · 'Work done' needs an agent

Always state

which force

does the work. By Newton's 3rd law the forces cancel, but the works

W_{12}

and

W_{21}

need not.

Kinetic Energy & Work–Energy Theorem

Kinetic energy

K = \frac{1}{2} m v^{2} = \frac{1}{2} m v \cdot v

m =

mass,

v =

speed; scalar,

\geq 0

; in terms of momentum

K = \frac{p ^{2}}{2 m}

p = 2 m K

Work–energy theorem:

K_{f} - K_{i} = W_{n e t}

For a variable force

K_{f} - K_{i} = \int_{x_{i}}^{x_{f}} F d x

F =

net force along motion; derived from

d K = F d x

Comparative:

K

p

links

Given	KE $K$	Momentum $p$
From the other	$K = \frac{p ^{2}}{2 m}$	$p = 2 m K$
Double the speed	$\times 4$	$\times 2$
Same K, $m_{1} > m_{2}$	equal	heavier has larger $p$
Same p, $m_{1} > m_{2}$	lighter larger $K$	equal

Notes

▸It is the $scalar$ form of Newton's 2nd law (an integral over displacement).
▸Holds in any inertial frame; in a non-inertial frame include the pseudo-force in $W_{n e t}$ .
▸ $W_{n e t} > 0 \Rightarrow$ speeds up; $W_{n e t} < 0 \Rightarrow$ slows down.

🚫 Examiner Trap · Work–energy theorem

(1)

W_{n e t}

is the work of

every

force (incl. friction, gravity, normal) — not just the applied force. (2) KE depends on the

frame

, so

W_{n e t}

does too; only fix one frame per problem. (3) Equal momentum

\neq =

equal KE: lighter body carries more KE for the same p. (4)

K \geq 0

always — a 'negative KE' answer means an arithmetic slip.

⚡ Shortcut · Skip the kinematics

Need only speed/KE and not time? Use

K_{f} - K_{i} = W_{n e t}

directly — variable forces, rough patches, or unknown force but known energy change. Stopping distance:

\frac{1}{2} m v^{2} = f d \Rightarrow d = \frac{v ^{2}}{2 μg}

🎯 Exam · When to use it

Best when you need only speed/KE and not time — variable forces, rough patches, or when the force is unknown but the energy change is known. Stopping distance:

\frac{1}{2} m v^{2} = f d

Potential Energy & Conservative Forces

Potential energy: Energy stored by virtue of position/configuration, defined for a

conservative

force:

Δ U = - W_{co n s}

Gravitational PE (near surface)

U = m g h

m =

mass,

g =

grav. field,

h =

height above the chosen zero level

Elastic (spring) PE

U = \frac{1}{2} k x^{2}

k =

spring constant,

x =

deformation from natural length

Spring force F=kx linear and PE U = half k x squared parabola — Spring: $F = k x$ (linear), $U = \frac{1}{2} k x^{2}$ (parabola).

Force from PE

F (x) = - \frac{d U}{d x}; U_{i} - U_{f} = \int_{x_{i}}^{x_{f}} F d x

F =

conservative force; the force points 'downhill' on the

U (x)

curve

🚫 Examiner Trap · Potential energy & forces

(1) PE belongs to a

system

, not a body, and only for a

conservative

force — there is no 'friction PE'. (2)

F = - d U / d x

has a

minus

sign; force is toward

decreasing

U. (3) Spring PE

\frac{1}{2} k x^{2}

uses deformation from natural length, not its total length. (4) Only

Δ U

is physical; the zero level is arbitrary.

Comparative: conservative vs non-conservative

Property	Conservative	Non-conservative
Examples	gravity, spring, electrostatic	friction, viscous drag
Work along path	path-independent	path-dependent
Round-trip work	$0$	$\neq = 0$
PE definable?	yes ( $Δ U = - W$ )	no

Conservative vs non-conservative

▸ $Conservative$ (gravity, spring, electrostatic): work is path-independent; round-trip work $= 0$ ; a PE can be defined.
▸ $Non-conservative$ (friction, viscous drag): work is path-dependent; round-trip work $\neq = 0$ ; no PE.

⚡ Shortcut · Conservative test

A 1-D force is conservative iff it depends only on position x (then

W = \int F d x

is path-free). Velocity- or path-dependent (friction

μ N

, drag

- b v

)

\Rightarrow

non-conservative — no PE exists.

★ Remember · PE is relative

PE is defined only up to an additive constant — choose the zero level conveniently. Only

Δ U

has physical meaning.

Conservation of Energy & PE Curves

Mechanical energy

E = K + U = const

K =

kinetic,

U =

potential energy; if only conservative forces act:

K_{i} + U_{i} = K_{f} + U_{f}

With non-conservative forces

E_{f} - E_{i} = W_{n c}

W_{n c} =

work by non-conservative forces; e.g. energy lost to friction

= W_{f r i c t i o n} < 0

Potential energy curve U(x): unstable maximum and stable minimum, F = -dU/dx — Equilibrium from U(x); $F = - \frac{d U}{d x}$ .

