Work, Energy & Power: theory

CONCEPT

Work by a Constant Force

In one line
Work is done when a force moves a body: only the part of the force along the displacement counts.

DEFINITION

5.1.1 Work by a constant force

A force F at angle θ has two parts: F cos θ along the displacement and F sin θ across it.
Only F cos θ does work.

Force F moves the body through s

W = Fs cos θ

W = F·s = F_xs_x + F_ys_y + F_zs_z

Work is a scalar, the dot product of force and displacement.
Given two positions: s = final position − initial position.
Several forces: add them first, then take the dot product with s.

NATURE OF WORK

5.1.2 Positive, negative and zero work

The angle between F and s decides the sign

Angle	Work	Example
0° ≤ θ < 90°	positive, W_max = Fs at 0°	lifting force on a body; pull on a lawn roller; stretching a spring
90° < θ ≤ 180°	negative, W_min = −Fs at 180°	gravity on a body lifted up; friction on a sliding body
θ = 90°	zero	coolie walking on a level platform; centripetal force; magnetic force on a charge

Lifting force and stretching force do positive work; gravity and friction here do negative work

Zero work has three cases: F ⟂ s, or s = 0, or F = 0.
Pushing a wall that does not move: W = 0.
Holding a weight up: W = 0. Only lifting it takes work.

Two ways to get zero work

GRAVITY AND FRICTION

5.1.3 Work against gravity and friction

against gravity: W = mgh

h is the vertical rise. The path does not matter.

Same rise h on all four paths: same work mgh

against friction: W = μmg s

Force at an angle: R changes. Pull at θ: R = mg − F sin θ.
On an incline: f = μmg cos θ.
Uniform velocity: applied force along the motion = friction.
Total work on a slope = mgh + work against friction.

UNITS

5.1.4 Units and frame of reference

[W] = [ML²T⁻²]

Unit	Value
1 J	1 N × 1 m = 10⁷ erg
1 kg-m	9.81 J
1 g-cm	981 erg
1 eV	1.6 × 10⁻¹⁹ J
1 kWh	3.6 × 10⁶ J
1 cal	4.18 J

Work depends on the frame, because displacement does.
Porter climbing stairs: the work of his lifting force is zero in his own frame and mgh for a person on the ground.

Porter with a suitcase on a staircase

Practise PYQs on this topicWork by a Constant Force · 25 questions with solutions

CONCEPT

Work by a Variable Force

In one line
When the force changes with position, integrate: the work is the area under the F-x graph.

INTEGRAL FORM

5.2.1 Work as an integral

Small work over each small step ds

dW = F·ds

W = ∫_x₁^x₂ F dx

Put the force as a function of x and integrate between the two positions.
A constant term adds to the answer as constant × distance.

GRAPH

5.2.2 F-x graph

Each strip has area F dx

W = area under the F-x graph

Split the area into rectangles and triangles.
Area below the x-axis counts as negative work.
Starting from rest: K at the end = total area.

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Corner

Force in two or three dimensions

W = ∫F_x dx + ∫F_y dy + ∫F_z dz

Integrate each component over its own limits.
A force with only F_x(x) and F_y(y) gives the same work on every path.

Retardation or friction that varies

Retardation a = kx: loss in K = ∫mkx dx, so loss ∝ x².
μ = μ₀ − αx: replace μ in f = μmg and integrate over x.
Average of a linear force = half the sum of its end values.

Practise PYQs on this topicWork by a Variable Force · 26 questions with solutions

CONCEPT

Kinetic Energy & Momentum

In one line
Kinetic energy is the energy of motion. It is tied to momentum by K = p² ÷ 2m, the most used relation of this chapter.

KINETIC ENERGY

5.3.1 Kinetic energy

Energy a body has because of its motion: flowing water, wind, a moving hammer, a bullet.
The work done to bring a body from rest to speed v is stored as K.

