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Work, Energy and Power

Work-energy theorem, potential energy, conservation of energy, collisions.

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Reading time~7 min
Revision time~2 min
Last updated2026-07-19
1 Read the chapter ~7 min

🎯 Key Points

  • Work-Energy theorem: W_net = ΔKE; Power P = W/t = F·v·cos θ
  • Conservative forces (gravity, spring): path-independent work, zero work in closed loop, F=−dU/dx
  • Elastic collision: momentum AND KE conserved; Inelastic: only momentum conserved, KE is lost
  • Coefficient of restitution e = (velocity of separation)/(velocity of approach); e=1 perfectly elastic, e=0 perfectly inelastic
  • Equal-mass elastic collision (1D): velocities are completely EXCHANGED
Energy Conservation: KE ⇄ PE on a SwingA: PE maxKE = 0B: KE maxPE minC: PE maxKE = 0

As a pendulum swings, total mechanical energy (KE+PE) stays constant: at the highest points A and C, all energy is potential (zero speed); at the lowest point B, all of that energy has converted into kinetic energy (maximum speed).

Work

  • W = F·d·cos(θ) (scalar product of force and displacement)
  • Work done by conservative forces is path-independent
  • Work done against friction = heat generated

Kinetic and Potential Energy

  • KE = ½mv²
  • Gravitational PE = mgh
  • Elastic PE (spring) = ½kx²
  • Work-Energy Theorem: W_net = ΔKE

Conservation of Energy

  • Total mechanical energy (KE + PE) = constant for conservative systems
  • Power: P = W/t = F·v

Collisions

  • Elastic collision: both momentum and KE conserved
  • Inelastic collision: only momentum conserved; KE lost
  • Perfectly inelastic: objects stick together; maximum KE loss
  • Coefficient of restitution e = relative velocity of separation / relative velocity of approach

Conservative and Non-Conservative Forces

  • Conservative force: work done depends only on initial and final position, not on path (gravity, spring force, electrostatic force); work done in a closed path = 0
  • Non-conservative force: work done depends on the path taken (friction, air resistance); work done in a closed path is not zero, mechanical energy is dissipated as heat
  • For a conservative force, F = -dU/dx (force is negative gradient of potential energy)

Power

  • Average power: P_avg = W/t (total work done divided by total time)
  • Instantaneous power: P = dW/dt = F·v·cos(θ) (theta is angle between force and velocity)
  • SI unit: watt (1 W = 1 J/s); commercial unit: 1 horsepower = 746 W; 1 kWh = 3.6 × 10⁶ J

Elastic Collisions in One Dimension

  • For masses m₁ (velocity u₁) and m₂ (velocity u₂) colliding elastically, final velocities:
  • v₁ = [(m₁ - m₂)u₁ + 2m₂u₂] / (m₁ + m₂)
  • v₂ = [(m₂ - m₁)u₂ + 2m₁u₁] / (m₁ + m₂)
  • Special case: if m₁ = m₂, velocities are exactly exchanged
  • Special case: if m₂ is initially at rest and m₁ >> m₂, then v₁ ≈ u₁ and v₂ ≈ 2u₁ (light target shoots off at twice the incoming speed)
  • Special case: if m₁ >> m₂ and target at rest, the heavy mass continues almost undisturbed

Potential Energy Curve

  • Slope of U-x graph gives force: F = -dU/dx; positive slope means force acts in -x direction
  • Points where dU/dx = 0 are equilibrium points (stable if U is minimum, unstable if U is maximum)
  • A particle oscillates between turning points where total energy E equals U(x) (KE becomes zero there)

Work Done by a Variable Force

  • When force changes with position, work = ∫F·dx = area under the Force–displacement (F–x) graph
  • Spring force F = −kx is variable; work done in stretching from 0 to x = ½kx² (area of the triangle under the F–x line)
  • For a force varying in direction too, W = ∫F·ds (line integral of the dot product along the path)

Motion in a Vertical Circle

  • For a body on a string looping a vertical circle, minimum speed at the TOP: v_top = √(gr) (tension just zero, gravity supplies centripetal force)
  • Minimum speed at the BOTTOM to complete the loop: v_bottom = √(5gr) (from energy conservation between top and bottom)
  • Tension at any point: T = mv²/r − mg·cos(θ) contribution; tension is maximum at the lowest point, minimum at the highest point
  • Difference in tension between lowest and highest points = 6mg (a standard result)

Two-Dimensional (Oblique) Collisions

  • In 2D collisions, momentum is conserved SEPARATELY along two perpendicular directions (x and y)
  • m₁u₁ = m₁v₁cos(θ₁) + m₂v₂cos(θ₂) along x; 0 = m₁v₁sin(θ₁) − m₂v₂sin(θ₂) along y
  • For elastic collision of equal masses (one at rest), the two bodies move off at 90° to each other

Types of Equilibrium

  • Stable equilibrium: potential energy is minimum; a small displacement produces a restoring force pushing the body back (d²U/dx² > 0)
  • Unstable equilibrium: potential energy is maximum; a small displacement pushes the body further away (d²U/dx² < 0)
  • Neutral equilibrium: potential energy is constant; body stays in the new position (d²U/dx² = 0)

🚀 JEE Advanced Edge

Oblique elastic collisions: When two equal masses collide elastically and one is initially at rest, after the collision the two velocity vectors are always perpendicular to each other (90° apart) — a powerful shortcut for billiard-ball-style 2D collision problems, derivable from simultaneously conserving momentum (vector) and KE (scalar).