Comparative: equilibrium types from

U (x)

Type	$U (x)$	$U^{''} (x)$	On displacing
Stable	minimum	$> 0$	force restores
Unstable	maximum	$< 0$	force pushes away
Neutral	flat	$= 0$	no net force

Equilibrium from

U (x)

▸ $F = - \frac{d U}{d x} = 0$ at equilibrium.
▸ $Stable$ : U minimum, $\frac{d ^{2} U}{d x ^{2}} > 0$ . $Unstable$ : U maximum, $< 0$ . $Neutral$ : U flat.

Spring–block oscillation

\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} k x_{m}^{2}

x_{m} =

amplitude; speed max at

x = 0

v_{ma x} = k / m x_{m}

🚫 Examiner Trap · Conservation of energy

(1) Mechanical energy is conserved

only

if non-conservative work is zero — friction/drag break it (

E_{f} - E_{i} = W_{n c}

). (2) Normal force and tension usually do

no

work (perpendicular to motion), so include them as forces but not in energy. (3) On a U(x) curve,

minimum

U is stable — many flip it. (4) A body is confined where

E \geq U (x)

; turning points are where

E = U

⚡ Shortcut · Energy beats force

For variable forces (springs, curved tracks) the energy method gives speeds directly — no need to integrate the force. Read motion off the

U (x)

curve:

K = E - U

, so the body speeds up in

U

-valleys, slows on

U

-hills.

🎯 Exam · Energy beats force

For variable forces (springs, curved tracks) the energy method gives speeds directly — no need to integrate the force.

Power & the Vertical Circle

Power

P_{a v} = \frac{W}{t}; P_{in s t} = \frac{d W}{d t} = F \cdot v

W =

work,

t =

time,

v =

velocity; scalar; dimension

[M L^{2} T^{- 3}]

Units

▸ $1 W = 1 J s^{- 1}$ ; $1 hp = 746 W$ .
▸ $1 kWh = 3.6 \times 1 0^{6}$ J is a unit of $energy$ , not power. $(36 km/h = 10 m/s)$ .

🚫 Examiner Trap · Power

(1)

P = F \cdot v

uses

instantaneous

v — at constant power, force

falls

as the body speeds up (

F = P / v

). (2)

1 kWh

energy

, not power. (3) Average power

\neq =

power at the average speed unless motion is uniform. (4) Use

P = F v cos ϕ

when

F

and

v

are not parallel.

⚡ Shortcut · Constant-power motion

Engine at constant power P from rest:

v = \frac{2 Pt}{m}

and

x = \frac{8 P}{9 m} t^{3/2}

(so

v \propto t^{1/2}

x \propto t^{3/2}

). Max speed when drag balances:

v_{ma x} = P / F_{d r a g}

Vertical circle (energy method)

Vertical circle: v_top = sqrt(gr), v_bot = sqrt(5gr), diameter 2r — Critical speeds on a vertical circle of radius $r$ .

Speed & critical values

v^{2} = u^{2} - 2 g l (1 - cos θ); v_{b o t} = 5 g l, v_{s i d e} = 3 g l, v_{t o p} = g l

u =

speed at the bottom,

l =

radius,

θ =

angle from the lowest point

Comparative: string vs rod (loop the loop)

Constraint	Complete loop	Oscillates if
String / track	$u \geq 5 g l$	$u \leq 2 g l$
Light rod	$u \geq 4 g l$	$u \leq 2 g l$

Conditions (from the bottom)

▸Complete the loop (string): $u \geq 5 g l$ (at the top $m g = \frac{m v ^{2}}{l}$ ).
▸Light rod: $u \geq 4 g l$ ; oscillates if $u \leq 2 g l$ .

🚫 Examiner Trap · Vertical circle

(1) String needs

v_{t o p} \geq g l

(tension

\geq 0

); a

rod

can push, so it only needs

v_{t o p} \geq 0

(

u \geq 4 g l

). (2) For

2 g l < u < 5 g l

a string goes slack and the body leaves the circle — it does NOT just oscillate. (3) At the top, gravity supplies (part of) the centripetal force; don't set

T = m v^{2} / l

alone.

Collisions — One Dimension

Momentum (always conserved)

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

u =

before,

v =

after; true in every collision (no external impulse during the short contact)

1D collision before and after: m1 with u1 hits m2; after m1 v1 and m2 v2 — Head-on collision, before and after.