Force F takes the body from rest to speed v over distance s

K = 12 mv²

K is a scalar and can never be negative.
K depends on the frame: a passenger has K = 0 in the train and K = ½mv² for a person on the ground.
Unit: J. Dimension: [ML²T⁻²].

RELATION WITH MOMENTUM

5.3.2 K and p

K = p²2m = 12 pv

p = √(2mK)

A body cannot have K without p, or p without K.

How K changes with v, p and m

COMPARING TWO BODIES

5.3.3 Comparing two bodies

Same	Result	Which is larger
K	p₁p₂ = √(m₁m₂)	heavier body has more p
p	K₁K₂ = m₂m₁	lighter body has more K
m	K₁K₂ = (p₁p₂)²	more p, more K

PERCENTAGE CHANGE

5.3.4 Percentage change

K ∝ p², so find the new K as (new p ÷ old p)².
Momentum up by 20%: K × 1.44, a rise of 44%.
K × 36: p × 6, a rise of 500% (not 600%).
Speed left when 5% of K remains: v√0.05.

FORCE, MOMENTUM AND K

5.3.5 Force, momentum and K

area under the F-t graph = Δp

Get the final momentum from the area, then find v = p ÷ m and K if needed.

Practise PYQs on this topicKinetic Energy & Momentum · 21 questions with solutions

CONCEPT

Work-Energy Theorem

In one line
The net work done on a body equals its change in kinetic energy.

THEOREM

5.4.1 Work-energy theorem

W_net = ΔK = 12 mv² − 12 mu²

Valid for all kinds of forces: conservative or not, internal or external.
K rises: net work is positive. K falls: net work is negative.
Whole motion from rest to rest: ΔK = 0, so the works add up to zero.

STOPPING

5.4.2 Stopping a vehicle

Stopping force F acts over distance x

Fx = K ⇒ x = KF

Ft = p ⇒ t = pF

Same	x₁ : x₂	t₁ : t₂
F, v	m₁ : m₂	m₁ : m₂
F, p	m₂ : m₁	1 : 1
F, K	1 : 1	√m₁ : √m₂

BY FRICTION

5.4.3 Stopped by friction

x = v²2μg = p²2μgm²

t = vμg

The mass does not matter for a given v.
Speed × n: stopping distance × n².

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Work from a given motion

Given v(x) or x(t): find v at the two ends, then W = ΔK.
Given F(t): integrate F ÷ m to get v, then W = ΔK.
v = a√s means acceleration = a²2, a constant.

Air resistance

W_gravity + W_air = ΔK

W_air is always negative.
Thrown up: energy lost = K at start − mgh at the top.

Bullet and planks

Loses 1n of its velocity in a distance x: K used = 2n − 1n² of K, so further distance = x(n − 1)²2n − 1.
Number of planks to stop it = n²2n − 1.
Falls h, then sinks s in sand: mg(h + s) = Fs.

Practise PYQs on this topicWork-Energy Theorem · 32 questions with solutions

CONCEPT

Potential Energy & Equilibrium

In one line
Potential energy belongs to conservative forces, and the force is the negative slope of U.

CONSERVATIVE FORCES

5.5.1 Conservative forces

Conservative force: same work on paths I, II and III

Conservative	Non-conservative
work does not depend on the path	work depends on the path
work over a closed loop = 0	work over a closed loop ≠ 0
gravity, spring, electric	friction, viscous, air drag

Round trip on a rough table: W = 2μmgs

POTENTIAL ENERGY

5.5.2 Potential energy and force

ΔU = −W_conservative

F = −dUdx

Defined only for conservative forces.
Work done against a conservative force raises U.

EQUILIBRIUM

5.5.3 Equilibrium

Points where the curve is flat have zero force

equilibrium: dUdx = 0

Type	U	Second derivative
stable	minimum	positive
unstable	maximum	negative
neutral	constant	zero

Marble in a bowl, on a bowl and on a table

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Three dimensional force

F = −(∂U∂x î + ∂U∂y ĵ + ∂U∂z k̂)

∂U ÷ ∂x: differentiate with y and z held constant.