Energy loss in perfectly inelastic collisions: KE lost = ½ × (m₁m₂)/(m₁+m₂) × (relative velocity)² — this "reduced mass" formula directly gives the energy dissipated as heat/deformation without needing to separately calculate initial and final KE.

Worked problem: A 4 kg block moving at 6 m/s collides perfectly inelastically with a stationary 2 kg block. Find the velocity after collision and the KE lost. Approach: Momentum conservation: 4×6 = (4+2)v → v = 4 m/s. KE before = ½(4)(36) = 72 J. KE after = ½(6)(16) = 48 J. KE lost = 72−48 = 24 J (matches the reduced-mass formula: ½ × (4×2)/6 × 36 = 24 J).

2 Revise ~2 min before the exam

📐 Formula Sheet

  • Work: W = F·s·cosθ  |  variable force: W = ∫F·dx (area under an F–x graph)
  • Kinetic energy: KE = ½mv²  |  in terms of momentum: KE = p²/2m
  • Work–energy theorem: Wnet = ΔKE = ½mv² − ½mu²
  • Gravitational PE (near Earth): U = mgh  |  Spring PE: U = ½kx²
  • Conservation: KE + PE = constant when only conservative forces act
  • Power: P = W/t = F·v  |  1 hp ≈ 746 W
  • Elastic collision (1-D, equal masses): the two velocities simply swap
  • Coefficient of restitution: e = (relative speed after) / (relative speed before); e = 1 elastic, e = 0 perfectly inelastic
3 Practice apply it

✍️ Worked Examples

Example 1 — Work done against friction
Q: A 5 kg block is pulled 10 m along a rough floor by a 30 N force at 60° above the horizontal. Friction is 8 N. Find the net work done.
Step 1 — Work by the applied force: only the horizontal component does work. W₁ = 30 × cos60° × 10 = 30 × 0.5 × 10 = 150 J.
Step 2 — Work by friction: friction opposes motion, so W₂ = −8 × 10 = −80 J.
Step 3 — Gravity and normal force: both act perpendicular to the displacement ⇒ zero work.
Answer: Wnet = 150 − 80 = 70 J. Trap: using the full 30 N instead of its horizontal component gives 300 J.

Example 2 — Speed from a compressed spring
Q: A spring of k = 200 N/m is compressed 10 cm and released, pushing a 0.5 kg block on a frictionless surface. Find the block's speed as it leaves the spring.
Step 1 — Energy stored: U = ½kx² = ½ × 200 × (0.1)² = 1 J.
Step 2 — All of it converts to KE: ½mv² = 1.
Step 3 — Solve: ½ × 0.5 × v² = 1 ⇒ v² = 4 ⇒ v = 2 m/s.
Answer: 2 m/s. Trap: writing x = 10 instead of 0.1 m inflates the energy 10,000×.

Example 3 — Power of a lift motor
Q: A lift of total mass 800 kg rises at a steady 2 m/s. What power must the motor deliver? (g = 10 m/s²)
Step 1 — Steady speed means zero acceleration, so the motor force just balances weight: F = mg = 800 × 10 = 8000 N.
Step 2 — Use P = F·v: P = 8000 × 2 = 16,000 W.
Answer: 16 kW. Note: because the speed is constant, no extra power goes into changing KE.

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Frequently Asked Questions — Work, Energy and Power

What are the key concepts in Work, Energy and Power?
Work-energy theorem, potential energy, conservation of energy, collisions.
Is Work, Energy and Power important for NEET & JEE?
Yes. Work, Energy and Power is part of the Physics Class 11 NCERT syllabus and is directly tested in NEET and JEE examinations. StudyHub provides structured notes, diagrams, and practice questions covering all exam-level subtopics.
How can I practice Work, Energy and Power questions on StudyHub?
Open StudyHub and select Physics → Work, Energy and Power. Choose Easy, Medium, or Hard difficulty. Hard-tier questions are at NEET & JEE level with full step-by-step explanations.

References

  1. NCERT Class 11 Physics Textbook — Chapter: Work, Energy and Power
  2. CBSE Curriculum — Physics (Class 11)
  3. NTA NEET UG Official Syllabus — subject-wise topic list
  4. NTA JEE Main Official Syllabus — subject-wise topic list