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Physics → Work, Energy & Power
TG EAPCET 2027
Chapter progress
Layer 1 of 5
Physics High Weightage β˜…β˜…β˜…β˜… Class 11

Work, Energy & Power

Work-Energy theorem is the Swiss Army knife of mechanics β€” use it when Newton's Laws become messy. Expect 3–4 EAPCET questions every year.

3–4Questions in EAPCET
~3%Paper Weightage
8Core Formulas
4Mistake Traps

Concept Core

Work, kinetic energy, potential energy, conservation β€” and when to use each.

Work Done by a Force

Work = Force Γ— displacement component in direction of force.

W = F Β· d = Fd cosΞΈ

ΞΈ is angle between F and displacement. Work is a scalar.

W > 0: force aids motion. W < 0: force opposes motion (friction, braking). W = 0: force βŠ₯ displacement (normal force, centripetal force).

Kinetic Energy & the Work-Energy Theorem

KE = Β½mvΒ². The Work-Energy Theorem is one of the most powerful tools in mechanics:

W_net = Ξ”KE = Β½mvΒ² βˆ’ Β½muΒ²

Net work done on an object = change in its kinetic energy. Use when forces are complex β€” no need for acceleration.

Potential Energy

Gravitational PE: U = mgh (h = height above reference)

Elastic PE (spring): U = Β½kxΒ² (k = spring constant, x = compression/extension)

PE is the energy stored by virtue of position or configuration. It depends on the reference level β€” only changes in PE are physical.

Conservation of Mechanical Energy

When only conservative forces do work (gravity, spring β€” NOT friction):

KE + PE = constant Β½mv₁² + mgh₁ = Β½mvβ‚‚Β² + mghβ‚‚

This is the "no-friction" shortcut. Height and velocity trade off smoothly.

Power

Rate of doing work:

P = W/t = FΒ·v = Fv cosΞΈ

Unit: Watt (W) = J/s. 1 HP = 746 W.

When a machine operates at constant power: P = Fv. As speed increases, the force the engine can exert decreases (for fixed P).

Collision Types

Elastic: Both momentum AND KE conserved. e = 1 (coefficient of restitution).

Inelastic: Only momentum conserved. KE lost as heat. e < 1.

Perfectly inelastic: Bodies stick together. Maximum KE loss. e = 0.

m₁u₁ + mβ‚‚uβ‚‚ = m₁v₁ + mβ‚‚vβ‚‚ (always)
Spring Problems β€” Elastic PE

Spring compressed by x stores energy U = Β½kxΒ². When released, all PE converts to KE:

Β½kxΒ² = Β½mvΒ² β†’ v = x√(k/m)

For two springs in series: 1/k_eff = 1/k₁ + 1/kβ‚‚ (softer combined). In parallel: k_eff = k₁ + kβ‚‚ (stiffer combined).

Formula Vault

All Work-Energy-Power formulas β€” exam-ready.

Work Done
W = Fd cosΞΈ
ΞΈ = angle between F and d
Kinetic Energy
KE = Β½mvΒ²
Always β‰₯ 0
Work-Energy Theorem
W_net = Ξ”KE = Β½mvΒ²βˆ’Β½muΒ²
Net work = change in KE
Gravitational PE
U = mgh
h from reference level
Spring PE
U = Β½kxΒ²
k = spring constant; x = deformation
Conservation of Energy
KE₁ + PE₁ = KEβ‚‚ + PEβ‚‚
No friction; conservative forces only
Power
P = W/t = Fv cosΞΈ
1 HP = 746 W
Elastic Collision Velocities
v₁ = (mβ‚βˆ’mβ‚‚)u₁+2mβ‚‚uβ‚‚)/(m₁+mβ‚‚)
Similarly for vβ‚‚
Perfectly Inelastic
v = (m₁u₁+mβ‚‚uβ‚‚)/(m₁+mβ‚‚)
Bodies stick together
KE lost in P.I. Collision
Ξ”KE = m₁mβ‚‚(uβ‚βˆ’uβ‚‚)Β²/2(m₁+mβ‚‚)
Always positive (lost)

Worked Examples

5 problems showcasing when Work-Energy beats Newton's Laws.