Reading a U-x curve

K = E − U, and K can never be negative: the particle stays where U ≤ E.
|F| = size of the slope. Fastest at the lowest U.

Two atoms: U = a ÷ r¹² − b ÷ r⁶

r⁶ = 2ab, U_min = −b²4a

Dissociation energy D = b²4a

Practise PYQs on this topicPotential Energy & Equilibrium · 16 questions with solutions

CONCEPT

Spring & Elastic Energy

In one line
A stretched or compressed spring stores U = ½kx², whichever way it is deformed.

HOOKE’S LAW

5.6.1 Restoring force

Restoring force is opposite to the extension

F = −kx

k is the spring constant: the force for unit extension.
A stiff spring has a large k.
Unit: N m⁻¹. Dimension: [MT⁻²].

STORED ENERGY

5.6.2 Elastic potential energy

U = 12 kx² = 12 Fx = F²2k

Stretch by the same x: U ∝ k. The stiffer spring stores more.
Stretch by the same F: U ∝ 1 ÷ k. The softer spring stores more.
Series springs carry the same force, so U ∝ 1 ÷ k.
U ∝ x²: x × 3 gives U × 9.

STRETCHING FURTHER

5.6.3 From x₁ to x₂

W = 12 k(x₂² − x₁²)

From 0 to x: W₁. The next x: W₂ = 3W₁.

From	To	Work by the spring
natural	stretched or compressed	−12kx²
stretched or compressed	natural	+12kx²
stretched by x	compressed by x	0

ENERGY GRAPH

5.6.4 Block on a spring

Block oscillating between −a and +a

U = 12 kx²

K = 12 k(a² − x²)

E = 12 ka²

U and K are parabolas; their sum stays constant

USING IT

5.6.5 Spring problems

Situation	Equation
block of speed v hits a spring	12mv² = 12kx²
with friction f	12mv² = 12kx² + fx
ball falls h on a spring, compresses x	mg(h + x) = 12kx²
block hung and released	x_max = 2Mg ÷ k
constant force F on a spring	v_max at x = F ÷ k
spring launch to height h	12kx² = 12Fx = mgh

The force used to compress a spring is not constant, so stored energy = ½Fx, not Fx.
Largest momentum of a block on a spring: p = x√(km).

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Spring with its own mass

K = 16 mv²

Free end pulled at speed v, other end fixed; speed grows along the length.

Two blocks on one spring

Pushed apart at v each: total K goes into the spring.
Rolling body on a spring: use the whole K. Solid cylinder: K = ¾mv².

Practise PYQs on this topicSpring & Elastic Energy · 34 questions with solutions

CONCEPT

Conservation of Energy

In one line
With only conservative forces K + U stays constant. Friction takes energy out as heat, but total energy is never lost.

THE LAW

5.7.1 Conservation of mechanical energy

K + U = constant

ΔK + ΔU = 0

If K rises, U falls by the same amount.
Use it for any path: only the start and end heights matter.

Body thrown up: K falls, U rises, E stays fixed

FREE FALL

5.7.2 Free fall

K = nU at height h ⇒ h = Hn + 1

v = √(2gh)

Fall through h: K = mgh, so K ∝ m for the same fall.
Dropped from H: K at a depth x below the top = mgx.

PROJECTILE

5.7.3 Projectile at the top

K_top = K cos²θ

U_top = K sin²θ

θ is measured from the horizontal. Only the horizontal speed is left at the top.
Angle α from the vertical: U_top = K cos²α.
Vertical throw and throw at α from the vertical: U ratio = 1 : cos²α.

TRACKS AND PENDULUM

5.7.4 Tracks, pulleys and pendulum

Smooth track: v = √(2g × net fall). The shape does not matter.
Bodies on a pulley: loss in U of one − gain in U of the other = total K.
Pendulum at angle θ: h = L(1 − cos θ).
Small amplitude A: U = mgA²2L.
Slowly pulled sideways by a force at 45°: F = mg(√2 − 1) by the work method.