Coefficient of restitution

e = \frac{separation}{approach} = \frac{v _{2} - v _{1}}{u _{1} - u _{2}}

e = 1

elastic (KE conserved),

e = 0

perfectly inelastic (stick together);

0 \leq e \leq 1

Elastic, head-on (

m_{2}

at rest)

v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}; v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}

m_{1}, u_{1} =

incident mass & speed;

m_{2}

initially at rest

Comparative: elastic vs inelastic

	Elastic	Perfectly inelastic
$e$	$1$	$0$
Momentum	conserved	conserved
KE	conserved	max loss
After	separate	move together

Elastic special cases (

u_{2} = 0

)

▸Equal masses: $v_{1} = 0, v_{2} = u_{1}$ — they $exchange$ velocities.
▸ $m_{1} ≫ m_{2}$ : $v_{1} \approx u_{1}, v_{2} \approx 2 u_{1}$ .
▸ $m_{1} ≪ m_{2}$ : $v_{1} \approx - u_{1}, v_{2} \approx 0$ — light body bounces back.

Perfectly inelastic

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}; Δ K = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}

v =

common final velocity;

Δ K =

maximum possible KE loss

🚫 Examiner Trap · Collisions (1D)

(1) Momentum is conserved in

every

collision; KE only if

e = 1

(elastic). (2) Momentum is a

vector

— keep signs by a chosen

+

axis (a body moving

- x

has

- m v

). (3)

e = 0

means they move together, NOT that they stop. (4) 'KE lost' goes to heat/sound/deformation — total energy is still conserved.

⚡ Shortcut · With restitution

e

KE lost

Δ K = \frac{1}{2} μ (1 - e^{2}) (u_{1} - u_{2})^{2}

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

(reduced mass). Relative velocity simply reverses and scales:

v_{2} - v_{1} = - e (u_{2} - u_{1})

🎯 Exam · KE loss with restitution

e

Δ K = \frac{1}{2} μ (1 - e^{2}) (u_{1} - u_{2})^{2}

, where

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

(reduced mass).

Collisions — 2D & Special Cases

Oblique (2D) collision

2D collision: m1 hits stationary m2, fly off at angles theta1 and theta2 — $m_{1}$ hits a stationary $m_{2}$ ; they fly off at $θ_{1}, θ_{2}$ .

Momentum components conserved

m_{1} u_{1} = m_{1} v_{1} cos θ_{1} + m_{2} v_{2} cos θ_{2}; 0 = m_{1} v_{1} sin θ_{1} - m_{2} v_{2} sin θ_{2}

θ_{1}, θ_{2} =

deflection angles above/below the incident line; resolve

∥

and

⊥

u_{1}

Equal-mass elastic

▸After a glancing elastic collision of equal masses (one at rest), the two fly off at $θ_{1} + θ_{2} = 9 0^{\circ}$ .

🚫 Examiner Trap · Collisions (2D)

(1) Conserve momentum

component-wise

(two scalar equations); you cannot add the speeds as scalars. (2) The

θ_{1} + θ_{2} = 9 0^{\circ}

rule holds

only

for equal masses, elastic, one target at rest. (3) 2D problems usually need a third relation (KE if elastic, or e) — momentum alone has too many unknowns.

Ballistic pendulum & bouncing

Ballistic pendulum

(M + m) v = m u; v = 2 g y \Rightarrow u = \frac{M + m}{m} 2 g y

m, u =

bullet mass & speed;

M =

block;

y =

rise. Momentum for the hit, energy for the swing

Ball bouncing off the floor

v_{n} = e^{n} v_{0}; h_{n} = e^{2 n} h_{0}; H_{t o t a l} = h_{0} \frac{1 + e ^{2}}{1 - e ^{2}}

v_{0}, h_{0} =

initial speed/height,

n =

bounce number,

e =

restitution

🚫 Examiner Trap · Ballistic pendulum & bouncing

(1) Ballistic pendulum is

two stages

: momentum (inelastic embed) THEN energy (swing). Do NOT use energy conservation through the embedding — KE is lost there. (2) After embedding,

v = m u / (M + m)

, not u. (3) Bounce heights scale by

e^{2}

per bounce, speeds by e.

🎯 Exam · Neutron moderation

Fractional KE transferred in a head-on elastic hit

= \frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}}

— maximum (

= 1

) when

m_{1} = m_{2}

(why light moderators slow neutrons best).

More JEE Main Physics formula sheets

Frequently Asked Questions

What are the most important Work, Energy and Power formulas for JEE Main?

This Work, Energy and Power formula sheet covers all the high-yield Physics formulas, definitions and theorems you need for JEE Main, across Work done by a constant force and a variable force, Kinetic energy, Work-energy theorem, Power, Notion of potential energy — each shown with the key result and, where useful, a worked example.

Is this Work, Energy and Power formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Work, Energy and Power formulas?

Blurt the Work, Energy and Power formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Also useful: all formula sheets · JEE Main previous-year papers · most important chapters.