EasyFind velocity of 2 kg block after 10 J of work done, starting from restβ–Ύ
A 2 kg block starts from rest. Net work done on it = 10 J. Find its final velocity.
1
W_net = Ξ”KE β†’ 10 = Β½(2)vΒ² βˆ’ 0 β†’ vΒ² = 10 β†’ v = √10 β‰ˆ 3.16 m/s
βœ“ v = √10 m/s
MediumBall dropped from 10 m β€” find velocity just before hitting groundβ–Ύ
A 1 kg ball is dropped from height 10 m. Using conservation of energy, find velocity just before impact. (g = 10)
1
At top: KE = 0, PE = mgh = 1Γ—10Γ—10 = 100 J. At bottom: PE = 0.
2
Conservation: Β½mvΒ² = 100 β†’ vΒ² = 200 β†’ v = 10√2 β‰ˆ 14.1 m/s
βœ“ v = 10√2 m/s (same as v = √(2gh) = √200)
MediumSpring compressed 0.2 m launches 0.5 kg block β€” find speedβ–Ύ
A spring (k = 800 N/m) is compressed by 0.2 m and releases a 0.5 kg block on a frictionless surface. Find the block's speed after release.
1
Spring PE β†’ KE: Β½kxΒ² = Β½mvΒ² β†’ kxΒ² = mvΒ²
2
vΒ² = kxΒ²/m = 800Γ—0.04/0.5 = 32/0.5 = 64 β†’ v = 8 m/s
βœ“ Speed = 8 m/s
EAPCET LevelPerfectly inelastic collision β€” find KE lostβ–Ύ
A 4 kg block moving at 6 m/s collides and sticks to a 2 kg block at rest. Find the KE lost.
1
Momentum conservation: 4Γ—6 + 2Γ—0 = (4+2)v β†’ v = 24/6 = 4 m/s
2
Initial KE = Β½Γ—4Γ—36 = 72 J
3
Final KE = Β½Γ—6Γ—16 = 48 J
4
KE lost = 72 βˆ’ 48 = 24 J
βœ“ KE lost = 24 J
Trap QuestionEngine at constant power β€” does velocity keep increasing forever?β–Ύ
An engine of power P drives a car against drag force f. What is the maximum speed? ⚠️ Conceptual trap.
1
The trap: Students think constant power β†’ constant acceleration β†’ speed increases forever. Wrong.
2
P = Fv. As v increases, the available engine force F = P/v decreases.
3
Maximum speed when engine force = drag force: P/v_max = f β†’ v_max = P/f
4
At v_max, net force = 0 β†’ acceleration = 0 β†’ constant speed. The car reaches terminal velocity.
βœ“ Maximum speed = P/f where f = drag force

Mistake DNA

4 errors from distractor analysis β€” conceptual and computational.

πŸ“
Work Done by Normal Force and Gravity While Moving Horizontally
Students sometimes add mgh to a horizontal motion problem or claim N does work on a horizontal surface.
❌ Wrong
Block moves 5 m horizontally:
W_gravity = mgh β‰  0 βœ—
W_normal = Fd β‰  0 βœ—
βœ“ Correct
Horizontal motion, h=0:
W_gravity = mgΓ—0 = 0 βœ“
W_normal = NΓ—dΓ—cos90Β° = 0 βœ“
Work = Fd cosΞΈ. Normal force βŠ₯ motion β†’ ΞΈ=90Β° β†’ W=0. Gravity does work only when there's vertical displacement.
πŸ”‹
Using Conservation of Energy When Friction is Present
Conservation of mechanical energy requires no non-conservative forces. Friction violates it.
❌ Wrong
With friction ΞΌ on incline:
mgh = Β½mvΒ² βœ—
(ignores friction work)
βœ“ Correct
mgh βˆ’ W_friction = Β½mvΒ²
W_friction = ΞΌmgcosΞΈ Γ— d βœ“
Energy lost to friction
With friction: use W_net = Ξ”KE (Work-Energy Theorem). W_net = W_gravity + W_friction = Ξ”KE. Never ignore friction in energy problems.
πŸ’₯
Assuming KE is Conserved in All Collisions
KE is conserved only in elastic collisions. Momentum is always conserved.
❌ Wrong
Inelastic collision:
Using Β½m₁u₁² = Β½(m₁+mβ‚‚)vΒ²
to find v βœ—
βœ“ Correct
For inelastic: only
momentum conserved:
m₁u₁ = (m₁+mβ‚‚)v βœ“
Always identify collision type first. "Sticks together" = perfectly inelastic. Use momentum equation to find v, then compute KE if asked.
⚑
Power = Force Γ— Velocity Confusion with Direction
When force and velocity are at an angle, P = Fv cosΞΈ. Students forget the cosΞΈ.
❌ Wrong
Force 100 N at 60Β° to
motion, v = 5 m/s:
P = 100 Γ— 5 = 500 W βœ—
βœ“ Correct
P = Fv cosΞΈ
= 100 Γ— 5 Γ— cos60Β°
= 500 Γ— 0.5 = 250 W βœ“
Power is the dot product F⃗·v⃗ = Fv cosθ. Only the component of force along velocity contributes to power.

Chapter Intelligence

Work-Energy is a bridge chapter β€” it connects kinematics to everything ahead.

EAPCET Topic Weightage (2019–2024)
Work-Energy theorem
~7
Conservation of energy
~6
Collisions
~5
Spring problems
~4
Power calculations
~3
High-Yield PYQ Patterns
Speed using energy conservation KE lost in inelastic collision Spring releases block β€” find speed Max speed at constant power Work done against friction Ball on incline β€” energy method
Exam Strategy
  • When a question involves height and velocity β€” try energy conservation first. It's faster than Newton's 2nd Law + kinematics.
  • For collision questions: identify elastic or inelastic first. Perfectly inelastic (stick together) β†’ momentum conservation only.
  • Maximum speed at constant power: set F_engine = F_drag β†’ v_max = P/F_drag. This appears almost every year.
  • Spring questions: energy stored = Β½kxΒ². If the block slides on a frictionless surface and the spring releases, all spring PE β†’ KE.
  • Work done by friction is always negative. W_friction = βˆ’ΞΌmgcosΞΈ Γ— d. This energy is "lost" (converted to heat).
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