WITH FRICTION

5.7.5 Energy lost to friction

E_start = E_end + work against friction

The lost energy appears as heat.
Slides down h, stops on a rough floor: L = hμ.
Upper half of a slope smooth, lower half rough: stops at the bottom if μ = 2 tan θ.

CHAIN

5.7.6 Chain on a table

Mass of hanging part = M ÷ n; its centre of mass is L ÷ 2n below the table

W = MgL2n²

Find the mass of the hanging part, then raise its centre of mass.

Chain leaving the table

v = √(gL(1 − 1n²))

Rolled carpet: mass ∝ R². A roll of half the radius has mass M ÷ 4.

OTHER FORMS

5.7.7 Energy in other forms

Mass-energy: E = mc².
Heat: bullet stops, fraction of K goes to heat: ΔT comes from msΔT.
Soap bubble: work ∝ surface area.
Efficiency: useful energy = efficiency × energy supplied.

Practise PYQs on this topicConservation of Energy · 44 questions with solutions

CONCEPT

Power

In one line
Power is the rate of doing work. At any instant it is force dot velocity.

AVERAGE POWER

5.8.1 Average power

P_av = Wt

Same work in less time means more power.
Lifting: P = mght.
Motor raising water: P = (mt) gh.
Turbine with loss: output = efficiency × input.
Also P_av = ΔK ÷ t.

INSTANTANEOUS POWER

5.8.2 Instantaneous power

P = dWdt = F·v = Fv cos θ

Power is a scalar.
Constant force, from rest: P = F²tm, so P ∝ t.
Rotation: P = τω = Iαω.

Unit	Value
1 W	1 J s⁻¹ = 10⁷ erg s⁻¹
1 hp	746 W
1 kW	10³ W
1 kWh	3.6 × 10⁶ J (a unit of energy)

[P] = [ML²T⁻³]

GRAPHS

5.8.3 Graphs

Slope of the W-t graph is P; area under the P-t graph is W

FORCE WITH TIME

5.8.4 Force given as a function of time

Find a = F ÷ m for each component.
Integrate to get v, starting from rest.
Then P = F·v = F_xv_x + F_yv_y.

LIFTS AND CARS

5.8.5 Lifts and vehicles

P = (mg + f) v

Moving	Power at constant speed v
level road, friction f	P = fv
up a slope θ	P = (mg sin θ + f) v
down a slope θ	P = (mg sin θ − f) v

Constant speed means no extra force for acceleration.
On a road that rises 100 m in 1 km: sin θ = 0.1.

Practise PYQs on this topicPower · 51 questions with solutions

CONCEPT

Constant Power, Conveyors & Jets

In one line
With constant power, K grows in step with time, so v ∝ t^1/2 and s ∝ t^3/2.

CONSTANT POWER

5.9.1 Automobile with constant power

Pt = 12 mv²

v = (2Ptm)^1/2

s = (8P9m)^1/2 t^3/2

Speed and distance under constant power

F = Pv = (mP2t)^1/2

Force falls with time: F ∝ t^−½.
If a force is inversely proportional to speed, then Fv is constant: this is constant power, and K ∝ t.
Starting from rest is assumed in all of these.

CONVEYOR

5.9.2 Sand on a conveyor belt

Sand falls on a belt moving at constant speed

F = v dmdt

P = v² dmdt

P here is twice the rate of gain of K of the sand: the other half is lost as the sand slips on the belt.

JETS AND WIND

5.9.3 Water jet and wind

dmdt = λv = ρAv

P = 12 dmdt v² = 12 λv³ = 12 ρAv³

λ is the mass per unit length of the jet.
Wind: the power a turbine can get ∝ v³.

AVERAGE AND END VALUE

5.9.4 Average and final power

Constant force, from rest: P_av = half of P at the end of the time.
Heart: P = pressure × volume per second.

Practise PYQs on this topicPower · 51 questions with solutions

CONCEPT

Collisions

In one line
In every collision momentum is conserved. Kinetic energy is conserved only in an elastic one.

TYPES

5.10.1 Types of collision

Type	K	e
perfectly elastic	conserved	1
inelastic	not conserved	0 < e < 1
perfectly inelastic	largest loss, bodies stick	0

Head-on collision: motion stays on one line

RESTITUTION

5.10.2 Coefficient of restitution

e = v₂ − v₁u₁ − u₂

Speed of separation ÷ speed of approach.
Use it together with momentum conservation for an inelastic collision.

REBOUND

5.10.3 Ball hitting the ground

Each bounce is lower

e = √(h₁h₀)

Loss of a fraction of K: new height = fraction left × old height.

STICKING

5.10.4 Bodies sticking together

Move as one body after the collision

v = m₁u₁ + m₂u₂m₁ + m₂

K lost = 12 m₁m₂m₁ + m₂ (u₁ − u₂)²

Moving at right angles: conserve momentum along each axis.

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Head-on elastic collision: results

v₁ = m₁ − m₂m₁ + m₂ u₁ + 2m₂m₁ + m₂ u₂

v₂ = m₂ − m₁m₁ + m₂ u₂ + 2m₁m₁ + m₂ u₁

Case	After collision
m₁ = m₂	velocities are exchanged
m₁ ≫ m₂, u₂ = 0	v₁ = u₁, v₂ = 2u₁
m₁ ≪ m₂, u₂ = 0	v₁ = −u₁, v₂ = 0

Energy given to a target at rest

ΔKK = 4m₁m₂(m₁ + m₂)²

Equal masses: 100% is transferred.
Larger mass difference: less is transferred.

Oblique elastic collision

Momentum is conserved along x and y separately

Equal masses with one at rest: the two move off at 90°, so u² = v₁² + v₂².

Ball bouncing

hₙ = e²ⁿ h₀

total distance = h₀ 1 + e²1 − e²

total time = 1 + e1 − e √(2h₀g)

Loses 50% of K on each bounce: e² = 12, total distance = 3h.

Bullet into a suspended block

Block and bullet rise together

V = mum + M, V = √(2gh)

u = m + Mm √(2gh)

K lost = 12mMm + Mu². On a rough floor: V² = 2μgs.

Explosion and splitting

Momentum is conserved, K rises.
Body in flight: find its velocity at that instant first, then use momentum.

Practise PYQs on this topicCollisions · 14 questions with solutions

EXAM-DAY CARD

Just see this before the exam

W = F·s = ∫F dx

work

K = p²2m

kinetic energy

W_net = ΔK

theorem

U = ½kx²

spring

K + U = constant

conservation

P = F·v

power

FORMULA SHEET

Whole chapter on one page

Work

W = Fs cos θ = F·s

W = ∫F dx = area

against gravity: mgh

against friction: μmg s

K & momentum

K = ½mv² = p²2m

p = √(2mK)

same K: p ∝ √m

same p: K ∝ 1m

Theorem & stopping

W_net = ΔK

Fx = K, Ft = p

x = v²2μg

x = p²2μgm²

Potential & spring

F = −dUdx

stable: U minimum

U = ½kx² = F²2k

W = ½k(x₂² − x₁²)

Conservation

K + U = constant

K_top = K cos²θ

mg(h + x) = ½kx²

chain: W = MgL2n²

Power & collisions

P = Wt = F·v

P const: s ∝ t^3/2

conveyor: P = v² dmdt

e = √(h₁h₀)

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

K × 36 means p up by 600%

Spring compressed by F: U = Fx

K at the top of a projectile = 0

Same K: lighter body has more p

CORRECT ✓

p × 6, a rise of 500%

Force grows from 0: U = ½Fx

K_top = K cos²θ

Same K: heavier body has more p